L I B ^AHY OF THE U N IVLR.SITY OF ILLINOIS <* 510-7 v-9 Digitized by the Internet Archive in 2012 with funding from University of Illinois Urbana-Champaign http://www.archive.org/details/highschoolmathem09univ HIGH SCHOOL MATHEMATICS Unit 9. ELEMENTARY FUNCTIONS: POWERS, EXPONENTIALS, AND LOGARITHMS UNIVERSITY OF ILLINOIS COMMITTEE ON SCHOOL MATHEMATICS MAX BEBERMAN, D/recfor HERBERT E. VAUGHAN, Editor UNIVERSITY OF ILLINOIS PRESS • URBANA, 1962 HIGH SCHOOL MATHEMATICS Unit 9. ELEMENTARY FUNCTIONS: POWERS, EXPONENTIALS, AND LOGARITHMS UNIVERSITY OF ILLINOIS COMMITTEE ON SCHOOL MATHEMATICS MAX BEBERMAN, Director HERBERT E. VAUGHAN, Editor UNIVERSITY OF ILLINOIS PRESS • URBANA, 1962 © 1962 by the Board of Trustees of the University of Illinois. Manufactured in the United States of America. [9-iii] PREFACE There is a number b such that 10 = 100. Do you think that there is a number a such that 10 a = 75? A number c such that 10 C = 125? If there are such numbers, what can you say about a, b, and c? Here is a problem whose solution depends on being able to find such exponents: A certain isotope of the element strontium [Sr 89 ] decays [into yttrium eighty nine] according to the formula: A(t) = A«10~ 2t [Here, for each t, A(t) is the number of grams of the isotope which remains in a sample which, t years earlier, contained A grams. ] A sample which now contains, say, 100 gm of yttrium, contains 1 gm of Sr , Supposing that all of the yttrium resulted from the decay of Sr 89 , how many years ago did the sample contain 75 gm of the isotope? To solve this problem, we need to find the number t such that 1 = 75*10~ 2t 2t --that is, such that 10 = 75. Evidently, t = a/2, where a is the number such that 10 = 75. One of the things you will learn in this unit is how to solve such equations. In Unit 8 you learned the meaning of expressions such as: 10 2 , 10" 1 , (a 2 +3)~ 2 , (b + 5) 59 , and established some useful laws of exponents [Theorems 152a and 154- 161]. In this unit you will discover how to assign meanings to expres- sions like: In l. 87506 ln -J ,vT . ,77 ,0.8,f 10 , 10 4 , 5 and: (a - 3 ) 3 , in such a way that these laws of exponents still hold. To do this, you will need to learn one more basic principle for real numbers- -the least upper bound principle. [Except for definitions, this is the last basic principle you will need in order to develop all the properties of the real numbers.] [9-v] TABLE OF CONTENTS Preface 9.01 Definite description Principal square roots An existence condition A uniqueness condition A defining principle for l v ' The uniqueness theorem for 4 v ' A typical proof Exercises on square roots Rationalizing denominators A defining principle for ' v ' 3 — The uniqueness theorem for ' V Exercises on the absolute value operation Miscellaneous Exercises ;9-iii] 9-1] .9-1] 9-3] !9-3] 9-5] 9-5] :9-s] 9-6] 9-10 9-12 9-12 9-14 9-19 9.02 The need for a new basic principle The system of rational real numbers Closure theorems for the rationals A theorem which requires a new principle Explorations Exercises Upper bounds and greatest member Lower bounds and least member Least upper bound 9-22; 9-22; 9-22 9-23; 9-24; ;9-25; ;9-26; 9-26 9.03 The least upper bound principle The least upper bound principle [iubp] The cofinality principle is a theorem Exploration Exercises Inverses Increasing and decreasing functions The power functions and the restricted power functions ;9-3o; 9-32 9-33 9-34; .9-34 9-35 [9-36] [9-vi] [CONTENTS] Monotonic functions Functions whose domains are segments Continuous functions Subsets of continuous functions The range of a continuous monotonic function Miscellaneous Exercises 9.04 Principal roots Operators and functions The defining principle (PR) for * V ' Two theorems about the principal nth root function Radical Expressions Radical, radicand, and index Exercises on principal roots Theorems about principal roots Roots of negative numbers The defining principle (PR') Theorems about odd principal roots Exercises on transforming radical expressions Removing factors from a radicand Introducing factors into a radicand Reducing the index Miscellaneous Exercises 9.05 The rational numbers A definition (R) for the set of rationals Using (R) in proving that numbers are rational Using (R) to prove closure theorems Using (R) in proving that numbers are irrational Exercises on rationality Roots of integers are integral or irrational Rational-linear combinations The rationals are dense in the reals The reciprocating operation 9-37 9-38 9-42 9-43 9-45 9-45 9-48 9-48 9-49 9-49 9-50 9-50 9-51 9-52 9-53 9-55 9-55 9-56 9-58 9-58 9-58 9-60 9-62 9-62 9-62 9-63 9-63 9-64 9-65 9-66 9-67 9-67 [CONTENTS] [9-vii] 9.06 Rational exponents Integral exponents- -parts (1) and (2) of a definition Extending [the domain of] a function Looking for a definition Maintaining the same laws for exponents Consistency of a proposed definition Parts (3) and (3') of a definition of exponentiation Exercises on rational exponents Testing our definition Laws of exponents for the case of rational exponents Proofs for the laws of exponents More exercises on rational exponents Rational exponents and negative bases Exploration Exercises An increasing function 9.07 The exponential functions More about rational exponents Some properties of exponential functions with rational arguments Irrational exponents Definition of 2 Justifying a definition The exponential functions Parts (4 1 ), (4 2 ), and (4 3 ) of a definition of exponentiation Some properties of exponential functions Exercises on irrational exponents Exploration Exercises 9.08 Computing with inverses of exponential functions Obtaining better approximations 9-68 9-69 9-70 9-71 9-71 9-72 9-73 9-75 9-75 9-76 9-79 9-84 9-86 9-87 9-89 9-89 9-90 9-90 9-91 9-91 9-92 9-92 9-93 9-93 9-95] 9-100] 9-100] h Vlll [CONTENTS A more convenient choice of base A table for the inverse of the exponential function with base 10 Exercises on computation Linear interpolation Exercises on interpolation and computation Exploration Exercises The exponential function with base 5--exp The logarithm function to the base 5- -log 9.09 The logarithm functions The defining principle ( L) for 'log ' 3. Theorems about logarithm functions Scientific notation Characteristics and mantissas Exercises on logarithmic computation Exercises on graphing Exercises on exponential and logarithmic equations Relating logarithms to two bases Miscellaneous Exercises Exploration Exercises Sentences whose solutions are functions 9.10 Some laws of nature Gay-Lussac's law for gasses A case whose value-differences depend on argu- ment-differences Homogeneous linear functions Gay-Lussac's Law Absolute temperature Isopiestic conditions and isothermal conditions General Gas Laws [9-102 9-103 9-108 9-110 9-112 9-112 9-113 9-114 9-115 9-117 9-120 9-121 9-123 9-128 9-129 9-131 9-134 9- 137 9- ■138 9- 139 9- 139 9- 141 9- 142 9- 142 9- 143 9- 143 9- 144 [CONTENTS] Decay of a radioactive substance A case where value-ratios depend on argument- differences A characterization of exponential functions Half-life Another approach to radioactive decay The natural logarithm function The decay constant l/x The function whose values are those of '(1 + x) ' '■ the number e Newton' s law of cooling Transient currents in simple circuits Compound interest A diabatic compression of gasses A case where value-ratios depend on argument- ratios A characterization of power functions The number e Tangents to logarithm functions Another characterization of e [9-ix] [9-147] 9-148] 9-150] 9-151] 9-152] 9-152] 9-153] 9-155] 9-158] 9-159] 9-162] 9-164] 9-165] 9-165] 9-168] 9-169] 9-171] The natural logarithm function The slope of a tangent to log Exercises on tangents to logarithm and exponen- tial functions Exercises on 'inx/x' Exercises on the inequation 'x^ > y ' Exercises on 'xlnx' and 'x * 9-172] 9-173] 9-173] 9-175] 9-176] 9-176] Summary Review Exercises Miscellaneous Exercises 9-177] 9-181] 9-186] [9-x] Appendix A - - The simplest functions Operators, functions, and definite descriptions Monotonic functions Inverses of monotonic functions- - Theorem 184 Exploration Exercises Continuous functions Definition Each positive integral power function continuous Theorem 186 A theorem on continuous monotonic functions To find the range of a monotonic function The main theorem on inverses- - Theorem 187 Appendix B- - Irrational numbers The irrationality of roots Infinite sets One-to-one correspondences Definition of finiteness for sets Definition of 'infinite' Exercises on one-to-one mappings Countably infinite sets Some sample theorems Counting the rationals Uncountably infinite sets The diagonal argument Uncountability of the reals Uncountability of the irrationals Decimal representation of real numbers Exercises on density The density of the rationals in the reals Appendix C- - The exponential functions More on rational exponents Uniform continuity [CONTENTS] [9- 190] [9- 190] [9- 194] [9- 197] [9- 201] [9- •210] [9- 211] [9- ■218] [9- 220] [9- •220] [9- •228] [9- 231] [9- ■231] [9- ■233] [9- •234] [9- •234] 19- ■235] [9- -235] [9- •237] [9- ■237] [9- ■238] [9- -241] [9- -241] f c-. • 242] [9- -242] [9- •243] [9- -248] [9- ■249] [9- •250] [9- ■251] [9- ■255] [CONTENTS] [9-xi] Justification of the introduction of irrational exponents Proof of Theorem 202 Monotonicity of the exponential functions Continuity of the exponential functions Range of the exponential functions Proofs of Theorems 203-211 A uniqueness theorem on continuous functions Three standard theorems on continuity Product of continuous functions Reciprocal of a continuous function Composite of continuous functions Continuity of power functions [9-256] [9-258] [9-258] [9-259] [9-261] [9-261] [9-262] [9-267] [9-267] [9-268] [9-268] [9-269] Appendix D- - Volume - measures Area-measure from Unit 6 Informal assumptions about solids Points, lines, and planes Convention on variables A line contained in a plane A line intersecting a plane A line parallel to a plane Parallel planes Definition of 'parallel' A line perpendicular to a plane Distance between a point and a plane Distance between parallel planes Coplanar sets Skew lines Projection of a point on a plane Dihedral angles and their measures Perpendicular planes Bisector of a dihedral angle Parallel planes and proportional segments Some simple solids Prisms [9- ■270] [9- 271] [9- 272] [9- 272] [9- 272] [9- •273] [9- •273] [9- •273] [9- ■273] [9- •2V4] [9- •275] [9- •275] [9- •276] [9- ■277] [9- •278] (9- •279] [9- •279] [9- •280] [9- •280] [9- ■281] (9- 281] [9-xii] [CONTENTS] Lateral face, surface, and edge; bases Parallelepipeds Right prisms Altitude of a prism Pyramids Vertex and base of a pyramid Lateral faces Tetrahedra Altitude of a pyramid Regular pyramid; slant height Cylinders and cones Cross- sections Cross- sections of a prism [cylinder] are congruent Cross- sections of a pyramid [cone] are similar Ratio of the area-measure of the cross- sections of a pyramid [cone] Exercises on sketching solids Numerical exercises An axiom on volume-measure Cavalieri's principle Application to prisms and cylinders Volume formulas Another axiom- -V = abc Volume-measures of prisms and cylinders Volume-measures of triangular pyramids Volume-measures of pyramids and cones Numerical exercises Inclination of the axis of a circular cylinder Prismatoids Bases, midsection, and altitude The prismoided formula Frustums Volume-measures of frustums Solid spheres Spherical region--ball 9-281 9-281 9-281 9-281 9-282 9-282 9-282 9-282 9-282 9-282 9-282 9-282 9-282 9-283 9-285 9-285 9-286 9-287 9-289 9-289 9-290 9-290 9-290 9-291 9-292 9-292 9-29 3 9-295 9-295 9-296 9-298 9-299 9-300 9-300 [CONTENTS] [9-xiiij Applying Cavalieri's principle Volume-measures of solid spheres Surface area formulas Rectangular parallelepiped Regular pyramid Right circular cone Right circular cylinder Spherical region Summary of mensuration formulas Numerical exercises Solid of revolution Locus problems 9-300 9-301 9-302 9-302 9-302 9-303] 9-303] 9-304] 9-306; 9-308] 9-309; 9-311 Appendix E - - Some functional equations Solution of a functional equation The idpms and Theorems 205, 215, and 218 A queer function A theorem on homogeneous linear functions Proof of Theorem 221 Three other theorems Proof of Theorem 222 Proof of Theorem 223 Proof of Theorem 224 An application of the four theorems 9-313] 9-314] 9-315] 9-319; 9-322 9-326; 9-326; 9-329; 9-332; 9-333] Basic Principles and Theorems Basic principles for the real numbers Theorems from Unit 2 Basic principles for P--basic principle for > Theorems about > Basic principles for I + Theorems about positive integers Cofinality principle and basic principle for I Theorems about integers 9-337 9-337] 9-337 9-341 9-341 9-343. 9-343 9-344] 9-344] [9-xiv] [CONTENTS] Definitions of the greatest integer function and the fractional part function Theorems involving '[[... ]}' and '{[ . . . }} ' Definition of the divisibility relation Theorems about | Definition of Z-notation Theorems about continued sums Definitions of a difference- sequence, an arithmetic progression, and the common difference of an AP Theorems about arithmetic progressions Summation theorems involving inequations Definitions of II- notation and the factorial sequence Theorems about continued products Definitions of exponential sequences and negative integral exponents Theorems about exponential sequences Laws of exponents Definitions of a geometric progression and the common ratio of a GP Theorems about geometric progressions Definition of the absolute- valuing function Theorems about the absolute- valuing function Recursive definitions of the function C and the factorial sequence Theorems about the function C Counting principles (C x ), (C 2 ), (C 3 ), and(C 4 ) Theorems about combinatorial problems Theorem about binomial exponentials Theorems about prime numbers Least upper bound principle Definitions of increasing, decreasing, and monotonic functions Theorem about monotonic functions Definition of continuity Theorems about continuous functions Defining principle (PR) for principal root functions 9-345 9-345 9-345 9-345 9-346 9-346 9-348 9-349 9-349 9-349 9-350 9-350 9-350 9-351 9-352 9-352 9-353 9-353 9-353 9-354 9-354 9-355 9-355 9-356 9-356 9-357 9-357 9-357 9-357 9-358 [CONTENTS] [9-xv] Theorems about principal root functions Defining principle (PR') for odd principal root functions Theorem about principal odd root functions Basic principle (R) for the rational numbers Closure theorems for the rational numbers Theorem on irrational roots Definitions of 'finite' and 'infinite' Counting principle (C 5 ) Theorems about countability of sets Decimal representation principle Theorem on existence and uniqueness of decimal representations Theorem on the density of the rationals in the reals Definition of exponential functions Theorems about exponential functions Defining principle ( L) for logarithm functions Theorems about logarithm functions Theorems on continuity of power functions Theorems on functional equation Table of Squares and Square Roots Table of Trigonometric Ratios Table of Common Logarithm s 9-358] 9-358] 9-358] 359] 359] 359] 360] 9-360] 9-360] 9-361] 9-361] 9-361] 9-362] 9-362] 9-363] 9-363] 9-363] 9-364] 9-365] 9-366] 9-367] [9.01] [9-1] 9. 01 Definite description . --Consider these two questions about the greatest integer function: (1) What is the real number z such that [[z]) = 1. 5 ? (2) What is the real number z such that ([z]| =9 ? In order to answer either of these questions, you would have to name some real number. The first question is unanswerable because 1. 5 is not an integer. The second question is also unanswerable. Why? The underlined phrases are examples of definite descriptions . [A definite description is a phrase of the form ' the . . . such that * ; a phrase of the form 'a . . . such that * is an indefinite description. ] For a definite description to make sense it must be the case that (i) there ^ something which satisfies the description, and (ii) there are not two things which satisfy the description. [Of course, if there are more than two, there are two. ] PRINCIPAL SQUARE ROOTS Consider these questions about the squaring function: (3) What is the real number z such that z =4? (4) What is the real number z such that z = -1? (5) What is the real number z such that z ~ 1t? To answer any of these questions you must name some real number. Question (3) is unanswerable because there are two real numbers [2 and -2] whose squares are 4. [if each of these numbers were the number whose square is 4, then they would not be two numbers! In other words, it would be the case that 2 = - 2. ] Question (4) is unanswerable because squares of real numbers are nonnegative. [Theorem 97a and the pmO. ] Question (5) is also unanswerable. In the first place, as far as any- thing we have proved is concerned, there may not be a real number whose square is ir. But, if there is, there are two [Explain.]. Let's replace question (3) by a related question which is answerable: 2 (3') What is the real number z such that (z > and z = 4) ? What is the answer? {9-2] [9.011 If we make a similar change in question (4), we still have an un- answerable question. Why? If we make a similar change in question (5), the resulting question is answerable if and only if there is a real number whose square is 77. Explain. Let's go back to question (3') and see in more detail why it is an- swerable. In order for the definite description: the real number z such that (z > and z =4) to make sense, the description must refer to exactly one real number. In other words, it must be the case that two conditions are satisfied- - an existence condition : There is a number which is nonnega- tive and whose square is 4. and a uniqueness condition : There are not two numbers which are nonnegative and whose squares are 4. Since 2 > and 2 =4, the existence condition is satisfied. Let's con- sider the uniqueness condition. Suppose that (b > and b =4) and (c > and c = 4), and suppose that b ^ c. From the last assumption it follows that c > b or b > c. [Theorem 86a] Now, since b > 0, 2 2 if c > b then c > b if b > c then b 2 > c 2 and, since c >_0, J^> [Theorem 98c] Hence, c > b or b > c . 2 2 In either case [by Theorem 87], b fi c . Consequently, if b ^ c then b ^ c . 2 2 But, by hypothesis, b = 4 = c . Hence, b = c. There are not two num bers which are nonnegative and whose squares are 4. [9.01] [9-3] As you may recall from earlier units, we called the nonnegative number whose square is 4 the principal square root of 4 and abbreviated this to: V4 [You may also read this as 'radical 4*. ] In general, j— J (*) V >n vx = the real number z such that (z > and z = x). [Why 'x > 0' in the quantifier ? ] As in the case x = 4, in order for the descriptive phrase in (*) to make sense, two conditions must be satisfied- -an existence condition: (t,) V x > 3 z U>0 and z 2 = x) and a uniqueness condition: (t 2 ) V >Q V V z [((y > and y 2 = x) and (z > and z 2 = x)) => y = z] [Compare the instances of (t.) and (t 2 ) for x = 4 with the existence con- dition and the uniqueness condition on page 9-2. ] Again, as in the case x = 4, we need to ask whether (t x ) and (t 2 ) are satisfied. Let's begin with (t x ). We know that the instance of (t x ) for x = 4: 3 (z > and z 2 = 4) z is satisfied because we can prove that 2 > and 2 =4. And, in a simi- lar way, we can establish many other instances of (t 2 ). But, consider the instance: (1) 3 (z >0 and z 2 = 8) z — A few attempts will show you that this is not so easy to prove. [Note that it is not fair, at this point, to answer that, by definition, v8 is a nonnegative number whose square is 8. It is precisely in order to help show that 4 v8 ' makes sense that we need to prove the statement (1). ] As a matter of fact, it is not hard to show that the instance (1) of (t,) is not a consequence of the basic principles which we have adopted so far. [This is done on page 9-23. ] Consequently, the same holds of [9-4] [9.01] the existence condition (t x ). So, if we are to be able to prove theorems about principal square roots of all nonnegative numbers, we shall need a new basic principle. We might take (t^, itself: (t i } V x>0 3 z U -° and z2 = x) as such a basic principle and, in fact, this is what we did, implicitly, when we dealt informally with square roots in earlier units. However, this is a shortsighted policy. As we shall see, we shall want many theo- rems like (t,)--for example: V 3 y 3 = x, and: V ^ n 3 2 y = x x y x>0 y --and it is more economical to find a single basic principle which will allow us to prove all the theorems of this kind which we want. You will learn one such principle in section 9. 03. Now, let's consider the uniqueness condition: (t 2> V >0 V V t((y>0 and y 2 = x) and (z>0 and z 2 =x)) => y = z] Can we prove this now, or do we need still another basic principle? Fortunately, (t 2 ) is already a theorem. This should be clear from the proof, on page 9-2, of its instance for x = 4. To obtain a test-pattern for (t 2 ), all we need do is replace the *4's in this proof by, say, 'a*s. [Notice that the proof of (t 2 ) used only three theorems- -two general theorems about order and one theorem [Theorem 98c] relating to squar- ing. The other uniqueness conditions which we shall need we shall be able to prove in just the same way, using the same two theorems about order and, in each case, one theorem [like Theorem 98c] relating to an appropriate function- -for example, the cubing function and the expo- nential function with base 2. ] As remarked above, the existence condition (t ) can be proved with the help of one new basic principle, and the uniqueness condition (t 2 ) is already a theorem. Consequently [assuming that we have already adopted this new basic principle], we are entitled to speak, for any a > 0, of the real number z which is nonnegative and whose square is the number a. Rather than use such a lengthy description, we should prefer to abbre- viate it, as suggested in (*), by means of the operator *v *. However, if we are to prove theorems which contain this new operator, we need [9.01] [9-5] to have at least one basic principle in which the operator occurs. For this purpose, we shall adopt a defining principle for *v * : (* x ) V x>0 (Vx~>0 and (Vx~) 2 = x) [Compare (* x ) with (t x ).] The theorems (f x ) and (t 2 ) tell us that, for any a > 0, there is one and only one real number z such that (z > and z = a), The defining principle (*.), then, gives notice that this real number z is VT. Once we have adopted the defining principle (*,), we can use it and (t 2 ) to obtain a useful complement to (*,). This is the uniqueness theo- rem for W ' : ( *2> V x > V y [(y > and y = x) => y = Vx" ] 2 To prove (* 2 ), we proceed as follows: Suppose that b > and b = a. By (* x ), for a > 0, (VaT > and (vT) 2 = a). By (t 2 ), for a > 0, if (b > and b 2 = a) and (Va~ > and (Va") 2 = a) then b = Va". So, by the assumption and (*.), it follows that b = va. Hence, for a > 0, if b > and b = a then b = Va. 2 i — Consequently, V > n V [(y > and y = x) => y = vx ]. The generalizations (* ) and (* ) tell us all we need to know about the operator 'v *. To illustrate this, let's recall a typical proof about principal square roots- -the proof of: v x>o v y >o ^^Y = ^y For a > and b > 0, it follows from (* ) that Sa > 0, *> i-r t -y ^* (Va") = a, v'tT^ 0. and (Vb) = b. Since vT > and Vb" > 0, Sayfb>0. Moreover, (vTVb") 2 = (v / T) 2 (v / b") 2 and, so, since (vT) 2 = a and (v'b") 2 = b, (vTVtT) 2 = ab. Now, since, for a > and b > 0, ab > and since, for a > and b > 0, VaTVb" > and (vTVb") 2 = ab, it follows from (* 2 ) that vTVhT = Vab. Consequently, V > V >0 Vxvy = Vxy . In the following exercises, assume that (* x ) is either a theorem or [temporarily] a basic principle, and that, consequently, (* ) is a theorem. [9-6] [9.01] EXERCISES A. Prove. i. vT = i 2. Vo" = o 3. (VtT) = i\ B. Prove, or give a counter-example. 2. V >n 7? = x>0 1. V NX. = x X 3. V ^ n ^x x< 4. V Jx = Ixl [ Hint . V ^ A Ixj = x and V ^ rt |x| = -x] x ' ■ x>0 ' ' x<0 ' ' J C. True or false? 1. V r* 2 vx = x 2. V *^x = x x 4. V V Tx 2 * = x n n x l ww J 4n 2n 6. V V vx = x n x w / 6 10 3 5 V Nr x y = x y 8. VVMxy =xy x y 7 10. 7(1 - 2) 2 = 1 - 2 12. 7(3 - 5) 2 = |3 - 5| 14. ^9 X 4 =-3 X -2 6. 7(2 - 3) 2 = 2 - 3 18. V x<3 7(x - 3) 2 = 3 20 . V 7x 2 + 6x+9=x+3 3. V n" x. = -x 5. V V 77" = |x| n n x ' ' 7. V.V. nx y = x y x y 9. 7(2 - l) 2 = 2-1 11. 7(5 - 3) 2 = |5 - 3| 13. 7-9 X -4 = -3 X -2 fi 15. V ^(x - 3)" = x - 3 x 17 . V x>3 7(x- 3) 2 =x- 19. V 7(x - 3) 2 = |x - 3 21. 79 + 16 = 3+4 D. Practice manipulating square root expressions. Sample 1 . Simplify: "/50 + VT8 Solution . ^50 = V25 • 2 = >/25»vT = 5vT, VT8 = V ? »2 = V r ?~«vT= 3vT f V^O = vTi = 5vT + 3vT = 8vT [9.01] [9-7] Sample 2. Solution. Simplify: 3750 + 4771" - 7*7 + 748 3V50 + 4777 - VF + 748 = 1577 + 8vT- 2/i" + 477 = 137T + 12VT Simplify. 1. 777 + 77 2. 717 + 775 4. 5/108 + 2^243 - 717 + 2VT2 6. 4^24 - 4^54 + -W6" - ~VT50 2 3 4 2 8. Vl44 + Vl.44 + Vo.0144 10. V200 + 77 + V0.02 3. 7l25 - 720 5. 777 - 774 + 7l25 - V54 7. VbTT + 3V0.01 + 10V0. 001 9. ^14400 + Vl44 + 70. 000144 11. V2000 + V200 + 720 + 77 E. True or false? 1. 2 > 77 4. 0. 1 > TbTT 2. 0.01 > 7o.oi 5. 1 > 77 3. 7T > 7^" 6. 0.999 > 70.999 F. More practice manipulating square root expressions, 7l8 Sample 1 . Simplify: Solution. 717 77 77 = jm._^ = 3 Sample 2 . Simplify: 73 77 Solution . 77 7T = 73 '7 = TIT Sample 3 . Simplify: (72 ) Solution. (7T) 3 = (77j 2 77" = 277* Simplify. 771 1. 77 2. 577 77 3. 87To 77 4. n/TTTI [9-8] IT fib [9.01] ft ^ 0. V20 5. V2V5 6. 7. V8 Vo. 125 9. (VT) 3 10. (V0.01 ) 4 11. (Vo. 1 ) 4 12. (-VT) 3 13. m 14. (fr ! „ (&)-' 16. (O.OlVT) 2 17. 2 VI 18. (*r "• (wf - &r' G. Expand. Sample. (1 W| -t /4 + VI Solution. 1 - — ■f 16 + 8V2" + 2 4 9 + 4VT 2 1. (3 + V7) 2 2. (4 - VT) 2 3. (2 + VT) 3 4. u - VT) 9. (VT + 3)(VF- 4) 10. (3 - VT )(4 + V7) 11. (VT- Jz){>[5 + VT) 12. (2VT + : 3VT)(5VT - 2V2") 13. (2V5* - 4V7~)(V5~ + 2V7) 14. (VT2 + VT8)(V75 - Vi") 15. (3V7+ 2) 2 16. (5vT-2) 2 17. (VT-VT) 2 18. (x - Vy~) 2 19. (2a+VT) 2 20. (3VT - V^T) 2 H. Prove the following theorems, [Use (* L ), (* 2 )» and earlier theorems.] l - V x>0 V y>0 Vy = v^ 2 * V x>0 V y>0 7 y = y Complete. ir V - 3. V._^V.^ VxV = V r ~»v r ~ 4. V..^„V. 5 ' V x<0 V y<0 ^y x<0 y<0 y ~ ' v x<0 v y<0 ^ y " / /IT- v ^*y [9.01] [9-9] I_. The samples below suggest various techniques for transforming ex- pressions containing square root signs. Practice these techniques. Sample 1 , ^50x" Solution, n 50x = v(25x )x = 7(5x) 2 Vx" = |5x|Vx" = 5|x|Vx" For the expression W50x ' to make sense, the values of 'x' must be restricted to be nonnegative numbers. [Why?] So, the "final answer" is: 5xVx~ [x > 0] Sample 2 . ^1623 b Solution . '7l62a 7 b~ = »/(9at> ) 2 (2a) = 9|aV 3 |V2l For the expression W 162a b ' to make sense, the values of 'a' must be restricted to be nonnegative numbers, and those of 'b* must be restricted to be nonzero numbers. So, the "final answer" is: 9a 3 |b|~ 3 v / 2a [a > 0, b ^ 0] 1. V / 36y" 2. 4%l? 3. 5^ 21> 7J.V3 22< Va" 4 + vT 2 + vT Vx"- vV 23 . }^i_^ 24 . S_±J^ 25 . 3 VT+ 1 S ampl e 6. Find the rational approximation to -7= correct V3 - 1 to the nearest tenth. VF+ 1 • 1. 732+1 2.732 2732 Solution . —=z = = = = ? V3 - 1 1.732- 1 0.732 732 If you have a desk calculator available, the problem is easy to finish. But, if you wish to avoid doing "long division", it would be easier to transform the given expression as in the previous exercises: [9.01] [9-11] yT+ i _ (vT+ i)(^3"+ i) >/F- i (VT- i>(a/3~+ i) 4 4- zvT 2 = 2 + VT = 2 + 1. 732 = 3.7 Find rational approximations correct to the nearest tenth. [Use the table of square roots on page 9-365. ] VF+ 2 26. ./y 29. yf\5 - yfl yf\5 - VTo 27. 30. Vff- 2 1 + Vz v7~- VfF 28. 31. V6~- a/3~ 7F4 Wi J. Here are graphs of 'y = x' and 'y > 0' The existence and uniqueness conditions which justify our adopting the defining principle for principal square roots are: (t i } V x>0 3 z (z -° and ^ =x) (y V >n V V [ and < z >° ^d z 2 = x)) => y = z] 1. How does the figure suggest what (t,) says? 2. How does the figure suggest what (t 2 ) says? [9-12] [9.01] K. The cube root operator * v ' is defined by: (*') V "vT = the z such that z 3 = x x 1. State the existence and uniqueness conditions which justify the use of the definite description occurring in (*'): *x 3 z "• 'x V y V z (t 2 ') V„V„V„ ... 3 2. Draw a graph of 'y = x' and point out the properties of the graph that suggest what (t^/) says and what (t 2 ') says. 3. Complete. (a) vT = (b) 3 vT = (c) V^S = (d) 3/8/27 = (e) 3 vT vT = (f) y^^nz = (g) 3 V4o" = 2 3 y~ 4. State the defining principle and the uniqueness theorem for the cube root operator which bear the same relation to (*') as (* x ) and (* 2 ) on page 9-5 bear to (*) on page 9-3: (*x') V x ( ) 3 = <* 2 '> v x v y ... 5. Use (* 1 f ) and (* 2 ') in proving: V V vx" Yy ~ Txy x y JL. 1. The adoption of a descriptive definition like (*) commits one to two statements like (*.) and (* ). For example, if one were to adopt: 7 V Vx = the real number z such that (z>0 and z = x) he would be committed to: V (Vx" > and (Vx") 2 = x) So, in particular, he would be forced to accept a real number [9.01] [9-13] V^7 such that V^ > and (yPl ) Z - -1. Show that, because of a previous theorem, this would be an unfortunate position to be in. 2, Adoption, as a definition, of: V ^ n vx = the real number z such that z = x x >0 would commit one to the statement: V ^ n V [y = x => y = Vx ] x>0 y l7 J J Use the fact that 4 > 0, 2 =4, and (-2) =4 in showing that this commitment would be an unfortunate one. ■&, The two exercises of Part L point up the fact that it is desirable to be aware of one's commitments- - in particular, to be aware of what basic principles he reasons from. *v The rest of the exercises--Parts **M, N, and 0--are optional. M. Here are an existence condition and a uniqueness condition which, since they are theorems, justify the introduction of an operator [you studied this operator in Unit 7]: V 3 (z € 1 and z < x < z + 1) X z V V V [((y 6 I and y < x < y + 1) and (z e I and z y = z] 1. Let's consider how you might think of a proof for the existence condition. Given a number a, you are interested in the set of those integers which are less than or equal to a. The existence condition refers to a greatest such integer. Your problem is to prove that there is a greatest integer z such that z < a. What existence theorem about integers might be helpful in proving this? 2. The uniqueness condition can be proved by deriving it from three theorems- -one like Theorem 92, and Theorems 112and93. Do so. [9-14] [9.01] 3. What is the operator? 4. State the defining principle and the uniqueness theorem for this operator: V ( x V V [ => ] x y L J 5. The uniqueness theorem can be derived from the defining prin- ciple by slightly modifying the proof you gave in answer to Exercise 2. Make these modifications. ^6. (a) It is a trivial task to show that the existence condition is a consequence of the defining principle. Do this. (b) It is almost as trivial a task to show that the uniqueness condition is a consequence of the uniqueness theorem. Do this, A defining principle for the absolute value operator is: (*i") V x (|x| > and (|x| = x or |x| = -x)) * M. [continued] 7. State the existence and uniqueness conditions which would justify adopting the defining principle given above. 8. Using the principle for multiplying by and the 0-product theo- rem it is easy to prove: V V [(u = or v = 0) <=> uv = 0] u v l J This gives one a way of replacing alternation sentences of a cer- tain kind by sentences which do not contain the word 'or'. Use this method to restate, without using 'or', the defining principle given just before Exercise 7. [9.01] [9-15] 9. Use the restatement you gave in answer to Exercise 8 together with the uniqueness theorem (* 2 ) for square roots [page 9-5] to give a quick proof of: v x ixi = y? [What other theorem did you need to prepare the way for using (* 2 )?] N. In Part ^M, above, we gave a defining principle for the operator l | I': (*/') V x (|x| > and (|x| = x or |x| = -x)) In the following exercises and those of Part ^O you will see how this principle can be used to prove theorems about absolute values. [Most of these theorems you may already have proved, in another way, in Unit 8. Some of them are collected in Theorem 169. ] As an example of how this defining principle is used, let's prove: r ( I Theorem A . I V V [-y |x| < y] | J *1 I Suppose that -b < a < b. By the second part of the defining principle, |a| = a or |a| = -a. Suppose that |a| = a. In this case, it follows from our initial assumption that ja| < b. Sup- pose that |a| = -a. Since, from our initial assumption, -a < b [Why?], it follows, in this case also, that |aj < b. Hence, in any case, if -b < a < b then |a| < b. Consequently, V V [-y |x| |x| < y] I i i 1. Complete. Sample 1 . — 5 < a < 5 := > Solution . -5 lM^5 [Theorem B] [9-16] [9.01] Sample 2 . Solution. 3 < a < 5 3 < a < 5 <=> 3-4 (b) -2 < a < 2 => (c) 9 < a < 13 => (d) -6 < a < => (f)2 The converses of Theorems A and B are also theorems. To prove them it is useful to have a lemma: r V x -|x| < x< |x| I Let's prove this. By the first part of (*/'), |a|>0. Hence, 0<|a| + |a|. Consequently, - [a{ < |a|. Now, by the second part of t*^'), a=Jajora=-|a|. In the first case, -|a| < a = |a|. In the second case, -|a| =a < |a|. So, in either case, — |a | < a < |a|. [Explain. ] Now let's prove the converse of Theorem B: _ __ __ — _ ___ _ ( Theorem C. j V v V v [|x| < y => -y -y < x < y] i 2. Complete. Sample . | a + 3 | < 0. 5 «c=^> Solution . | a + 3 | < 0. 5 <=> -0. 5 < a + 3 < 0. 5 [Thm. D.] (a) | a | < 0. 1 <=> (b) |a + 2| < 0.05 <=^> (c) | a - 3 1 < 0.01 <=> (d) |a - 3| < d <=> (e) | a + 2 | < d <^=> (f) | a - a Q | < d <^=> (g) | 3a - 5 | < 1 <^> (h) |3a - 5| < c <=> As a consequence of the lemma for Theorem C, -(|a| + |b|)0 and(y = x or y = -x)) ^> y = |x|] x y Here is a proof of this uniqueness theorem. Suppose that b > and (b = a or b = -a). By the defining principle (*j") of Part N, either |a| = a or ja| = -a. In either case, since b = a or b = -a, it follows that (b = j a j or b = - | a | ). Suppose that b = - j a | . Since b > 0, it follows that - (a | > 0- - that is, that |a| < 0. Since, by (*/'), |a| > 0, it follows that |a| =0. Since b = -|a|, it follows that b = 0, also, and, so, that b = I a J . So, in eithe r case, b = j a j . Hence , if b > and (b = a or b = -a) then b = {a|. Consequently, V V [(y > and (y = x or y = -x)) => y = jxj]. As an example of how (* ") is used, let's prove: ! ! I Theorem G . V |-x| = |x| \ I * I According to (* "), what we need to prove is that, for any a, (j-a|>0and(|-a| =aor | -a j =-a)). That this is the case follows from (*.") and Theorem 17. 1. Use Theorems F and G to prove: r ; v v ! x y x| - |y| < |x +y| 2. Use Theorem A and the theorem proved in Exercise 1 to prove; , ! Theorem H. V V , x y i |x| - |y| < |x + y| 3. Use t*^') and (* 2 ") to prove: 1 Theorem I . V V | x | • | y | = j xy | j i ^ y ; i , [9.01] [9-19] MISCELLANEOUS EXERCISES 1. The sum of the first five terms of an arithmetic progression is one fourth the sum of the next five. What are the first two terms? 2. Solve these equations. (a) 84 + (a + 4)(a - 3)(a + 5) = (a + l)(a + 2)(a + 3) (b) y-(3y- ^1) = |(2y - 57) - f 3. If the sum of the first four terms of a geometric progression is 40 and the sum of the next four terms is 3240, find the first and second terms. (b) (27x + y) 2 - (26x - y) 2 3 3 (e) a + b + a + b 4. Factor. 2 (a) acx - bcx + adx - bd (c) (x+y) 4 - 1 (d) (x + y) m 2 2 -l (f) x - y + x - y (g) k 2 - k (h) t 2 - t (i) a 3 + 27b 3 (j) x 3 - 64y 3 5. The measures of the sides of a triangle are 13, 14, and 15. Use the formula: K = Vs(s - a)(s - b)(s - c) to find the area- measure of the triangle, and then find the sine ratios for the three angles. 6. Suppose that A, B, and C are three points on a circle of diameter d such that ZBAC is acute. Show that the sine ratio for ZBAC is BC/d. 3/— 4/— 7. Find a root of the equation: vx = 10 vx 8. Suppose that I is a line and Pisa point not on L If A , A , A , ...» A n are n points on i, what is the total number of triangles with P as one vertex and two of the points A , A_, A , . . . , A^ as the other vertices ? [9-20] [9.01] 9. Derive a formula for the area-measure A of a regular hexagon in terms of its perimeter P. 10. Five boys hire a ping-pong table for 1 hour. Only 2 boys play at a time. If each boy is to play the same amount of time, how many minutes can each play? 11. 12. Simplify. 1 (a) £ - £- s s - t Given : BD | | CE, m(ZB) = 140, m(ZC) = 150 Find: m(ZA) (b) s + 4 s + 1 13. If a man owns 2/3 of a piece of property and sells 3/5 of his holding for $2400, what is the value of the entire property? 2 14. Find a substitution for 'k' such that the equation 'x - 4x + 2k = 0' has two rational roots. 15. One kind of steel alloy is 0. 19% carbon. How much carbon is con- tained in 1730 pounds of this steel? 16. A regular pentagon has center (0, 0) and one vertex (3, 0). What are the coordinates of the other vertices? 17. Solve: (a) 2xV0. 05 = 5 (b) 1/x = V0. 16 18. Find the measure of an altitude of an equilateral triangle whose side-measure is 40. 19. Find three consecutive integers such that the product of the two smallest differs from the product of the two largest by the sum of the largest and the smallest. [9.01] [9-21] 20. Factor. (a) x + xy + xz + yz 2 2 (b) sx + tx + 2s + 2t 21. If a ball is thrown upwards with a velocity of v feet per second, the height reached at the end of t seconds is s feet, where s = vt - I6t . If the starting velocity is 100 feet per second, how long will it take the ball to reach its greatest height? 22. Simplify. (a) (b) , 2n . , 3n , _ 4n 6 x + ox - 1 2x n 3x 1+m l+2m l+3m , 1 + 4m X - x - X + X 3m 23. Suppose that the difference in area- measures of the circumscribed and inscribed squares of a circle is 72. What is the area-measure of the circle ? 24. A magic square is a table of numbers having as many rows as col- umns and such that the numbers listed in each column, each row, and each diagonal have the same sunn. Examples: 8 1 6 3 5 7 4 9 2 16 3 2 13 5 10 11 8 9 6 7 12 4 15 14 1 3- square 4- square What is the constant sum for a 50- square which lists the first 2500 positive integers? 25. How long is a circular arc whose chord is 16 feet long and whose "height" is 6 feet? [9-22] [9.02] 9. 02 The need for a new basic principle . --On page 9-3 we stated that the existence condition: (t x ) V x>0 a z U -° and * 2 = X) ' which is needed in order to justify our introduction of the square root operator, is not a consequence of our basic principles. In this section we shall substantiate this statement. The procedure will be to show, first, that all our basic principles continue to hold if they are reinter- preted as statements about only the rational real numbers, and, second, that (tj^) becomes false when it is interpreted as referring only to ra- tional real numbers. THE SYSTEM OF RATIONAL REAL NUMBERS Recall that, by definition, a real number is rational if and only if it is the quotient of an integer by a nonzero integer. For example, 2/9, 1, /TOO, 0, -3.125, 0.83, and 4%/Jl are rational real numbers. Be sure you understand why this is so [for example, vlOO = 10/1, and 10 and 1 are integers]. Then, give some other examples of rational numbers. If r and s are rational numbers then there are integers- - say, i, j, k, and I [with j / 0, and 1 / 0]--such that It follows that — and s = -T-. J * (1) r + s = li * kj . [Theorem 57] J* Since i, j, k, and SL are integers, ii + kj and ji are also integers [Theo- rems 1 lOd and 1 10b] and, since j / and I fi 0, ji fi [Theorem 55]. Hence, by (1) and definition, it follows that r + s is rational. Conse- quently, for each rational number r and each rational number s, r + s is a rational number. This result can be restated as: The set of rational real numbers is closed with respect to addition. Once you have understood this statement and its proof, you can restate the proof more briefly: [9.02] [9-23] Since the set of integers is closed with respect to multi- plication and addition, and since the set of nonzero real numbers is closed with respect to multiplication, it follows from Theorem 57 [and the definition of 'rational number'] that the set of rational numbers is closed with respect to addition. In just the same way you can prove that the set of rational real num- bers is closed with respect to the operations of multiplication, opposition, subtraction, and division [by a nonzero rational number]. These results on closure show that we can consider addition, multi- plication, opposition, subtraction, and division as operations defined on the set consisting of just the rational real numbers. Since, also, and 1 are rational, it is easy to see that our first 11 basic principles become true statements about rational numbers if we think of '+', ' • \ ' — *, l - \ and '-f' as referring to these restricted operations on rational real num- bers, and read l V x *» say, as 'for each rational number x'. The same holds for the basic principles (P 1 ) - (P 4 ) and (G) if we take 4 P' to be a name for the set of positive rational numbers and '>' as re- ferring to the greater-than relation for rational numbers. Since integers are rational numbers, the same holds of (I + ,) - U + 3 )' (C)» an< ^ (0« As a consequence of the preceding remarks every theorem which we could derive from our basic principles is, when reinterpreted as above, a true statement about the system of rational numbers. A person looking at our basic principles could not tell whether we are talking about the real numbers or only the rational numbers. Now, let's consider the statement (t,): V .„3 (z > and z Z = x) x>0 z Can this be derived from our basic principles? If it could, we would be able to show, since 8, say, is a nonnegative rational number, that there is a rational number z such that z > and z = 8. But, on page 4-48 of Unit 4, we proved that there is no such rational number. [in the proof we used only theorems which follow from our present set of basic prin- ciples.] A person who is told that (t 1 )is a basic principle [or, as it later will be, a theorem] knows that we are not talking about rational numbers. Consequently, (1^) is not a consequence of our basic principles, and we do need some additional basic principle to justify our dealing with square roots. [9-24] [9.02] EXPLORATION EXERCISES A. 1. Consider the set, N, of negative numbers. (a) Is there a number which is greater than or equal to each member of N? (b) Is there more than one number z such that, for each x € N, z > x? (c) Is there a z € N such that V [x € N => z > x]? 2. Repeat Exercise 1 for the set, I + , of positive integers. 3. Consider the set, N v^ {0}, of nonpositive numbers. (a) Is there a z such that V [x € N w {0}=> z > x]? (b) Is there more than one such z? (c) Is there a z £ N w {0 } such that V [x£Nw{0}=>z> x]? (d) Is there more than one such z? 4. (a) Is there a z such that, for each x, if x belongs to the set of positive rationals less than v2 then z > x? Is there more than one such z? (b) Is there a z in the set of positive rationals less than v2 such that, for each x, if x also belongs to this set then z > x? Is there more than one such z? 5. Consider the set M, where M = {y: -5 < y < —4 or y = or 1 < y < 2}. (a) Is there a z such that V [x £ M ^> z > x] ? I« there more than one such z? (b) Is there a z € M such that V [x £ M => z > x]? Is there more than one such z? [9.02] [9-25] o> The foregoing exercises deal with the ideas oi upper bound and greatest member . Here are definitions: Vc V [z is an upper bound of S <=> V, [x e S ^=> z > x]] VeV [z is a greatest member of S <^~> (z is an upper bound of S and z £ S)] -> y^ upper bounds of S The set S pictured above has upper bounds but no greatest member. What would be the easiest way to change S so that it would have a great est member ? 6. (a) Give two upper bounds for {t: 3t + 1 < 10 }. Does this set have a greatest member? (b) Give two upper bounds for {t: 3t + 1 < 10}. Does this set have a greatest member ? 7. Let R be the set of reciprocals of the positive integers. (a) Does R have an upper bound? (b) Does R have a greatest member? 8. Suppose that t is an upper bound of some given set S. What other numbers are sure to be upper bounds of S? Justify your answer. [9-26] [9.02] 9. Suppose that, for some given set S, S has a greatest member. Then, (t.) 3 (V [x e S => z > x] and z e S). (a) Check this by referring to the definitions. (b) State the conditions which says that S cannot have two greatest members: (t 2> V y V z [((V x [x e S => y > x] and y € S) and ( and )) => = ] (c) Prove the uniqueness condition (t 2 ). [ Hint . See Theorem 93.] (d) Fill in the blanks. Since, no matter what set of real numbers S is, the ? condition (t 2 ) is a theorem, it follows that, for any set S such that the ? con- dition (t,) is a theorem, we are justified in speaking of ? z such that (V [x e S => z > x] and z e S) --that is, of ? ? member of S. 10. Complete these definitions of lower bound and least member , (a) V Q V [z is a lower bound of S <=> V [x € S =5> z x]] (b) Vc V [z is a least member of S <= : > o 2> and )] 11. Fill in. (a) Just as in Exercise 9, one can prove that no set can have two least members. Hence, if a set has a least member, one is justified in calling it ? ? ? of S. (b) Suppose that U is the set of upper bounds of some set S. If U has a least member then this member of U is called ? ? upper bound of ? . [9.02] [9-27] 12. Fill in. (a) If U is the set of upper bounds of some set S then U is either a right half-line or a right ? or the whole ? . (b) If U is the set of upper bounds of some set S then each member of S is a ? bound of U. (c) Suppose that U is the set of upper bounds of some set S. If U has a least member then this member of U is called ? ? upper bound of ? . 13. Name a set which has an upper bound but does not have a least upper bound. [ Hint . There is only one such set. ] 14. Give an example of a set of real numbers which has (a) both an upper bound and a lower bound (b) a lower bound but no upper bound (c) an upper bound but no lower bound (d) neither an upper nor a lower bound (e) a greatest member but no upper bound (f) an upper bound which belongs to it (g) a least upper bound which does not belong to it (h) a lower bound but not a greatest lower bound 15. Which of the following sets has 3 as an upper bound? (a) {x: x < 3} (b) {x > 2: x < 3} (c) (x>3: x< 3} (d) {x>4: x< 3} (e) {x > 3: x < 3} (f) {x: x < 3} 16. "Which of the sets listed in Exercise 15 has 3 as its least upper bound? [9-28] [9.02] B. In Unit 5 you learned to picture a function [i. e. a mapping] by pic- turing its domain and range and drawing arrows. For Example, the adding 1 function "looks" like this: -3-2-10 1 2 3 4 ft 1. Make a similar picture of the function h such that h(x) = Z{ ** V , for x >0. (a) At first consider only the arguments 0, 0. 5, 1. 1, and 1. 2. [Carry out your computations correct to two decimal places, for if you don't, you may miss the point. ] fa (b) Your work in part (a) may suggest that, for any a > 0, h(a) > a. How does this show up in your picture? (c) To check the suggestion of part (b), here is a table listing more values of the function [correct to four decimal places]. h(a) 1. 3 1. 3939 1. 4 1.4118 1. 41 1. 4135 1. 414 1. 4141 Do these values tend to confirm the suggestion that, for any a >0, h(a) > a? [9.02] [9-29] (d) You can check the conjecture [that, for any a > 0, h(a) > a] more completely by carrying out the following manipulations: , . . 2 (a + 1) ? h(a) - a = — 2 — r-^r- - a = a + 2 a + 2 Do these manipulations, obtaining as simple a result as you can, and infer a theorem: 9 V ^ n h(x) - x = — 7—5- x>0 x + 2 (e) Does the theorem you proved in part (d) verify the conjecture? If not, what less sweeping conjecture does it verify? 2. You have seen in Exercise 1 that (* x ) V x>Q [x 2 < 2 =>h(x) > x] andV x>Q [x 2 > 2 -> h(x) + x]. [Complete this last. ] (a) Check this second generalization by drawing more arrows in the picture you drew for Exercise 1(a). Use arguments whose squares are greater than 2, such as those listed in the following table. a 1. 5 1. 7 2 3 h(a) 1. 428 1.459 1. 5 1. 6 (b) A more careful look at the figure you made in answering Exer- cise 1(a) may suggest another conjecture. Notice that the arrows corresponding to arguments of h whose squares are very close to 2 are very nearly vertical. This means that the value of the function, for such an argument, differs very little from the argument itself. This might suggest that (* 2 ) V x > [x 2 < 2^>[h(x)] 2 < 2]and V x ^ Q [x 2 >2 ^[h(x)] 2 > 2]. You can check these conjectures much as you did the conjecture in Exercise 1(d) by considering the expression: 2 - [h(a)] 2 Do so, thus proving the conjectures. [9-30] 19.03] C. Consider the set S such that S={x>0: x < Z}. 1. Show that S has no greatest member. [ Hint . You can do this by showing that, for any number in S, there is a larger number which is also in S. See the first generalizations in (-i^) and (4> 2 ) of Exercise 2 of Part B. ] 2. Suppose that b is a nonnegative number whose square is 2. Prove : (a) b is an upper bound of S. [ Hint . Use Theorem 98b. ] (b) V ^~ [x < b => xeS] [Hint. Use Theorem 98c. ] (c) No number less than b is an upper bound of S. [ Hint . Use Exercises 1 and 2(b).] (d) b is the least upper bound of S. 9.0 3 The least upper bound principle . - - In Part C of the preceding exer- cises you considered the set S such that S ={x > 0: x < Z) . You proved that if there is a nonnegative number whose square is 2 then S has a least upper bound. You did this by showing that any nonnegative number whose square is 2 is a least upper bound of S. Let's consider the converse problem. Is it the case that if S has a least upper bound then there is a nonnegative number whose square is 2? As we shall see, the answer is 'yes'. We shall prove this, as you might suspect, by showing that any least upper bound of S is a nonnegative num- ber whose square is 2. In doing so, we shall use the function h which you studied in Part B on pages 9-28 and 9-29. Recall that h(x) = * + 2^ ' forx ^°« and that you proved the generalizations: [9.03] [9-31] (*J V x>Q [x 2 < 2 =>h(x) > x] and V x > Q [x 2 > 2 => h(x) < x] U> 2 > V x > Q [x 2 < 2->[h(x)] 2 <2] and V x > [x 2 >2=>[h(x)] 2 >2] Suppose, then, that b is a least upper bound of S. We wish to show that b is a nonnegative number whose square is 2. That b > is obvious since c S and b is an upper bound of S. To show that b = 2 we shall 2 2 show that b ^ 2 and b f 2. Then, it will follow [by Theorem 86a] that 2 2 So, suppose, first, that b < 2. Since b > and b < 2, it follows, from (4^), that h(b) > b and, from ( 2 ), that [h(b)J < 2. So, since h(b) > 0, it follows that h(b) is a member of S which is greater than b. But, this is impossible because b is an upper bound of S. Hence, b f* 2. 2 2 Suppose, now, that b > 2. Since b > and b > 2, it follows, from (4> ), that h(b) < band, from (4> ), that [h(b)] > 2. Also, since b > 0, it 2 follows from the definition of h that h(b) > 0. Now, since [h(b)] > 2, [h(b)] is greater than the square of any member of S. So, [by Theorem 98b], since h{b) > 0, h(b) is an upper bound of S. But, since h(b) < b, and since b is the least upper bound of S, this is impossible. Hence, b 2 ^2. 2 2 2 Since b ^2 and b ~f> 2, it follows that b =2. Combining what we have just proved with what you proved in Part C, we see that the statements: 3 >n ■* = 2 z >0 and: {x>0: x <2} has a least upper bound are equivalent statements. More generally, the statements; and: (1) V y>0 3 z>0 z2= y (2) V >n (x>0: x 0 ^ valent? Explain. ] [Are '3 >fi z =0' and ' {x > 0: x < 0} has a least upper bound* equi- [9-32] [9.03] As was pointed out on page 9-4, our work with square roots could be justified by adopting one new basic principle: (3) V y>0 3 z >0 Z =Y [See (t 1 ) on page 9-22.] Since we can already prove that =0, we could also get by [as far as square roots are concerned] by adopting (1) on page 9-31 as a new basic principle. And, since (1) and (2) are equivalent, we could, instead, adopt (2). Now, by itself, (2) seems to have no advantage over (1). How- ever, we also wish to be able to deal with cube roots [see Part K on page 9-12] and, as we shall see, with the inverses of many other functions be- 2 3 sides those of {(x, y), x > 0: y = x } and {(x, y): y = x }. To do so we could, of course, adopt a basic principle like (1) for each such function. This would, obviously, be inelegant. The advantage of (2) is that, unlike (1), it suggests a basic principle which, as we shall see, covers all the cases we shall wish to consider. This new basic principle- -the least upper bound principle [iubp]--is: Each nonempty set which has an upper bound has a least upper bound. With this additional principle, it is easy to see that (2), for example, is a theorem. [Just show that, for b > 0, (x > : x < b} is nonempty and that it has an upper bound. ] Consequently, (3) is also a theorem, and our work with square roots rests on a firm foundation. EXERCISE Use the iubp to show that I + does not have an upper bound. [ Hint . Since 1 € I + , I + is nonempty. So, if I* has an upper bound then, by the iubp, I + has a least upper bound--say, b. Since b - 1 < b, it follows that b - 1 is not an upper bound of I + --that is, it follows that there is a positive integer p such that ... . [Complete the proof by using (l* 2 ) to obtain a contradiction. ] v»«. o* o, "i x *T» 'i v [9.03] [9-33] In Unit 7 we adopted the cofinality principle: (C) V 3 n > x x n as a basic principle. Now what (C) says is just that it is not the case that there is a number x such that each positive integer is less than or equal to x [Explain. ]. That is, (C) says that I* does not have an upper bound. Since, as you have seen, this follows from the iubp, (T ), (I*) and some [three] earlier theorems which do not depend on (C), we need no longer take (C) as a basic principle. With the adoption of the iubp, (C) has become a theorem. Moreover, it is possible to show that [except for definitions, such as (G) and (I), and recursive definitions] our first eleven basic prin- ciples together with (P x ) - (P 4 ), (1^) - (l\)t and the iubp, form a complete set of basic principles for the system of real numbers. No other basic principles will ever be needed [except definitions and defining principles like (* x ) on page 9-5]. In the Exploration Exercises beginning on page 9-34, we shall de- velop some general concepts one of which will give a new insight into the role of the iubp, and we shall state but not prove a theorem about inverse functions which will make it easy to settle such questions as the existence of square roots, cube roots, etc. [in your later study of mathematics, there will be other questions of this kind, and the theorem just referred to will enable you to settle each of them very easily. ] v^ 0- Oj. 'l" *T» 'C- For a more complete treatment of the topics covered in the following Exploration Exercises and in the intro- ductory paragraph of section 9.04, you may wish to study Appendix A [pages 9- 190 through 9-230]. If you do this, you should take up the text again with the subsection RADICAL EXPRESSIONS on page 9-50. [9-34] [9.03] EXPLORATION EXERCISES A. Here are graphs of several functions, each having the segment 0, 6 as domain. 1. Which of these functions do not have inverses? [Recall that a function f has an inverse if and only if [f( Xl ) = i(K^) =r> X x = x 2 J. ] V V Suppose that you pick two arguments of a function. If it turns out that, however you do this, the function value for the larger argument is larger than the value for the smaller argument, then the function is said [9.03] [9-35] to be increasing . In other words, a function f is increasing if and only if 2. Which of the nine functions pictured above are increasing? 3. Record your answers to Exercises 1 and 2 in the following table, f i * 2 f 3 f 4 f 5 f 6 f T f 8 f 9 No inverse v 7 Increasing n/ Decreasing 4. Complete: A function f is decreasing if and only if V V fx > x => V X X € ^ f V X 2 e k £ L 2 i 5. Complete the third line of the table in Exercise 3. 6. If you have done the foregoing exercises correctly, you will find that no column in the table of Exercise 3 contains two checkmarks. This illustrates the fact that [pick the right words] r e an inverse if a function { , } hav< v does not J then the function iss ... } increasing! /decreasing ^neither^ B Vnor-/ 6 7. If you did Exercise 6 correctly, you probably discovered that if a function is either increasing or decreasing then the function has an inverse. Which of the functions pictured on page 9-34 is a counterexample to the converse of this statement? [9-36] [9.03] 8. Sketch a graph of an increasing function, and on the same chart, sketch a graph of its inverse. Is the inverse increasing, de- creasing, or neither? 9. Is there an increasing function whose inverse is not increasing? 10. (a) Consider a function that has an inverse. Do any of its sub- sets have inverses? (b) Consider a function that does not have an inverse. Do any of its subsets have inverses? (c) Consider a function that is decreasing. Are any of its sub- sets increasing? B. For each n, the function {(x, y): y = x } is called the nth power func- tion . [The function {(x, y), x > : y = x } is the nth power function restricted to nonnegative arguments.] 1. Sketch, on separate charts, graphs of the 1st, 2nd, 3rd, and 4th power functions for arguments between -4 and 4. [You may find it convenient to use different scales for the horizontal and verti- cal axes. ] 2. Sketch, on separate charts, graphs of the converses of the first four power functions. 3. Which of the first four power functions have inverses? Which of all the positive -integral power functions have inverses? 4. Which of the positive-integral power functions are increasing? 5. Do your answers to Exercises 3 and 4 support your finding in Exercise 6 of Part A? 6. Consider the positive-integral power functions restricted to non- negative arguments. (a) Which are increasing? (b) Which have increasing inverses? [9.03] [9-37] In Part A of the preceding exercises you discovered that any func- tion which is increasing or decreasing has an inverse of the same type. It is convenient to introduce a word to cover both kinds of functions: Definition . A function is monotonic if and only if it is either increasing or decreasing. So, the theorem you discovered is: Theorem 184 . Each monotonic function has a monotonic inverse of the same type. In Part B you discovered that the positive -integral power functions restricted to nonnegative arguments are increasing- -that is: You also discovered that the odd positive-integral power functions are increasing [all the way!]- -that is: Theorem 185'. V V V [x„ > x."=> x n x, x„ l 2 i _^ 2n - 1 . 2n - 1 1 2 Applying Theorem 184 to the functions mentioned in Theorems 185 and 185' we see that (a) each restricted positive -integral power function has an increasing inverse, and (b) each odd positive -integral power function has an in- creasing inverse. [Proofs of Theorems 184 and 185 are given in Appendix A. Theorem 185' is discussed on page 9-54. ] [9-38] [9.03] s'.. «&» %.K 'I v "| N -f C. Here are graphs of several increasing functions each of which has • — • some subset of the segment 0, 6 as its domain. 1. The domain of all but one of the three functions pictured above is the entire segment 0, 6. Which one? 2. Look at the graph of g . Since g x is an increasing function, its • * range is bound to be a subset of the segment g x (0), g 1 (6). Explain. 3. From the graph, the range of g x is the entire segment g x (0), g 1 (6). For which of the other two functions can you make similar state- ments ? 4. Record the results you obtained in Exercises 1 and 3 in the fol lowing table. n 1 2 3 fc>n B n y D. The results recorded in the table of Exercise 4 of Part C might sug- gest that, for any increasing function g, the range of g is an entire segment if and only if the domain of g is an entire segment. Let's investigate this conjecture by considering some additional functions [9.03] [9-39] whose domain is the entire segment 0, 6. gjx) = x/2, < x < 4 x-1, 4 < x < 6 g.tx) = f x/2, < x < 4 Lx- 1, 4 < x < 6 1. (a) What is gJO)? g<6)? • • (b) The shortest segment which contains the range of g is ?, (c) What is the shortest segment which contains ft ? ft ? 2. (a) Is 3.98 an argument of g 4 ? If so, what is g 4 (3. 98)? (b) Is 4 an argument of g 4 ? If so, what is g 4 (4)? (c) Is 4. 1 an argument of g ? If so, what is g (4. 01)? 3. (a) Is 2 a value of g --that is, does there exist an x € ^ such that g 4 (x) = 2 g. (b) True or false? xe£ (1) 3._ g 4 (x) = 2.01 (2) & = gJO), g.(6) g. g 4. Complete. (a) ft = {y: < y g 4 (b) &g 6 = {y: (c) ft g = {y: *>5 2 or 3 y < 5} } ['<' or »<•] or or or [9-40] [9.03] J* O^ v'>. In Exercise 3 of Part D you found that, although g is an increasing function whose domain is the entire segment 0, 6, the range of g is not the entire segment g 4 (0), g 4 (6). So, the conjecture that the range of an increasing function is a segment if and only if its domain is a segment turns out to be false. [The conjecture has two parts--an if-part and an only if-part. For which part is g a counterexample?] We wish to find a sufficient condition that an increasing function whose domain is a segment have a segment as its range. To do so, let's see how g [and g and g, ] differ from the functions g x , g , and g which led us to our false conjecture. Your answers for Exercise 2 of Part D may suggest such a difference. You will explore this further in Part E. o, o^ ».•„ T" 'r 'i- E. 1. Complete the following table for g . X 3.9 3.98 3.998 4 4.01 4.001 4.0001 4.00001 g 4 <*> 1.99 2 3.01 |g 4 U)-g 4 (4)| 0.01 1.01 2. (a) Complete. You can find an x e £„ as close as you wish to the 64 . . argument 4 and such that |g 4 (x) - g 4 (4)| > (b) True or false? There is a number c > such that no matter how close you come to the argument 4 of g you can find an x e £ g such that |g 4 (x) - g 4 (4) | > c. 3. Look at the graph of the function g x of Part C on page 9-38. The function g. is defined by: f x- 1, 0 - gi< 4 H < o.oi. (b) No matter what xe^ g you choose, as long as |x - 4J < o 1 you can be sure that |g x (x) - g x (4)| < 0.002. (c) No matter what x e b you choose, as long as |x - 4| < you can be sure that |g x (x) - g 1 (4)| < 0.0002. (a) Explain. No matter what number c > you choose [however small], if x is any argument of g 1 which is sufficiently close to 4 then | gl (x) - g x (4)| < c. (b) In part (a), how close need the argument x be to 4? 7. True or false? (a) Choose any x Q e £• gi' No matter what number c > you choose, if x is any argument of g 1 which is sufficiently close to x then |g x (x) - gl (x )| 0. You can't find arguments x i of g x arbitrarily close to x for which Ig^x) - g (x ) j > c. O^ v>, o^. 'i- 'i- T* [9-42] [9.03] Each of the functions g L and g is an increasing function whose do- main is a segment- -the segment 0, 6. The range of g x is the entire segment g^Oh g x (6), but the range of g is only a part of the segment g (0), g 4 (6). The reason for this difference between g and g x is indi- cated in the exercises of Part E [Exercises 2 and 7}. In Exercise 2 you found that there are arbitrarily small changes in the argument of g which result in a change of at least 1 in the value of In Exercise 7 you found that this was not the case with g , You can be sure that the value of g x will change as little as you wish, if you take care to change its argument little enough- -a small change in the argu- ment of g x will result in a small change in the value of g x and, indeed, sufficiently small changes in the argument will result in arbitrarily small changes in the value. Functions which share this property with g x are said to be continu - ous at each argument. In contrast, g is dis continuous at its argument 4, Definition . A function f is continuous at x_ if and only if x Q e ^r and f(x) differs arbitrarily little from f(x Q ) for each x e ,5y which is sufficiently close to x Q . A function is continuous if and only if it is con- tinuous at each of its arguments. [A more formal definition is given on page 9-211, in Appendix A, pre ceded by additional Exploration Exercises.] F. 1. Look at the graphs of the functions f x through f of Part A on page 9-34. Which of these functions do you think are continuous? 2. (a) Which of the positive-integral power functions of Part B [page 9*36] do you think are continuous? (b) Which of the positive-integral power functions restricted to nonnegative arguments do you think are continuous? [9.03] [9-43] 3. (a) Suppose that a function f is continuous at one of its arguments x and suppose that f is a subset of f which also has x Q as an argument. Is f also continuous at x Q ? [ Hint . To answer this you need to consider the values of f Q at x and at nearby arguments of f . Think of f Q as what is left when you remove some ordered pairs from f. ] (b) True or false? Each subset of a continuous function is continuous. (c) Is the function g of Part C on page 9-38 continuous? G. 1. Complete the following table, referring to the increasing func tions of Parts C and D on pages 9-38 and 9-39. n 1 2 3 4 5 6 y ftg n ^n<°>*en< 6 > dp = 076 y V g is continuous 5 n y 2. If you have done Exercise 1 correctly, you will find that each column contains exactly two checkmarks. In particular, if some column has a checkmark in the middle row then the column con- tains just one other checkmark. This illustrates the fact that [pick the right words] if g is an increasing function whose domain . a segment a, b then &„ / g(a), g(b) if and only if g is continuous. 3. Complete: if g is a monotonic function whose domain is a seg- ment a, b then & g if and only if g is 4. (a) One of the functions of Part A is a discontinuous function whose domain and range are both entire segments,. Which one? (b) Can you sketch a graph of an increasing discontinuous func- tion whose domain and range are both entire segments? [9-44] [9.03] sly v»^ x»-« <•,> ^,v ^,V In Part E of the foregoing exercises you discovered the difference between continuous and discontinuous functions. In Part F you probably guessed that the positive-integral power functions are continuous --for any n, if | x - Xj_ | is small enough then | x 2 - x x | will be as small as you wish. This guess is correct: Theorem 186. Each positive-integral power function is continuous. [Theorem 186 is proved in Appendix A.] Since [Exercise 3 of Part F] each subset of a continuous function is continuous, it follows that each of the positive-integral power functions restricted to nonnegative argu- ments is continuous. You discovered in Part G that, for any monotonic function g whose domain is a segment a, b, (*) & e = g(a), g(b) if and only if g is continuous. Theorems 185, 186, and the if-part of (*) can be used to give an- other proof [besides the one given in section 9. 03] that there is a non- negative number whose square is 2. Briefly, by Theorems 185 and 186, the function ? {(x, y), x > 0: y = x } is an increasing continuous function. So, the same is true of its subset {(x, y), < x < 3: y = x 2 } . This latter function has the additional property that its domain is the segment 0, 3. So, by the if-part of (*), the range of this subset is the • . segment , 3 . So, since _< 2 < 9, 2 belongs to the range of this func- tion- -that is, there is an x such that < x < 3 and 2 = x . As you know from section 9.02, this result must depend, somehow, on the least upper bound principle. In fact, the proof of the if-part of (*) which is given in Appendix A is very similar to the proof in section 9.03 that the least upper bound of {x >0: x < 2} is a nonnegative num- ber whose square is 2. The iubp is used in the same way both in this [9.03] [9-45] proof and in the proof of the if-part of (*). Using Theorem 184 and both parts of (*) one can prove a fundamen- tal theorem about inverses which we shall need at several critical points in this and later units. The theorem is: Theorem 187 . Each continuous monotonic function f whose domain is a segment a, b has a continuous mono- tonic inverse of the same type whose domain is the segment f(a), f(b). The proof, also given in Appendix A, is very simple. The only tricky part is using the only if-part of (*) to show that the inverse of f is con- tinuous. Theorems 185, 186. and 187 can be used, just as we used Theorems 185, 186, and the if-part of (*) to prove that there is a non-» negative number whose square is 2--or, for that matter, to prove that, given any nonnegative number, there is a nonnegative number whose square is the given number. MISCELLANEOUS EXERCISES 1. Solve these systems. r a + 2b + 2c = 11 (a) I 2a + b + c = 7 3a + 4b + c = 14 x + 2y - 3z -1 (b) ( 5x + 4y + 6z -1 v. 7x~ l - By' 1 + 9z~* 1 24 14 2 2 4 4 3 3 2 2. Prove: V V (x + y ) (x + y ) x y 3. Factor. (a) (a + b) 2 - 4c 2 (c) 1 - (3a - 5b) 2 (e) t - s 2sr - r 2 Z Z 2 (g) y - 4by + 4b - a + 2ax - x U) 1 + d 3 (b) t 2 - (5p - 3q) 2 (d) (a - 7b + c) 2 - (7b - c) 2 2 2 2 (f) k - m + 2mx - x (h) t J - s J (j) 8t 3 + 27s 3 [9-46] [9.03] 4. Suppose that 5 less than twice a number is greater than 25, and 7 less than three times the number is less than 13 more than twice the number, If the number is prime, what is it? 5. Bill doesn't have enough money to buy a $14 baseball glove. But, if he borrows a third as much money as he now has, he could buy the glove and have more money left than he now lacks. How much money does he now have? 6. £ Hypothesis : ACDM and MEFN are squares, AM = MB, AGHN is a rectangle, NB = NH Conclusion : K(E)ACDM) = K(SMEFN) + K(EEOAGHN) 7. Prove. (a) V V (x 2 + xy + y 2 ) 2 - (x 2 - xy + y ) = 4xy(x + y ) x y ~> ? 2 2 (b) V V, V V [(ax + by) + (ay - bx) ][(ax + by) - (ay + bx) ] cL D x y .4 ,4,. 4 4 = (a - b )(x - y ) 8. Consider the four triangles into which the diagonals of a convex quadrilateral divide it. Prove that the centroids of these triangles are the vertices of a parallelogram. 9. Factor. (a) oax - 6a x (c) a + ab + ac + be (e) p - p + p - 1 (g) r 2 + 54r + 729 (i) a 2 - 32ab - 105b 2 (b) 25 + 15y (d) x 2 + 3x + xy + 3y (f) x 2 + 3x + 2 2 , 2 (h) t + 5tn + 6n (j) m 2 + 16m - 260 [9.03] [9-47] 10. Derive a formula for the number, C, of square feet of carpeting needed for a rectangular room p feet by q feet if a border r inches wide is to be left between the edge of the carpet and the walls. 11. Consider two circular cylindrical jars of water, the first of which has diameter t times that of the second. If p marbles of the same size are dropped in the first jar, the water level rises a distance a; if q marbles of another size are dropped in the second jar, the water level rises a distance b. Compute the ratio of the diameter of the marbles dropped in the first jar to that of the marbles dropped in the second jar. 12. Simplify, (a) 9x + 6x - 1 1 - 3x + 1 (b) 6x 13x + 5x + 3 3x - 2x - 1 13. If XY = 3(X - 3) and YZ = 3(Y - 3), show that ZX = 3( Z - 3). 14. Solve these equations. (a) 3(x- 7) + 5(x- 4) = 15 (b) 3(3 + x) - 3(2x - 5) = 6 - x - 2{3 - x) 15. If one side of a rectangle is 3 times as long as an adjacent side, how many times as long as the shorter side is the diagonal? 16. Suppose that a, b, and c are sequences such that a =120+2. 5n, b = 90 + 4n, and c =85 + 7. 5n. Find the smallest number p such n n that, for all q>p, a x 2 x + 2xy + y (b> 2ab a + b 2 K 2 a - b [9-48] [9.04] 9. 04 Principal roots . --We have seen that the justification of our work with the principal square root operator lies in two theorems about the squaring function . (t x ) V x > E! z U>0and Z 2 = x) and: (t 2> V x>0 V V z ^ y - ° and y2 = x) and (z - ° and z 2 = x)) => y = z] The first of these, (t,)> can be abbreviated to: (1) V y > 3 x > x 2 = y [V for 'x'; *x' for '*•] and the second, (t 2 ), is equivalent to: (2) V > Q V x > Q [x/ = x/ => x x = x 2 ] [*x x ' for 'y'; 'x 2 ' for ' 2 '] Looking at (1) and (2), we see [x > 0, * x > 0, x 2 > 0] that these are really statements about the squaring function restricted to nonnegative arguments . What they state is that [(!)] the range of this function con- tains all nonnegative real numbers and [(2)] this function has an inverse. Since all squares of nonnegative numbers are nonnegative, we may con- clude that the restricted squaring function has an inverse whose domain is precisely the set of nonnegative numbers, You have seen in section 9. 02 that the proof of (1) requires a new basic principle- -the least upper bound principle- -and you have seen at the end of the preceding Exploration Exercises how (1) can be derived using Theorems 185, 186, and 187, and that the i?ubp comes in in proving Theorem 187. Earlier in the Exploration Exercises you have seen that (2) follows from Theorems 184 and 185. Having proved (1) and (2), we are justified in introducing the opera- tor l v ' to refer to the inverse of the restricted squaring function, and we do introduce it by adopting the defining principle: i* L ) V x>0 (^x">0 and (Vx~) 2 = x) [See page 9-5. ] From (* x ) and (2) we have the uniqueness theorem: (* 2 ) V x > Q V y [(y > and y 2 = x) ^> y - vT ] [See page 9-5. ] The inverse, {(x, y), x > 0: y = vx }, of the restricted squaring func- tion is called the principal square root function. From Theorems 184 and 185 it follows that the principal square root function is increasing. [9.04] [9-49] Using Theorems 185, 186, and 187 it can be proved that the principal square root function is continuous. All that we have just said about the principal square root function was based on two very general theorems [Theorems 184 and 187] about inverses and two theorems [Theorems 185 and 186] about positive-inte- gral power functions. The discussion can be repeated with reference to any positive-integral power function restricted to nonnegative argu- ments. In particular, one can prove: < l J V V . n 5 . n x n = y [by Theorems 185, 186, 187] n n y>0 x>0 J and: (2 n> V n V x 1 >0 V x 2 >0 [x i n = x 2 n= ^ x i = X 2 1 [by Theorems 1 84, 185] In geometric terms (l n ) says, for each n, that each horizontal line with nonnegative intercept [y > 0] crosses the graph of the restricted nth power function; and (2 ) says that each horizontal line which crosses the graph does so in precisely one point. Since all nth powers of non- negative numbers are nonnegative we may conclude that, for each n, the restricted nth power function has an inverse whose domain is precisely the set of nonnegative numbers. Consequently, we are justified in intro- ducing an operator * V '» to use in referring to these inverse, by adopting the defining principle: (PR) V V . n ( n vT>0 and ( 'Vx" ) n = x) n x_^u From (PR) and (2 ) we have the uniqueness theorem: V n V x>Q V [(y>0 and y n = x) => y = n Vx"] [Theorem 188] And, calling the function {(x, y), x > 0: y = vx } the principal nth root function , we have, from Theorems 184-187: Each principal positive-integral root function is continuous and increasing [Theorem 189] on the set of nonnegative numbers. As in the past, we shall follow the custom of abbreviating ' V * to 'v '. Notice, also, that, for each x > 0, \/x~ = x. [To prove this we use Theorem 188: For a > 0, a > 0, and a 1 = a. So (by Theorem 188), a = vT.] [9-50] [9.04] RADICAL EXPRESSIONS The defining principle (PR) and Theorem 188 give us the necessary tools for manipulating radical expressions. A radical expression is one which contains a root sign. Here are some examples: 3/ r 5 2xV~3~7, VVn, ( 4 vT + vT) 4 , i-vT In particular, a radical expression of the form: n vr is called a radical . The expression under the root sign is called the radicand of the radical. The number n is called the index of the radical. Note that the defining principle restricts us to radicands whose values are nonnegative numbers. Example 1 . Show that >/54 = yfzf • yfz . Solution . By (PR), since 27 > and 2 > 0, \ll > and vT > 0. So, JIT ' /z > 0. Also, ( ^27 -\l f = ( JTl) ' ( \l ) = 27 • 2 = 54 > 0. 3/ — 3/ — 3/— So, by Theorem 188, V54 = V27 ■ V2 . Example 2 . Show that y 5 = v5 Solution . We wish to show that the principal sixth root of 5 is v5 . According to Theorem 188, to do this it is sufficient to show that V 5 > and 3/ — 6 2 3/ — ( V5 ) = 5 . By (PR), since 5 > 0, V5 > 0. Also, Again by (PR), ( VT ) - 5. So, ( vT) = 5 . Hence, by Theorem 188, "y5 = v5 . [9.04] [9-51] Example 3 . Show that \J JT = ^ . Solution. By (PR), since 7 > 0, v7 > 0, and, since (t^[(^)?=M 4 = 7 So, by Theorem 188, V Vf = -/T. Example 4 . Show that ( Js ) =\j 5 . 3/-. 2 Solution. Since squares are nonnegative, ( V5 ) > 0. So. <^) 2 =T?. 3 V6~ Example 5 . Show that ~\Itt - ~y~ Solution. i'-m-is-w-* A. Verify. 1. ^32 = 2%2~ 4. (W^ EXERCISES 12, 2 . v^ = V^ 5. d/z-r^V? 3 6.Vi- 3. -yvf . vr 7. \fl6 / -2 8. V ("W = and "vT = 1) [Theorems 190a, c] m l J B. True or false? v V x = x 1. V x>0 4. V tT 2 " X = x 2 - V x>0 ^ = x 2 j?~ = x 2 X i. vl 5. V.^Vx 1 * - « S 4. y x V^ = |x 3 C_. Complete. i. vT» vT2 = vToT 4. V49 = v~T :. ■?T r -?? 3. fr-V^* 5 ^ 5. V a/8~ = VT 6. ( ? ^) 3 = 3 7/ UNIVERSITY OF HUNOfc [9-52] [9.04] D, Prove. m 1. V V ^ rt Vx~ > [Theorem 190b] m x >0 L m* — m/ — m/ — r i 2. V V . n V ^ A vx vy = Vxy [Theorem 191a] m x>0 y>0 ' ' l J m nm , 3. V V V sn '/x = Vx 11 [Theorem 191b] m n x>0 -,n/ 4 - V V V ^ n V n ^"= m ^x [Theorem 191c] m n x >0 v l m- m 5. V-V V ^ n ( >/x") J = Vx J [Theorem 191d] j m x >0 *• E_. Show that each of the following generalizations is FALSE. 1. V >n "^f = 7 VT 2. V *v^ fc = Vx" x>0 v x>0 -4, x>0 WA ' A "" "' T x>0 v y>i 4/ — 3/ — 12> — \ / 4 4 „ n vGTV^= VT 4. V x>Q V Vx + y*=x + y F_. Expand. Sample . ( vT + 2) 4 Solution . ( vT + 2) 4 = ( vT) 4 + 4( vT) 3 ^ 1 + 6( vT) 2 '2 2 + 4( 3 V5 _ ) 1 -2 3 + 2 4 = 5 vT + 40 + 24 Iflb + 32 vT + 16 = 56 + 37 vT + 24 3 /25 i. (vT + D 3 2. ['if?. i)[if7+i) 3. <*r + tb> 4 4. ( 4 /7+2) 5 5. (I^T + n/x") 4 6. (V2~ + x) 3 G. Consider the sequence f defined recursively, for some integer m, by: mr— - . . m 1. Prove that each term of the sequence is positive. [9.04] [9-53] 2. Guess an explicit definition for f and prove your guess by mathe- matical induction. 3. Compute. 100 100 (a) ^Tf p , formal (b) ^T f , for m = 2 p=l p=l H. 1. Find the 10th term of the geometric progression 1, v2, v4, ... 2. Find the sum of the first ten terms of the geometric progression 1 vT V9" 3. Insert 4 geometric means between 1 and 2. 4. The 6th term of a geometric progression is 2 and the 16th term is 50. (a) Find the 26th term. (b) Find the 25th term. ROOTS OF NEGATIVE NUMBERS Up to now we have considered roots of nonnegative numbers only [see the defining principle (PR) on page 9-49]. On this basis we know, for ex- 28 ample, that the equation 'x =5' has a unique nonnegative solution, name- 28 2 2 ly» yfs» [And, since a = b if and only if (a = b or a = -b), it follows that the equation in question has exactly two solutions, \[$ and - vlf. ] TO On the other hand, we know that ' v'x" = -5' has no solution [Why?]. As another example, consider the equation 4 x J = 8'. Again, our knowledge of the principal root functions assures us that this equation has a unique nonnegative solution. Does it have a negative solution? Does the equation 'x = -8* have a solution? The preceding considerations suggest that it may be possible to ex- tend the domain of some of the principal root functions to include negative numbers. Since the domain of a principal root function includes --at most --the members of the range of the associated power function, we can in- vestigate this possibility by reconsidering the positive-integral power functions. As examples, let's look at the squaring function and the cub- ing function. [9-54] [9.04] x, y): y = x } {(x, y): y = x } As is clear from its graph, the converse of the squaring function is not a function [Explain,] It is for this reason that, in introducing the princi- pal square root function, we defined it as the inverse of the squaring function restricted to nonnegative arguments. However, as is clear from its graph [and as follows by Theorem 185'], the converse of the cubing function is a function- -the cubing function, itself, has an inverse. So, we might have defined the principal cube root function to be the inverse of the cubing function- -rather than, as we did, as the inverse of {(x, y), x > 0: y = x }. This important difference between the squaring function and the cubing function is expressed in two theorems: and: V (-x) = x 3 3 V x {-x) = -x Indeed, using the first it is not difficult to show that, since the squaring function is increasing on {x: x > }, it follows that it is decreasing on {x: x < }. Using the second we can show that, since the cubing function is increasing on {x: x > }, it is also increasing on {x: x < }. Conse- quently [since belongs to both {x: x > 0} and {x: x < }], the cubing function is increasing [everywhere]. It also follows from the second theorem that, since the range of {(x, y), x > 0: y=x } is {y: y > }, "5 that of {(x, y), x<0: y = x } is {y: y < }. Consequently, the range of the cubing function is the set of all real numbers. From these two results it follows that the cubing function has an in- verse—say, g--which is an increasing function whose domain is the set of all real numbers. Theorem 187 tells us that g is continuous. [9.04] [9-55] The difference we have noticed between the squaring and cubing functions is merely an instance of a difference between even positive- integral power functions and odd positive-integral power functions. In fact, using the theorem on squares noticed above it is easy to prove: . x 2n 2n V V (-x) = x n x and, using this, to prove: w w / \ 2n " * 2n - 1 V V (-x) = -x n x Using the last, we can repeat, for any odd power function, the remarks made above concerning the cubing function. This being so, it is con- venient to adopt an additional defining principle for principal odd roots, supplementary to (PR): (PRO V V (^'Vx") 211 " 1 = x n x We then have the theorems: and: V V V [y 2n " 1 = x => y = 2n " Vx~ ] [Theorem 188'] n x y ' ' J J Each principal odd positive-integral root function is continuous and increas- [Theorem 189'] ing on the set of all real numbers. Using (PR') and Theorem 188' it is easy to prove: V V 2n " V^x" = - 2n "Vx" [Theorem 190'] n x Moreover, the four parts of Theorem 191 now hold for all x in case m and n are both odd. For example, consider Theorem 191b. For m odd and for any number a, / m / — »nm , m/ — .mn r . m/ — >min n r , / .~ I -,/\i ( Va ) = ( Va ) = [( Va ) ] = a [by (PR')]. Now, if n is also odd, it follows that nm is odd and, so, by Theorem 188', for m and n both odd , nm. As just illustrated in the case of Theorem 191b, the proof of the supple- ment of each part of Theorem 191 is just like the proof of the part itself [9-56] [9.04] except that (PR') and Theorem 188' take the place of (PR) and Theorem 188. [For the resulting theorems, see Theorem 191' on page 9-359.] Theorem 191b can be supplemented in another way. Suppose that n is even. Then, for any a, a = |aj . Hence, since ja| > 0, Va = V |a| = \/l a l » b Y Theorem 191b. [The case n = 2, m = 1 should be familiar to you.] So, for n even , nm r — ni/ V V Vx" = YM • m x v vii Now, it is not hard to prove [using (PR') and Theorem 188] that, for m odd, m , v x Vw = l"^l. Consequently, for n even and m odd , nm Uiil / EXERCISES A. True or false? l - V x>o"v^ 3 = " *• v «ao"V^"-" 3 - \ X 5/ — e J>i — =- 5 r 4 - V x> V-^ 5 =x 5. V xf0 V-x 5 =x 6. V x V"^ 5 = -x 7 ' V x>0 Vx=X 8 ' V x<0 Vx=X 9 ' V x V/* = * 10. V x Vx 4 =|x| 11. V x£0 Vx 4 =|->c| 12. V x > V x4= 1*1 ~rT~T -vTIl ~\fy — 7 a 4 ; — 13. V Vx =xVx 14. V sn \/x =x'V7 15. V V x x v v x > v x V 14 3 = X • \/x -6/ 16, . V V x =x\/x 17. V V x = x 'Ux 18. V V x = x* Vx x v v x v iiv x v 10 w w"A 3 r~^7 z r\n~ ?n w w ~f/ 2z i4 5 3 r\ 4 /~2~^ 19. V x V V x y = xy • Vx y 20. V x V Vx Y = x y • V x Y \ / 14 28 2 4 \ / 2 4 \f~~\A 28 2 4 \ / 2 21, V V yx y =xy*Vxy 22.VV\/x y = x y • Vxy <•* 0„ «a« 'i" 'i"- *v» [9.04] [9-57] Study these examples of transforming radical expressions. ~A 3 / 5~T ~A 3 /~ ~\ 3 n> ""vVT Example 1 . V 54a b = V 54 • \J a *\J b = 3ab 2 ' V2a 2 b [Are any restrictions necessary on the values of 'a' and l b'?] Example 2 . ~V48a 6 b 10 =A/7l6a^b~ 8 )(3a 2 b 2 ) = 2 |a| b 2 -V3aV [Any restrictions needed?] Example 3 . V 48a 5 b 10 = V (l6a 4 b 8 )(3ab 2 ) = 2ab 2 ' V3ab 2 , [a > 0] Ar Example 4 . V 48a 5 b ? = V (I6a 4 b 4 )(3ab 3 ) = 2ab- V3ab 3 , [ab > 0] Example 5. 3, ^ = i^L = ^^ , [uv ^ 0] n 3 V 3 3 / S 2uv ,u v v< 2uv > 4/— :nr 3 2 x y Example 6 . ~\ / v = -\ / r * - , k/ 2xy 3 V2xy 3 2 3 x 3 y * /24x 3 y V(2xy) 4 a/ 24x 3 Y. 2|xy ^/24x 3 [xy > 0] :xy X , [xy > 0] [9-58] [9.04] 12 12- Example 7 . v 9a b c = '(3ab c ) 6 r 3ab c , _- V- • 2 [ac > 0] V2?v -V3xv 3 = Example 8 . V 2x y ' V 3xy "'(2x 2 y) 4 - 1 A/(3xy 3 ) 3 12 _ ^ r 74~3 11 13 = « 2 3 x y 12, = |y|- V432xy, [x>0 < y] o.. O- o- 'i v "p "i* _B. 1. Transform by reducing the radicand. [See Examples 1-4.] (a) Vl28y 3 k 5 m 12 (d) A/32a 5 b 8 c 17 5 (b) Vi6xy 3 z 7 (e) V 3 6x 36 (c) ClTaV (g) V x (J) (h) Vx 11 V77 2ab + b (k) J m + 2 m + 3 v x v (f) VoToOl: (i) V-y # (i) -6 -v/ 2m 3m V V V x y (m) Vx 2 (l+3x 2 y 2 ) (n) 3-V x 6 - x 4 y 2 (o) V64(x-y) 1 2. Transform to a radical. Sample . 3x\ A 2y = V(9x 2 )(2y) = V18x 2 y, [x > < y] (a) 4c *Vcf (b) 5k- Vk~ (c) x* Vx ... 2 3 n ; — (d) x y • Vxy (e) ^VlOt (f) 3a 2 |b|-V5a 2 b 2 (g) 0.4- 5 vTOO (h) -2-vT 3t 3. Transform by reducing the index. [See Examples 5 and 7.] (a) V49y^ 4 (e) 49a 225x 2 -\ / 4 -\ / 3 (b) V 9x (c) V27t^ (f) Va 2 +2ab+~b 2 (d) (g) l $] 2, 4 6 v a b c 7 ^r4irzn: V x v r [9.04] [9-59] 4. Transform to an expression which contains a single radical. [See Example 8.] (a) Vb-^JJ (b) a/^'V (c) 7Vx~ • 6 %T (i) A/2x 2 y 3 z -fA/3xyz 4 (g) 12x 2 y -Vl2x 2 y 4- (3xy 2 »V 2x y 3 ) 5. Transform to an expression in which the radicand does not con- tain a fraction. [See Example 6. ] -3/T" rr~ 5 r~r~ wVF < b »Vf «=>V^ 6. If f(x) = x 3 - 8x 2 + 19x - 14, show that f(3 - VF) = 0. "bC. Study the following true equations: ,f.y§ =V 4 -f - 2 Vf aAi^ J* V 4 H '•i-»-Vi '£ *h = 5 'Vh 1. Generalize . 2. Does the generalization you made in Exercise 1 include cases such as the following? If not, generalize further. yTt - 3 /t~ -\ 3 rr -vVr V z 7 " Z V 7 - V :5 26 3 V 26 ' V 4 63 " 4 V 63 * V*255 * V 225 ' -^5/ — r- -Srr- 10; ^ 10 f V 31 " V 31 ' V J 59048 V 59048 [9-60] [9.04] MISCELLANEOUS EXERCISES 1. Of the 75 boys in a small high school, 39 were on the football squad, 14 were on the basketball squad, and 27 did not take Dart in athletics at all. How many were on both the basketball and football squads? 2. Solve the equation: V^ 49x^ = 7 3. Simplify. 7 (a) ab + b' a 2 + ab (b) s + t s 2 - t 2 4. If a quart of water is approximately 60 cubic inches in volume and a cubic foot of water weighs approximately 60 pounds, about how many pounds does a gallon of water weigh? 5. Given : AB SBC, AEC is an arc of 120°, » « EB is a diameter Find: the ratio of AB to AD 3 1 6, Solve the equation: tx- t" x " * = *9 7. Simplify. (a) 2a b + 2 6 b + 2 i — - 3a (b) * y I y X (c) 9_ 25 r : A certain type of alloy is made of 7 parts copper to 3 parts zinc. How many pounds of zinc will be needed to make 830 pounds of this alloy? 9. Derive a formula for the area-measure K of a circle in terms of its circumference C. [9.04] [9-61] 10. Suppose that the sum of two integers is 45. If the ratio of the larger to the smaller is greater than 2 and the ratio of the smaller to the larger is greater than 7/16, what are the integers? 11. Find the dimensions of the rectangle with the largest area-measure that can be cut from a trapezoidal region whose bases are 3.5 and 1 and whose altitude is 10. 12. Solve the system: io m + n = 10000 io m " n = 100 13. If the measures of three angles of a convex pentagon are 100, 118, and 96, respectively, and if the other two angles are congruent, what is the measure of one of them? 14. Suppose that x and y are two-digit numbers. If the digits of one of them are the digits of the other and if their ratio is 5 to 6, what are the numbers ? 15. Hypothesis: Pe AABC, A f S PQ ± AB, PR l BC, PS ± AC r- i 2 j 2 j 2 v 2 j a 1 j. f 2 Conclusion: a + c +e =b + d + f 16. Find the first term and common difference of the arithmetic pro- gression whose 27th term is 186 and whose 44th term is 305. 17. The sum of the first four terms of an arithmetic progression is 44 and the fourth term is 17. What is the second term? 18. Expand and express without referring to negative exponents. I \ l 2 W~ 2 X 3 (a) (a b ) (b) (-a J b l c *y . , , -1. 2 3,-6 (c) (a b c ) [9-62] [9.05] 9. 05 The rational numbers . --In Unit 4 you learned a classification of real numbers: /"positive integers [i*] f integers [l] / f rationals [R] I (negative integers real numbers < I nonintegral rationals irrationals For example: 1, 2, 3, etc. are positive integers and — 1, — 2, —3 are negative integers; the positive integers, 0, and the negative integers make up the set of integers; the rational numbers are those which are quotients of integers by positive integers-- |, "7^* -y. vT'Vi", 6, 0.3, 0.9, and 0.83 are rational numbers; the remaining real numbers are called irrational numbers--v8 > v9, and V are irrational numbers. In Unit 7 we adopted three basic principles, (1^) - (I*) to character- ize the set I + of positive integers, and one basic principle (I) to charac- terize the set I of integers. We now need a basic principle to allow us to prove theorems about the set R of rational numbers. What we need is a principle which says which real numbers are rational. One common definition is the one mentioned above- -that a rational number is a real number which is the quotient of some integer by some positive integer. More briefly, V [x € R <=> 3, 3 x = -]. x k n n J By the principle of quotients, if x = — then xn = k. And, by the division theorem [Theorem 49], if xn = k then x = — . So, the foregoing possible definition is equivalent to: (R) V [x e R <=> 3 xn e I] x n J In our dealings with rational numbers, we shall use (R) as a basic prin- ciple. Now, how does (R) work? Let's suppose that someone has picked a real number and is wondering if it is rational. A mechanical procedure [9.05] [9-63] for testing it would be to multiply it by 1, then by 2, then by 3, etc. If one [or more] of the products is an integer then the number is rational. If the number is irrational, it will be impossible to get an integral pro- duct. Of course, such a mechanical procedure is never necessary. For 22 example, to show that the real number — is rational, think of the prin- 22 ciple of quotients. Since -=- • 7 = 22 and since 7 e I + and 22 e I, (R) tells 22 us that — is rational. Instead of 7, we can use 14 to test the rationality 22 of —zr . What other numbers can be used? Similarly, -5 is a rational number because -5*1= -5 and 1 € I + and — 5 6 I. What other positive integral multipliers can be used to test the rationality of -5? Sometimes, the problem can be more complicated. For example, consider — j=- . Is this number rational or not? The principle of quo- tients may suggest multiplying by 3v 2 , but since 3v2 f[ I + , doing so would be irrelevant. However, vT 2 3vT 3 [Why?] and -r is rational [Explain.]. We can also use (R) in proving closure theorems for the set of rational numbers. For example, let's prove that the set R is closed with respect to addition, that is, that each sum of rational numbers is rational. Suppose that r and s are rational numbers. It follows from (R) that there are positive integer s-- say, p and q--such that rp el and sq e I. Since lis closed with respect to multiplication, it follows that rpq e I and sqp e I, Since I is closed with respect to addition, rpq + sqp e I, that is, (r + s)(pq) € I. Since I* is closed with respect to multiplication, pq e I + . So, by (R), r + s is a rational number. So far, we have used (R) to prove that numbers are rational. Let's use it, now, to prove that v2 is not rational. To do this, let's suppose that v2 is rational- -that is [by (R)], suppose that there is a positive [9-64] [9.05] integer whose product with V2 is integral. Then, by the least number theorem [Theorem 108], there must be a smallest such positive integer --say q. Then, v2 # q is integral and, if p < q then v2 • p is not integral. 'y | "y -y Now, since 1 < (v2 ) < 2 , it follows [since the principal square root function is increasing] that 1 < vT < 2. Since q > 0, lq < yfl m q < 2q, and, so, {*) < vT- q - q < q. Now, since v2 • q and q are both integers and I is closed with respect to subtraction, v2 • q - q is an integer. So, from (*), v2 • q - q is a posi- tive integer less than q. Hence, by the choice of q, (1) vT (vT.q - q)Q VTeR (f) V Vr7e R n (b) V V r 2 + s 2 e R r s (d) V V [rx 6 R ^> x e R] r x J (g) V 3 . ^m" € R fe n m> 1 B. 1. Prove that v3 is irrational. 2. Guess which of these are rational, (a) vT (b) ^8" (c) 7 vT (d) V32 (e) V65 You have probably already guessed that, for each m and n, vm is either an integer or an irrational number. That is, vm is rational only if m is a perfect nth power [m is a perfect nth power if and only if it is the nth power of some integer]. In Appendix B [pages 9-231 through 9-249] your guess is proved correct: Theorem 193. n, — V V Vm is irrational unless n m m is a perfect nth power. [9-66] [9.05] 3. Which of these are rational? •7 7 / 3 / — (a)V2 14 (b)Vz 15 (c) ~y| (d) ^.00001 (e) ^64 Sample . Is >/F + VT rational ? Solution . Suppose that 4l + vT e R. Then, by (R), there is a positive integer, say q, such that (VF +vT)qe I. Since I is closed with respect to squaring, [(y/I + vT )q] 2 € I. That is, (5 + 2vT)q 2 € I. Since q € I + , it follows that 5 + zV6~ is rational. So, since 5eR and 2€R, so does v6 [Explain. ]. But, by Theorem 193, since 6 is not a perfect square, V6^R. Hence, VX^R. 4. Which of these are rational? (a) 7 - vT (b) 4vT - ^32 (c) ~- + ^- (d) 3VT2 + 2vT ^5. (a) Find rational numbers r and s such that rv2 + sv3 e R. How many pairs can you find? (b) [For any real numbers a and b, the rational -linear combi- ations of a and b are the numbers ra + sb, where r and s are rational numbers. ] You have seen that the only rational- linear combination of v2 and v3 which is a rational number is 0. Investigate the conditions on m and n and on r and s under which rvm + svn is irrational. C. 1. Prove that the real numbers are dense . That is, prove that be- ween each two real numbers there is another real number: VV^ 3 xx z [ Hint . How about the average of the given numbers?] [9.05] [9-67] 2. Prove that the rational numbers are dense. That is, prove that V V . 3 r < t < s. r s > r t 3. What theorem tells you that the positive integers are not dense? o^ o^ ^ 'r "t- 'i v It is perhaps not very surprising that the rational numbers are dense. But, since [as shown in Appendix B, see Theorem 200] there are many more irrational numbers than there are rational numbers, most people do find it surprising that the rational numbers are dense in the real numbers- -that is, that between any two real numbers there is a rational number: Theorem 201. V V . x y > a x r X < r < y For a proof of this, see Appendix B. j* o>- +)+ 'I- "f T 4. Use Theorem 201 to prove V V ^ 3 3 x x r s ' ^D. The reciprocating operation is defined by: /x = 1 -r x, for x f- Prove the following theorems. 1. V x/o x-/x=l 3 - VoVo /Wy = /Uy) 2 - vV/ rv x , /y = x ^y x y p 4. V^ //*=* 5. Reciprocating is not distributive with respect to addition. 6. V V V / n (x + y)'/z = X *A + y ' /'< x y z/= J ' ' ' ' 7. R is closed with respect to reciprocation. 8 ' VoVo /x ' /y:/(X ' yl 9 ' V x V y^0 V z^0 (xz) '/ ( y 2):=X, /y [9-68] [9.06] 9,06 Rational exponents. --In Unit 8 you learned about the use of inte- gers as exponents. We began with nonnegative integral exponents: (1) v x v k>o x = n x P =i and then introduced negative integral exponents: (2) x = x/0 v k<0 " -k ' x So, for example, (1) and the recursive definition of II-notation tell us 3 2 3 that 2 =2 • 2 = 2 # 2* 2, and that, for each x, x =1. Since 2=8 [and since 2/0 and -3 < 0] (2) tells us that 2~ 3 = l/2~~ 3 = l/2 3 = l/8. Using these two definitions we were able to prove a number of theo- rems which justified simple tricks for manipulating exponential expres- sions. For example, by Theorem 155, (vT) 5 (VT) 3 by Theorem 154, (VT)" 2 by Theorem 156, by Theorem 157, [(-1.5)" by Theorem 158, (nVF) by Theorem 160, by Theorem 159, (3/2) 3 = ( vT)- 5 + 3 = (vTf 2 , l/(vT) 2 = 1/3 [by (PR)], 6/ -4 77 JTl 2,3 6--4 10 77 = 77 (-1.5)°., (2/3) J = (3/2)", and, 3 3 /2 3 . In Unit 8, we spoke of (1) as defining, for each x, the expo- nential sequence with base x. Each such sequence is a function whose domain is the set of nonnegative in- tegers. H 1 1 h -I 1 1 1- [9.06] [9-69] Definition (2) shows how to extend each exponential sequence with nonzero base to a function whose domain is the set of all integers. These "exponential functions with integral arguments'* have the desirable properties, some of which are exemplified above, which are de- scribed in Theorems 154-160. Those with positive bases have the additional properties described in Theorems 152a and 161. In this section we shall discover meanings for expressions like: 4 i- ,24/10 which will enable us to extend, further, the exponential sequences with positive bases to "exponential functions with rational arguments" in such a way that these functions will also have the pro- perties described in Theorem 152a and Theorems 154-161. For example, it will be the case that 1/2,3/5 _ ,11/10 77 .-0. 83 = 2 and that ,-0.83,3 -2.5 177 ) ~ Tl Later, we shall see that each of these new functions is monotonic and continuous on its domain R, and we shall see how these properties can be used to define an extension of each such function to an exponential function whose domain is the set of all real num- bers. Theorems analogous to those previously mentioned will again hold, and these final exponential functions will turn out, also, to be monotonic and continuous --this time, on the set of all real numbers. [9-70] [9.06] LOOKING FOR A DEFINITION Our problem is to assign meanings to expressions like: 3 2 6 5 2 24/10 ff -0.83 in such a way that the laws of exponents which we have proved for inte- gral exponents will continue to hold. We faced a similar problem in Unit - 3 8 when we sought to assign meanings to expressions like '2 '. We solved that problem by, first, discovering what meanings would have to be as- signed to such expressions if one of the laws [the addition law] were to hold and, second, showing that if meanings were assigned in this way then this law and the others would hold. A similar procedure will be effective here. If the multiplication law for exponents is to hold when exponents are rational numbers, as well as when only integers ;.re allowed as expo- nents, it must be the case that (3 i/2 )2 = 3 i' 2 = 3 i = 3 1 2 --that is, it must be the case that 3 is a number whose square is 3. As e know, there are just two such numbers, v3 and — v3 . We could en- I . 7 I I I \ / *}\ "} o o o • that 13 ' = 3 by agreeing either that 3 = vT or that 3 =-vT. V ' 1 sure 4 Which agreement should we make? One way to decide is to consider 4 3 '. We must be able to assign such a meaning to this expression that -•2 i (3 1 / 4 ) 2 = 3 4 = 3 2 . To be able to do so, since squares of real numbers are nonnegative, 3 must be nonnegative. So, the only agreement as to ' 3 ' which has a chance of being satisfactory is that 3 2 = vT. 5 Now, let's consider l 6 '. If the multiplication law is to hold, [9.06] [9-71] (6 3/5 )5 =6 I' 5 = 6 3 2 5 3 --that is, 6 must be a number whose 5th power is 6 . There is only one such number, V 6 . [Explain why - V 6 is not such a number,] So, our only hope is to agree that I 5„ 5 B v 6" 24/10 24 Finally, let's consider v 2 ' '. Since jT) ' ^ ~ ^» ^e sarne kind of argument we have used before shows us that our only hope is to decide that 2 24/ 10 = »fyps This last example may suggest a possible source of difficulty. Not 24 24 only is — • 10 = 24, but, also, yr- • 15 = 36. So, the same line of rea- soning prescribes that we agree that 2 24/l0 = i "^36\ Clearly, our program will run into difficulties unless V7* = V?* Let's check: ^^ V, 2 12 ) 3 = V 2 12 = V (2 l2 ) 2 =V?* [Theorem 191b.] Apparently things are going to work out all right. If r is a rational number and m is a positive integer for which rm e I, we can, for any a > 0, agree that ' & m , r -W rm a = Va This makes sense because, for any a / and any integer rm, a is, by (1) and (2), a real number which, if a > 0, is, by Theorem 152a, a posi- m r~r — tive number. And, by the defining principle (PR), Va is, in this case, a real number. Moreover, if n is any other positive integer for which rn € I then n, mn , nm m Va™ = Va rnm = Va rmn = A4^™ Theorem 191b [9-72] [9.06] So, however we use this agreement, we shall always be led to as- signing the same value to any numeral of the form 'a '--the agreement is, we may say, self- consistent. Before adopting this agreement as a definition, we have still to check another consistency question. To see what this question is, no- tice that, since 2, say, is both a rational number and a positive integer, it follows by the definition (1) for exponential sequences, that 3 =3*3. Since 2* 1 e I [so that 2 is a rational number], it follows from the pro- ? / 7~ ~> posed agreement that 3 = V 3 , where the meaning of the second 4 3 ' is to be determined, as above, from (1). So, according to the proposed agreement, 3 = V 3 • 3. Since, for each x, vx = x, it follows that v3*3 = 3*3. So, both (1) and the proposed agreement assign the same value to '3 '. Generalizing on this argument shows that the proposed agree- ment is consistent with both the earlier definitions (1) and (2). The upshot of these checks for consistency is that no harm will come from adopting the proposed agreement as a supplement to (1) and (2). Vie shall do so, after using Theorem i91d to transform it into a slightly more convenient form: (3) V >n V V eT x r = ( m Vx") rm x >0 r m, rm el f [read 'with'] [The restriction on l m' is needed to ensure that (1) and (2) make sense of'ff)^.] Up to now we have rather neglected the exponential sequence with base 0. In Unit 8 we saw that it was futile to attempt to define negative- integral powers of 0, and the same argument applies to discourage us from defining other negative-rational powers of 0. However, it seems [and will turn out to be] reasonable to adopt: (3') 0° = 1 and V ^ n r = r >0 EXERCISES A. Definition (3) enables us to translate an expression for a rational power into an expression for an integral power. 2 .5 ,5/- v 2 AOnrA Examples : 3 = (V3~) = ( V3~) =... [9.06] (9-73] Complete each of the following. i. 2 ». ( 3 VT) ? . < 12 /F> ? = ( ? VF) 6 = (V?) 6 2. 8* = ( 5 Vi-) ? = ('W = ( 2( V8 ) ? - (VF) 7 3. ( 5 V7-» 2 = 7 ? 4. {ft . 9 ? = 3 ? B. Later we shall prove that definition (3) enables us to justify all of the familiar laws of exponents which we have already established for integral exponents. In this exercise, we shall justify the laws as they apply to special cases. 2 ~ 3 1 Sample 1 . Show that 3 = " ' / _! zl Solution. 3 =3 ■ M rerl- 1 Theorem 154 ,2/3 2 3 j_9 3 5 15 Sample 2. Show that 4*4 =4 2 3 i£ _9_ c i 4.- a* 4 5 „15. .15 Solution. 4*4 =4 '4 . (V) I0 -( 15 V4-) 9 = 4 19 15 [9-74] [9.06] 2 \3 Sample 3 . Show that \ 7 2\3 5 7 _6_ 35 Solution, \7 5 7 ^3 -2 I- 5 vrJ j 7 -6 Theorem 191d Theorem 157 35/— -6 „ 35 3 V7~r° = 7 2 2 2 Sample 4 . Show that (2 • 5) 3 = 2 3 «5 3 2 Solution . (2'5) 3 = ( 3 /2 _r 5) 2 2 2 = 2 3 -5 3 Theorem 191a Theorem 158 Establish each of the following, 2 j. " 5 1 1. 3 ,2/5 2. 4 3 = -1/3 3. 32 °' 8 = 0.0625 4 2 26 4. 3 7 -3 3 , 3 21 5. 64 3 »64 2 = 64 6 = 128 -I 1 2 5 ^ 5 6. 11 • 11 • 11 = 11 2\4 _8_ \3 5 i 7 = 3 35 10. 15 ' 1 ) - 1 , 5°- 01 12. (16' 27) 3 = 36' VT 14. ( 7 2 .n)^ 3 Vf 6 /T7 7, 7 9 f 7 8 = 7 4\2 n 72 9. \3 5 ) 7 = 3 35 3 3 3 4 4 4 11. (6-7)* = 6*7* ■ 4 luo i3. Uw) 1 19 4 = 2 y * 3 is. ( 2 7 * 3 5 ) 35 = 5 vf ♦ Vr ,m/ — . rm [9.06J [9-75] 2 2 2 _2_ 16. (|] 3 = (|) 3 17. 5 3 > 18. 5 3 > TESTING OUR DEFINITION We have seen that adopting: . . r m/ — . i x > u r m, rm e l will not get us into any trouble, and that (3) is the only definition from which we can hope to get satisfactory theorems. The preceding exer- cises are good evidence that it is possible to prove analogues of Theo- rems 152a and 154-161. We shall now state these analogous theorems and prove them by the methods you probably used in solving the exer- cises. [The theorem numbers listed are those which will be given to the final generalizations, to be proved later, in which *r' and *s' are re- placed by variables whose domain is the set of all real numbers. The bracketed versions are analogous to the unbracketed ones, but cover the case of base 0. In this case, as in the definition: (3') 0°= 1 and V r >Q r = 0, the exponents must be nonnegative. ] Theorem 203. V . n V x r > [V . ft V . n x r > 0] x >0 r L x>0 r>0 J Theorem 204. V ^ n V x~ r = — x > r r Theorem 205. V . n V V x r x S =x r+S [V . n V . n V ^ xV 5 = x r + S ] x>0 r s L x>0 r>0 s>0 J r Theorem 206. V ^ rt V V — =- = x x>0 r s x s Theorem 207. V . n V V (x r ) s = x rs [V ^ n V . ft V ^ n (x r ) S = x rs ] x>0 r s L x>0 y>0 s>0 J Theorem 208. V x> Q V y >Q V (xy) r = X r y r [V x > V y2Q V r > Q (xy) r = x r y r ] Theorem 209. V x>Q V y>0 ffi = £ f V x>0 V y > V r >0 (f )' = ^ ' Theorem 210. V x> Q V y >Q ^ (*)-* - g)' Theorem 211. V ^ n V V [x r = x S ^> (x = 1 or r = s)l x>0 r s L /J [9-76] [9.06] Here are proofs: Theorem 203 . Suppose that rm e I. Then, for a>0, by (3), a = ( >/a~ ) . By Theorem IVOb, for a > 0, Va~ >0 and, so, by Theorem 152a, since rm e I, ( v'a") > 0. Consequently, a > 0. [The bracketed form of Theorem 193 follows from that just proved together with (3') and the fact that 1 > 0. ] Theorem 204 . Suppose that rm e I. Then, since I is closed with res- pect to oppositing [and because of Theorem 21], -rm e I. So [by (R)], -r e R and, by (3), for a > 0, -r ,m/ — .— rm a = ( Va ) 1 Theorems 154 and 190b m/ — .rm rvr) r Theorem 205 . Suppose that rm e I and sn e I. Since I is closed with respect to multiplication [and because of the apm and the cpm], r(mn) € I and s(mn) € I. Since I is closed with respect to ad- dition [and because of the dpma], (r + s)mn € I. Since I + is closed with respect to multiplication, mn e I + . So [by (R)], r + s € R and, for a > 0, r s /Hin/ — ,r(mn) .mn/ — . s(mn) j a a = ( Va ) Va I ™, „ ice x \ Theorems 155 / , » ( and 190b ,mny — Jr+s)mn J r +s = a [The bracketed form of the theorem follows from that just proved together with: V r >0 V s>0° r ° S =° r + S This last is proved, using (3')> by considering two cases, (r > or s > 0), r = = s. In the first case, r S =0 = r + S ; in the second, =1=0 . ] [9.06] [9-77] Theorem 206 . [This follows using Theorems 205 and 203 in the same way in which Theorem 156 follows using Theorems 155 and 152b. Here' s how : Since R is closed with respect to subtraction, it follows, using Theorem 205, that, for a > 0, a a=a =a # Since, using Theorem 203, for a > 0, a / 0, it follows that, for a > 0, r - s r / s a = a /a . Theorem 207 . Suppose that rm € I and sn e I. For a > 0, by Theorem 203, a r > 0. So, by (3), for a > 0, /n , — \ sn v sn rm \sn Theorems 191d, 190b Theorem 191c \ rm\ sn ( m 7r) . w . ) Theorems 157, 190b \ (rm)(sn) I ' /rv,« \(rs)(mn) ,m 7r) Since rm el, sn € I, and I is closed with respect to multiplica- tion, it follows that (rs)(mn) e I. Since I* is closed with respect to multiplication, mn e I*. So [by (R)], rs e R and, by (3), for a > 0, . r.s rs (a ) = a [The bracketed form of the theorem follows from that just proved together with: V r>0*s>0 ( ° r > S " ° rS This last is proved, using (3'), by considering two cases, (r = or s = 0), r > < s. In the first case, (0 r ) S = 1 = rS ; in the second (0 ) =0=0 . ] [9-78] [9.06] Theorem 208 . Suppose that rm € I. Since, for a > and b > 0, ab > 0, it follows, for a > and b > 0, that (ab) r = ( m v^b) rm = ("VT"^" ) rm = ( m v^) rm ( m VbV m = a r b r . I t t t (3) Th. 191a Th. 158 (3) Th. 190b [The bracketed form of the theorem follows from that just proved together with: V y>0 V r>0 ( °V» r =°V *«*■■ V x >0 V r>0 (x0 > r = xr ° r The first follows from the second and the cpm. The second r v r follows from the pmO and 4 V ^ rt V v.^0 =x0 '. This last is r x >0 r >0 proved, using (3'), by considering two cases, r > 0, r = 0. In the first case, = = a0 = a0;in the second, =1 = 1*1 =a r r .] Theorem 209 . [This follows using Theorems 208 and 203 in the same way in which Theorem 159 follows using Theorems 158 and 152b, Here's how: For a > and b > 0, it follows, using Theorem 208, that (6) v - (5 •")'■-'■ Since, using Theorem 203, for b > 0, b p 0, it follows that, for a > and b > 0, (a/b) r = a r /b r . [The bracketed form of the theorem is proved in exactly the same way, using the bracketed form of Theorem 208. ] Theorem 210 . For a > and b > 0, a/b > and, using Theorem 203, a r £ and b r / 0. So, for a > and b > 0, (S\~ * . _1_ . _i_ . bl , (b\ r W , (a/b) r , a r /b r , a r , ^ a / ' Th. 204 Th. 208 Th. 73 Th. 209 [9.06] [9-79] r s Theorem 211 . Suppose, for a > 0, that a = a , where rm € I and sn c I. Then, by [closure properties and] (3), ( va I = [ Va"l and, by Theorem 161, either va = 1 or r(mn) = s(mn)--that is, either a = 1 or r = s. EXERCISES A. For each of the powers listed below, write a simple expression which contains neither an exponent symbol nor a radical. 3 2 Sample . 49 3 Solution . 49 2 = (V49) 3 = 7 3 = 3^3 3 3 2. 9 2 3. 16 4 4 1 4 5. 27 J 6. 1024 U * 8. 1024 10 9. 1024 0,3 5 3 11. 216 3 12. 81 3 2 8 14. 1000000 6 15. 1000000 6 17. ( 8 1 2 ) ? 18. 256 1 - 25 1. 4 16 2 4. 2 16 4 7. 243 ' 6 2 10. 52 2 I 13. 10000 2 16. (JY 19. (0.008) U66 22. 343" - 3 _ 1_ 25. (0.0001) 4 3 28. (16-81) 4 (625 5 ) 2\5 / 5\2 / 4 21. \625 4 / 5 I 1 ' 6 ) 3 24. 216 3 2 27. 216 3 (,.-) 3 3 J_ _l\ 4 ,81 ' 16, 29. (81*16) 4 30 [9-80] [9.06] JB. Simplify each of the following. 4. Sample. J. 1 5 5. 5 2 c i *• 5 ^>5 J c 2 3 5 2 ,15 Solution . — 2 * = 5 =5 5^5~ 7 2 .11 3 5 5 1. 3 '3 -3 _2_ A 11 _ 2_ 7 101 ! 5 .i8.i 102 8\2 2. 1 1 7 "* 3 .7 5 •7 1 2 3. 1 2 ^2 3 77 *77 1 1 77 • 77 5. 1 2 2 X 2 2 3 6. 1 2 3 2 + 3~ 3 J. 2 2 + 2 2 3 1 _i 2 3 3X3 8. M* 9. (64, 12 )^ ii. 1 5 1 * 5 • 1 6 1 6 77 13. 0 V y>0 ^ + y 3 x + y 26. V >n x x > x 2 [9.06] [9-83] 27. 3 = 27 31. 2 33. _1_ 27 1 1 = i 35. V4 1M = 4 4 2 28. -4 = 16^ 30. U)Mf 32. 2 3 5 71 + 77 = 77 34. ,- 3 5 2 ivrr 4 * 3 .1 i 36. 7 2 X 7 2 = 1 37. 0=0 -3 38. (77 + vT) -r (77 + \/2~) = 1 10 °' 3 ■ f IS 41. 10°' 3 = 5°- 6 1 / 4 40. 77 3 \77 3 + 77°/ = 7T + W 42. (0. 25) °* 5 = 2 E_. Expand. [Assume that the values of the variables are positive num- bers only. ] Sample . 2 2 x + y 2 2 I -,2 2 4 Solution . \x + y / = x + 2y x +y 2 , m + n 2 , 2 1 ' x + y 2 , 2 v x + y 1 lw 1 b 2 + c 2 JU 2 3 }_ .i\/ 2 1_ ,a 2 + 2a 2 - 4a 2 A2a 2 - 5a 2 5 3 lw 1 6. \2a 2 - 3a 2 + 4a 2 /\7a 2 - 3a [9-84] [9.06] F. Find the rational approximation correct to the nearest tenth. Sample . 10 Solution . Since 1 < 10 < 2 , 1 < 10 ' 4 < 2. So, a first approximation might be 1. 5. We check this by computing: 4 (1.5)* = 2 2 ,, _ v 2 (1.5) = (2.25) < 5. 1 1/4 Since 5. 1 < 10, 1. 5 < 10 ' , Our first approximation is too small. Try 1.8, and compute again: (1. 8) 4 = (3. 24) 2 > 10.4 So, 1. 8 is a bit too large. Try 1. 7. (1.7) 4 = (2. 89) 2 < 8.4. So, 1.7 < 10 ' < 1.8. Since we are looking for the ration- 1/4 al approximation to 10 ' correct to the nearest tenth, all we need to do is to decide between 1.7 and 1.8. One more calculation will tell us: (1.75) 4 = (3. 0625) 2 < 10 So, 1.8 is the rational approximation correct to the nearest tenth. I I 1. 10 3 2. 5 3 / 4 3. 2 5 4. 4 3,5 RATIONAL EXPONENTS AND NEGATIVE BASES The definition which we have been using of rational powers of posi tive numbers: (3) V >n V V , T x r = ( m Vx~) rm x> r m, rm € I can be supplemented by a definition of certain rational powers of nega- tive numbers. For example, ignoring the restriction 'x > 0*, we can use (3) to make sense of 4 (~8) 2 / 3 ': [9.06] [9-85] ( - 8) 2 /3 = (^rgjZ a ( _ 2) 2 = 4 In a similar manner, for any rational number r for which there is an odd positive integer m with rm 6 I, and for any a ^ 0, we can agree that r ,xr\, — .rm a = ( Va ) [if a > 0, this is merely a consequence of (3). The case in which a = is covered by (3').] A check of the proofs of Theorems 204-210 shows that the asser- tions made in these theorems for positive bases continue to hold for non- zero bases provided that the exponents are of this special kind. Although Theorem 203 cannot be generalized to this extent [(-8) ' = -2 ^ 0], its role in proving the later theorems is served as well by: V >n V x r ^ x >0 r ' and this does generalize to nonzero bases if the exponents are restricted as mentioned. Theorems 154-160, which are used in the proofs, are stated for nonzero bases; and the parts of Theorem 191 are paralleled by those of Theorem 191' [See page 9-359 of the theorem list.]. The role of Theorem 190 can be played by the readily proved theorem 4 V / n V c n m V5T £ 0' fsee Theorems 190 and 190']. The upshot of all this is that as long as you are dealing with nonzero bases you can perform the usual manipulations provided that the expo- nents are rational numbers which have odd multiples belonging to I. EXERCISES A. Which of the following expressions make sense? Sample 1. (-8) Solution . This makes sense because there is an odd posi 4 tive integer m [to wit, 3] such that 7-m € I. 4 4 So, (-8) 6 = (\/^8J 6 = 4. Note, however, that it makes no sense to say: ( — 8) = 1 V— 8/ [9-86] [9.06] _7_ Sample 2. (-8) Solution . This does not make sense because there is no odd positive integer m such that Tym € I. [How do you know there isn't?] J_ J_6 100 21 1. <-8) 18 2. (-9) 24 3. (-12) 50 4. (-2) 28 B. 1. On the same chart, sketch the graphs of the equations: I i 1 i 3 3 3,3 y = x, y = x, y = x, and: y = x 2. Repeat Exercise 1 for the graphs of the equations: 111 £ 4 4 4 4 y = x , y = x , y = x , and: y = x EXPLORATION EXERCISES A. You know that 2° = 1. Also, 2 1 ' 4 = 1.1892, 2 1 ' 2 = 1.4142, and — o /a 2 =1. 6818. Use these data to make a table of powers of 2 from — 4 4 2 to 2 , changing the exponent by 1/4 from one entry to the next. [ Hint . 2 5 ' 4 = 2 1 ' 4 -2 1 ; 2*" 1 ' 4 = 2 3 ' 4 * 2 X . Your table should have 33 entries. ] J3. Round off the entries in your table to two decimal places, and use the results to plot points on a graph of the exponential function with base 2 and with rational arguments, from -4 to 4. [ Hint . Use as large a scale as your paper allows, and a sharp pencil. Do as accu- rate a job as you can. ] C_. Draw as smooth a curve as you can through the points you plotted in Part B. [ Hint . Turn your paper more or less topside down, and use your elbow as a pivot. ] Do you think that, if you plotted other pairs (r, 2 r ), for rational numbers between -4 and 4, these points would lie on your curve? [9. 06] [9-87] D. Use your graph to find an approximate solution for each of the fol- lowing equations. 1 4 Sample 1 . 2 = x Solution . The ordinate of the point of the graph whose abscissa is 1.4 is approximately 2.6. So, the solution is approximately 2.6. [Check: 2 1 * 5 = l4l = 2 • 1. 4 = 2. 8] Sample 2 . 2 =10 Solution . The abscissa of the point of the graph whose ordinate is 10 is approximately 3. 3. So, the solution is approximately 3. 3. [Check: 2 3 = 8] Sample 3 . 4=6 2 x | 2 1 x Solution . Since 4=2,4 = 6 if and only if 2 = 6--that is [by Theorem 197], if and only if 2 =6. From the graph, 2 * =6. So, the solution is approxi- mately 1. 3. 1. 2 2 ' 3 = x 2. 2 X = 5 3. 2~ ' 3 = x 4. 2 X = 3.5 5. 2 2x " 3 = 3.5 6. 2 3x + 4 = -2 7. 8 2x = 4. 2 8. 4 X = 2 X " 2 9. 2 2 = x E_. 1. Your work with the exponential function with base 2 and rational arguments has probably suggested to you that it is an increasing function- -that is, that V V [r_ > r. ^> 2 2 > 2 i], t r 2 l J 1 2 Here is an outline of a proof that this is the case. Your job is to [9-88] [9.06] fill in the blanks. Since 2 > and R is closed with respect to , it follows by Theorem 20_, that 2 r 2 = 2 r 2~ r i» 2 r i. So, since 2 r i = 1 • 2 r i , it follows that 2 r 2 > 2 r i if and only if • 2 r i > '2 r i. Since, by Theorem 20_, 2 x _, it follows that 2 r 2 -r i • 2 r i > 1 • 2 r i if and only if 2 r 2~ r i > 1, Suppose, now, that r 2 > r 1# Since R is closed with respect to subtraction, there is, by (R), an m such that (r - r 1 )m _ _. Since r 2 - r x is , it follows that (r 2 - r 1 )m is some positive integer-- say, p. By definition (3) on page 9-72, it follows that 2 r 2" r i = , and, of course, 1 = {"VT ) P . Hence, 2 r 2 _r i > 1 if an d only if > . Consequently, assuming that 2 r 2>2 r i if and only if {"Vl ) P > ("VT ) P . Now, since 2 > 1 and since the mth principal root function is an function [Theorem 189], it follows that "vT "vT. From this, by (PR) and the fact that the pth restricted to nonnegative arguments is , it follows that ^^^^ > . Hence, r r Consequently, if r 2 > r x then 2 2 > 2 x 2. Complete: (a) For a 1, the exponential function with base a and rational arguments is increasing. (b) For a , the exponential function with base a and rational arguments is decreasing. [9.07] [9-89] 9.07. The exponential functions . - -In the preceding section we extended the domain of the exponential functions with positive bases to include all rational numbers, and we proved a number of important theorems [Theo- rems 203-211 on page 9-75], In this section we shall complete the defini- tion of these functions by indicating how to interpret irrational exponents. MORE ABOUT RATIONAL EXPONENTS The last numbered theorem of the preceding section was Theorem 211 for rational exponents: V ^ n V V [x = x' x>0 r s (x = 1 or r = s)] This theorem tells us that, for < a ^ 1, the exponential function with base a and rational arguments has an inverse [Explain. ]. Since most of the functions we have met which have inverses are also monotonic, Theorem 211 suggests that these functions may be monotonic . Also, work with the one with base 2, in Exercise 1 of Part E on pages 9-87 and 9-88, bears out this conjecture. The conjecture is verified in Appendix C [pages 9-250 through 9-269]. {(r, y): y = a r } < a < 1 1 < a Your work with the exponential function with base 2 and rational arguments may also have suggested that exponential functions with posi- tive bases and rational arguments are continuous - -that is, sufficiently small changes in argument result in arbitrarily small changes in value. This conjecture is also correct, and is verified in Appendix C. Finally, Theorem 203 tells us that the range of each exponential function with positive base and rational arguments is a subset of [9-90] [9.07] {x: x > }. [if you have studied Appendix B you will know that the range must, then, be a proper subset of {x: x > }. ] Your work with base 2 probably suggests that [as long as the base is not 1] an exponential func- tion has arbitrarily large values . Consequently [Theorem 204], such a function has arbitrarily small positive values. Collecting these results, we have: For 1. For < a < 1, given M > 0, there is an N such that, for each r > N, a"~ r > M and < a r < l/M. For a > 1, given M > 0, there is an N such that, for each r > N, a r > M and < a~ r < l/M. The proof of monotonicity is like that given for the case a = 2 in Part E on page 9-88. The proof of continuity makes use of Bernoulli's Inequali- ty [Theorem 162], The last two results also depend on Bernoulli's Ine- quality. For details, see Appendix C. IRRATIONAL EXPONENTS rx" How should we define '2 '? As with any definition, we are free to choose- -but, we shall be stuck with whatever consequences follow from — 3 3/4 our choice. In defining '2 ', say, and '2 ' *, we attempted [success- fully] to control the consequences by choosing definitions which would enable us to justify the laws which we already knew to hold for nonnega- tive integral exponents. Here, similarly, we shall do well to pay attention to what we already know about the exponential function with base 2 and rational exponents. As it turns out, the simplest property to consider in looking for a ■JT definition of *2 ' is monotonicity. Monotonic functions are especially simple- -in particular, they have inverses- -and, as you proved in Part E on page 9-88, the exponential function with base 2 and rational argu- ments is an increasing function. In extending the definition of the expo- nential functions to include irrational arguments, we shall try to preserve [9.07] [9-91] J? this property- -in particular, we shall try to define '2 "in such a way that, for each rational number r, (a) if r < VT then 2 r < 2 2 and (b) if VT < r then 2 Z < 2 r . It will be of help to restate the requirement (a) in terms of a set, <* : 3 r- The requirement (a) is merely that 2 is to be greater than each mem- ber of this set- -in other words, it is to be an upper bound of the set which does not belong to the set. Now, it is not hard to show that, given any member of the set, there is a greater member- -that is, no member of the set is an upper bound of the set, [Proof: Given a rational number r < v 2 then, by Theorem 201 on page 9-67, there is a rational number s such that r < s < v2. Since the exponential function with base 2 and rational arguments is increasing, it follows that 2 > 2 . ] So, require- ment (a) will be satisfied if and only if 2 V ^ is some upper bound of - Now, let's look at requirement (b). Our chance of satisfying this requirement is better the smaller 2 is. So, our best chance of satis- fying both (a) and (b) is to agree that v? r 2 = the least upper bound of {y: 3 < JT Y = 2 }. However, before adopting this agreement, we had better make sure that {y: 3 < jy y = 2 } does have a least upper bound. According to the iubp, all we need do is show that the set in question is nonempty and that it does have some upper bound. This is easy. There is a rational num- ber less than v2 --for example, the number 0. So, the set is not empty --for example, 2 belongs to the set. On the other hand, there is a rational number greater than v2 --for example, the number 1.5. So, if r < v2 then r < 1.5. Since the exponential function with base 2 and ra- /— r 1 5 tional arguments is increasing, it follows that if r < V2 then 2 < 2 ' 1 5 Hence, 2 * is an upper bound of the set. Consequently, the set in question does have a least upper bound. [9-92] [9.07] THE EXPONENTIAL FUNCTIONS The definitions we have so far adopted: k w v x v k>o* k= 77 x ' P =i (2) V x^0 V k<0 xk = "=k» and: (3) V >n V V c T x r = ("fit)*" 1 x>0 r m, rm € I define, for each x > 0, the exponential function with base x and rational arguments- -that is, {(u, y), u € R: y = x }. We shall now remove the restriction c u c R*, by adopting: (4 X ) V >t V x = the least upper bound of {y: 3 < y = x }, ^ V 0 1 and any b, {y: 3 , y = a } does have a least upper bound. [Once this is done, we are sure that (4 X ) makes sense. That (4 ) makes sense then follows from Theorem 164; and there is no question about (4 ). ] Second, we must show that the new definitions are consistent with those already adopted. For example, we must show that, for any a > 1 and any beR, the least upper bound of{y: 3 1 and any b, {y: 3 < L.y = a }i s nonempty and has an upper bound. That the set in question is nonempty follows from the fact that, for any b, there is a rational number less than b [Theorem 201, on page 9-67, or the cofinality principle]. That the set has an upper bound follows from the fact that, for a > 1, the exponential function with base a and rational exponents is increasing, and the fact that, for any b, there [9.07] [9-93] is a rational number greater than b [again, Theorem 201 or the cofi- nality principle]. The second and third tasks are somewhat more difficult, and are carried out in Appendix C. For the third, we shall make use of: Theorem 202. For each x > 0, the exponential function with base x is a continuous function whose domain is the set of all real numbers; if x fi 1, its range is the set of all positive numbers; it is decreasing if < x < 1 and increasing if x > 1. which is proved in Appendix C. EXERCISES A. Use the graph you made in Part C of the Exploration Exercises on page 9-86 to estimate each of the following. [Use Theorems 205 and 207 as suggested in the Hints. ] 1. 2 vT 2. 2 77 3. 2 -77/4 4. 4 VT 5. 8^ [Hint. 2 3/F = 2< 3VT - 1 >.2] 6. 3 3 [Hint. If 2 X = 3 then 3 3 = (2 X ) 3 .] j3. True or false? [Use Theorem 202.] 1. 2 > 2 3 77VT vT 3. 3 ' > 3'3 2. (77/4) 2/3 "> (TT^/ 11 4. (1/vT) 77 > (1/vT) 22 / 7 C. Simplify. 1. 3 VT /3^ 2. (S 77 ) 2 /* 3. [(-6)"] -M*1 2 A [9-94] 4.0 3.5 3.0 — 2.5 — 2.0 1.5 1.0. I L 0.5 J L J L [9.07] J I L -0.2 -0.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 I.I 1.2 [9.07] [9-95] EXPLORATION EXERCISES A. On page 9-94 is a graph of part of the exponential function with base 77. Use it to find approximate solutions to the following equations. Sample , 77 = 2 Solution . This problem amounts to finding a number x such that (x, 2) belongs to the exponential function with base 77, Just find an approximation to the first component of the point whose second component is 2. An approximate solution is 0. 61. 2---V-- 1. 77 = 3.14 x 2. 77 = 3 3. 77 = 1 4. 77 X = 277 5. 7T X = 3.57T 2 J, O* vl. The exponential function with base 77 has [since 77 > 1, and by Theo- rems 202 and 184] an inverse which is an increasing function whose domain is the set of all positive numbers and whose range is the set of all real numbers, [Explain, ] j- «.i^ «^ 'i- "I s T B, 1, Sketch the inverse of the exponential function with base 77, 2. Use your sketch to find approximate solutions to these equations. Sample , V K = 2 ?" <- - Solution . This problem amounts to finding a number x such that (2, x) belongs to the inverse of the exponential function with base 77, Just find an approximation to the second component of the point whose first component is 2, An approximate solution is 0.61. 3x (a) 77 X = 3.3 (b) 77 X = 1 (c) 77 X = VF (d) 77 X = 0.8 (e) 77 = 2.7 [9-96] [9.07; [9.07] [9-97] C» The chart on page 9-96 shows a graph of part of the inverse of the exponential function with base 5. Use it to help you find approximate solutions to these equations. Sample . 25 X = 75 x Solution. 25 =75 (5 2 ) X = 5 2 -3 5 2x /5 2 = (5 2 '3)/5 2 5 2x " 2 = 3 ? --< From the graph, we see that 2x - 2 is approximately 0.68. So, a solution is approximately 1. 34. 1. 25 = 15 2. 125 = 20 3. 5* = -2 4. 5~ X = 10 5. 25 X = 5 X " 2 6. 625 2x- 3 = 18 x 7. 5 = 6250 8. 5 A = 55. 5 9. 5 X = 5~ X 10. 5* = 2.6 11. 5* = 3.7 12. 5 = 2.6 X 3.7 o, „•„ *l„ "C 'i* T Exercises 10, 11, and 12 suggest an interesting use for the inverse of the exponential function with base 5. Let us suppose that we wish to compute 2.6 X 3.7. We know that there are numbers a and b such that 2.6 = 5' and 3. 7 = 5 . So, 2.6X3.7 = 5 a X5 b = 5 a + b From the chart, a = 0. 59 and b = 0. 81. So, a + b = 1. 4- -that is, 2. 6 X 3.7 = 5 1.4 = 5°' 4 X5. Now, from the chart, 5 - 4 = 1.9. Hence, 2.6 X 3.7 = 1.9 X 5 = 9.5. [9-98] [9.07] Here is another example of how the chart on page 9-96 can be used to approximate the results of computations. Suppose that we wish to compute 7. 2 f 3. 5. Now, 7. 2 is not in the range of arguments covered by the graph. But 7.2 = 1.44 X 5, 0. 23 and, from the graph, 1. 44 = 5 c _ 7 . .0.23 1 .1.23 So, 7. 2=5 X5 =5 and, from the graph, 3. 5 = 5 . Consequently, _ - ^ _ , .. ,.1. 23 ^ 0.78 -1. 23-0.78 = 5°« 45 = 2. 1 . As a third example, let's use the graph to approximate (0. 39) . 0. 39 = 3.0 X 10" 1 (0.39) 2 = (3. 9 X 10" 1 ) 2 = (3. 9) 2 X 10~ 2 = (5 * 85 ) 2 X 10~ 2 17 -2 = 5 ' X 10 7 -2 = 5 ' ' X 5 X 10 C = 3.08 X 5 X 10~ 2 = 0. 154 D. Use the graph on page 9~96 to obtain an approximation to each of the following. 1. 7.2X3.5 2. 0.45X3.5 3. 20.2-^6.4 4. (9. 7) 3 5. V60 [Hint. 60 = ? X 25] 6. 40.5-^38.6 7. * 2 ' * 2 * 9 8 ' 5 8. 2. 9tt *t^ O^ -A. ■V '?• "I" [9.07] [9-99] In Part D you used a graph of the inverse of the exponential function with base 5 to make computing easier. You can achieve similar results using a graph of the inverse of any exponential function. You can carry out the work directly on the graph without reading off the ordinates. Here is a graph of the inverse of an exponential function. Let's use it to find an approximation to something simple-- say, 2X3. If the base of the exponential function is b, then AB So, 2 = b and 3 = b 2-3 = b AB .b CD CD = b AB + CD If DE = AB, then CE = AB + CD = GF, where b GF = 6. Hence, 2X3 = 6. Do you have to know the base b when using this geometric method for computing? J«. O^ vl* T" 'i- *(- E_. Use geometric methods and the graph given above to compute each of the following. ,3/2 1. 1.5 X 3 2. 5 4-2 3. 2 4. 24X2.8 [ Hint . Since 24 = 2. 4 X 10, if 2. 4 = b a and 2. 8 = b C then 24 X 2. 8 = b a+C X 10. ] [9-100] [9.08] 9.08 Computing with inverses of exponential functions . --You have seen how the inverse of an exponential function can be used to replace multi- plication, division, and exponentiation steps in computations by additions, subtractions, and multiplications, respectively. Using a graph of one such function- -the inverse of the exponential function for the base 5-- you have carried out some such computations. The results you obtained were probably not very accurate, but this was to be expected, since the approximations you could read from the graphs to solutions of equations of the form *x = 5 y ' were not very accurate. If we could improve the accuracy of these approximations, we would expect the accuracy of our results to be improved. OBTAINING BETTER APPROXIMATIONS For an example, consider the problem o* using the inverse of 7 / D {(x, y): y = 5 } to find (19. 5) . Since 19. 5 is not one of the arguments covered by the graph on page 9 - 96, let us begin by dividing by 5-- 19. 5 = 3. 9 X 5. Now, from the graph, 3.9 = 5°- 85 . So, 19.5 = 5 1 - 85 and (19.5) 7/8 = ( 5 K85 ) 7 / 8 5 1.85X(7/8) ^ 5 1.62 = 5°- 62 X 5. Since, from the graph, we know that 5°- 62 = 2.71, (19. 5) 7 / 8 = 2. 71 X 5 = 13.55. This approximation is not bad, since, correct to the nearest hundredth, (19. 5) 7 ' 8 = 13.45. [Since 13.45 - 13. 55 = -0. 1, the error is about -0. 1. Since -0. 1/13.45 = -3/400, the relative error is about -3/400. The percentage error is about -3/4%.] However, we can do better if we use better approximations than those we can read from the graph. [9.08] [9-101] In fact, it can be shown [later, you will see how] that 3.9 = 5 ' 8456 . Using 0.8456, instead of our earlier 0.85, we find on repeating the pre- ceding steps, that (19. 5) 7 / 8 = 5°* 6149 X 5 [1.8456 X | = 1.6149], A more accurate way than our graphical one of estimating 5 * gives So, we know that 5 ' 6149 = 2.69. (19. 5) 7 / 8 = 2. 69 X 5 = 13.45 --our better approximations have given us the rational approximation to 7 /ft (19. 5) ' ' which is correct to the nearest hundredth. [The percentage error is between -0.05% and 0. 05%. ] One way of obtaining the two approximations we used in the course of the computation is to have a very large-scale graph of the inverse of the exponential function with base 5. The scale would have to be very large for us to be able to read, for example, that the value of this in- verse function for the argument 3. 9 is 0. 8456, correct to the nearest ten- thousandth. Pretty clearly, such a graph would be impractical. Another way of obtaining the approximations we need is to have a table which lists approximations to the values of the inverse function for lots of arguments. Such a table would also be impractical if it had to list values for all the arguments we might ever need. Fortunately, this is not necessary. For one thing, as illustrated both at the beginning and the end of the computation in the example, we need consider only arguments between 1 and 5. All other positive numbers can be brought into this range by multiplying [or dividing] by an integral power of 5. This still leaves us with a lot of arguments, but, as it turns out, we can get sufficiently good approximations to the numbers we are interested in if the table contains approximations to the values of the inverse func- tion for a relatively small number of arguments which are evenly spaced between 1 and 5. [That we can do so is a consequence of the fact that the inverse function—as well as the exponential function itself--is con- tinuous. ] [9-102] [9.08] A MORE CONVENIENT CHOICE OF BASE Such a table would be a great help in carrying out computations but, very probably, none has ever been made. To see why, let's look again at the way such a table would be used. To begin with, unless the num- bers one starts with are between 1 and 5, one must replace them by numbers in this range by multiplying by the appropriate integral powers of 5. This can be a fair amount of work in itself. For example, before using the table to compute an approximation to 0. 0543 X 2536, you would have to multiply the first factor by 25 and divide the second by 625. A similar problem arises at the end of the computation. If, for example, you have found that the number you are computing is approximately c 4 # 6832.. ,. ., . .. . ( . , .. -0. 6832 5 then, after using the table to find an approximation to 5 , 4 you have still to multiply this by 5 . Now, there is an easy way to clear up the first difficulty. Although multiplying and dividing by integral powers of 5 is not too easy, multi- plication and division by integral powers of 10 is a trivial matter. This is, of couse, because we use the base- 10 system of numerals. Each positive number is the product of a number between 1 and 10 [or l] by some integral power of 10. For example, 0. 0543 = 5. 43 X 10 and 2536 = 2. 536 X 10 . So, to approximate 0. 0543 X 2536, we can, instead, approximate 5.43 X 2. 536 and multiply the result by 10. Our table will have to give us approximations for arguments between 1 and 10, rather than merely between 1 and 5. Let's assume that it does. Now, how about the second problem- -multiplying by a power of 5 at the end? This is a little more difficult. We would find, for example, that 5.43 X 2.536 = 5 1,6274 . Since 1, 6274 is beyond the values given in our table [5. 43 X 2. 536 > 10], we must, as at the beginning, proceed indirectly. One way is to find an approximation to 5 " [from the table] and multiply this by 5. This 4 isn't too hard, but if we had, instead, had to multiply by, say, 5 , there wouldn't have been much point in the whole procedure. However, there is another way out. As our table might tell us, 10 i 5 1 ' 4306 . So, since 1.6274 - 1.4306 = 0. 1968, [9.08] [9-103] 5 1.6274 5 0. 1968 x 5 1. 4306 = 5°- 1968 X10. Hence, a better procedure would be to approximate 5 " from the table and multiply by 10. Fortunately, there is a still better procedure. All our difficulties have arisen from the fact that, in making use of limited tables of the inverse of the exponential function with base 5, it is most convenient to have the arguments we must deal with ** reduced" to the interval from 1 to 5 by multiplying and dividing by integral powers of 5- -but, because we use the decimal system of notation, it is not always easy to carry out this reduction. Although we have seen how to get around this trouble, it should be evident that the most simple way out is through it. Either change to the base- 5 system of numeration- -in which case, multipli- cation and division by integral powers of 5 would be as simple as multiplication and division by integral powers of 10 is in the decimal system- -or use the inverse of the exponential function with base 10. Since the decimal system of numeration is well entrenched, we shall adopt the latter alternative. This requires that we have a table of ap- proximations to the values of the inverse of the exponential function with base 10 for closely spaced arguments between 1 and 10. There is such a table on pages 9-367 through 9-369. To see how to use it, let's recal- culate (19. 5) 7 / 8 . From the table, ,0. 2900 So, since 19. 5 = 1. 95 X 10* , and 1.95 = 10 19.5 = 10 1 ' 2900 (19. 5) 7 / 8 = (lO 1 ' 2900 ) 7 / 8 .. 1Q 1. 1288 = 10°' 1288 x 10 _ Now, 0. 1288 is not listed in the table among the approximations to the values of the inverse of the exponential function with base 10. However, from the table, we see that 10°- 1271 = 1.34 and 10°" ' M3 = 1. 35. [9-104] [9.08] Since 0. 1288 is about midway between 0. 1271 and 0. 1303, we shall not 1288 be far off in assuming that 10 ' is midway between 1. 34 and 1. 35: 1Q 0.1288 . x 345> Hence, (19. 5) 7 / 8 = 1. 345 X 10 = 13.45. [As previously remarked, this result is correct to the nearest hundredth.] To check your understanding of the table, use it to verify the follow- ing statements. 2.00 = 10°- 3010 2.01 = lO ' 3032 2.02 = lO ' 3054 4.37 = lO * 6405 lO ' 9390 = 8.69 lO ' 0128 = 1.03 1Q 3. 5527 ^ 3 5? x 1Q 3 85300 = io 4 * 9309 Use two of the approximations you have checked and another one from the table to obtain an approximation to 8. 59 X 1.0 3. EXERCISES A, Use the table on pages 9-367 through 9-369 to find approximations to the roots of the following equations. , in 0.3979 , ln 1.3979 1. 10 = x 2. 10 = x „ in 4. 3979 in -0.6021 3. 10 = x 4. 10 = x [ Hint for Exercise 4. lOx = 10 1 " 0# 6021 ] 5. 10 y = 848 6. 10 y = 84800 7. 10 y = 8.48 8. 10 y = 0.848 [9.08] [9-105] Q in 0.9978 ln ..0. 7042 9. 10 = x 10. 10 = x 11. 10 y = 136 12. 10 y = 4360 13. 10 y = 110 14. 10 y = 101 15. 10 y = 1080 16. 10 y = 1800 Sample 1 . 10 y = 9.5492 The number 9. 5492 is not listed in the table as an argu- ment. However, 9.55 is one of the listed arguments and 9.5492 = 9.55. From the table, 9.55 i 10°- 9800 . So, the root of the given equation is approximately 0. 98. c . _ In 0. 7453 Sample 2 . 10 = x The number 0. 7453 is not listed in the table as a value. However, 0. 7451 is one of the listed values and 0.7453 = 0. 7451. Hence, 1Q 0.7453 . 1Q 0.7451 and since, from the table, 10 0.7451 . 5>5fe> the root of the given equation is approximately 5. 56. ._ in 0.6308 1Q in 0.7610 17. 10 =x 18. 10 =x 1Q .1.0056 yn in 2.4899 19. 10 = x 20. 10 = x 21. 10°' 4769 =x 22. 10 y = 6.3592 23. 10 y = 4.3152 24. 10 y = 521.62 25. 10 y = 6382419 26. 10 y = 1000010 [9-106] [9.08] B. Use the table for the inverse of the exponential function with base 10 in carrying out the computations required in the following problems, [The volume and area formulas you need to set up some of the prob- lems are discussed in Appendix D, pages 9-270 through 9-312. ] Sample 1 . Find the area of a rectangle which is 5. 64 feet wide and 18. 3 2 feet long. Solution . We need to find the product of 18. 32 by 5. 64. 18.32 X 5.64 = 18. 3 X 5. 64 = 1.83 X 5. 64 X 10 = io°- 2625 xio°- 7513 xio 10 (0. 2625 +0.7513 + 1) 1Q 2. 0138 = io°- 0138 xio 2 = 1.03 X 10 2 = 103 The area is approximately 103 square feet. Sample 2 . Find the area of a circle whose radius is 6. 08 inches long. 2 Solution . We need to compute 77(6.08) . 77(6. 08) 2 = 3. 14 X (6. 08) 2 , = ^0.4969,^0.7839^ = io^^xio 1 - 5678 1Q 2.0647 = io°- 0647 xio 2 = 116 The area is approximately 116 square inches. [9.08] [9-107] 1. Compute the area of a rectangular region which is 32. 6 feet long and 17. 2 feet wide. 2. Find the perimeter of the rectangle of Exercise 1. 3. What is the area of a circular region whose diameter is 29.8 centi- meters long ? 4. Compute the circumference of the circle of Exercise 3. 5. /NWWWW If d = 38. 6 and t = 27. 4, what is the area-measure of the square region? [Hint. Factor 'd 2 - t 2 '.] 6. Find the area-measure of the region bounded by a circular sector whose central angle is an angle of 172° and whose radius is 7. 36. 7. The side-measures of a triangle are 38. 2, 46. 3 and 68. 7, respec- tively. Compute the area-measure of the region bounded by the tri- angle. 8. The hypotenuse of a right triangle is 25. 3 inches long and one of the acute angles is an angle of 72°. Compute the measures of the legs and the area-measure of the region bounded by the triangle. 9. How many cubes 2 inches on a side can be stacked in a rectangular box whose inside dimensions are 8 feet 6 inches, 11 feet 4 inches, and 17 feet 10 inches? 10. Compute the area-measure of the circular ring whose inner radius is 3. 27 and whose outer radius is 5.68. 11. Solve the system of equations: 9.82x + 7.63y = 11. 2 4. 83x + 9.21y = 19. 3 12. Compute an approximation to 11 8!/ 114!. [9-108] [9.08] LINEAR INTERPOLATION You have seen one way in which, given an argument [of the inverse of the exponential function with base 10] which is not listed in the table, you can still use the table to get an approximation to the value of the in- verse function for the given argument. Just take the closest approxima- tion to the given argument which is listed, and use the listed approxima- tion to the corresponding value. For most purposes one needs better approximate values than are given by this method. Linear interpolation is a method of getting much better approximations. Here's how it works. Suppose that we want an approximation to the root of: (1) 10 y = 3.443 Now, 3. 443 is not listed as an argument in the table, but 3.44 < 3.443 < 3.45 and, according to the table, 3.44 = 10 0.5366 and 3.45 = 10 0.5378 Drawn to a large scale, a piece of a graph of the inverse of the expo- nential function with base 10 looks something like this: (3.443, (3.44,0.5366) (3.45, 0. 5378) Up to now, we have used the ordinate of A, 0.5366, as an approxi- mation to the root of (1). However, from the sketch, it is clear that the ordinate of the point P whose abscissa is 3. 443 and which lies on the chord AB is a better approximation to the root of (1). Using similar right triangles-- specifically, the similarity APD ♦— ABC--it is not hard to find DP. Adding this to 0. 5366, we have the ordinate of P. Here is how one works it out. [9.08] [9-109] Due to the similarity, we have DP/CB = AD/ AC and, so, AD DP = CB AC = (0.5378 - 0. 5366)' 3.443 - 3.44 3.45 - 3.44 = 0.0012 X °q °q 3 = 0.0012 X 0.3 = 0.00036 = 0.0004. Hence, the ordinate of P is approximately 0. 5366 + 0.0004 and, so, 0. 5370 is an approximation to the root of (1). Here is a convenient form in which to arrange the work. [With practice, you may find that you can do it all in your head. ] Notice the neglect of decimal points. [They are usually omitted, also, from tables like the one we are using.] 10 3450 5378 3443 C?370 3440 5366 12 12X^ = 3.6 = 4 [5366 + 4 = 5370] Explanation : The argument we are interested in is 3 tenths of the way from the smaller to the larger of the listed arguments. So, the value we want is about 3 tenths of the way from the value for the smaller listed argument to that for the larger. The difference between the two values is 12 [really, 0.0012] and 3 tenths of this is about 4. So, add 4 to 5366-- 10°- 5370 = 3.443. The same procedure works backwards to find an approximation to the root of: _ co _ n 10 = x 10 3450 5478 344 3^) 5370 3440 5366 12 l s iS0.3 12 3 Since the value we are interested in is about 3 tenths of the way between the listed values, the argument we want will be about 3 tenths of the way between the corresponding listed arguments. [9-110] [9.08] EXERCISES A. Use linear interpolation to approximate the missing component. 1. (6.387, 3. (3.483, 5. (0.8256, 7. (58.32, ) ) 9. (7.805, ) [ Hint for Exercise 10 . 10 2. , 0. 8440) 4. , 0.9935) 6. , 0.8256) 8. , 1.5121) 10. , -0.3723) -0.3723 _ l0 °' 6277 X 10 _1 ] Note : From now on, use linear interpolation to find approximations to arguments and values not listed in the table. B. Use the table for the inverse of the exponential function with base 10 in carrying out the computations required in the following problems. [The volume and area formulas you need to set up some of the prob- lems are discussed in Appendix D. ] Sample . Solution. Find the area of the triangle whose base is 0. 006523 feet long and whose height is 0.05326 feet. -|- X 0.006523 X 0.05326 = (5 X 10" 1 ) X (6. 523 X 10~ 3 ) X (5. 326 X 10~ 2 ) = 5 X 6.523 X 5. 326 X 10" 6 = io°- 6990 xio - 8144 xio°- 7264 xio- 6 k 2. 2398- 6 = 10 = 10 0. 2398- 4 = 1.737 X 10 The area of the triangle is approximately 0.0001737 square feet. [9.08] [9-111] 1. Find the area of the region enclosed by a parallel og- am which has a base 0.005624 feet long and a height of 0.07893 feet. 2. Find the volume of a rectangular solid whose dimensions are 7. 80 2 meters, 0.9905 meters, and 3.01 meters. 3. Find the volume of a cone whose base has a radius of 9. 86 meters and whose height is 17. 2 centimeters. 4. Find the volume of a solid sphere whose radius is 0.9124 feet long. 5. Find the surface area of the sphere in Exercise 4. 6. Find the volume of a cylinder whose base has a radius of 0. 2745 feet and whose height is 0, 8183 feet, 7. Find the lateral area of the cylinder of Exercise 6. 8. Find the total area of the cylinder of Exercise 6. 9. Find the area of a trapezoidal region whose bases are 3. 768 and 0. 5892 centimeters long and whose height is 60. 14 centimeters. 10. Find the volume of a cube of edge-length 0. 5185 centimeters. 11. Find the area of the lateral surface of a right circular cone whose radius is 1. 247 feet and whose slant height is 4, 687 feet. 12. Find the total area of the cone of Exercise 11. 13. Find the weight of an iron washer whose outer diameter, inner diameter, and thickness are 1 3. 5 inches, 8. 74 inches, and 0. 215 inches, respectively, [iron weighs about 488.8 lbs. /cubic foot.] 14. Find the total surface of a solid hemisphere whose diameter is 6. 875 inches. [9-112] [9.08] EXPLORATION EXERCISES A. Let's use 'exp,' to name the exponential function with base 5. Then we can use functional notation. For example: 2 1 exp,- (2) = 5 = 25 and exp c .(— 3) = r ? v Complete. 1. exp & (3) ~ 2. exp 5 (1) = 4. exp 5 ( ) = 0. 04 5. exp 5 (y) 3. exp ( ) = 0.2 vT 6. exp 5 ( ) = — 7. exp & (1.5) = 8. exp (u) = 9. exp ( ) = 1 10. exp 5 = {(x, y): y } 11. exp.* ) = 12. exp (-2) = 13. exp^ ( ) = -2 14 - *e*p 5 " < x: 15. ft = {y: exp iy 16. exp 5 (3)* exp 5 (4) = exp 5 ( ) 17. exp,-(18) = exp,-(12) • exp 5 ( ) 18. exp (12) = exp 25 ( ) 19. 25- exp 5 (7T) = exp 5 (2+ ) 20. Suppose that exp^a) = b. It follows that b • exp_(zr) = exp_( ). B. By Theorems 202 and 184, exp- is monotonic and has an inverse. Sample . [the inverse of exp, ](1 25) = ? Solution . We are looking for the number a such that exp- (a) = 125. Since expj.(3) = 125, it follows that [the inverse of exp,-](125) = 3. 1. [the inverse of exp e .](25) = ? 2. [the inverse of expjj(5) = ? 3. [the inverse of exp,](0) = ? 4. [the inverse of exp-K ? ) =-1 * [9.08] [9-113] The inverse of exp,. is usually called the logarithm function to the base 5_, and this is abbreviated to 'log '. Thus, log,. (125) = 3. [Read this as 'the logarithm to the base 5 of 125 is 3', or as 'the logarithm of 125 to the base 5 is 3', or as 'log sub-five of 125 is 3'. ] 5. log 5 (625) = ? 6. log 5 (0) = ? 7. log 5 (0.04) = ? 8. log 5 ( ? )=j 9. log 5 (5 6 )=? 10. log 5 < ? ) = 1.5 11. log 5 (exp 5 (2)) = ? 12. exp 5 (log 5 ( ? )) = 4 13. [log 5 oexp 5 ](17) = ? 14. [exp 5 olog 5 ]( ? )=? 15. [log 5 »exp 5 ](-9) = ? 16. [exp 5 °log 5 ](-9) = ? 17. log 5 ( ? ) = -l 18. log 5 (0)=? 19. log 5 (-l)=? 20. The domain of log is {x: }. 21. The range of log- is {y: }. 22. log 5 = {(x, y): } 23. exp 5 (log 5 (7)) = ? 24. 5 5 = ? log 5 (24) log (6) log«-(4) 25. 5 b = ? 26. 5 5 = ? 27. 5 5 = ? 28. 5 < lo g5< 6 > +1 °g5^) = ? 29. log 5 (6) + log 5 (4) = log 5 ( ? ) 30. log 5 (48) = log 5 (12) + log 5 ( ? ) = log 5 (l6) + log 5 ( ? ) 31. log 5 (22) = log 5 (ll) + log 5 ( ? ) = log 5 (44) - log 5 ( ? ) 32. log- (22) = log. (66) + log. ( ? ) [9-114] [9.09] 9.09 The logarithm functions .- -In the preceding section you learned how to use the inverse of an exponential function to translate problems in- volving multiplication, division, or exponentiation into problems which require only addition, subtraction, or multiplication, respectively. These "inverse exponential functions'* are called logarithm functions . Specifically, for < a f 1, the inverse of the exponential function with base a is the logarithm function to the base a. Since, for a > 0, the ex- ponential function with base a is {(x, y): y = a X }, it follows that, for < a ^ 1, the logarithm function to the base a is {(x, y): a y = x}. These graphs show an exponential function with base a > 1 and the logarithm function to the same base. Make a sketch to illustrate the case in which < a ^ 1. Also, sketch the exponential function with base 1 and its con- verse. That exponential functions have inverses --that is, that there are loga rithm functions- -follows from Theorem 202 and Theorem 184. [Why isn't there a logarithm function to the base 1? To a negative base?] Here are pairs of equivalent statements [fill in the missing state- ment]: 4 4 = the logarithm to the base 2 of 16 2 =16 1 = the logarithm to the base 7 of 7 . 7 =7 3 = the logarithm to the base 4 of 64 0. 3010 = the logarithm to the base 10 of 2 ... 10°" 301 ° = 2 [9.09] [9-115] = the logarithm to the base 77 of 1 77 = 1 — 4 -4 = the logarithm to the base 1/2 of 16 . . . (1/2) = 16 [The statements in the first column are customarily abbreviated to: 4 = log 2 16, l = log ? 7, 3 = log 4 64, 0.3010 = log 2, ... Note the fourth of these abbreviated statements. 'log. ' is usually ab- breviated to 'log', and the logarithm function to the base 10 is often called the common logarithm function . Tables like the one you have been using on pages 9-367 and 9-369 are called tables of common logarithms .] As illustrated, for < a ^ 1 and b > 0, the number log b is a the number z such that a = b. In order to prove theorems about the logarithm functions we adopt the defining principle: log x < L > V 00 a a " * As for any inverse function, we have a uniqueness theorem. Here is a proof: Since, by Theorems 202 and 184, the exponential function with base a has [for < a f- 1] an inverse, we know that V 0y = z] - So [taking 'log x* for 'z' and using (L.)], we obtain the uniqueness a theorem: V n s /,V sn V [a y = x => y = log x] [Theorem 212] 00 y L 7 6 a J L J Notice thatamore complicated way of stating what (L) says is: 'n ^ 1 i V ^ n V [y = log x -> a y = x] 00y 7 & a J So, we can think of Theorem 212 as the converse of (L). Together they tell us that, for example, the sentences: 5 3 = 125 and: 3 = log 125 are equivalent. So, a sentence of the form of either of these can be translated into a sentence of the form of the other. [9-116] [9.09] EXERCISES A. Here are sentences of the forms illustrated above. Translate each into a sentence of the other form. 1. 6 3 = 216 2. 3 6 = 729 3. 10 1 = 10 4. log 1() 100 = 2 5. Iogl000=3 6. log 5 3. 5 = 0. 78 7. 10°- 477 = 3 8. 2 10 = 1024 9. 10°' 0781 = 1.197 10. log x 17 = 2.3 11. log 5 y=13 12. log 1. 19 = 0. 0755 13. 10 a = 50 14. 3 a = b 15. b a = c 16. log 4 5 = a 17. log 3 b = a 18. log c = b 6 a 19. 5°= 1 20. 8 = 21. 1 6 =! 22. log Q 3 = \ 23. log 3 9 24. log 1/5 3.89 25. 4 V8l= 3 26. 4 81 = 3 27. 4 2 = 2 4 2 5 _ 5 / A 28. 27 3 = 9 29. 64 6 = 32 30. 64 6 = ±- 32 13. Simplify. 1. log 4 2. io lo s 4 log Q 5 4 3. 0.5 U#!> 4. (log 2 3+log 2 5) 5. 3 J -log 3 2) 6. J 6 log 6 ,)3 7. 31og,7T 6 b 51og Q 2 8. 9 V 9. log 5 4 * 10, -log 7 5 log 5 7 11. 5 3 ' C, Use Theorem 212 to complete each of the following. 1. Since 5 =1, it follows that = log, . [9.09] [9-117] 2. Since 77 = *• it follows that = n 3. Since (log 2 3 + log 2 5) — 3*5, it follows that Since log 2 3 + log 2 5 (log 3 7 - log 3 2) 4. 7 2' it follows that log 3 7 - log 3 2 = • 5. Since it follows that Since 31og A 77 6 b = 3 m 6. -log 7 5 1 7' it follows that • THEOREMS ABOUT LOGARITHM FUNCTIONS Using (L) and Theorems 184, 187, and 182 one can prove: Theorem 21 3 . The domain of each logarithm function is the set of positive numbers and its range is the set of real numbers. Each such func- tion is continuous and monotonic- -decreasing if its base is between and 1 and increasing if its base is greater than 1. Since, for any a, a = 1 and a = a, it follows from Theorem 212 that 1t\s l i (l°g 1=0 and log a = 1) < a/£ 1 & a 6 a [Theorem 214], In a similar way [the same way in which the theorems about the princi- pal root functions were proved], you can prove theorems which justify the computational uses you made of logarithms. For example: V ft ^ y i V v.r.V ^«log (xy) = log x + log y [Theorem 215] 00 y>0 6 a ' to a 6 a l J [9-118] [9.09] Proof . For b > and c > 0, by Theorem 205 [on the exponential functions], (log b+log c) l°g a b lo g a c a = a • a Since b > and c > 0, it follows, by (L), that log b log c a a. a • a = b'c. Consequently, (log b + log c) a a. • a = be. Since b > and c > 0, be > and, so, by Theorem 212, log o (be) = log b + log c. ci a a EXERCISES A. Prove. 1 - V 00 lo Sa(f) = l0g a x * log a y 2 - V 00 V a 1 °6a« xU > = u,1 °8a X 3 - V 00 lo «a (1 / x) = - lo Sa X [Theorem 216] [Theorem 217] [Theorem 218a] * 4 ' V 0 1, log (x - V x 2 - 1 )= -log (x +V X 2 - 1 4. Show that, for x > 1 and for x < 0, log(l- I)._tog(l + » . C^. Use the table of common logarithms on pages 9-367 through 9-369 to find an approximation to the solution of each of the following equations Sample 1 . log 672. 1 = x Solution . Since 672. 1=6. 721 X 10 , log 672. 1 = log 6. 721 + log 10 2 = log 6. 721 + 2 . From the table, using linear interpolation, log 6.721 = 0.8275. So, the root of the equation is approximately 2. 8275. Sample 2 . log x = -2. 7 Solution , log x = —2. 7 if and only if log(x» 10 3 ) = -2.7 + 3 = 0. 3 . From the table we see that this is the case if and only if x* 10 3 = 1. 995. So, the root is approximately 0.001995. 1. log 3 = x 3. log 34. 55 = x 5. log 0.003455 = x 7. log x = 2. 3990 9. log x = -1.4425 11. log 0.000001067 = x 13. log 1000 = x 2. log 3.455 = x 4. log 345500 = x 6. log 0.0618 = x 8. log x = 0.8386 10. log x = 3.6618 - 5 12. log 0.5959 = x 14. log x = [9-120] [9,09] SCIENTIFIC NOTATION As you have seen while using common logarithms for computation, you can find log u, for any u > 0, if you know the values of log for argu- ments from 1 up to 10, Just express the number in scientific notation and use Theorem 215, For example, suppose that u is 984. 7, Then, log u = log 984. 7 = log (9. 847 X 10 2 ) = log 9. 847 + log 10 2 . Similarly, > 0, is it possible to find a number v such that 1 < v < 10 and an inte- ger k such that , u = v X 10 K ? As you might suspect, the answer is 'yes' and we can use our knowledge of the greatest-integer function and of the common logarithm function to prove it. Recall that, for each x, x = ffxfl + tfx}}, where [[xj, the integral part of x, is the integer k such that k < x < k + 1, and {[x]} , the fractional part of x, is x - j[x]], and, so, is nonnegative and less than 1. Examples : 46.38 = J46.38]] + {[46. 38]} = 46+0.38 25 25 3 €f» - 8 + i 3JJ 3 971 = [[97r]l + {[97r}} = 28+0.2744" -1.765 = [[-I. 765]] + {[-1. 765}} = -2 + 0. 235 Now, suppose that we are given a number u > 0. Since u is a posi- tive number, it has a common logarithm. And, of course, this common logarithm has an integral part and a fractional part. (1) logu = Hlogu]] + «loguJ, [9.09] [9-121] where {[log u]j is the integer k such that k < log u < k + 1 and, conse- quently, < {{log ul < 1. Now, by (L), (2) u = 10 6 . So, from (1) and (2), u = 1Q (|[logu]l + {[loguJ) = io^g^x 10 tt lo g u B. Since < {[logu}} < 1, and since log is increasing, (3) 10°<10 l[lo g U ^<10 1 . Hence, u = v X 10 k , where v = I0* logu -" and k = [[logu]]. By (3), 1 < v < 10, and, since k is the integral part of a number, k is an integer. CHARACTERISTICS AND MANTISSAS It is customary to refer to the integral part of log u [for u > 0] as the characteristic of log u and to refer to the fractional part of log u as its mantissa . What we have shown in our discussion of scientific notation is that, given u > 0, the characteristic of log u is the greatest integer k such that 10 < u and the mantissa of log u is, for this k, The simplest shortcut for finding the characteristic and mantissa of the common logarithm of a number u is to express the number in scientific notation: u = v X io k T / r I 'characteristic of log u mantissa of log u is log v [9-122] [9.09] For example, since 243. 1 = 2.431 X 10 2 , the characteristic of log 243. 1 is 2 and the mantissa of log 243. 1 is log 2.431. Also, since 0.002431 = 2.431 X 10~ 3 , the characteristic of log 0. 002431 is -3 and the mantissa of log 0.002431 is, again, log 2.431. So, log 243. 1 = 2 + log 2. 431 = 2 + 0. 3858 = 2. 3858 and logO. 002431 = -3+log 2.431 = -3 +0.3858 = -2.6142. A word of warning about negative characteristics ; Since, for example, log 3.725 = 0.5711, it follows that log 372.5 = 2 + 0. 5711 and log 0.3725 = -1 +0.5711. The warning is to note that, while 2 + 0.5711 = 2.5711, -1+0.5711 £ -1.5711. [Explain.] In fact, in standard decimal notation, log 0. 3725 = -0.4289. Since it is usually desirable to keep track of the mantissas of one's logarithms, it is usually better not to use the standard decimal notation in the case of a negative characteristic. What does often turn out to be convenient is to replace ' — 1', say, by some such equivalent numeral as 4 9 - 10', *2 - 3', or '19 - 20', and simplify. The suggested replace- ments lead to: log 0. 3725 = 9. 5711 - 10 = 2. 5711 - 3 = 19. 5711 - 20 [Some texts abbreviate *— 1 + 0.5711' to '1.5711', putting the oppositing sign over the *1'. Our advice to you is: Don't.] [9.09] [9-123] EXERCISES A. Use the tables of common logarithms to find approximations. 1. log 5. 42 2. log 5420 3. log 0.000542 4. log (5. 42 X 10~ 4 ) 5. log 32. 56 6. logO. 3256 7. logO. 007652 8. log 763. 5 9. 10°* 4771 10. 10 2 - 4771 11. 10 - 2+0 - 4771 12. ICT 1 ' 5229 Note : The exponential function with base 10 is sometimes called the antilogarithm function . So, for example, antilog 3. 6921 = 10 ' . 13. antilog 1. 5569 14. antilog (9. 5569 - 10) 15. antilog 2. 4807 16. antilogO.0837 17. antilog (7. 8125 - 10) 18. antilog 4. 9409 19. antilog (2. 5308 - 3) 20. antilog (9. 0000 - 10) B. 1. What can you say about a number if the characteristic of its common logarithm is negative? Zero? Positive? 2. Suppose that g is the function such that, for each x > 0, g(x) is the characteristic of log x, (a) What is g(19)? g(29)? g(309)? g(3090)? g<0.002)? (b) Complete: V, V x>Q g(10 k x) = + (c) Does g have an inverse? (d) What are the domain and range of g ? (e) Find b given that g(a) = log b and 10 < a < 100. 3. A digital computer computed powers of 2 and kept track of the number of digits in their decimal representations. It turned out that 2 has 302 digits. Use this to find the decimal repre- sentation of log 2 correct to 3 decimal places. 4. How many digits are in the decimal representation of 3 ? [9-124] [9.09] C, Use common logarithms to carry out the indicated computations. Sample 1 . 0. 52 x 13. 7 Solution . log (0. 52 X 13. 7) = log 0. 52 + log 13. 7 log 0. 52 = <).7/ o3 4*7400 log 27 = 1.43/4 (X) 3 3 log 27 = 4*??4X log 0.362 = \ m 5$d7 - 2 (X) 1/2 \ log 0.362 = 0. 770 log b X= Toi~b which gives us the change-of-base formula. Carry out the proof along the lines indicated above, [a slight vari- ation on the proof goes this way: log b c / log c\ b = c; so, log \b /= log c. to a i b a Hence [by Theorem 217], log.c • log b = log c. Consequently, .. . ] d ^l a Using Theorem 218b [which is, in fact, an instance of Theorem 219], the formula of Theorem 219 can be rewritten: log b x = log a x«lo gfc a Since a -f 1, log, a £ 0. So, we see that each logarithm function can be obtained from any other by multiplying the latter by a nonzero constant. Conversely, each nonzero multiple of a logarithm function is, again, a logarithm function. [9-132] [9.09] [9.09] [9-133] EXERCISES A. 1. Prove the last statement made in the text. [ Hint . Given < a fi 1 and c / 0, for what number b, if any, is c = log, a? If you can find such a number b, is it positive and different from 1?] 2. Use a table of common logarithms to approximate each of the following: (a) log 3 5 (b) log u 2 (c) log ? 265 (d) log 9g 36 (e)log 0>5 13 (f) log l/3 64 3. Solve for 'x': log, a* log c • log x = a b a c 15. Several logarithm curves are shown on page 9-132. 1. Why do all the curves pass through the point (1, 0)? 2. Fill in the blank so as to make the following sentence true. For each positive abscissa, the ordinate to the logy — - curve is the ordinate to the log -curve. 3. Why is it the case, for each of the log curves drawn on page 9-132, that points on the curves which are to the right of (1, 0) are above the x-axis? 4. State a sentence like that in Exercise 2 but with 'log- ' in place 5. Sketch, on page 9-132, the log -curve. C_. The chart on page 9-132 illustrates the fact that all log curves have much the same shape. One can distinguish between two log curves by comparing how steep they are at (1, 0). Fill the blank in such a way as to make the following sentence true. For each two numbers a and b, both greater than 1, the log -curve is steeper at (1, 0) than the log -curve if and only if . [9-134] [9.09] MISCELLANEOUS EXERCISES 1. Simplify. / \ a ~ 4 (a) 7 (b) 3t t - 1 t + 1 2. Solve these systems. (a) r4x + 7y,29 1 x + 3y = 11 (b) 9y + 10x - 290 = 12x - lly - 130 = 3. Suppose that ship A is heading north at 1 2 knots and ship B is head- ing east at 9 knots. If A sights B 10 nautical miles directly north of A at 12 noon, at what time will they be the least distance apart? 4. Consider the semicircle AXB with AB as diameter. Let N x , N 2 , N , ... be points of AB which divide it into congruent segments such that AN 1 < AN 2 < AN 3 < . . . . Let P^ P 2 , P 3 , ... be points of AXB such that P 1 N 1 , P 2 N 2 , P 3 N 3> . . . are perpendicular to AB. Show that -, -, -, AP AP, AP, 5. Factor. (a) 28 - x - 2x 2 (c) m p 3mpq - 10q' (b) 6z + 7z - 3 (d) k 2 j 2 - 16kj + 39 (e) 4m - 1 (h) b - by (f) 1 - 49a (i) 7r + r (g) y - 9 / x 6 6 A (j) s t - 4 6. Show that 20 21 100 > 100. 7. Simplify. (a) log -g- - 2 log - + log 243 (b) log \-3 7-5 7- 4 .5 3 ^ 7- 1 >5 5 2. 7 -3 -5 8. Solve for 'x' x+ 1 . , x - 1 a - b 2x [9.09] [9-135] 9# Solve these systems. f 9x~ l - 4y~ 1 = 1 f 5x~* + y " l = 79 < a M -1 -1 (b) -i -i k I8x + 20y = 16 Ll6x - y = 44 10. Solve these systems for *x* and *y*. ( ax - by + ex = be f xa + yb = c "» V^ x -a =b - y v. xb = ya y - * 11. Solve the system: < 3 2 3 x = y r 3x 10. 2y [a = m b ' 12. Solve for *x' and *y*: \ 2 «i 3 L a x = m b y p a + b = 6 b 13, Solve the system: < . , . . ^ 3 a-l = 2 b + l 14. Simplify. (a) (3x 4 - 10x 3 y + 22x 2 y 2 - ?2xy 3 + 15y 4 ) -r (x 2 - 2xy + 3y 2 ) (b) (x 7 - 5x 5 + 7x 3 + 2x 2 - 6x - 2) -r (1 + 2x - 3x 2 + x 4 ) 15. Solve these equations. (a) 8(x - 1) - 17(3 - x) = 4(4x - 9) + 4 (b) 8(x - 3) - 2(3 - x) = 2(2 + x) - 5(5 - x) (c) 5a - (3a - 7) - [4 - 2a - 3(2a - 1)] = 10 16. Consider a convex quadrilateral one of whose diagonals bisects the region bounded by the quadrilateral. Prove that this diagonal bi- sects the other diagonal, 17. If, in AABC, AC = 8 and ^A is an angle of 30°, find the measure of the altitude CD. [9-136] [9.09] -=- . If A is multiplied by 9 and D is a constant, what change takes place in B ? 19. What is the greatest common divisor of 2160, 1344, and 1440? 20. If a first number is five times a second then jr of the second is % of the first. 21. Suppose that the ratio of the length-measure to the width-measure of a rectangle is 4 to 1. If the rectangle has the same area- measure as that of a square with side-measure 8, what are the dimensions of the rectangle? 22. Expand. (a) (3x 4 - 2x 3 + x - 7)(3x 2 - x + 1) (b) (5a 3 b 3 - 3a 2 b 2 + 2ab - l)(2a 2 b 2 + ab - 5) 23. If 360 is divided by a certain number, this divisor will be 1 more than the quotient and 1 more than the remainder. Find the number. 24. Solve the equation: 0.0000358 = 3. 58 X 10 X 25. The perimeter of this regular hexagram is 120. What is the area-measure of the region it bounds? Hypothesis : QP J- RS at T, T € QP r> RS , I and II are square regions, III and IV are rectangular regions Conclusion : Kj x l< n > (K m + Kj y ) [9.09] [9-137] 27. Suppose that O is the sequence of odd positive integers and E is the sequence of even positive integers. Prove: n n 2n n (a > 1°p + Z e p= Z p ,b) E °p = n *- (m - l » p=l p=l P = 1 p = m 2 n n m - 1 '"P p = m P = 1 P = l -2 -2 28. If a = -2 and b = 3 then a b - ab = 29. If a = 16, b = 8, c = 4, and d = 2 then (a - bHcd)"" 1 EXPLORATION EXERCISES Consider the tripling function. Is the triple of the sum of two numbers the sum of their triples? Since the answer to this question is 'yes*, the tripling function satisfies [that is, is a solution of] the open sentence (1) V V f(x +y) = f(x) +f(y) x y In other words, if you replace *f* in (1) by a name for the tripling func- tion, the resulting statement is a theorem. To prove this theorem, suppose that g(x) = 3x, for all x --that is, suppose that g is the tripling function. Then, by definition, g(a + b) = 3(a + b), and, since g(a) = 3a and g(b) = 3b, g(a) + g(b) = 3a + 3b. Since, by the idpma, 3(a + b) = 3a + 3b, it follows that g(a + b) = g(a) + g(b). Consequently, V V g(x + y) = g(x) + g(y). x y So, g satisfies (1). [9-138] [9.09] Here are some open sentences which are satisfied by various func- tions. (1) V x V f(x + y) = f(x) + f(y) (2) V x>0 V y>0 f(x y ) = f W + W (3) V x V y f(x + y) = f(x)f(y) (4) V V f(xy) = f(x)f(y) x y (5) v x>o v y >o f(x y ) = f(x)f( y } Does the tripling function satisfy (2)? Let's see. For a > and b > 0, g(ab) = 3ab, g(a) = 3a, and g(b) = 3b. Is it the case that, for each a > and each b > 0, 3ab = 3a -!- 3b? No. So, the tripling function does not satisfy (2). Does the tripling function satisfy (3)? Let's check. Is it the case that, for each a and each b, 3(a + b) = (3a)(3b)? No. So, it doesn't satis- fy (3). Does it satisfy (4)? [if it did satisfy (4), would it satisfy (5)?] Does it satisfy (5)? A. You have seen that the tripling function satisfies (1). For each of (2), (3), (4), and (5), find a function which satisfies it. 13. In each of the following exercises you are given a function. For each function, tell which one(s) of the five sentences it satisfies. 1. the doubling function 2. h(x) = x 3. log 4. log ? 5, the linear function defined by 'y = 7x + 3' 6. y = 7Tx 7. {(x, y): y = 2} 8. the constant function 1 9. 10. the identity function [y = x] 11. g(x) = 71" 12. h(x) = x , for x > 13. the reciprocating function 14. the exponential function with base 3 1 5. y = Vx 16. y = (1/2) X 17. {(u, v): v = 2 _U } 18. y = X 19. y=log(3 X ) 20. y=(log3) X 21. y = 1 X [9.10] [9-139] 22. a function f such that, for some c £ 0, f(x) = ex 23. {(u, y): y = a }, where < a ^ 1 24. {{x, y), x > 0: y = x }, for some c different from and 1 25. the function g<»log, where g is a function which satisfies (1). C. Consider the function g which is defined for all real numbers as fol- lows. For each x, if there are rational numbers r and s such that x = r + sv2 then g(x) = r; otherwise, g(x) = 0. So, for example, g(5) = g(5 + 0vT) = 5, g(3 - VT) = g(3 + -lVF) = 3, g(0) = g(0 +0V7) = 0. 1. Show that g(v3 ) = 0, [ Hint . If there were rational numbers r and s such that r + sVT = VT then (r + svF) = ... . ] 2. Compute g(l + v3 ). 3. Does g satisfy (1)? 4. Does g have a subset- -say, f--such that v x £ > f V> £ f(x + y) = f(x) + £(y>? 9. 10 Some laws of nature . --In this section you will see one way in which mathematics helps in discovering physical laws. In the process you will learn more matheinatics and become acquainted with a number which crops up in many parts of mathematics and in many of its applications. In fact, this number is so ubiquitous that, like H, it has been given a letter name, 'e'. GAY-LUSSAC'S LAW FOR GASSES In an earlier unit we talked about Boyle's Law for gasses which says that the volume V of a gas sample whose temperature is constant is in- versely proportional to its pressure P. Gay-Lussac's Law describes the behavior of a gas sample whose pressure is constant. Let's see how such a law might be discovered. [9-140] [9.10] A physicist wishes to determine whether the volume V of a gas sample whose pressure is kept fixed is a function of its temperature T and, if so, what function V is of T„ To do this he might make use of an apparatus like that shown below. The pressure on the gas in the gradu- ated tube can be kept fixed by raising or lowering the mercury reservoir gas mercury so that the difference in the levels of the two surfaces of the mercury remains the same. The volume of the sample is read from the gradu- ations on the tube, and the thermometer gives the temperature of the gas in degrees centigrade. A series of experiments is then carried out. The physicist varies the temperature of the gas sample by applying a flame to the tube or packing the tube in some cooling agent. Each read- ing of temperature and volume gives a pair (t, v) of numbers. Here is a table of some of the ordered pairs he obtains: t -4 20 16 76 44 48 28 16 60 12 52 84 28 V 49.3 53.7 52.9 63.9 5 8.1 5 8.8 55.1 52.9 61.0 52.2 59.5 65.4 55.1 The physicist finds that any two experiments which yield the same value for the temperature also yield the same value for the volume --more correctly, the differences are small enough to be laid to experimental errors [errors in adjusting the height of the mercury reservoir or in making the two readings]. So, he concludes that the volume is, indeed, [9.10] [9-141] a function of the temperature --that is, that there is a function g such that V = gcT. In order to discover the function g, he again examines his data. What he finds is that, to within experimental errors, whenever the dif- ference in the temperatures he measured in two experiments is the same as the difference in temperatures for another two experiments, it is also the case that the differences in volumes for the two pairs of ex- periments are the same- -changing the temperature by a given number of degrees always results in the same change in volume, no matter what temperature he starts from [as long as the pressure is kept constant]. Now, what this means is that there is a function f such that, for any two temperatures t x and t 2 and corresponding volumes g(t x ) and g(t_), (1) g(t 2 ) - g^) = f(t 2 - t x ). In particular, taking t ± to be and replacing 't * by *x', (2) V x f(x) = g(x) - g(0). Consequently, if we can find what the function f is then, knowing the vol- ume g(0) of the gas sample at 0°C, we can find the function g and, so, learn how the volume of the sample depends on its temperature. To discover f we begin by noting that, by (2), for any t x and t 2 , f(t 2 ) - f(t x ) = [g(t 2 ) - g(0)] - [g(t 1 ) - g(0)] = g(t 2 ) - g(t x ) ^ (1) = f(t 2 - t x ), J that is, f(t 2 ) = f(t 2 - t x ) + f(t 1 ). This result becomes more illuminating if we replace l t 2 - t x * by 4 x', *t 1 ' by *y*, and, of course, *t 2 ' by 'x + y'. We then see that f must satisfy: (3) V V f(x +y) = f(x) +f(y). x y The generalization (3) may suggest to you the idpma. In fact, if f is a function such that, for some c, (4) f(x) = ex then, by the idpma, f satisfies (3). [Such a function f for which c f- is [9-142] [9.10] called a homogeneous linear function. ] Since the physicist's function f cannot well be the constant function whose value is 0, this suggests the idea that the physicist's function f may be a homogeneous linear function. We could be sure that this is the case if we knew that each function which satisfies (3) is a homogeneous linear function. [Notice that this is quite different from what we just discovered- -that each homogeneous linear function does satisfy (3). ] As a matter of fact, (3) has many solutions which are not homogeneous linear functions. However, as is shown in Appendix E [Theorem 221], except for the constant function whose value is 0, these extra solutions are very queer functions indeed. Each such function f has the surprising property that, given any ordered pair of real numbers, there is an ordered pair (x, f(x)) which belongs to the func- tion and is as close to the given ordered pair as you wish. Now, such a function — one which contains ordered pairs arbitrarily close to each ordered pair in the whole number plane- -certainly cannot be the physi- cist's function. [For one thing, as the physicist may observe, his func- tion is an increasing one, and none of these queer solutions of (3) can be as simple as that. ] So, since each solution of (3) is either a homogeneous linear function or a ''physically impossible" function, the physicist's function f must be linear and homogeneous. Returning now to the function g, we see from (2) and (4), that, for any temperature t, g(t) = g(0) + £(t) = g'(0) + ct, for some number c -f 0. The more usual form of writing this is: g(t) = g(0)[l +at] in which, of course, a = c/g(0). Since g is the function such that V = g<»T, we see that, if v is the volume-measure of the gas at 0°C then (5) V = v Q [l +aT]. [Gay-Lussac's Law] [Surprisingly enough, it turns out that the constant a is nearly the same for all gas samples- -no matter what size the sample is, no matter what kind of gas, and no matter at what value the pressure is fixed, the con- stant a is approximately 0.00367. ] [9.10] [9-143] EXERCISES A. A certain gas sample has a volume of 120 cubic centimeters at a temperature of 0°C. If, without changing the pressure, the temper- ature is reduced to -273°C, what is the new volume of the sample? [Actually, any gas sample will liquefy at sufficiently low tempera- tures and, being no longer a gas sample, Gay- Lus sac's Law will not apply. Since l/a = 273, any gas sample which remained a gas at about -273"C would have volume-measure 0! This, and other evi- dence, suggests that nothing can be colder than — 273*C (approxi- mately), and this temperature is called absolute zero . The absolute temperature of an object is its centigrade temperature plus 273. ] O^ vl* o, *T* "i v T* According to Gay-Lussac's Law, if the pressure of a gas sample is held constant and its volume at 0°C [at the given pressure] is v Q then its volume v at t°C is given by the formula: v = vjl +at] = v a[t + I] By definition [see above], t + — is the absolute temperature. So, Gay- Lussac's Law says that, under isopiestic conditions [fixed pressure], the volume V of a gas sample is proportional to the absolute temperature T. Of course, the coaetant of proportionality depends on the given pressure. Hence, what Gay- Lus sac's Law tells us is that, for each gas sample, there is a function g such that (1) V = (goP)'T. [For each pressure p Q , g(p ) = v Q a. ] Now, according to Boyle's Law, under isothermal conditions [fixed temperature], P is inversely proportional to V. In this case, the con- stant of proportionality depends on the fixed temperature. Hence, Boyle's Law tells us that, for each gas sample, there is a function b such that (2) PV = boT. Let's see if we can find out what function bis. [9-144] [9.10] Since, from (1), (3) PV=P-[(g.P)'T] = [P-(g°P)]"T, it follows from (2) and (3) that boT = [P-(g°P)]'T, that is, that (4) (b.T) v T = P'(g°P). Now, (4) tells us a surprising thing. Choose some value of T. Accord- ing to (2), we can get any value of P we like by varying V. By fixing a value of T, we have fixed the value of the function (b« T) t T. So, the function P*(g°P) is some constant function, say, c. Hence, from (3), (5) PV = cT. [General Gas Law] [Of course, the function (b°T) f T is also a constant function. Why? So, what kind of a function is b?] Since for different gas samples we have different functions g and b [for example, looking at (1) we can see that for a sample containing twice as much gas as another, the function g for the first sample is twice the function for the second], it is reasonable to expect that the function c will vary from one sample to another. This is the case, but interestingly enough, c depends only on the number of molecules in the sample and not on the kind of gas. In fact, c is proportional to the num- ber N of molecules in the sample. If the values of P and V are in metric units [dynes per square centimeter and cubic centimeters] then the con- stant of proportionality is approximately 1. 38 X 10 . So, (6) PV = 1. 38 X 10~ 16 NT. It should be remarked that while Gay-Lussac's Law holds with con- siderable accuracy over a large range of volumes and temperatures, the same cannot be said for Boyle's Law. For most gasses, the latter holds only for moderate ranges of values of P, V, and T. So, the same applies to (6). Example . The volume of a gas sample is 137.9 cubic inches at 294°C and a pressure of 28.6 pounds per square inch. What is its volume at 137°C and a pressure of 35.6 pounds per square inch? [9.10] [9-145] Solution , In view of formula (5), we see that P 2 V 2 *2 Pi v. So, Hence, 35.6 X v 2 137 + 273 28.6 X 137.9 294 + 273 ' 28.6 X 137.9 X 410 v. 2 35. 6 X 567 By logarithms, v 2 = 80.12. So, the volume is about 80. 12 cubic inches. kV Ox Ox "i" "i x "r B. Solve the following problems. 1. A gas sample under a pressure of 38.2 pounds per square inch has a volume of 183. 5 cubic inches at a temperature of 86. 3°C. What is its volume under a pressure of 46. 3 pounds per square inch and at a temperature of 47. 5°C? 2. The pressure on a gas sample is doubled and its absolute tem- perature is doubled. What change takes place in the volume? 3. If the pressure on a gas sample is increased by 25% and the temperature is kept fixed, what is the %-change in the volume? 4. If the pressure on a gas sample is increased by m% and the ab- solute temperature is decreased by n%, what is the %-change in the volume ? 5. A gas sample under a pressure of 43, 6 pounds per square inch has a volume of 216. 3 cubic inches at a temperature of 74. 3 C C. What is its volume under a pressure of 53.6 pounds per square inch and at a temperature of 74. 3'C? 6. Suppose that the temperature of a gas sample is increased from 50°C to 75°C without a change in pressure. What is the %-change in the volume? [9-146] [9.10] 7. A gas sample had, at one time, a pressure of 1 5 pounds per square inch and a volume of 1 cubic foot. At a later time its pressure was 20 pounds per square inch and its volume was 1. 5 cubic feet. (a) At which time did the gas have the higher temperature? (b) What is the ratio of the absolute temperature-measures of the gas at the two time s ? (c) What is the ratio of the centigrade temperature-measures of the gas at the two times? (d) If the temperature of the gas at the earlier time was — 10*C, what was its centigrade temperature at the later time? C. Pretend that someone is trying to discover what function the peri- meter P is of the length-measure L of a rectangle of constant width. He accumulates some data: 1 15 13 1 18 j 30 25 23 13 18 p 50 46 I 56 ! 80 70 66 46 56 He knows that there is a function g such that P = g°L. Examining the data further leads him to suggest that the difference in perimeters is a function, f, of the difference in corresponding length-measures. That is, for any £ > and i, > 0, g(i 2 ) - g U x ) = f( ? ). In particular, taking £ x to be, say, 30, { * } V x>0 f(x " 30) = gU) " ? - Now, to discover f, he notes that, for any £ > and 1^ > 0, f(i 2 - 30) - i(£ 1 - 30) = [g(i 2 ) - _J_] - [gUJ - _?_] = gtf 2 ) " _L_ = f( ? ), [9.10] [9-147] that is, f(i 2 - 30) = f( ? ) + f(i x - 30) Replacing *i 2 - i^ by 'x' and 'i 1 - 30' by 'y\ and, of course, 4 i g - 30' by * ? ', we see that f must satisfy: V x> _ 30 V y> _ 30 f<_^) = f(_?_) +£(_?_) Aside from certain queer functions, the only functions which meet this condition are subsets of the constant function and of linear homogene- ous functions. From the nature of the problem [for example, f is an increasing function], we can rule out the queer functions and 0. Hence, f is a subset of a homogeneous linear function. So, there is a number c -f such that v x> . 30 f(x) =_L_. Substituting in (*), we have: v x>0 c •(_!_) = g(x) - _JL_ In other words, for some number c, and for any length-measure St > 0, (**) gtf) = ? +c'( ? ). We can tell what number c is by using any ordered pair from the table. Take the pair (15, 50), and substitute in (**): 50 = ? + C (15 - 30) Hence, c = ? , and g is the function such that, for each i > 0, g(i) = 20 + _?_. Since P = g » L, we see that P = ? DECAY OF A RADIOACTIVE SUBSTANCE Now, let's see how a physicist might tackle the problem suggested in the Preface [page 9-iii]. There it was said that a sample which, at one time, contains A(0) grams of a certain isotope of strontium will, t years later, contain A(t) grams of the isotope, where A(t) = A(0)« 10" 2t . [The reason for the loss of weight is that some of the atoms of the [9-148] [9.10] strontium isotope have "decayecT'into less heavy atoms of another element. ] Our problem is to see how the above formula might be dis- covered. To begin with, the physicist needs some experimental data. For a start he may collect several samples from the same source and make a chemical analysis of some of them. This analysis can tell him how much strontium isotope he has to start with in one of the remaining samples. Suppose that this is A(0) grams at time 0. Then, at various times t, he can, from the weight of the sample, calculate the weight of the strontium isotope still left in the sample at these times t. In this way he builds a table of approximations to values of the function A for various arguments. In order to discover what the function A is, the physicist examines the data he has accumulated. In doing so he discovers that, to within experimental error, whenever the time interval between two weighings is the same as that between two others, the ratio of the weights of stron- tium isotope for the first two times is the same as the ratio for the other two- -during all intervals of the same duration, there is the same per- centage weight loss in the strontium isotope. Now, what this means is that there is a function f such that, for any times t x and t 2 , with t 2 > t x > 0, A(t 2 ) (1) "^ = f(t 2 - t x ). In particular, taking t x to be and replacing 't 2 ' by l x', (2) V x>Q f(x) = A{x)/A(0). Consequently, if we can find the function f then, knowing A(0), we can find the function A. To discover f we begin by noting that, by (2), for t 2 > t^ > 0, f(t 2 ) A(t 2 )/A(0) Wj " A(t x )/A(0) A(tJ A( tl ) S (1) = f(t 2 - tj. [9.10] [9-149] that is, f(t 2 ) = f(t 2 - t 1 )f(t 1 ). It is convenient to replace, here, 4 t 2 - t x ' by l u', *t x ' by V, and, of course, *t ' by l u + v'. Doing so, we see that f is a function such that (3) V u>0 V v>0 f(u + v) = f(u)f(v) « The generalization (3) may suggest to you Theorem 205: w w w u +v u v V .. A V V x = x x x>0 u v and, so, suggest that the function f is an exponential function restricted to nonnegative arguments. This is, in fact, the case. To prove it, we make use of the result on homogeneous linear functions described on page 9-142. We begin by noting that if f is any function which satisfies (3) then, for any u > 0, f(u) = f(| + f) = [f(f)] 2 > 0, and since, in particular, f(0) = [f(0)] 2 , either f(0) = or f(0) = 1. Also, if f(0) = then, for any u > 0, f(u) = f(u + 0) = f(u)f(0) = f(u)'0 = --that is, f is the constant function whose value is and whose domain is {x: x > }. Using a slightly more complicated argument [for which, see Appendix E] it can be shown that if f(0) = 1 then either f is the ex- ponential function with base or, for each u > 0, f(u) > 0. To discover what f may be in this last case we consider the function F where F = log °f. By (3) [since, for each u > 0, f(u) > 0], it follows that, for each u > and v > 0, log (f(u + v)) = log (f(u)f(v)) = log (f(u)) + log (f(v)) --in other words, (4) V u>0 V v>0 F(U + V) = F(u) + F(v) * Now, just as in the case of (3) on page 9-141, it can be proved that any function F whose domain is {u: u >0} and which satisfies (4) is either a homogeneous linear function, or the constant function 0, restricted to nonnegative arguments, or a very queer function- -so queer that there are pairs (u, F(u)) belonging to the function which are arbitrarily close [9-150] [9.10] to any given ordered pair whose first component is nonnegative. Since, whatever F is, f(u) = 10 F(u) [for u >0], it follows that, in the case we are considering, either there is a number c such that, (5) f(u)=10 CU [foru>0] or there are pairs (u, f(u)) arbitrarily close to any point in the first quadrant. Summarizing, the only functions which satisfy (3), above, are these: (i) the constant function whose value is and whose domain is {x: x > 0}, (ii) the exponential function with base 0, (iii) the exponential functions with base [10 C ] greater than restricted to {x: x > }, and (iv) some very queer functions. Since physical considerations show that the function f which we are seeking [see (3), above] cannot be one of those described under (i), (ii), or (iv), it must be an exponential function with positive base. So, by (3) and (5), (6) V t > Q A(t) = A(0)-10 ct , for some number c. The number c can be determined if we know the value of A for some argument in addition to 0. For example, c = log (A(1)/A(0)). [Explain.] Measurements made on strontium isotope samples of the kind referred to in the Preface give c = — 2 [when, as we have been assuming, time is measured in years]. EXERCISES A. The function A such that, for all t > 0, (*) A(t) = A(0)' 10" 2t describes the decay of a certain strontium isotope. It tells us that if we start with a sample containing A(0) grams of this isotope, [9.10] [9-151] ~2t there will be A(0) • 10 grams left at the end of a t-year period. 1. The value — 2 of *c* was determined on the basis of using one year as the unit of measurement for time. What is the value of l c' if the time -unit is one day? One century? 2. How many years will it take for a sample containing this strontium isotope to decay to a sample which contains just half as much of the isotope? [ Hint . Suppose that x is the required number of years. Then, A(x) = [2- A(x)]« 10~ 2x . So, ... .] Since radioactive substances do decay in the manner described by (6), one can record the facts concerning the decay of such substances by merely listing, for each substance, the experimentally-determined value of 'c 1 . A more common practice is to give the time required for half the isotope to decay. This is called the half-life of the substance in question. 3. In terms of 'c\ what is the number of years in the half-life of any radioactive substance ? 4. What is the half-life in days of the strontium isotope referred to in ( *) ? 5. The strontium isotope in Exercise 4 is Sr 89 . Unfortunately, the stron- tium isotope which people worry about is Sr 90 and its half-life is 25 years. What is its law of decay? 6. The half-life of uranium- 232 is 70 years. How long will it take 3/4 of the U 232 in a sample to decay? 5 7. The half-life of uranium- 234 is 2. 48 X 10 years. How long will it take 1/4 of the U 234 in a sample to decay? — 2t 8. (a) Draw a graph of l y = 10 ', plotting points for t = 0, 0. 1, 0. 2, . .. , 1. (b) Relabel the vertical scale so that what you have is a graph of the decay law of Sr 89 . [9-15Z] [9.10] B. Solve these problems. 1. The half-life of radium-226 is 1620 years. What fraction of the Ra 226 in a sample remains after 810 years? 2. Due to radioactive decay, the neptunium-237 content of a sample decreases by 75% in 4.4 million years. What is the half-life of Np 237 ? 3. One of the isotopes of silver decays to 75% of its initial amount in about a half-hour. What is its half-life ? ANOTHER APPROACH TO RADIOACTIVE DECAY The appearance of the '10' in generalization (6) on page 9-150 [and in the earlier formula (5)] was purely fortuitous- -had we, for example, defined the function F to be log ? °f rather than log <> f, we would have at found that, for t > 0, A(t) = A(0) '2 , for some number a [a = c/log 2. Why?]. The function A can, evidently, be thought of as the product of the constant A(0) and a function determined by composing the exponential function with any positive base different from 1 with a properly chosen homogeneous linear function restricted to nonnegative arguments. There seems no very compelling reason to prefer one base to another. [The use of 10 as a base might simplify computations and, for base 2, the number a is related in a particularly simple way to the half-life [How?]. But the fact that we use base- 10 numerals and that we choose to speak of half-lives [rather than, say, third-lives] doesn't have much to do with the nature of radioactive decay. ] However, approaching the problem of decay in a different way will lead us to discover a number which is, in a real sense, the most "natural" base to use. This number occurs in many parts of mathematics and its applications and is denoted by the letter 'e'. The function log is called the natural logarithm function [and 'log ' is abbreviated 'in'], and the exponential function with base e is called the exponential function. To discover what e is, let's consider what a second physicist might do when confronted by the first one's table of values of the function A. To this second physicist it seems likely that, for time intervals of the same duration, say h, the number of atoms which decay should be [9.10] [9-153] proportional to the number of undecayed atoms present at the beginning of the interval. So, if N(u) is the number of undecayed atoms present at time u then the ratio N(u) - N(u + h) N(u) should be the same for all u > 0. Since, also, A is proportional to N, it should be the case that the ratio A(u) - A(u + h) A(u) is the same for all u > [Explain. ]. The physicist checks this conjecture, using the values of A listed in his table, and finds that, to within experimental errors, it is con- firmed- -so, there is a function g such that, for h > and u > 0, A(u) - A(u + h) _ , aTuI " g (h) - In checking his conjecture, the physicist has accumulated a table of values of g and finds, naturally enough, that g is an increasing function --the longer the sample sits around, the more of its atoms will have decayed. The physicist now makes another conjecture- -that g(h) is pro- portional to h. If this is correct, there should be a positive number \ ["lambda"] such that, for h > 0, g(h) = \h. To check this, he adds a new line to his table in which he lists values of 'g(h)/h*. This time, he is not so lucky. The entries are not the same. However, he does notice that, for small values of 'h', the values of 'g(h)/h' are all fairly close to 4.6, and are closer the smaller the value of l h\ So, he modifies his conjecture and guesses that, for small posi- tive values of 'h', g(h) = \h where X = 4. 6. Summarizing, his data suggest that, for small h > 0, and all u > 0, <»> A ' U> hAfa" + ^ = X I" 4 " 61 ' and that the error can be made as small as he wishes by taking h suf- ficiently small. [\ is called the decay constant . ] This suggests a way of computing approximations to the values A(t) [9-154] [9.10] of A. For any t > 0, if n is a large positive integer then t/n is a small positive number and, choosing this for h, one can, knowing A(0), use (1) with u = to compute an approximation to A(t/n). Then, using this approximation, one can take u = t/n in (1) and again using t/n for h, compute an approximation to A(2t/n). After n such steps, one would obtain an approximation to A(nt/n), that is, to A(t). In order to do this, let's rewrite (1) as: A(u +h) = (1 - Xh)A(u) and replace 'h' by l t/n' to get: (2) A(u +r) = (1 For u = 0, this tells us that n U n )A(u) A{ ±) i ( i . M) A (0). n n Using (2) again, this time for u = — , we find that A( iL) i (i . M )A (i) n n n = (1 -) 2 A(0). n Since t = nt/n, repetition leads eventually to the formula: (3) A(t) = (1 - ^) n A(0) n Of course, since (2) yields only approximations, its repeated use may result in rather large errors. However, the basic error can be reduced as much as one wishes by choosing a large enough value for l n'. Al- though doing so will increase the number of steps and, so, might leave us as badly off as before, it's worth trying. Let's see what happens to the values of '(1 - — ) n ' as n gets larger [for given values of 4 \' and l t']. n To do this, it is convenient to transform the expression '(1 - — ) '. By the principle for subtraction and Theorem 207, (1 - ^) n = n So, since — — is the reciprocal of \t ^ n 1 + - n n -\ Xt \t -\t (4) (1 ^) n = [(l+x) 1 /*] 1/xl-Xt w here x = — \t/n. Since, for large n, \t/n is a small positive number, [9.10] [9-155] our problem is to investigate the values of '(1 + x) 1/x, for negative values of 4 x' close to 0. Here is a table of approximations to some l/x values of the function f for which, for x fi 0, f(x) = (1 + x) X -1/2 -1/3 -1/5 -ID" 1 -lo- 2 -lo' 3 -lo' 4 -10" 5 f(x) 4 3.38 3.05 2.87 2.73 2.7196 2.7184 2.71829 The evidence is that the function f is a decreasing function and that the values of f for negative arguments have a greatest lower bound which is not much different from 2.718. More evidence for this is furnished by- considering values of f for some positive arguments. X lO" 6 lo" 4 lO" 3 io~ 2 10" 1 f(x) 2. 71827 2. 71815 2. 717 2. 70 2. 59 -5 -5 Comparison of the values for the arguments -10 and 10 suggests that the greatest lower bound of the values for negative arguments is between 2. 71827 and 2. 71829. As a matter of fact, this is the case. The greatest lower bound of these values [which is, also, the least upper bound of the values of f for positive arguments] is approximately 2. 718281828459. This is the number e which was mentioned earlier. So, we have seen that, whatever \ and t may be, (1 - — ) is the n ( — \t)th power of a number which, for n sufficiently large, is as close to e as we wish. Since the power function with exponent -\t is continuous at e [see Appendix C, Theorem 220], we conclude that, for n sufficiently \t large, (1 - — ) is as close to e as we wish. Looking back at (3), this suggests that v t>o A(t) = A <°)' e ~ Xt > where, as we saw earlier, \ = 4. 6. To check this with the original formula: V t>0 A(t) = A(0) * 10 ct [c = -2] we need only check whether e = 10 ' or, equivalently, whether e 4 ' 6 =10 2 . Since e = 2. 718, log e = 0. 4343. Since 4. 6 X 0. 4343 = 1. 99778 = 2, the formulas are in good agreement. [9-156] [9.10] Example . Find the half-life [in years] of Sr 89 . Solution . The law of decay of Sr 89 is given by the formula: A(t) = A(0)'e *'° l We are looking for the t such that A(t) is half of A(0)--that is, for the t such that (*) e~ 4 ' 6t = 0.5. Using natural logarithms we see that -4. 6t = inO. 5 = -in 2. [Recall that l in' is an abbreviation for 'log '. ] From a table of natural logarithms we find that in 2 = 0. 69315. So, . . 0. 69315 . n ._ t = 4<6 =0.15. Hence, the half-life of Sr 89 is about 0. 15 years. Of course, if you don't have a table of natural logarithms available, you can still solve the problem using common loga- rithms. From (*), it follows that -4.6t log e = log 0.5 = -log 2. Solving this for *t* amounts to using common logarithms to compute in 2 and then dividing by 4. 6. Explain. EXERCISES A. 1. Plot a graph of 'y = e ' from x=— 2tox=2. [Hint. Compute -2 -1.5 -1 1.5 2 rr , approximations toe ,e ,e ,...,e > e . For example, since log e -1 ' 5 = -1. 5 log e = -1. 5 log 2. 718 = -1. 5 X 0. 4343 = -0.6515 = 9.3485-10, e" 1,5 = 0.223. Of course, a table of natural logarithms makes the job easier. ] 2. Sketch a graph of in. 15. Use Theorem 219 and common logarithms to find approximations to values of in. 1. in 2 2. in 10 3. in 20 4. in 1.35 5. in(e 2 ) 6. in 0. 95 7. inl 8. in(-2.4) [9.10] [9-157] C_. Find approximate solutions. 1. 2n x = 0. 7894 2. in y = 2. 7643 3. in z= -1.4786 D. 1. Two functions which occur frequently in applications of mathe- matics are the functions c and s for which x -x X —x c(x) = y and s(x) - ^ . Sketch graphs of c and s. [ Hint . First sketch graphs of 'y = e and 'y = e 'on the same chart. ] *>* The graph you drew of the function c suggests, by its shape, a rope or chain whose ends are fastened to two posts. As a matter of fact, it can be proved that a rope or chain of uniform weight per unit length does take this shape when it hangs freely between two points. More pre- cisely, with respect to properly chosen coordinate axes, "the equation of such a chain" is 'y = a , c(x/a) 1 , for some a > 0. For this reason [see the Latin for 'chain'], the curve you drew is called a catenary . For numerically small arguments, 2 c(x) = 1 + "y . Consequently, near its lowest point a catenary is very nearly a parabola. As a matter of fact, a chain which is loaded in such a way that the total load per unit horizontal distance is constant has the shape of a parabola. Ox 2. Use elementary algebra to show that, for each x, (a) [c(x)] 2 - [s(x)] 2 = 1, and (b) 2'c(x)s(x) =: s(2x). 3. Which of c and s has an inverse? For the one which does, show that its inverse is the function f such that f(x) = 4n(x + Vx 2 + 1 ). [ Hint . See Exercise 10 of Part H on page 9 - 130. ] [9-158] [9.10] NEWTON'S LAW OF COOLING As another example of exponential decay, consider the cooling of a warm body to room-temperature. If, at various times t, one measures the difference , T(t), between the temperature of the body and the [con- stant] temperature of the room, he finds that, to within experimental error, the decrease in T during an interval of short duration h is nearly proportional to the product of h by the value of T at the beginning of the interval. [Compare with (1) on page 9-153. ] So, as in the case of radioactive decay, the law of cooling is such that, for some k > 0, — let V t>Q T(t) = T(0)*e . [Newton's Law] [See the derivation of (5) on page 9-155 from (1) on page 9-153. ] You can perform an interesting experiment in class or in the chem- istry laboratory to check Newton's Law of Cooling. Heat some water to about 70°C and pour it into a beaker [about 250cc] which is set on a ring- glass stirrer ^) thermometer stand. Record the room-temperature. Then, make a temperature- time record in which you list water -temperatures at the ends of 1- minute intervals. Keep stirring the water with a glass rod while the water is cooling. Plot the difference between water-temperature and room-temperature against time. [T(0) is the temperature difference at the start, T(l) is the temperature difference at the end of the first min- ute of cooling, etc. ] Do you get what looks like an exponential curve? [9.10] [9-159] TRANSIENT CURRENTS IN SIMPLE CIRCUITS The growth and decay of an electric current in certain simple cir- cuits provide further examples of the occurrence of 'e' in statements of natural laws. As a basis for understanding such laws, let's consider a situation in which the current is steady. Consider a flashlight. Schematically, it looks like this: y switch 'if v i .M.h r bulb , battery The battery can exert a certain electromotive force in pushing electrici- ty around. The measure of such a force is usually given in volts [after Volta, one of the earlier experimenters with electricity]- -the cells in a 3-cell flashlight make up a battery of about 4 volts. When the switch is closed, the battery pushes electricity around the circuit — through the closed switch, through the lamp filament, back through the battery, and around again, [it may be helpful to think of the battery as a water-pump pushing water around a closed circuit of pipe. The switch is a valve, and the lamp filament is a rough section of pipe. ] In pushing the elec- tricity around, the battery has to overcome resistance [just as the water-pump has to work to overcome frictional resistance between the water and the wall of the pipe and the parts of the pump]. This resist- ance is usually measured in ohms , a unit named after the discoverer of Ohm's Law, who found that the rate at which electricity flows through the circuit is directly proportional to the electromotive force of the bat- tery and inversely proportional to the resistance of the circuit. If E is the volt-measure of the electromotive force, and R is the ohm-measure of the resistance, then electricity flows at the rate of i amperes , where E l= R* [Ampere, like Ohm and Volta, was a physicist who discovered many of the fundamental facts about electricity. ] For example, a 6-volt battery can push electricity through a resistance of 3ohms at a rate of 2 amperes, [9-160] [9.10] The unit used in measuring quantities of electricity is the coulomb - - a rate of 1 ampere is 1 coulomb per second. Now, let's consider another set-up. A condenser is one kind of device for storing electricity. In a simple case, a condenser may be a pair of metal plates which are fastened close together but not touching one another. Suppose that a battery is connected to the plates of a con- denser, as indicated in ';he figure, by closing a switch. The battery E J will pull electricity from one plate and push it onto the other. As it does this, its job gets harder. [Think of a water-pump lifting water from one reservoir and pushing it through a hole in the bottom of an- other. The higher the level of the water in the second reservoir gets-- and the lower the level in the first reservoir- -the harder it is for the pump to do its job. "When the difference in levels becomes too great, the pump is no longer strong enough to lift any more water.] Finally, the battery will have moved as much electricity as it can, and nothing more happens. Coulomb discovered that the amount the battery can move is proportional to its electromotive force. The constant of pro- portionality, C, depends on the design of the condenser and is measured in farads [after Faraday] when the electromotive force E is measured in volts and the amount Q of electricity is measured in coulombs. With these units, Coulomb's Law is: Q = CE Suppose that we now disconnect the battery by opening the switch, and then switch the condenser into another circuit containing a resist- R< E M ance of R ohms. The electricity which has been stored on one plate of [9.10] [9-161] the condenser will flow back through the resistance to the other plate. The rate at which current flows in the circuit will vary- -"the higher the electricity is piled up, the faster it will flow away**- -the less of the electricity which remains on the charged plate, the slower it will flow away. It can be shown that, t seconds after the switch is closed, the ampere -measured L of the current in the circuit is given by: t E "RC l t * R C Comparing this with Ohm's Law, we see that the discharge of the con- denser creates a current like that due to a battery whose electromotive force t seconds after the switch is closed is Ee" '' , Let*s consider a third set-up. If current flows at an increasing rate through a coil of wire, there is generated in the coil a counter - electromotive force which tends to slow up the rate of flow of electricity. The amount of this counter- electromotive force is proportional to the rate at which the flow of current is changing. The constant of propor- tionality, L, measured in henries is called the self- inductance of the coil. Suppose a battery of E volts is connected to a resistance of R ohms and an inductance of L henries. At the instant the switch is closed, the cur- rent in the circuit is amperes. As this increases, a counter-electro- motive force is set up in the coil which, effectively, reduces the electro- motive force of the battery. So, the current cannot immediately build up, as it otherwise would, to E/R amperes. What happens is told by the formula: Rt E(, ~ L From this we see that the current gradually builds up to the limiting value E/R which Ohm's Law leads us to expect. [9-162] [9.10] COMPOUND INTEREST Although this is not an example of a natural phenomenon, the growth of a "sample" of money on which interest is compounded bears some resemblance to exponential growth. Let's investigate this. In this case, we don't have to do any experimental work to find the law of growth. By definition, a principal P invested for a given interest period at a simple interest rate k [that is, 100k% for the period] grows to an amount A such that A = P + Pk = P(l + k). If you leave this amount A invested for a second period at the same rate k then the amount B due you at the end of the second period is B = A(l + k) = P(l + k) 2 . In general, if you invest a principal A(0) for the duration t of m periods, the amount A(t) due you is A(t) = A(0)(1 +k) m . Suppose that h is the length of one period. Then, A(t + h) = A(0)(1 + k) m + 1 . So, the growth during this (m + l)th period is A(t + h) - A(t) = A(0)[(1 + k) m + l - (1 + k) m ] = A(0)(1 +k) m [(l +k) - 1] = A(t)*k. Hence, A(t + h) - A(t) A(t) That is, the growth during an interest period is proportional to the amount at the beginning of the period. [Compare with ( 1 ) on page 9-153.] Now, as it actually occurs, interest rates are given on an annual basis and the interest is compounded one or more times per year. If the annual interest rate is r and interest is compounded n times per year, then the interest period is l/n years, the rate for that period is r/n, and the number of periods is nt. So, if A (t) is the amount due at [9.10] [9-163] the end of t years, (1) A (t) = A <0)<1 + ^) nt n n n [Compare this with (3) on page 9-154. ] As in the case of radioactive decay, for sufficiently small values of 'r/n', (1 + ^) nt = e rt n [Explain. ]. So, for a given annual interest rate r, A (t) = A (0)-e rt n n (2) for large values of 'n'. The formula (2) is sometimes useful in getting an estimate of A (t). The larger n is, the better is the estimate. To see how good an esti- mate this is, let's consider money invested at 6% for 5 years, com- pounded for various periods, and compare formulas (1) and (2) % Compounded n annually 1 semi-annually 2 quarterly 4 bimonthly 6 monthly 12 semi-monthly 24 (1 + Q.Q6 yn5 n 1. 3382 1. 3439 1.3469 1. 3478 1. 3489 1. 3494 0.06»5 e 1.3499 1.3499 1.3499 1.3499 1.3499 1.3499 difference 0,0117 0. 0060 0.0030 0.0021 0.0010 0.0005 So, for example, the interest on $100 at 6% compounded annually for 5 years is $33. 82. The approximation formula (2) gives you an estimate which is too large by $1. 17. But, if the interest were compounded semi- monthly, the estimate would be too large by only 5£. If the interest were compounded weekly, by how much do you think the approximation formula would be in error? The amount A(t) computed by the formula: A(t) = A(0)«e rt is said to be the amount a(0) grows to in t years at the annual rate r compounded continuously . If interest on an investment were compounded continuously at the annual rate of 6%, how many years would it take for the investment to double? [9-164] [9.10] ADIABATIC COMPRESSION OF GASSES As you may know from using a bicycle pump, compressing a gas heats it. Suppose that the outlet of the pump is closed so that, in effect, you have a closed cylinder with a piston. At the beginning of a stroke [see Figure 1], the pressure, volume, and absolute temperature have Pi v , '//////.. 7 ■/ p 2 v„ *2 >t l P 3 P 3 t„ = t, >/>;>'. '.;/,.} Fig. 1 Fig. 2 Fig. 3 given values p., v x , and t,. /vt the end of the stroke [see Figure 2], the values are p , v , and t . By the General Gas Law [see page 9-144], Pp v 2 2 1" 1 [Explain.]. 2 1 Since compressing the gas raises its temperature, t > t . So, P 2 V 2 > Pl V l« Now, suppose [see Figure 3] that you wait for the gas to cool down to the initial temperature t , holding the piston in the same position as you wait. How will the pressure p compare with p 2 ? How will p 3 v com- pare with p v ? The answer to this last question is predicted by Boyle's Law which says that PV is a constant when T is a constant. That is, under iso- thermal conditions, PV is a constant. But consider the situation illus- trated by Figure 2. Since t /■ t , Boyle's Law doesn't apply. However, if the stroke is very fast so that there is little loss of heat from the gas during the stroke [or, better, if the cylinder is insulated against heat loss], there is another law which relates the initial and final pressures and volumes. Processes [compressions or expansions] during which the heat content [rather than the temperature] is constant are called adiabatic . [9.10] [9-165] processes . Let's see how one might discover the adiabatic gas law. When measurements are made of a gas sample under adiabatic con- ditions, one obtains pairs (p, v) of measures of pressure and volume. It turns out that if {p 1$ v x ) and (p 2 , v ) are pairs of measures for which v 1 = v then p x = p 2 ~-that is, P is some function g of V. It also turns out that if (p , v ) and (p , v ) are other pairs of measures such that v /v = v /v x then p 4 /p 3 = Pp/Pi - "*-* 1 ^ * s > there is some function f whose value for the ratio of any two volume-measures is the ratio of the corresponding pressure-measures. [if Boyle's Law held, this f would be the reciprocating function. ] Our problem is to find f and g. Now, by the descriptions given above for f and g, it follows that, for any positive numbers [volume-measures] v x and v 2 , m f(^\ g(Vg) [Compare with (1) 1 ' \vj g(v x )' on page 9-141.] In particular, taking v x to be 1 and replacing 'v ' by *x', U) V x fU) = g(x)/g(l). Consequently, if we can find what the function f is then, knowing the pressure g(l) when the volume of the gas sample is 1 unit, we can find g and, so, learn how the pressure of the sample depends on its volume. To discover f we begin by noting that, by (2) and (1), for any posi- tive v and v ? , f(Vg) g(v 2 )/g(l) g(v 2 ) ( v 2 f(v x ) giv^/gd) g( Vl ) l \v that is, f(v 2 ) = f -? f( Vi ). 21 :] ' V 2 ' Replacing — by 'x', *v 1 ' by 'y', and *v ' by *xy', we see that f must satisfy: v i < 3) V x>0 V y>0 f( ^ = iMi{ y ] The generalization (3) may suggest to you Theorem 208: v x>oVo v u<*y> u = xV and, so, suggest that the function f is a power function restricted to positive arguments. This is, in fact, the case. Let's prove this. [9-166] [9.10] Although there are many solutions of: (3) V x>0 V y>0 £Uy) = f(x)£(y) which are not power functions, none of these can fit the physical situ- ation. To begin with, by inspection, the constant function [defined on the positive numbers] is a solution of (3) and is not a power function. But, obviously, this constant function does not fit the physical conditions, Now, it is easy to see that each other solution f of (3) is such that, for each x > 0, f(x) > 0. For, to begin with, for x > 0, f(x) = f(Vx"'Vx") = f(vx~)f(>/x~) >0. Suppose now that, for some x Q , f(x Q ) = 0. Then, for each x > 0, f(x) =f(^.xJ = f(^)f(x o ) = f(^)0 = 0. x o So, if f satisfies (3), either the values of f are all or all positive. In the latter case, (3) tells us that (3') V x>0 V y>0 in(f(xy)) = in(f(x)) + in(f(y)). If in (3') we had 4 x + y* instead of 4 xy\ (3') would tell us [see Theorem 221] that in <>f is a subset of a homogeneous linear function [or a function too queer to fit the physical siutation]. Knowing what function in <>f is could enable us to find f. This suggests that we try to find another func- tion F which is simply related to in of and such that (*) V V F(x + y) = F(x) +F(y). x. y If, for x > and y > 0, we let a = in x and b = in y then x = e and y = e . It follows from (3') that, for all a and b, in(f(e a e b )) = in (f(e a )) + in(f(e b )). „ . a. b a+b _ But, e • e = e . So, V a V b in(f(e a + b )) = in(f(e a )) +in(f(e b )). Comparing this result with (#), we see that if F is the function such that, for each a, F(a) = in(f(e a )) then F satisfies (*). Since we know that f cannot be a queer function, neither can F. So, in view of (*), F is a homogeneous linear function or the constant func- tion whose value is 0. Hence, for some c, [9.10] [9-167] Sxi (f(e )) = ca, for all a. Since x = e and a = in x, we have that, for some c, So, for some c, in(f(x)) = c inx = in(x C ), for all x > 0. f(x) = x C , for all x >0 In other words, f is a power function restricted to positive arguments. In the case of adiabatic processes, the exponent of the power func- tion f turns out to be a negative number, -y. [For air, v = 1.4.] So, returning to (1), we see that if (p , v.) and (p , v ) are pressure-volume measurements made on the sample under adiabatic conditions then p 2 g(v 2 ) /v 2 \ /"v^Y Vl Y vj" Ui) = f — = that is, Y Y P 2 V 2 = Pi v i • In other words, the pressure P of the gas sample is inversely propor- tional to the vth power of its volume V: PV^ = k [Adiabatic Gas Law], Example . A sample of air has a volume of 182 cubic inches when it is under a pressure of 16. 3 pounds per square inch. If the pressure is increased to 21.8 pounds per square inch under adiabatic conditions, what is the new volume? Solution . If v is the resulting volume then 14 14 21. 8v = 16. 3* 182 ' . 1.4 . 16. 3 . 109 1. 4 = 2TT8 182 -*«-(iH) l/l - 4 log v = log 182 + log 16. 3 -log 21. 8 , 2 . 2601+ 9^8737^0 « 2.2601 + 13.8737-14 1. 4 = 2. 1699 So, the new volume is about 148 cubic inches. [9-168] [9.10] EXERCISES Solve these problems. 1. A sample of bromine gas [v = 1. 32] occupies 140 cubic inches of volume when under a pressure of 23. 4 pounds per square inch. If the pressure is reduced to 18. 6 pounds per square inch, to what does the volume increase assuming adiabatic conditions? 2. Repeat Exercise 1 assuming isothermal conditions. 3. If you double the pressure on a sample of neon gas [v = 1. 64] under adiabatic conditions, what change takes place in the volume? 4. Under adiabatic conditions, by what factor must you multiply the pressure of a sample of propane gas [y = 1.13] to reduce its volume by 50%? 5. Repeat Exercise 4 assuming isothermal conditions. THE NUMBER e In discussing radioactive decay we discovered the number e, one of whose definitions is: e =£ub{y:3 x>0 y = (1 +x) l/x } We shall now arrive at this number by a different approach. Consider, for some a > 1, a graph of log , and choose a point P on cL the graph. It is easy to guess that, among all the straight lines you [9.10] [9-169] might draw through P, there is just one such that each point other than P on the graph of log lies below this line. This line is a graph of a cl linear function which is said to be the tan ge nt to the function log at P. a [What change needs to be made in this discussion if0 1] is characterized by the fact that (i) t is a linear function, and (ii) t a (i) = o, (UL) V 0 log a (x). Let's try, now, to compute m . We could, of course, get an approx- a imation by making measurements on a graph. But, there is a better way. Consider any other logarithm function for base greater than 1, say, log, . By our change-of-base formula [Theorem 219], for each x > 0, [9-170] [9.10] log b x = log a x-log b a. Also, since b > 1 and a > 1, log. a > 0. Hence, since, for < x / 1, t (x) > log x [see (iii), on p. 9-169], a a it follows that, for each such x, (*) t a (x)"log b a > log a x-log b a = log b x. Now, since t a is a linear function and log b a is a constant, it follows that Also, So, Finally, by (*), t * log, a is a linear function [with slope m * log, a], a 6 b L v a 6 b J t a (l)'log b a = 0'log b a = 0. t 'log, a has the value for the argument 1. V 01 °gb X - Thus, the function t 'log. a meets the three conditions which charac a °b terize the tangent at (1, 0) to the function log K . So, t a *log K a is the tangent t, . Hence, m, = m 'log, a. b a b b Now, for a > 1 and b > 1, it follows that m. m • log, a b , a 5 b = b I v = ( b l0g ^) m a ! Theorem 207 ^ m a = a 1 J (L) In other words, the function f such that, for a > 1, m f(a) = a a is a constant function. It will turn out that the value of this constant function is the number e. [When we discover this, we shall know that m = l/in a. Explain. ] However, since we do not yet know the value a of the constant function, let's call it * c'. So, for each a > 1, a a = c. [9.10] [9-171] A glance at the graph of log [for a > l] will convince us that, for a a > 1, m > 0. So, since an exponential function whose base is greater than 1 is increasing, it follows that, for a>l, a > a =1. Conse- quently, c > 1. So, c is an argument of f and- -lifting ourselves by our boot-straps-- m c c = c. Since the exponential function with base c has an inverse, it follows that m = 1. c Now, m is the slope of the tangent at (1, 0) to log . So, for each x, t (x) = m *{x- 1) = x - 1. c c Hence [see (iii) on page 9-169], for < x ^ 1, (1) log x < x - 1. to c It is now not difficult to show that c = e. Keep in mind that, by definition, e =iub{y: 3 x>Q y = (1 +x) l/x }. We proceed as follows. Since, for < x ^ 1, < — -f 1, it follows from (1) that, for each such x, i l < l i 1 - x log — < 1 = . °C X X X However, log — = — log x and, so, i *■ 1 ~ x - log x < & C X --that is, for < x ^ 1, (2) x^l< logcX . Replacing 'x' in (1) and (2) by ' 1 + x\ it follows that, for -1 < x ^ 0, (3) jf^ < log c (l +x) 0, ■ log (1 +x) X < 1. 1 + X B c Since c > 1, the exponential function with base c is increasing. So, it [9-172] [9.10] follows that, for x > 0, 1 1 (4) c 1+X < (1 + x) X < c. From this result, it follows at once that c is an upper bound of {y: 3 x>0 y = d+x) 1 ^}. So, by definition c > e. Suppose that c > e. In this case, log c > 1, and there is a positive number x^ such that 1 + x^ = log c. It follows from Theorem 218b that o o =>e and, so, Since, by definition, it follows that 1 , 1 + x Q - '-c C 1 1 +x c u = e. 1 ,x, > (1 + x )^o , c 1+X o > (1 + x ) X o . But, by (4), this is not the case. Consequently, c ^ e and, since c > e, c = e. THE NATURAL LOGARITHM FUNCTION We have seen that the function log --for short, in- -is the logarithm function whose tangent at (1, 0) has slope 1. It is this property of in which makes it, for many purposes, simpler than any other logarithm function and which accounts for its being called the natural logarithm function . The simplicity of in in comparison with the other logarithm func- tions can best be shown if we return to the considerations, on page 9-169, which led us to our re-discovery of the number e. /. s in that discussion, given any x, > 0, there is just one linear function- -the tangent to in at (x., in x.)- -which contains the point (x., inx.) and whose value, for each x such that < x ^ x,, is greater than in x. In particular, we have seen that when x. = 1, this linear function is the function t where, for each x, [9.10] [9-173] t e (x) = l-(x - 1) = x - 1 --that is, we have seen that, for < x ^ 1, (1) in x < x - 1. It is now easy to find the tangent to in at any point (x x , in x x ). For, using (1), for < x./x. x / 1, in — < — - 1 x i x i and, so, by Theorem 216, for < x ^ x x , (2) inx < — - 1 +in x.. X 1 Since the function t such that, for each x. t( x ) = — - 1 + in x. is a linear function which contains (x,, In x x ), it follows from (2) that t is the tangent to in at (x x , in x x ), In particular, for each x x > 0, the slope of the tangent to in at (x x , in x x ) is l/x 1 . EXERCISES A. The tangent to log at (x,, log x ) was defined on page 9-169 [for a > 1 and x x > 0] to be the linear function t such that t(x x ) = log a x x and, for each positive x fi x x , log a x < t(x). [in adopting this definition, we made the rather large assumption that, given a > 1 and x x > 0, there is one and only one such function t. This assump- tion can be justified, but the proof makes use of concepts which we shall not develop here.] In the case a = e--that is, for the function in- -we proved, above, that, for x x > 0, the tangent to in at {x lt fnx 1 ) is such that (*) t(x) = -J-(x - x.) +inx x . 1. Use (*) and the definition to find the tangent to log a at (x x , log a x 1 ) for a > 1 and x x > 0], [ Hint. See the discussion on the top half page 9-170. ] [9-174] [9.10] 2. How should the descriptions of the tangent to log a at (x L , log a x 1 ) be modified to fit the case < a < 1 ? 3. Does the formula you found for t in Exercise 1 [where a > l] also work if < a < 1? 4. The tangent to log a at (x 1 , log x^ is the linear function which con- tains (x 1( log x x ) and whose graph lies [except at this one point] above the graph of log if a > 1, and below the graph of log if a a < a < 1. Find a single "directional phrase" [like, but different from, 'above' and 'below'] which can be used to describe the tan- gent in both cases. 5. For < a ^ 1, exp [that is, the exponential function with base a] is the inverse of log . Guess a description of the tangent to exp a a at (x x , exp a x 1 ). 6. By (*) [and the definition of the tangent to in at (1, 0)], for < x ^ 1, in x < x - 1. From this infer that V / n e X > 1 + x. x^ 7. Use the result obtained in Exercise 6 to prove that, for < a / 1, V , n exp (x) > 1 + xJln a. x/0 r a 8. Use the results of Exercises 5 and 7 to show that the tangent to :tion t, where, f( t(x) = 1 + xina. exp at (1, 0) is the function t, where, for each x, a 9. Use the result of Exercise 6 to prove: V V / e X > e X i + e X Mx - x.) x 1 x^x x * [ Hint . Consider using Theorem 206. ] 10. From Exercises 5 and 9 argue that the slope of exp [that is, of the exponential function with base e] at (x x , expx x ) is expx^ [9.10] [9-175] Sin x B. Consider the function f where f(x) = , for x > 0. 1. Sketch what you think a graph of f might look like. [One ordered pair you should plot is (e, f{e)). ] 2. We know that, for x x > and for < x / x 1 , in x < — (x - x x ) +ln x 1 x i and, so, (*) f(x) < — - - + ^f(x,). ' x i x x Check (*). Then, use it to show that, for < x / e, f(x) < l/e. [This result, together with the fact that l/e = f(e), shows that f takes on its maximum value at e. Does you graph show this?] 3. Show that f is decreasing on {x: x > e }. [ Hint . From the result of Exercise 2 it follows that all you need to show is that if e < x x < x g then f(x 2 ) < f(x x ). To do so, suppose that e < x x < x and that f(x„) > f(x ). Use this assumption together with (*) [with 4 x 2 ' for 'x'] to infer that ffx^ < l/x x . Use (*) again [with 4 e' for l x'] to obtain a contradiction. ] 4. Show that f is increasing on {x: < x < e}. 5. Show that, for all sufficiently large arguments, f has arbi- trarily small positive values. [ Hint . Use (*) to show that, given any positive number x,, f(x) < Z/x 1 if x is sufficiently large. ] 6 t Assuming [as is the case] that f is continuous, your work in Exercises 3 and 5 shows that the range of the decreasing part of f is {y: < y < l/e}. Explain. 7. What is the range of the increasing part of f ? 8. If you have not done so, modify the sketch you made in answer to Exercise 1 in order to show the information you now have about f. [9-176] [9.10] e 7T C_. 1. Which is larger, U or e ? [Hint. V x >Q V y >Q [xY > y X <=> f(x) > f(y)]]. 2. Complete. (a) For e < y < x x , x y y . (b) For < y < x < e, x^ y . (c) For 1 = y < x, x y y . 3. Arrange in order of magnitude. V7^, 1.4^, 1.4 1 - 4 , VF 1 - 4 4. Represent the results of Exercise 2 graphically. [ Suggestion. Mark with 4 + 's the parts of a graph of {(x, y): < y < x} for v x which, by Exercise 2, x J > y and with *-'s those for which, by Exercise 2, x y < y . ] v x 5. Of course, x 7 = y if x = y > 0. But there are also points (x, y) with < y < x for which x y = y x . Study the graph of f which you made in answering Exercise 8 of Part B and, on your answer for Exercise 4 of Part C, sketch {(x, y), < y < x: x^ = y }. 6. Complete the graph you started in Exercise 4 to show, for x > and y > 0, the points where x > y , x = y , and x < y . D. 1. Note that, for each x > 0, f(l/x) = -xinx. Use this and your graph of f to sketch a graph of the function g, where g(x) = xin x. Where does g assume its minimum value? 2. The function exp°g [exp is the exponential function with base e] is the function h such that, for each x > 0, h(x) = x . Study your graph of g and use your knowledge of exp to sketch a graph of h. 3. Sketch a graph of exp°f. [9-177] SUMMARY In this unit you have learned more about logic [definite description], completed your set of basic principles for real numbers [the least upper bound principle], discovered and learned to use two important properties which some functions have [continuity and monotonicity], defined and studied three important kinds of functions [power functions, exponential functions, and logarithm functions], and learned some ways in which functions of these kinds "occur in nature". You also learned more about rational numbers [and, if you studied Appendix B, about irrational num- bers and infinite sets], about mensuration formulas [if you studied Ap- pendix D], about using logarithms to simplify computations, and [if you studied .appendix E] about functional equations. [Some of this last you learned, anyway, in seeing how laws of nature may be discovered.] Final- ly, you learned something about an important number- -the number e. In studying definite descriptions you learned that in using a phrase of this kind-- say, 'the so-and-so' --you imply that there i_s a so-and-so and that there are not two so-and-sos. Hence--you learned- -before using such a description, it is the better part of valor to make sure that the implied existence and uniqueness conditions are satisfied [or, at least, are not in contradiction with your previous commitments]. Once you have done this, it is then quite safe to use the definite description in question, and you can formalize this use by adopting a defining prin- ciple. For example, once you have proved: (* ) V 3 z 3 = x 1 x z and: (* ) V V V [(y 3 = x and z 3 = x) => y = z], 2' x y z you can safely adopt the defining principle: V t 3 /^) 3 = x x Incidentally, from (* ) and the defining principles you can infer the theorem: _ VV [y =x=>y= -/x - ] x y ' J Many of the applications you made of definite descriptions were, as in the preceding example, for the purpose of introducing inverses of functions with which you were already acquainted. [Many, but not all. [9-178] For example, you saw that the absolute value function and the greatest integer function- -which, since they don't have inverses, are not inverses of other functions --could most simply be introduced by defining prin- ciples which arise out of the possibility of naming each of these functions by means of a definite description. You will presently be reminded—if you haven't already been- -of another important example of this kind. ] In cases involving the inverse of a known function, satisfaction of the relevant uniqueness condition is established by proving that the known function has an inverse. The search for a standard way of prov- ing such theorems led to the concept of monotonicity and to the basic re- sult that monotonicity implies the existence of an inverse. The attempt to guarantee existence led to the adoption of the least upper bound prin- ciple; and the search for a standard way to prove the relevant existence theorems led to the concept of continuity. [The iubp is an existence principle which asserts that the "least upper bound function" whose arguments are sets of real numbers and whose values are real numbers, has for its domain the set of all nonempty sets of real numbers which have upper bounds. The corresponding uniqueness principle follows from the theorem according to which no set has two least numbers. Does the "least upper bound function" have an inverse?] The general considerations which have just been summarized formed the basis for introducing and studying power functions, exponential func- tions, and logarithm functions. In Unit 8 we had already defined the integral power functions- - {(x, y): y = x }, for some integer k [Sketch those for k = 0, 1, 2, 3,-1 and -2. ] --and the exponential sequences-- {(k, y): y = a k }, for a / 0, and {(k, y), k >0: y - k } [Sketch those for a = 2, l/2, -2, and 0. ]. The fact that each positive- integral power function is continuous and is increasing on the set of non- negative numbers [and that those with odd exponents are increasing on the set of all real numbers] justified the introduction of the [continuous and increasing] principal root functions- - {(x, y): y = Jx } and {(x, y), x>0: y = vx }, for some n. [9-179] Using these root functions we were able to define the exponential functions- - {(x, y): y = a X }, for a > 0, and {(x, y), x > 0: y = }. Those with base a > are continuous and, of these, those with base a / 1 are monotonic. Their inverses [which are, of course, continuous and monotonic] are the logarithm functions- - {(x, y), x > 0: y = lo gfc x}, < b ^ 1. It turned out that the exponential and logarithm functions for a cer- tain base--the number e--are particularly important. e = iub{y: 3 x>Q y - (1 + x) 1//x } = 2.71828 The exponential function with base e is the exponential function, and is often denoted by 'exp'. The logarithm function to the base e is the natural logarithm function, and is often denoted by 'in'. The function in is distinguished from other logarithm functions by the fact that its slope at (1, 0) is 1 and, so, at (x , lnx ) is l/x Q , for x > 0]. In terms of the exponential function we can define the power func- tions with positive arguments-- {(x, y), x > 0: y = x }, for some u. These functions are continuous and monotonic. When u is an integer, the power function with positive arguments whose exponent is u is a restriction of the corresponding integral power function of Unit 8. When u is the reciprocal of a positive integer n, the power function is a res- triction of the principal nth root function. In general, for u / 0, the power functions with exponents u and l/u are inverses of one another. [Using Theorem 217 and the defining principle (L) it is easy to see that the power function with positive arguments whose exponent is u is {(x, y), x>0: y= e U * n *}. This is often used as a definition of the power functions. ] Clearly, the exponential functions, the logarithm functions, and the power functions with positive arguments are closely related. If we con- sider the set of ordered triples {(x, y, z), x > 0: y = x }, [9-180] we see that the exponential function with base a > corresponds to the "slice" of this set consisting of all its members for which x = a, and that the power function with exponent u and positive arguments corres- ponds to the slice consisting of all members with z = u. Also, using Theorems 217 and 219, the slice consisting of all members with y = b, for < b / 1, corresponds to the reciprocal of the logarithm function to the base b. The basic nature of the exponential, logarithm, and power functions is brought out strikingly in Appendix E. There, as corollaries of more general results, it is shown that functions of these three kinds are the only monotonic functions which satisfy: (2) V x V y f(x + y) = f(x)f(y), and: (3 » V x>0 V y>0 f(xy) = £(x) + £(y). (4) V x>0 V y>0 f(xy) = £(x)£(y), respectively. The proofs given in Appendix E for these results depend on another one of a similar nature: The only monotonic functions which satisfy: (1) V V f(x + y) = f(x) +f(y) x y are the homogeneous linear functions- - {(x, y): y = ex}, for some c ^ 0. [9-181] REVIEW EXERCISES 1. Critizize the following *' defining principle" for square roots: For each x > 0, l — 2 Vx is the number z such that z = x. 2. Criticize the following "defining principle" for function composition: For all functions f and g, f«g is the function h such that h(x) = f(g(x)), for each x € <& . g 3. Simplify. (b) Va 2 - 2ab + b 2 (c) V 3 * 2 + Vl2x 2 id>vi + "v| (e) {z ^ + iyfT)Z ■ (2> ^ - ZyfI) ' 7 3a be . , 5 - VT ,f, VT?T (g) 777T 4d e U) r- (j) (h) vr + vr b (k)" J 2 *' 2 3 y v ,* -3 y VT - l l - VT + VT 4. Which is larger, V7~+VTo or VT + Vl7 ? 5. Find the least upper bound of {x: 3(x - 4) < x}. 6. Find the greatest and least members of {x: | 3x - 6 | < 12}. 7. Find the least member of {x: 3 < x }. 8. Which of these functions are monotonic ? (a) {(x, y): y = x 2 } (b) {(x, y), x > 4: y = x 2 } (c) {(x, y): y = ffx]]} (d) «x, y), x > 0: yx = 1 } [9-182] 9. Consider the function f where f(x) = x - -, for x > 0. x (a) Draw a graph of f. (b) Prove that f is monotonic. (c) Draw a graph of the inverse of f. (d) You have probably guessed that the domain of the inverse of f is the set of all real numbers. Prove that it is. [ Hint . One way is to argue from the continuity of f--you may assume that f is continuous --using Theorem 187. An easier way is to find a formula for computing values of the inverse. ] (e) You can introduce an operator-- say, *Q'--such that, for each x, Qx is the value of the inverse of f at x. For example, since f(l) = 0, Q0 = 1; and Q-4 = VF - 2. State the appropriate de- fining principle: V (Qx > and ) x (f) From Exercise (b) it follows that f has an inverse and, so, that v sn^ ^n [y " — = z " — ^ > y = z ]« y>uz>0 y z From this and the defining principle of part (e), deduce the uniqueness theorem: VV. n [y--=x=>y = Qx] x y >0 L/ y J J (g) Use the defining principle and the uniqueness theorem to prove: V Q-x = 1/Qx x ' 10. Suppose that f and g are monotonic functions such that there are m numbers x for which f(x) = g(x). How many numbers x are there such that f~ L (x) = g _1 (x)? 11. Which of these functions are continuous? x/3, x < 6 fx/3, x < 6 (a) f(x) = (b) g(x) = / x/2, 6 ? 15. Show that the system of rational numbers is closed with respect to oppositing. 16. Which of these are irrational numbers? (a) ^64 (b) ^28 (c) V? + V7 (d) V7 17. Simplify. (a) 49 1 / 2 (b) 0. 49°* 5 (c) 0.0081 °* 5 (d) (3 12 ' ] i)" 3 <*m~ l " wW 1 " «(-* A/ y -1/5 5 3 l\ / 3 1 2' rn v 3 4 \ 6 4 2 2 6 3 U) \xy z/\x y z / \x y z _I I / 1 J. J, 2 /•» 3 3 3 3 3 3 (j) x J y \x y + x y 1 2 4\ (k) [x 3 - 2y 3 ] [x 3 + 2x 3 y 3 + 4y 3 J 18. Use the fact that, for all x > and y > 0, (>/x~ + Vy~) = x + y + 2\/xy , to find the square roots of 23 + 4vl5 . 1\ 19. Suppose that g(x) = x - x - 7x + 7x + 10x. Show that g 2 j = g 5 [9-184] r-S2 20. Which of these numbers is smaller than V2 ? (a) 1.4^ (b) vT^ (c) vT 1 * 4 (d) vT^ (e) 1.4 1 * 4 21. Show that log b* log c • log, a = 1. 22. Use logarithms to compute an approximation toy(28.39) - (12.71) , 23. Compute the length of a radius of a solid sphere whose volume is 726.3 cubic inches. 24. Solve these equations. (a) 5 2x ~ 3 = 10 (b) 9 X = 100 (c) 2 X =:3 X " 1 (d) log (x + 1) - logx = log 3 (e) log x - log (x + 1) s 2 25. Suppose that f is log and g is log- . Complete each of the following sentences. (a) V x>0 g(x) = ~ = '§< > (b) V x>0 [J(x) = g(x) <=> x = ] (c) V x>0 «x 2 ) = 2- (d) V y>Q g(V7) = (e) V x>0 V y>0 f(x y> ■ + 26. Suppose that f is the exponential function with base 3 and g is the linear function with slope 6 and intercept —3. Find, by inspection, all numbers x such that f(x) < g(x). [ Hint . Sketch the graphs of f and g. ] x 2 27. Given the functions g and f where, for each x, g(x) = 2 and f(x) =x . How many numbers x are there such that f(x) = g(x) ? [9-185] 28. In a room in which the temperature is 20°C a piece of iron cools from 350°C to 75°C in 15 minutes. Assuming that Newton's JLaw of Cooling applies [page 9-158], find how long it took the metal to cool from 350° to 100°C. 29. The intensity of a light as seen through fog diminishes with distance according to the formula: 1(d) = l(0)e~ kd Suppose that the intensity diminishes by 50% at 15 feet from the source. (a) Compute k. (b) How far from the source can the light be seen by a man who is capable of perceiving a light whose intensity is one ten thou- sandth that of the source? 30. To say that a function has an inverse is to say that it matches the members of its domain with the members of its range in a one-to- one manner- -that is, it determines a one-to-one correspondence between the members of its domain and the members of its range. Examples . The function g where g(n) = n + 1, for all positive inte- gers n, determines a one-to-one correspondence between the positive integers and the positive integers greater than 1. Any exponential function with positive base different from 1 determines a one-to-one correspondence between the real numbers and the positive numbers. (a) Consider the function f, where f(x) = T7-, for x >0. 1 + x Does f have an inverse ? What is the range of f ? (b) Your answers to part (a) tell you that f determines a one-to-one correspondence between all positive numbers and the members of some set of real numbers. What set? (c) Use what you have discovered in (b) and the fact that the expo- nential function [y = e ] determines a one-to-one correspond- ence between the real numbers and the positive numbers to find [9-186] a function which determines a one-to-one correspondence be tween all real numbers and all numbers between and 1. (d) Show that the function g, where x g(x) = 1 + |x| ' determines a one-to-one correspondence between all real num- bers and those between -1 and 1. [Note: V g(— x) = — g(x). ] (e) Starting with the function g of part (d), find a constant function c such that g + c determines a one-to-one correspondence between all real numbers and all numbers between 3 and 5. (f) Starting with the function g of part (d), define a function h which determines a one-to-one correspondence between all real num- bers and all numbers between and 1. (g) Find a formula for computing values of the inverse of g. [ Hint . You need to solve the equation: x = i + |y| for 'y 1 . Consider two cases: x >0 and: x < 0] MISCELLANEOUS EXERCISES 1. Factor. (a) (3x + l) 2 - (2x - l) 2 (b) x 2 - 4xy + 4y 2 - 9x 2 y 2 (c) r 2 + 8r 3 (d) 8a 3 - b 3 (e) a 4 + 8a 2 + 7 (f) b 2 - 29ba + 54a 2 (g) t 4 - 16s 2 (h) 25 - 64a 2 (i) t 2 + t - 56 (j) m 2 - m - 20 2. Solve. , . 7 x 2 + 8 x 2 x 2 + 4 ... x + 3 1 - 2x (a) = — + (b) = 21 3 8x 2 - 11 7 - 2x x - 3 3. Suppose that a is a geometric progression. Show that V V V [(p < n and q < n) => a a = a a 1. n p q " n+pn-p n + qn-q [9-187] 4. Solve: < . 5. Solve: x = 36 6. Solve: log(2x + 4) = 2 J2 X = 3 y 7. Solve: x X /x~ = x X 8. Solve: log x 6 = 12 9. Solve: 4 2x ' 3 = 2 2x + 6 10. Show that if a, b, and c are consecutive terms of an AP then so are 2 2 2 2 2 2 b + be + c , c +ac + a, and a + ab + b . Is the converse true? 11. Show that if each term of a nonconstant geometric progression is subtracted from the next term, the successive differences are the terms of a geometric progression. 12. If A 2 + B 2 = 7AB, show that log^(A + B) = y(log A + log B). 13. Solve for 'x': (a 4 - 2a 2 b 2 + b 4 ) X " 1 = (a - b) 2x (a + b)~ 2 14. Solve: x = (xvx ) 15. Factor. (a) 1000x 2 y - 40y 3 5 2 3 (b) x D - 8x y (c) 5a 4 - 15a 3 - 90a 2 (d) 1 - (x 2 + y 2 ) + 2xy (e) -\ - 27y 6 x° (g) (a + 5b) 2 - 4b 2 27 (f) -¥-x - l x J y J (h) (2a - 3b) 2 - 9b 2 (i) 16a 2 - (2b - 3c) 2 (j) (m + n) 2 - (p + q) 2 2n 16. Complete: V n V 2 P = P-l 17. Solve these equations. (a) 3(k - 1) 16 ^(k-4) = f(k-6) + ^ (b) T " 17 (t + 10) ■ (t ' 3) = ~5l 4 [9-188] 18. Find the arithmetic progression a with common difference 7 such that m n V V m f n > a = m > a . n ^-i P Z^ P p=l p=l 19. Expand, and express without referring to negative exponents. (a) (x + 5 + 6x~ 1 )(l + ox" 1 + 8x~ 2 ) (b) (3x - 8)[x - 1 - (1 - x[4 + x]" 1 )" 1 ]"" 1 20. Simplify. (a) a + b + a - ab (b) x 3 + xy 2 + ? Xy ? 21. Simplify. (a) 2 ,. u 2 a + b ab ab + b 2 a 2 + ab (b) ^^ + 2 y ? - 2x v x + y k - y d 22. Sintiplify. (a) 17.4 X 8.6 - 7.4 X 8. 6 (c) 24* 13 + 9' 26 - 4-39 7 X23 - 14 X 5 (e) 13 X 18 + 26 X 12 (b) 18. 8 X 12.4 - 17. 6 X 6. 2 (d) 12' 14 + 4* 28 - 5'28 39 X 7 + 21 X 21 (f) 9 X 34 - 12X17 23. If n arithmetic means are inserted between 1 and n , what is the smallest of these means? 24. If the square of A varies as the cube of B, and A has the value 3 when B has the value 4, what value of B corresponds with the value vT/3 of A? 25. A certain number of persons shared equally in the cost of a party. If there had been 10 more, each would have paid $1 less; if there had been 5 fewer, each would have paid $1 more. How many per- sons shared the cost? [9-189] 26. Two numbers are in the ratio 7: 12. Find the smaller if the larger exceeds it by 275. 27. Solve for the indicated variable. (a) t = 2„\/§£ ; T (b) s = __76v_ . v 28. A wagon wheel of radius r picks up a piece of paper from the road and carries it round l/lZ of a revolution. How high is it above the ground? How high will it be when it has gone 7/12 of the way round? 29. Simplify. f-ti - 2 Y Z - (b) 2 ab , + — £— 2x + z 2x z + 5xz + 2z^ 9a^ - I6b^ 6a - 8b 30. If the ratio of x to y is 3 to 4, what is the ratio of 7x - 4y to 3x + y? 31. Find a root of '27X 1 * 5 = 1\ 32. Express 93. 105003 as an indicated sum of multiples of powers of 10. f4'3 2 " X " y = 3-4 1_ y 33. Solve the system: \ -> , , 3-2 =2*3' 3 34. Solve: h- 4 l°g* + /4\logx^ 23 35. How many ordered pairs of integers (k, j) are there such that < j < 2 k and k < 5 ? 36. Complete; 100 99 (a) YS l + { " 1)P J = (b) X [l +( ' 1)P] = p=l p=l 2n n (c) V n ^[1 + (-D P ] = (d) V n ^[1 +(-D P ] p=l p=l [9-190] [App. A] APPENDIX A [This Appendix deals in a more detailed manner with the material cov- ered in section 9.04 and the Exploration Exercises which precede that section. In particular, it contains proofs of the fundamental Theorems 184 and 187 on the positive- integral power functions.] The simplest functions . --We have seen in section 9.01 that the justifica- tion of our work with the principal square root operator lies in two theo- rem s about the squaring function : (t i> V x>0 3 z (z - >0andz2 =x > and: (t s ) V x>0 V y V 2 £ ((y -° and yZ = x) and < z >° and * = x » => y = x] The first of these can be abbreviated to: U) V y>0 3 x>Q x 2 = y [Vfor'x'; *x'for'z'] and the second is equivalent to: (2) V x >0 V x >0 [x i 2 = x 2 2r>x i = x ^ [V £or V.V£orV] [Although (2) is simpler than (t 2 ) it is not hard to see that, in view of the fact that squares of nonnegative numbers are nonnegative, (2) and (t 2 ) are equivalent. For, in the first place, because of the fact just cited, the restriction *x>0' in (t 2 ) is unnecessary [according to the antecedent of (t 2 )» x is the square of the nonnegative number y and, so, is bound to be nonnegative]. And, in the second place, the 4 x's in the 2 2 antecedent of (t 2 ) serve merely to guarantee *y = z '. So, in view of the fact about squares, (t 2 ) is equivalent to: V V [(y>0 and z>0 and y 2 = z 2 ) => y = z ] And, (2) is merely an abbreviation of this.] We shall develop a better insight into how an operator like ' v* [for example, * S7 ', * V"" *, and other operators which you have not yet heard of] can be justified by concentrating on a related function [for *v\ the squaring function; for the examples mentioned, the cubing function, the 18th power function, and, for one you probably haven't heard of, the exponential function with base 2]. [App. A] [9-191] Looking at (1) and (2), we see [x > 0, x x > 0, x 2 > 0] that these are really statements about a subset of the squaring function- -namely, the function sq* defined by: sq + ={(x, y). x >0: y = x } \ \ / / / / \ v (x, y) : y > and y = x) In terms of sq*, statements (1) and (2) become: and' |1 ' ) Vo 3 xa M y Wsy sq (2') V. V sq x,e£ + x„ e £ + sq [ sq + (x^ = sq + x 1 = x 2 ] In geometric terms, ( 1') tells us that each horizontal line with nonnega- tive intercept [y> 0] crosses the graph of sq + - -there are no holes in the graph. (2') tells us that each horizontal line that crosses the graph does so in precisely one point If we recall that a function has an inverse if no two of its ordered pairs have the same second component [that is, if its converse is a function], we see that (2') says merely that sq* has an inverse. As soon as we know [by (2')] that sq* has an inverse, we are justified in speaking of the principal square root of any number in the range of sq*. Doing so would not be of much use unless we knew what numbers belong to the range of sq + , that is, to the domain of its inverse. (1') tells us part of the story- -each nonnegative number [V > Q . . . ] belongs to the range of [9-192] [App. A) sq + . And, since we know that V ^ n sq + (x)>0, [Theorems 15 and 97a] Xj> it follows that £ + is (y: y>0}. sq Once we have proved (1') and (2'), we know that sq + has an inverse whose domain is the set of nonnegative numbers. Consequently, we may then introduce an operator, W , to use in referring to this in- verse. We may, as we have seen in section 9-01, do so by adopting a defining principle: V ^^ (Vx~e b + and sq + (Vx ) = x) x >0 sq [ Compare this with (* ) on page 9 - 5. ] From this defining principle and (2') we can deduce a uniqueness theorem : v x>o V» J-i*=* ^y-^] — ' sq [Compare this with (* ) on page 9-5.] Since sq + is a subset of the more familiar squaring function, we may reformulate the defining principle and the uniqueness theorem in + * 2 * more familiar terms by eliminating 4 sq ' in favor of the operator Recalling that £ + = {x: x >0| we obtain the defining principle: sq i* x ) V x >Q (Vx" >0 and (Vx~) 2 = x) and the uniqueness theorem: V ^ n V . n [y 2 = x =>y = Vx"] x >0 y > J J Since each number belongs to the domain of the squaring function, this uniqueness theorem is merely an abbreviation of: ( *2> V x>Q V y [(y>0 and y 2 = x) ^> y = Vx" ] The preceding discussion illustrates a general procedure for justi- fying the introduction of an operator [like ,- v ] to abbreviate a definite description ['the real number z such that z>0 and z = ']. Such a description refers to a function f [the function sq + ] which is a subset [ l z>0*] of a known function f ["z --the squaring function]. The pro- cedure is to show that this subset f , whose domain is some subset D of [App. A] [9-193] <>£, has an inverse whose domain is a certain set R [\x: x>0}]. This is done by proving, first, that R is contained in the range of f : (I) V R 3 xeD f(x) = y [Compare with (1) on page 9-190.] and that the range of f is contained in R [V >Q sq*(x) >0]. [In most cases this second inclusion is already known- -just as we know that all squares of real numbers are nonnegative, we shall usually know that the range of the entire function f is contained in R.] Secondly, one proves that f has an inverse: <"> V x a £D V Xs£D [f(x 1 )=f(x a ) *x 1 = x 2 ] [Compare with (2) on page 9- 190* J Having done these things, we are entitled to introduce a name- -say, *g* --[or an operator] for the inverse of f Q » and we do so by adopting a de- fining principle: (« x ) V x€R * y = gU) * J Now that we have seen that the problem of justifying the introduction of an operator often amounts to proving [(II)] that a subset of a related function has an inverse and to establishing ((I)] what the domain of this inverse is, we shall proceed to discover a large class of functions- -the monotonic functions - -each of which has an inverse. This will give us an easy way to prove the theorems we wish to of the form (II). Later in this Appendix we shall show how to determine the ranges of many of these monotonic functions--namely of those which are continuous and which [like sq + ] have "unbroken" domains [^ + is a ray; rays, segments, sq intervals, half-lines, and the set of all real numbers are examples of what we mean by 'unbroken domains*]. This will give us an easy way to prove the theorems we wish to of the form (I). [9-194] [App. A] MONOTONIC FUNCTIONS To discover a property of functions which insures that they have in- verses, let's return to the example of sq* and take note of one way of proving (2) on page 9- 190: [You may already have proved (2), probably in a different way, in Unit 7. For, as you probably have noticed, (2) is Theorem 98a.] Our proof de- pends heavily on an order theorem about the squaring function. By Theorem 98c, for a>0, and, for b >0, 2 2 if b > a then b > a 2 2 if a > b then a > b . Suppose, now, that a / b. It follows [from Theorem 86a] that b > a or a > b. For a > and b > 0, it then follows that b > a or a > b . 2 2 In both cases [by Theorem 87], a / b . Consequently [by contra- position], for a > and b > 0, 2 2 if a = b then a = b. So, using the general properties of order formulated in Theorems 86a and 87, the fact that sq + has an inverse follows from Theorem 98c. This latter theorem is, in view of the definition of sq + [and transitivity], equivalent to: V* ,V t > Jx 2 >x x => s q *( X2 ) > sq^x,)] i sq 2 sq For short, we say that the function sq + is an increasing function. Definition. A function f is increasing on a set . D if and only if DC1 and V x l£ D\ eD [ x 3 >x i =* f(x 2 ) > f(x !>]■ A function which is increasing on its domain i is called (merely) an . increasing functio n. [App. A] [9-195] Notice that an increasing function is increasing on each subset of its domain--in other words, each subset of an increasing function is an increasing function. The same procedure we used above in proving Theorem 98a can be used to prove that each increasing function has an inverse. [See Part B on page 9- 196. ] f is increasing on {x: s < x < t} For the function f shown above, name a set on which f is decreasing . Name another. EXERCISES A. 1. Formulate a definition of a function decreasing on a set D: A function f is decreasing on a set D 9 D C ? a nd V T-, V t-n [x„ > x, ^ x eD x^D L 2 l 2. Formulate a definition of a decreasing function. A function which is either increasing on D or decreasing on D is said to be monotonic on D. A function which is either an increasing function or a decreasing function is said to be ? [9-196] [App. A) B. 1. Prove that each increasing function has an inverse. Suppose that f is an increasing function. By definition, for a eb f and b e ? , if b > a then f(b) > f(a) and if a > b then ? Suppose, now, that a / b. . . *- [ Complete the proof. ] Therefore, V N V . [Hx x ) = f(x 2 ) =s> x, = x 2 J. 2. How would you change the foregoing proof to a proof that each decreasing function has an inverse? 3. Suppose that f is an increasing function. By Exercise 1, we know that f has an inverse, say, g. What kind of function is g? [Make some sketches to test your discovery.] 4. Prove that if f is an increasing function then V^^f 1 "* 2 ' >f(Xi> *** >x * ] - 5. Prove that the inverse of each increasing function is also in- creasing. [ Hint Suppose that f is an increasing function, that g is the inverse of f, and, for c and d in & , that d > c. Now, g use the defining principle (•& x ) on page 9-193 (with R = £ and D = & f ) and the theorem of Exercise 4. J 6. How would you modify your work in Exercises 4 and 5 to show that the inverse of a decreasing function is also a decreasing function ? vi* +** a* T* *f* *P The work you have done in Part B amounts to proving: [App. A] [9-197] Theorem 184 . Each monotonic function has a monotonic inverse of the same type, O*. O/ x»- *¥> 'f- 'C C. 1. (a) Use the definition of a monotonic function to show that the squaring function is not monotonic. (b) Use the theorem that each monotonic function has an inverse to show that the squaring function is not monotonic. 2. Is the function sq~ monotonic? If so, what is its inverse? 3. Suppose that r is the reciprocating function- - that is, that r = {(x, y), x / 0: y = x }. (a) Sketch a graph of r. (b) Does r have an inverse? (c) Prove that r is not monotonic. (d) Are there sets on which r is monotonic? (e) Is the donnain of r the union of two sets on each of which r is monotonic ? 2 3, 4. Repeat Exercise 3 for the functions r and r 5. The functions which you have considered in Exercises 3 and 4 are the first three negative- integral power functions. On the basis of your work in these exercises make general remarks on monotonicity and existence of inverses of the negative- integral power functions. 6. Investigate the positive- integral power functions. [The second of these is, of course, the squaring function.] [9-198] [App. A] D. 1. Prove, again, that sq + is an increasing function. That is, prove: Do this by using the algebra theorem: 2 2 V x V x x 2 - x 2 = (x 2 - x 1 )x 2 + and y > 0, xy > and x + y > 0. From this, together with the pmO and the paO, it is easy to see that ( *> v x >o v y>o (xy - ° andx + y >0 >- The theorems just mentioned, together with the algebra theorem and Theorems 84 and 92 should suffice for your proof.] Prove that the positive- integral power functions, restricted to nonnegative arguments, are increasing functions. That is, prove (by mathematical induction]: \\>o\>0 [k ^ X i =* x 2 n>x i^ [Theorem 185] Do this by using the algebra theorem: Vwwr n +1 n+1. . n , , n n. i n V X V X [X 2 ■ X 1 = U 2 ' X 1 )X 2 + (X 2 " *1 )X J [ Hint . Your answer for Exercise 1 should suggest part (ii) of the inductive proof. In this part you will also need Theorem 152a.] «&» o^ «.«,. T" *-,>. *-,-» In the investigation you carried out in answering Exercise 6 of Part C, you probably discovered that the odd positive- integral power functions are increasing functions, but that the even positive-integral power func- tions are increasing on{x: x >0| and decreasing on {x: x<0}. It is not hard to derive these results by using Theorem 185. The difference between the behavior of the even power functions and that of the odd power functions is due to the fact that ... >_, w // \2n 2n , , ,2n - 1 2n - 1. (*) V V ((-x) = x and (-x = -x ), n x [(*) is an easy consequence of Theorems 28, 158, and 150c] Here is a [App. A] [9-199] proof, consisting of two parts, that each odd positive-integral power function is increasing: Suppose, for a x <0 and a 2 <0, thata 2 >a 1 . It follows that -a 2 >0, -a 1 >0, and-a 1 >-a 2 . Hence, by Theorem 185, (-a 1 ) 2 P- 1 >(-a 2 ) 2 P" 1 . So, by <*), -a/P " * > -a/? " 1 --that is [Theorem 94], a/ p ' l >a* p ~ l . Hence, for a x <0 anda 2 <0, if a 2 > a x then a/P " l >af P ' 1 . Consequently, each odd positive- integral power function is increasing on {x: x < 0}. Now, since each odd positive- integral power function is increasing on (x: x < 0} and, by Theorem 185, is also in- creasing on {x: x > 0}, it follows [ Theorem 86c] that each such function is an increasing function [Explain. Hint. If a g > a x then, either a 2 and a x are both nonpositive or a and a x are both nonnegative, or a > > a . ] So, we have proved: V n V x V x K >x i ^x* n ~ l >xl n - 1 ] [Theorem 185'] ■J* o.'^ •«.»* "I* *"t v "\- *E_. 1. Repeat the first part of the proof of Theorem 185' with'Zp - 1' replaced by '2p' and, so, obtain a theorem about even positive- integral power functions restricted to nonpositive arguments. 2. If, for some p, the function g is the inverse of the 2pth power function restricted to nonnegative arguments, what is the in- verse of the 2pth power function restricted to nonpositive arguments ? 3. Recall [Exercise 3 of Part C] that the reciprocating function, r, is decreasing on {x: x < 0) and on (x: x > 0} , but not on (x: x/0). But, why cannot the second part of the proof of Theorem 185' be modified to prove that r is decreasing on {x: x / 0} ? [9-200] [App. AJ On page 9-193 we reached the conclusion that the introduction of an operator can often be justified by proving that a certain function has an inverse and establishing what is the domain of this inverse- -that is, what is the range of the function. So, it became our goal to discover generalizations from which, for each of many functions, it would be easy to deduce that the function in question has an inverse and easy to establish what the range of the function in question is. Theorem 184 is one of the sought-for generalizations. It tells us that if a function f is monotonic on a set D then the function obtained by restricting f to the set D has an inverse. In other words, if f is mono- tonic on D then [ see page 9- 193] (in / x 1 «DV DlfUJ * f j ] that (b) ^x^O V x 2 >of X x n = x 2 n ^ x i = * 2 ]« A theorem of the form (II) justifies our introducing an operator which gives us, for each number in the range of the function obtained by restricting f to arguments in D, the value of the inverse of this function at this number. For example, the theorem (b) gives us the right to introduce an operator [it will be v '] which, for each n, gives us the values of the inverse of the restricted nth power function. However, before we can use this operator properly, we must find out for which numbers this inverse is defined- -that is, we must determine the ranges of the restricted power functions. The major step in doing so consists in proving theorems of the form [see page 9-193]: (I » VR 3 xeD f <*> = y Using the notion of continuity, which you will discover in the Exploration [App. A] [9-201] Exercises which follow, we shall be able to prove theorems of this kind. In particular, we shall be able to prove: (c) V V 3 v.,. x n = y n y>0 x>0 7 This, together with the rather easy theorem: V V >A x n >0, n x >0 — shows that the range of each of the restricted positive- integral power functions is {y: y > 0) . So, the operator' v ' can [like 'v '] be applied to each nonnegative number- -for each n, and for each x>0, we are, by the theorems (b) and (c), entitled to speak of the principal nth root of x. [Similarly, the theorem (a), together with another theorem like (c): w w zi 2n - 1 V V d x = y, n y x ' (which we shall prove in the same way) shows that the operator v can be applied to each real number, nonnegative or not.] EXPLORATION EXERCISES A. Consider the function f defined by: x, x < 2 ( jx + 1 , 2 < x < 2x - 8, x > 6 Complete. 1. f(-8) = 2. f(75) = 3. f(4.44) = 4. f{ ) = 1.99 5. f(2.02) : 6. f( ):=2.01 7. f(3) = 10. f(201) = 13. f(7) =_ 16. f(20) = 5. f(2.02) = 8. f(4) = 11. f(54) = 14. f(6.5) = 17. f( ) = 20 9. f(5) = 12. f(9) = 15. f(6.2) = f( ) = 18. 3 [9-202] [App. A] (As a partial check on your computations, compare your completed Exercises 18 and 8.] 19. f(2) = 20. f(2.2) = 21. f(2.4) = 22. f(2.6) = 23. f( ) = 2.4 24. f( ) = 2. 5 [Compare your completed Exercises 24 and 7.] 25. f( ) = 2.6 28. f( ) = 3.3 31. f( ) = 3.7 26. f( ) = 2, 27. f( ) = 3.2 29. f( ) = 3.4 30. f( ) = 3.5 32. f( ) = 3.9 33. f( ) = 4. 1 [Compare your completed Exercises 33 and 15.] 34. The domain of f is 35. The range of f is B. Consider the function g defined by: r x, x < 2 g(x) = / 3x - 4, 2 < x < 6 ^x+ 10, x > 6 Complete. 1. g(-8) : 2. g(4) = 5. g(6.3) = 4. g(6. 1) = 7. g(1.99) = 10. g(2.01) = 13. g(4.2) = 16. g( ) = 8 19. g( ) = 11 11. g(3) = 14. g( ) = 6 17. g( ) - 8.6 3. g(5.9) 6. g(6.5) g( ) = 1.99 9. g(2) = 12. g( ) = 5 15- g(f)=_ g< ) = 9 20. g( ) = 11.6 21. g( ) = 11.9 [App. A] [9-203] 22. g( ) = 12.5 23. g( ) = 13. 1 24, g ( ) = 13. 7 25. g( ) = 14.3 26. g( ) = 14.9 27. g( )=15.5 28. g( > = 16. 1 29. g( ) = 16.3 30. g( ) = 16.5 [Compare Exercises 4, 5, 6 with Exercises 25, 26, 27 and with Exercises 28, 29, 30.] 31. The domain of g is . 32. The range of g is . C. If you have not yet done so, draw the graphs of the functions f and g of Parts A and B. [Use a separate sheet of paper for each graph.] 1. (a) Is each real number an argument of f ? That is, is it the case that, for each real number x r there exists a real number y such that f(x) = y? Justify your answer in terms of the defi- nition of f and also in graphical terms. (b) Is each real number a value of f ? That is, is it the case that, for each real number y, there is a real number x such that f(x) = y? Justify your answer. (c) True or false ? (1) V 3 f(x) = y (2) V 3 f(y) = x (3) V 3 f( x ) = y x y x y y x (4) 3 V f{x) = y (5) 3 V f(x) = y (6) V V f(x) = y xy yx ' x y * 2. (a) Now look at the graph, and the definition, of g. Is each real number an argument of g ? That is, is it the case that V 3 y = g( X ) ? x y J 6 Justify your answer. (b) True or false? v y a x y = g(x) (c) Which of the exercises in Part B justify your answer to (b) ? (d) Interpret your answer to (b) in terms of the graph of g. [9-2C4] [App. A] D. [Refer to the graph, and the definition, of f, concentrating your atten- tion on the point (4, f(4)).] 1. Complete, (a) f(3.4) = (b) f(4) = (c) f(4.6) = (d) {t: f(t) >Z.7} = {t: t > } (e) (t: f(t) < 3.3} = {t: t< } 2. Verify the following statement: {t: 2. 7 < f(t) < 3. 3} = {t: 3. 4 < t < 4. 6} 3. True or false ? (a) For each x, 2. 7 < f(x) < 3. 3 if and only if 3. 4 < x < 4. 6. (b) For each x, if 3. 4 < x < 4. 6 then 2. 7 < f(x) < 3. 3. (c) V [ 4 - 0. 6 < x < 4 + 0. 6 => 3 - 0. 3 < f(x) < 3 + 0. 3] (d) V [-0.6 < x - 4 < 0. 6 => -0. 3 < f(x) - f(4) < 0.3] (e) V [ |x - 4 | < 0.6 => |f(x) - f(4) | < 0. 3] [See Fig. 1. ] H, f(4)) Fig. 1 (f) V [ |x - 4| < 0.4 (g) V |x - 4| < 0.4 (h) V [ |x - 4| < 0.01 (i) V [ Jx - 4| < 0.6 |x - 4| < 0. 6] |f(x)- f(4)| < 0.3] > |f(x) - f(4) | < 0.3] |f(x) - f(4)| < 157] [See Fig. 2. ] [App. A] [9-205] 4. Complete. (a) f(4) = (b) f( ) = 3-0.07 (c) f( ) = 3 + 0.07 5. Verifying the following statement: {t: 2.93 -0.07 < f(x) - f ( 4) < 0.07] (c) V [ |x - 4| < 0. 14 => |f(x) - f(4) | < 0.07] (d) V [ |x - 4| < 0. 1 => |f(x) - f(4) | < 0.07] 7. Complete. V [ |x - 4 | < =J> |f(x) - f(4)| < 0.01] [ Hint . Compare this with Exercises 3(e) and 6(c).] 8. Verify your answer to Exercise 7. 9. Complete. (a) V [ |x - 4| < => |f( x ) - f(4) | < 0.002] (b) V [ |x - 4 | < => |f(x) - f(4) | < 10~ Z °] (c) V x [ |x - 4 | < => |f(x) - f(4) | < 10" 10 °] (d) V x [ |x - 4 | < (e) |6.5 - 4| = (f) f(6.5) = (g) |f(6.5) - f(4)| ? True or false? [Draw figures like those in Exercise 3.] (h) V [|x- 4| < 3 => |f(x) - f(4)| < 1.5] (i) V x [|x-4|<2.9 => |f(x) - f<4)| < 1.5] (j) V x [|x-4|<2.7 ^> |f(x) - f(4)| < 1.5] [9-206] [App. A] (k)V[|x-4| <2.5 => |f(x) - f(4)| < 1.5] (JL) V [ |x - 4 | < 2. 5 => |f(x) - f(4) | < 1. 49] (m) V [ |x - 4 | < 2 => |f(x) - f(4) | < 1. 49] (n) V [ |x - 4 1 < 2 => |f(x) - f(4)| < 1.0001] 10. (a) Choose a very small positive number c. Find a positive num- ber d such that V [ |x - 4 1 < d => |f( x ) - f(4)| < c]. (b) Choose a number c > c. With the same positive number d you found in part (a), is it the case that V x [ |x - 4| < d ^> |f(x) - f(4)| < cj? (c) Write a formula for computing a suitable number d, given c, in case (i) < C < 1; d = (ii) c > 1; d = [See Exercise 9(n).] 11. True or false? (a) For each positive number c [ no matter how small], there is a positive number d such that V [ |x - 4 1 < d => |f(x) - f(4)| < c]. (b) V c>0 3 d>0 V l X * 4 I < d * l f(x) " f(4) I < c] E_. [Refer, again, to the graph and definition of f, concentrating your attention, now, on the point (6, f(6)).] 1. Complete. (a) f(6) = (b) f( ) = 4-0.2 (c) f( ) = 4 + 0.2 2. Verify the following statement: {t: |f(t) - 4 | < 0.2} = {t: 5.6 < t < 6. l} 3. True or false ? (a) For each x, if 5. 6 < x < 6. 1 then |f(x) - f(6) |<0.2. (App. A] [9-207] (b) V [-0.4 < x - 6< 0. 1 => |f(x) - f(6) | < 0.2] (c)V [-0.4 -0.4 -0.4 < x - 6 < 0.1] (e) V [ |x - 6| < 0. 1 -> |f(x) - f(6)|< 0.2] 4. Complete. V [ |x - 6 | < => |f(x) - f(6) | < 2 .10" 12 ] 5. True or false ? V c>0 3 d>0 V x [ l X " 6| 0 3 d>0 V h [|h| " f(6) l 0 V x [ l X ' 4 l |g(x) - g(4)| |g(x) x 0 V[ |x- 6 | |g(x) - g(6)| < 0.3] (b) V[5.9 |g(x) - g(6) | < 0.3] (c) V x [|x-6|<0.1 => |g(x) - g(6)| < 0.3] (d) V x<6 [ |x - 6| < 0.1 ^ |g(x> - g(6)| < 0.3] 7. True or false ? (a)3 d>() V x [|x- 6 < d ^> g(x) - g(6)| < 0. 3] < b > 3 d>oVl x - 6 < d :=> g(x) - g(6)| 3 d>oVI*- 6 < d =£> g(x) - g(6)| < 1.9] (d > 3 d>o V xN- 6 < d => g(x) - g(6)| <2] < e ) 3 ,>n V J|x- 6 < d => g(x) - g(6) < 2.0001] (f) Starting at 6, there are arbitrarily small changes in the argu- ment of g which result in changing the value of g by at least 2, [App. A] [9-209] H. Suppose that h is a function such that i Mx) = < x, for x < 2 x + 1, for 2 < x < 4 -3x + 17, for 4 < x < 5 1, for 5 < x < 10. 1. What is the domain of h? 2. Draw a graph of h for the segment -4, 10 of its domain. 3. True or false ? (a > V O0 3 d>0 V x<10 [ l* + M0 3 d>0 V x<10 [ l x " 2 I - h < 2 >l 0 3 d>0 V x<10 [ ' X - '""I 0 3 d>0 V x<10 1 I"" 2 - 011 0 V x<10 ( l X " 4 I < d * ' h(x) " h<4) l * C J (£ > V O0 3 d>0 V x<10 [ l x - H V c >0 3 d >0 V x <10 [ l X " ? l < d * I K(X) " h (7) | < c] (h) V c>0 3 d>0 V x<10 [ l X - '°l 0 3 d>C V x<10[l---ol0 3 d>0 V x'-''ol0 3 d>0 V x<10 ( l x - "ol 0 and Vjjx - J | < (b) > and V [ |x - 3| < ^>|f(x)-f(|)| |f(x) - f(3)| < 1] 2. True or false ? o v x !|x - 3 I - f < 3 >l * c ) 3. Complete V x < V c>0 3 d>0 V xN X - X ol t f(x Q ) I < c] CONTINUOUS FUNCTIONS In the preceeding exercises you have seen that, given a function f and an argument a of f, it may be the case that, starting at a , any sufficiently small argument-change results in an arbitrarily small value- change - -that is, V C>0 3 d>0 V x6^l X - a ol< d => I «*> - «»0> I < C]. If [and only if] that J_s the case, the function f is said to be continuous at a . — o [App. A] [9-211] Definition. A function f is continuous at a Q if and only if a o °f andV c>o 3 d>o V xe* f [ l x " a o |f(x) f(a Q )| v O0 3 d>0 V xe* f ( tb) Starting at . . . 2. State in two ways that f is continuous at 6. 3. State in two ways that f is continuous. 4. Tell why the following statements are false. (a > V c >0 V d> V xO £ ( I* " 2 I < d "* l fW " f(2 » I < c l < b > 3 d>0 V O0 V xO £ l l«" 21 <- * l*W - « 2 >l < ^ 5. (a) State in two ways that sq is continuous. (b) State in two ways that sq* is continuous. (c) Explain why, if sq is continuous at 2, sq* must be. (d) Explain why, if sq is continuous, sq* must be. JB. 1. Suppose that f is such that f(3) = 5 and, for x / 3, f(x) = 2x. (a) Graph the function f. (b) True or false? (i) For each x , f is continuous at x Q . (ii) f is continuous at 3. (c) In part (b), you should have answered 'false' for (ii) and, consequently [Explain.], for (i). (i) Put a restriction on 'x' in (i) of part (b) to obtain a true statement, (ii) Change the definition of f at 3 so that (ii) of part (b) will be true. [App. A] [9-213] Z. The signum function is defined by: f-l, for x < sgn(x) = / 0, for x = L I, for x > (a) Graph sgn. (b) True or false? (i) sgn is continuous, (ii) sgn is continuous at 0. (c) In part (b), you should have answered *false* for (ii) and, conse- quently [Explain.], 'false* for (i). (i) Complete: sgn is continuous except at . (ii) Can you change the definition of sgn at so that (ii) of part (b) will be true ? 3. Here is part of the graph of a function g whose domain is {x: |xj< l) . The whole graph consists of the origin and the points on the legs of an infinite number of isosceles triangles. [The base of each tri- —I l ' angle is one of the segments , - — ■ — r or one of the segments • — ? j — • B n n + 1 B — ■ — r» — » for any n, and the altitude of each triangle is 1,1 By definition, g(0) = 0. True or false ? < a > v oo 3 d>o v xe* f [ l x l |f(x) - f 0, it follows that, for c > 0, c / \ -3 | > 0. Consequently, for each c > 0, there is a number d > [c / | -3 | is one such number] such that Vjlx- a | |f(x) - f(a Q )| < c]. Hence, f is continuous at any number a--that is, f is continuous, *•- vl- O* 'i x *v 'i v C. 1. Suppose that, for some numbers m / and b, f (x) = mx + b, for all x. Prove that f is continuous. [That is, prove that each function is continuous. ] [ Hint. Follow the Example above.] [App. A] [9-215] 2. Suppose that g is the absolute value function. [Graph it.] Prove: Vc>0 3 d>0 V x l l X ' X ol < d * le (x) - 8 (x oH < c] --that is, prove that g is continuous. [ Hint . Note that V x V y ||x| - |y||< |x-y|. ] 3. Consider the principal square- rooting function h: h(x) = Vx", for x > (a) Graph h. (b) Complete [by putting the same expression in both blanks]. V c>0 ( >0 and V X > Q [ |x - 7 | < ^|h(x) - h(7)\ < c] ) [ Hint . Notice that, for any a > 0, |a - 7| = |(vT+ vT)<>/a~ - VT)| = |vT + VT| • |vT - vT| = (Va" + V7)|vT - VTj > /r|vT -VT | [Explain.] and, so, i 7 i |h(a)-h<7)| < l^I» .] (c) True or false ? (i) The function h is continuous at 7. (ii) The function h is continuous [that is, h is continuous at each x Q >0]. (d) Support your answer for (ii) of part (c). [ Hint . You may want to consider the cases x Q > and x = separately.] 4. Consider the squaring function sq: 2 sq(x) = x , for all x (a) Graph sq. (b) Do you think that sq is continuous ? [9-216] [App. A] Let*s try to prove that sq is continuous at, say, 2- -that is, that, given any c > 0, we can find a d > such that Now, for any a, and, so, and that is, 'x[ |x - 2 | < d =t>\ sq(x) - sq<2) | < cj. 2 2 sq(a) - sq(2) = a - Z = (a + 2)(a - 2) sq{a) - sq(2) | = ja + 2 J • ja - 2 J [Explain.] sq(a) - sq(2) | < c if |a + 2 | • |a - 2 | < c if |a - 2 I < |a+2|" One's first impulse, here, is likely to be to say that the problem is finished- -for any c > 0, we can take for d the number c/ja + 2J. The trouble with this is that, given a number c, the value of the expression 'c/ja + 2 J' depends on the value we choose for 'a'. And, what we need is one number d > such that, for any a, |sq(a) - sq(2)| < c if |a - 2| < d. If we could find a positive number d such that, for any a^ |a + 2| we'd be all set. For, for such a d > 0, if | a - 2 | < d then |a - 2| < and, as we have seen, if I a - 2 j < - — - — 1 ' a + 2 |a + 2| then |sq(a) - sq(2) | < c. To see what is happening, let's graph the values of 'c/|a + 2|'. [App. A] [9-217] [Here is a table used in drawing the graph: a -1 1 2 3 4 c/| a + 2| c c/2 c/3 c/4 c/5 c/6 Note that, since we have not indicated the scale on the vertical axis, the same graph will do for any number c > 0. ] Evidently, for larger values of 4 a*, one gets smaller values of *c / |a + 2 I '--in fact, if a is sufficiently large, c / |a + 2 | is an arbi- trarily small positive number. It turns out that there is no d > such that, for any a, ~ a + 2 | However, there is a way out. We aren't interested in all numbers a--just in those near 2. And, from the graph, if, say, |a - 2J < 1 [that is, if 1 < a < 3], £ < £ ■ 5 ~ |a + 2| So, if |a - 2 | < 1 and |a - 2J < c/5 then |sq(a) - sq(2) | < c. Hence, for d, we can take the smaller of the two positive numbers 1 and c/5. If the number c we choose is greater than 5, we take d to be 1 ; if not, we take d to be c/5. So, in any case, if c > then there is a d > such that V x [ |x - 2 | < d => |sq(x) - sq(2)| < c] --that is, sq _is continuous at 2. Now, let's see how our knowledge of absolute values could have been used to show that this way of choosing d would work. Recall that we showed, rather easily, that Now, and, since and, so Hence, sq(a) • sq(2) | < c if |a + 2 | • |a - 2 | < c. [What theorem ?] [What theorem ?] |a + 2| < ja| + |a| - |2| < ja - 2 |a| < |a - 2 | a | + | 2 | < | a - 2 la + Z\ < la - 2 + |2|. + 2|2|. + 2|2|. [9-218] [App. A] Consequently, for ja - 2 J < 1, |a + 2| < 1 + 2|2|. So, for |a - 2 | < 1, |a + 2| • |a - 2| < (1 + 2|2|)|a - 2|. Hence, for |a - 2 | < 1, if (1 + 2|2|) |a - 2 | < c then |sq(a) - sq(2) | < c - -that is, if |a - 2 j < 1 and ja - 2 | < . ~ -, then | sq(a) - sq(2) j < c. c So, again, we take for d the smaller of the numbers 1 and ? ? . . This procedure will work to show, for any number a Q , sq is con- tinuous at a Q . [Just replace most of the '2's by 'a 's. ] 4. [ continued] *(c) Prove that sq is continuous *(d) Prove: Theorem 186 Each positive-integral power function is continuous. [ Hint . One way to proceed is by induction. Clearly, the first positive- integral power function is continuous [Explain. ]. Suppose, then, that the pth power function is continuous- - say, at a . That is, assume that <* )V O0 3 d>0 V x [ l X - a °l [ Explain. ] J -.i i . , rr is a positive number 2 d ♦ l a oD and, so, by the inductive hypothesis (*), there is a positive number --say, d.-- such that if |a-a | |xP + 1 -a P +1 |0 3 x> 0*" = *" As pointed out there, we need this result in order to know that, for each n, the domain of the principal nth root function is the set of all non- negative numbers. More generally, our problem is to discover a method for showing, of as many numbers as we can, that they belong to the range of a given function f. That is, given a function f, we wish to have a method for proving existence theorems like: V cR 3 £ . f(x) = y [See (I) on page 9-193.] y t X\ X t rJr for as "large" a set R as possible. Actually, our present interest is in functions which have inverse s- - that is, such that V . V [f(x ) - f(x 2 ) =>Xj = x 2 ] [See (II) on page 9-193.] X l f X 2^ f and, more particularly, in monotonic functions. So, we might reformulate our problem as follows: (P) Find a method for determining the range of a monotonic function. If we solved this problem, we should have a method for determining the domains of the inverses of monotonic functions. Since there are many monotonic functions which— like the positive-integral power functions, restricted to nonnegative arguments — have useful inverses, this would be a good thing to be able to do. As a matter of fact, all the monotonic functions which will interest us will—again like the [restricted] power functions — be continuous. Moreover, the domain of each of these functions will be either the set of all real numbers, or a ray [for example, (x: x > O} ], or a half- line [{x: x > 0}], or an interval [{x: |xj < l}], or a segment [{x: |x| < l}]. So, it will be sufficient to solve a subcase of the problem (P): [App. A] [9-221] (P ) Find a method for determining the range of a continuous monotonic function whose domain is either the set of all real numbers, or a ray, or a half-line, or an interval, or a segment. Since what can be done for increasing functions can be done for de- creasing functions, we shall begin by considering continuous increasing functions,, And, because it is the simplest case, we shall consider, at first, continuous increasing functions whose domains are segments. It will turn out that once we have solved this case of problem (P c ), each of the remaining cases will be easy to handle c As a dividend, we shall dis- cover that the inverse of each function of the kinds described in (P c ) is not only--as we already know- -monotonic, but is, itself, continuous. To begin with, consider the functions g and h whose graphs are shown in Fig. 1 and Fig. 2. Both are increasing functions, each has the segment a, b as its domain, and g(a) = c = h(a) and g(b) = d = h(b). d •-- Fig. 1 Fig. 2 Since both functions are monotonic, it follows that the range of each is a subset of the negment c, d. But, the range of g is the whole of this segment, while the range of h is only part of this segment. What dif- ference between g and h do you think accounts for this difference in their ranges ? It is easy to see that if f is any increasing function whose domain is the segment a, b then the range of f is a subset of the segment f(a), f(b). And, as you have seen, the range may or may not be the whole of this [9-222] [App. A] segment. For simplicity, if the range is the whole segment, let us say that the function f has maximal range . [Why 'maximal'?] The examples g and h may have suggested to you that an increasing function whose domain is a segment has maximal range if and only if it is continuous. This is the case. We shall prove it to be so, beginning with the if-part. This part amounts to the following statement: (a) A continuous increasing function whose domain is a segment has maximal range. Before proving (a), let's illustrate how it will help us in solving our original problem. This was, you remember, to show that each non- negative number has, for each n, a nonnegative nth root. We shall illustrate, in a small way, the use of (a) by using it to prove that the nonnegative number 5 has a nonnegative 4th root- -that is, that there is a nonnegative number whose 4th power is 5. This is, now, very easy to do: From Theorem 186 we know that the 4th power function is continuous. Consequently, the restricted power function f, where f = {(x, y), x > 0: y = x 4 }, is also a continuous function. Furthermore, by Theorem 185, this function is increasing. Consider, now, the subset f Q , where f = {(x, y), < x < 2: y - x 4 } . As a subset of the continuous increasing function f, f is, also, a continuous increasing function. Moreover, the domain of f Q is a • • segment--the segment 0, 2. Consequently, by (a) [and the meaning » — — * of 'maximal range'], the range of f is the segment f o (0), f Q (2). Since f Q (0) = and f Q (2) = 16, it follows that the range of f Q is { y: < y < 16). Consequently, any number between and 16 is the 4th power of some number between and 2. Since 5 is be- tween and 16, 5 has a nonnegative 4th root. It should now be clear how (a) [together with Theorems 185 and 186] can be used in proving that, for each n, each nonnegative number has a nonnegative nth root. [App. A] [9-223] In proving that 5 has a nonnegative 4th root, we needed, besides (a) and Theorems 185 and 186, to know that there is a number whose 4th power is not less than 5. One such number is 2, and we used this in describing the function f Q . We could as well have used 1.5 [Explain.]. In proving the general result we shall need to know that *„ V y>0 3 x>0 xn ^- This follows easily from Bernoulli's Inequality [Theorem 162]. Now that we have seen how (a) can be used to solve (P c )» let*s find a way to prove (a). [We shall see that the least upper bound principle will be of help to us.] Consider a continuous increasing function f whose domain, fy, is a segment {x: a < x < b} . We know that the range of f is a subset of (O.f(b)) "^ (0,f0: x < 2} is 2. In the latter proof we used the monotonicity of sq' and two general- izations [( ) on page 9-3l] concerning a special function h. In the proof of (ii) we shall use the monotonicity of our function f [instead of the monotonicity of sq + ] and the continuity of f [instead of (•&>,) and( )]. Now, finally, here is a proof of (a): Suppose that f is an increasing function whose domainis {x:a y Q . Suppose that f(x ) < y . It follows that if c = y Q - f(x ) then c > 0. So, since f is continuous at x , there is a d > such that V x£^ f [ |x - x Q | < d =i> |f(x) - f(x Q ) | < c]. Now, for each x, ^ ^ , ^ i i ^ , x Q |x - x Q | < d and |f(x) - f(x Q )| < c =>f(x) - f(x Q ) < c. [App. A] [9-225] Consequently, for each x€b f> it follows, since c = y Q - fC* )» that if x Q < x < x + d then f(x) - f(x Q ) < y Q - f(x ). o o X^ - d Xp x 1 Now, since, by hypothesis, y < f(b) and, by assumption, f(x ) < y Q , it follows that f(x Q ) < f(b) and, since f is increasing, that x_ < b. Hence, there is an x e& f such that x Q < x < x Q + d. [Both b and x + d are greater than x . For x one might take the midpoint of the segment whose end points are x and the smaller of the numbers b and x + d. ] As we have seen, for such a number x , f(x ) - f(x ) < y - f(x )--that is, f(x ) < y . Hence, by definition, x eE. So, x is a member of E which is greater than x Q . But, this is impossible because x is an upper bound of E. Hence, f(x Q ) ^ y Q . Suppose, now, that f(x Q ) > y Q . It follows that if c = f(x Q ) - y Q then c > 0. So, since f is continuous at x , there is a d > such that V xe ^( |x - x | |f(x)-f(x )| j x - x Q I < d |f(x) - f(x )| < c ==>-c f(a) and, by assumption, f(x Q ) > y_, it follows that f(x ) > f(a) and, since f is increasing, that x > a. Hence, there is an x. € £ f such that x - d < x < x [ Explain. ]. As we have seen, for such a number x,, y - f(x Q ) < ffx^ - f(x Q )--that is, y < f(x 1 ). Since, for each xeE, f(x) < y Q , it follows that, for each x€E, f(x) < f(x.) and, since f is increasing, x < x . So, x. is an upper bound of E which is less than x . But, this is impossible because x is the least upper bound of E. Hence, f(x Q ) ^ y Q . So, we have proved (a). [9-226] [App. A] Before applying (a) [in the way we have illustrated] to prove the existence theorem on page 9-220: V n V y>0 3 x>0 xn = y let*s recall that (a) was only the if-part of a conjecture you made on page 9-223. This conjecture was that an increasing function whose do- main is a segment has maximal range if and only if it is continuous. To complete the proof of this conjecture we still have to prove: (b) An increasing function whose domain is a segment and whose range is maximal is continuous. [Aside from checking on our conjecture, this theorem will be useful to us because it [together with the if-part] will allow us to conclude at once that the inverse of a continuous increasing function whose domain is a segment is also continuous. ] Here is a proof of (b) : Suppose that f is an increasing function whose domain is the segment (x: a < x < b} and whose range is maximal- -that is, the range of f is { y: f (a) < y < f(b)} . We wish to show, for each x^ e /5y» that, given c > 0, there is a d > such that V x€^- [|X - X ol -fU )| and c 0 which works for this c. For such a number c, c < f(x Q ) - f(a) and c < f(b) - f(x Q ) --that is, f(a) < f(x Q ) - c < f(x ) < f(x Q ) + c < f(b). [App. A] [9-227] Since the range of f is maximal, it follows that both f(x ) - c and f{x Q ) + c belong to the range of f--that is, there are arguments- - say, x Q ~ and x Q + --of f such that f(xr) = H* ) - c and f(xj ) = f(x Q ) + c. Since f is increas- ing and since f(x ") < f(x n ) < f(x *) , it follows that x~ < x Q < x * . Moreover, since f is increasing, it follows that, for eachx€-5y, that is, if x " < x < x Q + then f(x -) < f(x) < f(x Q + ) if x" < x < x Q + then f(x Q ) - c < f(x) < f(x Q ) + c. Since x " < x Q < x + , it follows that both x Q - x Q and x Q + - x are positive, Let d be the smaller of these numbers. Then d > and d ^ x o " x o" and d < x + - x Q that is, x " < x - d and x Q + d < x Q + . Consequently, for each x, if x - d < x < x Q + d then x ~ < x < x + . So, for each x€^- f , if x Q - d < x < x Q + d then f(x Q ) - c < f(x) < f(x Q ) + c. In other words, x € £ [|x - x Q | < d =» |f(x) - f(x Q )| < c] --f is continuous at x Q . [9-228] [App. A] The preceeding discussion shows that f is continuous at each x_ such that a < x Q < b. To prove that f i6 continuous, it remains to be shown that f is continuous at a and at b. Since £- = {x: a < x < b} and since f is increasing, showing that f is continuous at a amounts to showing that, given c > 0, there is a d > such that (*) V £V [a f(a) < f(x) < f(a) + c]. As before, it is sufficient to consider numbers c such that < c and satisfies (*). The proof that f is continuous at b is carried out in a similar manner. This completes the proof of (b). Combining (a) and (b), we have the result that an increasing function whose domain is a segment has maximal range if and only if it is continuous. Since, if f is a decreasing function whose domain is a segment, -f is an increasing function whose domain is a segment, and since f is continuous if and only if -f is continuous, and has maximal range if and only if -f does, it follows that the word 'increasing* in our result can be replaced by 'decreasing'--and, hence, by 'monotonic*. Rather than use (a) to establish our fundamental existence theorem on principal nth roots, it will be more convenient to use the result just obtained to prove an important supplement to Theorem 184 and then use this to establish theorems about the principal nth root func- tions. The supplement we wish is: Theorei Eac domain monotoi main is n 187. h continuous monc >tonic has a same function f w l continuous type whose hose do- is a segment a, b lie inverse of the the segment f(a), f(b). To prove Theorem 187, we use Theorem 184 and the result we have just established. Suppose that f is a continuous monotonic function whose domain is a, b. By Theorem 184 [since f is monotonicj f has an inverse, [App. A] (9-229] g, which is also monotonic of the same type. By the if-part of the theo- rem just established it follows ( since, also, f is continuous and its domain is a, b J that the range of f is the segment f(a), f(b). Since this is also the domain of g, it follows that g is a monotonic function whose domain is the segment f(a), f(b). Since the range of g is the domain of f, which is the segment a,b [and since a = g(f(a)), b = g(f{b))j, it follows that g has maximal range. Consequently, by the only if-part of the theorem just established, g is continuous. Let's now see how Theorems 184-187 can be applied in the case of the positive-integral power functions, restricted to nonnegative argu- ments. The nth such function is the function f such that f = {(x, y), x > 0: y = x D }. As we have seen [Theorems 185 and 186], this function, whose domain is the ray {x: x > o} , is continuous and increasing. Consequently, by Theo- rem 184, f has an inverse g which is also increasing and whose domain is the range of f. Because f is an increasing function and = 0, it follows that the values of f are nonnegative. So, the range of f is a subset of the ray {y: y > O} . In showing that each nonnegative number belongs to the range of f, as well as that g is continuous, we shall make use of Theo- rem 187 and of a simple lemma about the function f: *n\>0<* ******* | 1 This lemma can be proved in several ways. One easy proof uses Bernoulli's Inequality [Theorem 162]. According to this, for a >-l, (a + 1) > 1 + na. Since n > 1, it follows that, for a > 0, na _> a. So, for a > 0, (a + 1) > a + 1. [It is also easy to prove this lemma by induction. ] I I Now, as to the range of f, suppose that y > and consider the function f Q such that f Q = {(x, y), < x < y Q + 1: y = x"). [Why *y + V? Hint . Recall the proof on page 9-222 that 5 has a nonnegative 4th root, and see the lemma. J Since f Q is a subset of f, f Q is also a continuous increasing function, and the domain of f Q is the segment 0, y Q + I. By Theorem 187, f Q has a continuous increasing inverse g ( a subset of [9-230] [App. Aj the inverse g of f] whose domain is the segment f Q (0), f (y ♦ !)• Since f o {0) = < y 1, it follows that the domain of g Q contains all arguments y of g such that |y - y D I — *• Since the domain of g Q contains all arguments of g which are sufficiently close to y Q and since g Q is continuous at y , so is g. Consequently, the domain of the inverse g of f is {y: y > 0} , and g is continuous at each of its arguments. These results on positive-integral power functions justify ac- ceptance of the descriptive definition: V V ^^ Vx* = the z > such that z = x n x >0 — Our actual course will be to adopt the defining principle for principal roots: (PR) V V . n ('Vx">0 and (>/x") n = x n x>0 The effect of this principle is to introduce the operator * v ' in terms of which we can define the function g of the preceding paragraph- - the principal nth root function --as follows: g = {(x, y), x > 0: y = ^x } Since g is the inverse of the function f, we have the theorem: V q V >() V((y>0andy n = x) => y = ^ ] (Theorem 188] And, as proved above: Each principal positive- integral root function is continuous and increasing on the set of nonnegative numbers. [Theorem 189] As in the past, we shall follow the custom of abbreviating ' v 'to 'v\ Notice, also, that, for each x > 0, vx = x. To prove this, we use Theorem 188: 1 l/~ For a > 0, a > and a = a. So [ by Theorem 188], a = V a . [App. B] [9-231] APPENDIX B [This Appendix fulfills some promises made in section 9.05. ] Irrational numbers . --In section 9. 05 the main emphasis was on the rational numbers, but some results were obtained concerning the ir- rationals. In this Appendix, the main emphasis is on the irrational numbers, but we shall also learn more about rational numbers. THE IRRATIONALITY OF ROOTS In section 9. 05 we proved that v2 is irrational and, from some examples, conjectured: Theorem 193 . V V vm is irrational unless m is n m a perfect nth power. In other words, if a principal root of a positive integer is a rational number then it must be an integer: V V [7m" € R => 7m" el] n m To say that Vm e R is to say that, for some rational number r, Vm - r--that is, that, for some rational number r, r > and r = m e I + . So, the theorem last displayed can be restated as: VV,Jr n £r^>r£l] n r > l J While we are about it, we shall prove the stronger theorem: VV [r n £l^>rel] n r J To do so, it is sufficient to prove its contrapositive : V V [r i I =S> r n i I] n r L * * J or, more briefly: [9-232] [App. B] We shall prove (*) by a kind of mathematical induction. What we shall do is to show that, for any given nonintegral rational number r t and any positive integer m > 1, (*>:<) if, for each p < m, r P g 7 I then r ^ I. When we have shown this, (*) follows easily by the least number theorem [Theorem 108], For, suppose (*) is not the case--that is, suppose that there is an r f[ I and a positive integer n such that r el. Then, by the least number theorem, there is, for this r, a least such positive integer --say m. By the choice of m, r el and, for each p < m, r" ^ I. Now, since r/l and r € I, it follows that m ^ 1- -hence, m > 1. Since m > 1 and, for each p < m, r P i I, it follows from (**) that r { I. But, by the choice of m, r el. Hence, the assumption that (*) is not the case leads to a contradiction. Consequently, (*). Since, as we have seen, Theorem 193 is a consequence of (*), all that remains to complete the proof of Theorem 193 is to complete the proof of (*) by proving (#*). Suppose, then, that r ^ I and that m > 1, and suppose that, for each p< m, r*g I. Since R is closed with respect to multiplication [and r e Rj, it is clear that, for each p, r" e R. So, by (R), for each p, there is a positive integer q such that r" • q e I. Since I + is closed with respect to multiplication, the product of the numbers q is a positive integer- - say q--which has each of the numbers q as a factor. Consequently, q is a positive integer such that, for each p < m, r ' q £ I. Hence, by the least number theorem, there is a least such positive integer- -that is, a positive integer q such that "> V p 1, m - 1 is a positive integer less than m and, so, by hypothesis, r m ^ I. Hence, there is an integer k such that k 1, (**) if, for each p < m, r p f( I then r i I. INFINITE SETS The proof in section 9.05 that v2 is irrational showed, in particular, that there is at least one irrational number. Theorem 193 goes con- siderably further by showing how to find lots of irrational numbers. Our purpose, now, is to go still further and show that there are more irra- tional numbers than there are rational numbers. In fact, we shall prove that there are more irrational numbers between and 1 than there are rational numbers altogether. To make sense of what has been said we must first discover how to tell when one set has a greater number of members than another set. For a start let's recall the counting principle (C x ) of Unit 8: Two sets have the same number of members if and only if the members of one set can be matched in a one-to-one way with those of the other. This suggests that we agree that a first set has a greater number of members than a second set if, when you match members of the first set in a one-to-one way with all those of the second, you end up having [9-234] [App. Bj members of the first set left over. For example, if students enter a classroom and begin sitting, one in each seat, and when the seats are all filled there are students still standing, then there is a greater num- ber of students than of seats. This seems simple enough. However, in the example just given, the two sets [students and seats] were both finite sets. For each set, you could [if you were in the classroom] give a positive integer [or 0] as the number of its members. The set of irrational numbers and the set of rational numbers are not finite sets- -for short, are infinite sets- -and, as you will see, infinite sets are tricky. Notions you have formed by thinking about finite sets may not work when you try to apply them to in- finite sets. For example, consider the finite set {l, 2, 3, 4, 5, 6} and its sub- set {2, 4, 6}. There are many ways of matching some of the members of the first set, one-to-one, with the members of the second set. For example, we might match each of the numbers 2, 4, and 6 with itself. Or, we might match 1 with 2, 3 with 4, and 5 with 6; or 1 with 2, 2 with 4, and 3 with 6. However we do the matching, we find that there are 3 members of the first set left over. Now, for contrast, consider the set 1 + of all the positive integers, and the subset E whose members are the even positive integers. As before, there are many ways of matching some of the members of I*, one-to-one, with all the members of E. To begin with, we can match each member of E with itself. In this case, the members of I* which are left are just the odd positive integers. On the other hand, we can match with each member of E the positive integer which immediately precedes it: 1, <-» -3, *, Of • . . , cxx — 1, ... (1) \, \. \ \. ... This time it is the even positive integers which are left over, rather than, as in the first matching, the odd ones. This is not very surprising because, as this second matching shows, there are, according to the counting principle (C x ), the same number of even positive integers as there are of odd positive integers. With each of these two ways of match- ing, there is the same number of members of I + left over. Finally, let's [App. B] [9-235] try a third matching. This time, match with each member of E its half: 1, 2, 3, 4, 5, 6, 7, . . . , n, ... (2) \ 2, 4, 6, , 2n, Which members of I + are left over this time? Since, given any n e I + , n is matched with the member 2n of E, there is no member of I* left over! We have discovered a characteristic difference between finite sets and infinite sets. Given a finite set [such as {l, 2, 3, 4, 5, 6}], it is im- possible to map all of it in a one-to-one way on any of its proper subsets, [A proper subset of a set is one which does not contain all the members of the set. ] On the other hand, an infinite set [such as I* ] always has a proper subset on which it can be mapped in a one-to-one way- -each in- finite set has the same number of members as does some one of its proper subsets. [This property of infinite sets is often used as a defi- nition. So, the matching (2) shows that, according to this definition, 1 + is infinite. ] Evidently, we need to revise our idea as to when a first set has a greater number of members than does a second set. /according to this idea, I + has a greater number of members than E does, because when we match members of I + with all those of E [according to the scheme (1)], there are members of I + left over. But, according to the same idea I + does not have a greater number of members than E does, because when we match members of I + with all those of E [according to the scheme (2)], there are no members of I + left over. In view of this situation, we shall adopt a counting principle: < C 5>< A first set has a greater number of members than a second set if and only if members of the first set can be matched in a one-to-one way with all those of the second set and, no matter how this is done , there are members of the first set ^ left over. Notice that to say that a set A has more members than one of its subsets, B, is likely to be confusing. It may mean that B is a proper subset of A [A has other members than those of B], Or, it may mean [9-236] [App. Bj that A has a greater number of members than B does. If A is a finite set, this confusion may not be important [Explain. ]. But, if A is in- finite then A can contain other members than those contained in a given subset B and still not have a greater number of members than B. EXERCISES A. To show that there is a one-to-one matching between the members of a set A and those of a set B it is sufficient to find a function whose domain is A, whose range is B, and which has an inverse. [For example, the matching (2) on page 9-235 between the positive inte- gers and the even positive integers is determined by the function f for which f(x) = 2x, for x e I + --and is also determined by its inverse f , for which f (x) = x/2, for x e E. ] Such a function is called a one-to-one mapping of A on B ['on' means the same as 'onto all of'.], 1. Define a one-to-one mapping g of I + on the set O of odd positive integers. 2. Define a one-to-one mapping of I* on {n: n > 1 }. 3. On the set of nonnegative integers. 4. On the set of negative integers. B. Recall that if f is a mapping of A on B and g is a mapping of B on C then, by definition, g»f is the mapping of A on C for which [g°f](a) = g(f(a)), for each a e A. 1. Show that if f is a one-to-one mapping of A on B and g is a one- to-one mapping of B on C then g»f is a one-to-one mapping of A on C. [ Hint . You must show that g°f has an Inverse- -that is, that, for a x e A and a 2 e A, if g(f(a 1 )) = g(f(a ? )) then a x = a 2 . Since, by hypothesis, g has an inverse, if g(f(a x )) = g(f(a 2 )) then . . . . ] 2. Suppose that A x and J\ 2 are sets such that A x r\ A 2 = and that there is a one-to-one mapping h x of A x on I + and a one-to-one mapping h of A on I + . Show that there is a one-to-one mapping [App. B] [9-237] h of A x w A 2 on I*. [ Hint . There is {see Part A) a one-to-one mapping f of I* on E and a one-to-one mapping g of I* on O. By- Exercise 1, f ° h x is a one-to-one mapping of . . . . If h{x) = [foh^x), for x € A x , and h(x) = [g°h 2 ](x), for x € A 2 , then, since . . . , h is a mapping of on E w O. Since ...» his one- to-one. ] COUNTABLY INFINITE SETS A set is said to be countably infinite if and only if there is some one-to-one matching of its members with those of I*, I* and E are ex- amples of countably infinite sets. Give some other examples. From the discussion in Part A of the preceding exercises, it follows that a set A is countably infinite if and only if there is a one-to-one map- ping of A on I + and, also, if and only if there is a one-to-one mapping of I* on A. By Exercise 1 of Part B, a set [the set A] which has the same num- ber of members as a countably infinite set [the set B; with C = I*] is also countably infinite. By the same exercise, two sets [A and C] which are countably infinite [B = I + ] have the same number of members [Theo- rem 194], By Exercise 2 of Part B, if each of two disjoint sets is countably infinite then their union is also countably infinite [Theorem 195], So, for example, by Exercises 3 and 4 of Part A, the set I of all integers is countably infinite [Theorem 198]. Although we shall not give the proof, it can be shown [Theorem 196] that each infinite subset of a countably infinite set is countably infinite. So, for example, each subset of I* either is a finite set or has the same number of members as I* does. It also follows that each infinite set whose members can be matched, one-to-one, with some [or all] of the members of a countably infinite set is, itself, countably infinite. We now have what we need in order to prove [Theorem 198] that the set R of all rational numbers is countably infinite --that is, that the ra- tional numbers can be matched in a one-to-one way with the positive integers. To establish this somewhat surprising result, we shall begin by considering--not the rational numbers--but the fractions whose numerators and denominators are decimal numerals for positive integers. [9-238] [App. B] We shall prove that this set is countably infinite. To see how, imagine the members of this set arranged in a ''square". 1 2 3 4 5 1 1 1 1 1 1 2 3 4 5 2 2 2 2 2 1 2 3 4 5 3 3 3 3 3 1 2 3 4 5 4 4 4 4 4 1 2 3 4 5 5 5 5 5 5 Although there would be infinitely many fractions in each row and in each column of such a table, each diagonal line sloping up from the left would contain only a finite number of fractions, [in fact, the nth dia- gonal, counting from the top, would contain n fractions. ] This suggests the following matching: til t 1 * *■>» » 1 * *">' «1* » 1 • »">* »■>' * A* • 1 » » "> » * O ' */*» » C » 1 • (*) 1 II A J i \ V > J 1 V 3 < > 6 It r 5 2' 4 Us. \f 2 4v 1 z 8 10 11 12 13 14 15 16*** From the figure, it is clear that there is a one-to-one matching, as suggested above, of all the fractions in question with at least some of the positive integers--to each fraction there is assigned a positive integer, and different integers are assigned to different fractions. [Since people do sometimes doubt this and since, in any case, what is *'clear from a figure'* is not always so, we shall prove this. But, for the moment, let*s accept it.] Since there is a one-to-one matching of all the fractions [whose numerator -numbers and denominator -numbers are positive inte- gers] with at least some of the positive integers, and since the set of fractions is an infinite set, it follows that the set of fractions is a count- ably infinite set. Now, each positive rational number is listed many times in our se- quence of fractions. But, if we match each such number with its first [App. B] [9-239] listing in the sequence, we have a one-to-one matching of all the positive rational numbers with some of the fractions. Since the set of positive rational numbers is an infinite set, and since the set of fractions is a countably infinite set, it follows that the set of positive rational numbers is countably infinite --that is, there is a one-to-one matching of all the positive rational numbers with all the positive integers. We can extend any one-to-one matching of the positive rational num- bers with the positive integers to a one-to-one matching of R with I [Explain. ]. So, since I is countably infinite, so is R. Although each positive integer is a rational number, and although there are infinitely many rational numbers which are not integers, there is exactly the same number of rational numbers as there is of positive integers! I 1 Before going on, let's check that the matching (*) described on page 9-238 actually does match all the fractions whose numerator-numbers and denominator-numbers are positive integers in a one-to-one way with some of the positive integers. [Actually, each positive integer is used in the matching, but we don't need to use this fact. ] Consider the fraction whose numerator -number is m and whose de- nominator-number is n. Inspection of the "square" on page 9-238 shows that this fraction is in the (m+n- l)th diagonal and that, counting up from the left, it is the mth fraction in this diagonal. [For example, the fraction — is the 4th term in the 6th diagonal. Since, for each p, the pth diagonal contains p fractions, there are m +n - 2 P p=l fractions in the diagonals above the fraction in question and, as we have seen, there are m - 1 fractions in its own diagonal which precede it. So, the number of fractions which precede it in the suggested matching is m +n - 2 p + (m - 1) p=l --that is, (m + n - 2)(m + n - 1) , .. r _ _ . ■, J " L + (m - 1). [Explain. J I [9-240] |>pp. B] Hence, the fraction whose numerator -number is m and whose denomi- nator-number is n is to be matched with the positive integer (m + n - 2)(m + n - 1) , 2 + m ' Our problem is to prove that the matching so-described actually does have the property suggested by the figure — that no two fractions are matched with the same positive integer. This amounts to proving that ;f (m+n-2)(m+n- 1) (p + q - 2)(p + q- 1) (*) lf 2 + m = 2*^ + P then m = p and n = q. For simplicity, suppose that m + n - 1 = j and p + q - 1 = k. Then, m < j and p < k, and what we want to show is that if (j - l)j + 2m = (k - l)k + 2p then m = p and n = q. To do so, suppose that (j - l)j + 2m = (k - l)k + 2p. Since p < k, it fol- lows that (k - l)k + 2p < (k - l)k + 2k = (k + l)k. Hence, (j - l)j + 2m < (k + l)k, and [since m > 0] (1) (k + l)k > (j - l)j. Now, the function f for which, for each i € I + , f(i) = (i - l)i is an increas- ing function. [Suppose that i 2 > i . It follows that i 2 - 1 > i - 1 and, since i 2 > 0, that (i 2 - l)i 2 > (i x - l)i 2 . Also, since i 2 > i 1 and i x - 1 > 0, it follows that (i 1 - l)i 2 > (i x - l)i x . Hence, if i 2 > i x , (i 2 - l)i 2 > (i 1 - 1)^, ] And, (1) says that f(k + 1) > f(j). Since each increasing function has an increasing inverse, it follows that k + 1 > j--that is, that k > j. Similarly [since m < j], j > k and, so, j = k. Since, by hypothesis, (j - l)j + 2m = (k - l)k + 2p, it follows that m = p. And, since j = m + n - 1 and k = p + q - 1, it follows, because j = k, that m + n = p + q. So, since m = p, n = q. Consequently, (*). I _l Notice that, although we have spoken of fractions and their numer- ator-numbers and denominator-numbers, we might as well have spoken of ordered pairs of positive integers. So, what we have shown is that the Cartesian product I + X I + * s countably infinite. It is only a small step from this to the result [Theorem 197] that the Cartesian product of any two countably infinite sets is countably infinite. [App. B] [9-241] UNCOUNTABLY INFINITE SETS You may have begun to wonder whether there are any infinite sets which are not countably infinite. There are, and, as we shall see, the set of irrational numbers is one such set. To prove that this is the case, it is sufficient to prove that the set of all real numbers is not countably infinite. For, as we have proved, the set of rational numbers is count- ably infinite and, if the set of irrational numbers were also countably infinite, it would follow that the set of real numbers- -being the union of two countably infinite sets--was countably infinite. So, once we prove that this is not the case, it will follow that the set of irrational numbers is not countably infinite, either. In order to prove that the set of real numbers is not countably in- finite, we need a result about real numbers which depends on the least upper bound principle. What we need to know is that each "infinite decimal" represents a real number [Theorem 199a], and that different decimals all of whose digits are 'O's and 'l*s represent different num- bers [Theorem 199b], If you accept this, the proof that the set of real numbers is not countably infinite is very simple. Here it is: Consider the set of all nonnegative real numbers less than 1 which can be represented by infinite decimals using only the digits *0* and l l\ If the set of all real numbers is countably infinite, then [by Theorem 196] this subset is also countably infinite. We shall show that it is not. To do so, suppose we have any one-to-one matching of some of these "0, 1-numbers** with all the positive integers. The following figure suggests such a matching: 1 — * °.a 11 a 12 a 13 . . . 2 — 0. a 21 a 22 a 23 . . . 3 «— *■ 0.a 31 a 32 a 33 . . . [Each of the 'a * stands for either a '0' or a '1'. ] Now, con- mn J sider the real number which is represented by the decimal whose nth digit after its decimal point is a *0' if a =1 and is ° r nn a *1* if a =0. This real number is a 0, 1-number which is nn [9-242] [App. B] different from all of those which have been matched with positive integers. [For each n, it is different from the number matched with n because the decimal representations of the two numbers have different digits at their nth places. ] So, no matter how we match 0, 1-numbers, one-to-one, with the positive integers, there will be some 0, 1-number left over- -that is, there is no one-to- one matching of all the 0, 1-numbers with the positive integers. Hence, the set of 0, 1-numbers is not countably infinite and, even more so, the set of all real numbers is not countably infinite. Notice that we have also shown [Theorem 200a] that the set of real numbers in the segment 0, 1 is not countably infinite [Explain.]. So, since the set of rational numbers in the segment is countably infinite [Why?], it follows that the set of irrational numbers between and 1 is not countably infinite. We can now prove, as we promised, that there is a greater number of irrational numbers between and 1 than there are rational numbers altogether [Theorem 200b]. What this means, you recall, is that there are enough irrational numbers between and 1 to match, one-to-one, with all the rational numbers, but that, however this is done , there will be some irrational numbers left over. Now, there is at least one irrational number between and 1. [One such number is v2 /2. Another can be found by enumerating the rational 0,1- numbers which are nonnegative and less than 1 and applying the diagonal procedure" used, above, in the proof that the set of real numbers is not countably infinite. ] If a is such an irrational number then, for each n, a/n is another [Explain. ] In fact, the matching: n *-* a/n is a one-to-one matching of some of the irrational numbers between and 1 with all the positive integers. So, since there is a one-to-one matching of all the positive integers with all the rational numbers, there are enough irrational numbers between and 1 to match, one-to-one, with all the rational numbers. But, since the set of rational numbers is countably infinite and the set of irrational numbers between and 1 is not, however such a matching is carried out, there will be irrational numbers left over. [App. B] [9-243] To put the results just obtained on a firm basis, we need to know that each decimal of the form: (*) 0. a x a 2 a 3 . . . , where each *a ' stands for either a *0* or a '1*, "represents" a non- m r negative real number less than 1, and that two such decimals [which differ in some decimal place] represent different real numbers. To begin with, we need to know what is meant by saying that a decimal re- presents a real number. What real number? In the following discussion, we shall, for completeness, consider decimals of the form (*) in which each 'a ' may be any of the ten digits *0', '1', . . . , *9'. The meaning of a repeating decimal has already been discussed in Unit 8 in connection with sums of infinite geometric progressions. Also, that each nonnegative real number less than 1 can be represented by a decimal like (*) [and what this means] follows from an optional section in Unit 8 on base-m approximations of real numbers. The following discussion is, however, independent of this material from Unit 8. Given any decimal (*), each of whose digits is l 0', ' 1', . . . , or '9', consider the sequence whose mth term is the number a • 10~ m . For this sequence, consider its continued sum sequence s, whose nth term s„ , is n n 1 a p «10 P. p=l Since a > 0, s > 0. Since, for each p, a < 9i so that a • 10 < 9 * 10 , r P it follows [see Theorem 143] that, for each n, n = 1 > n < y 9* io p . By Theorem 167c on the continued sum sequence of a geometric pro gression, it follows that, for each n, n I 9 . 10 -P = 9'IQ l (\ ■ 10~ n ) 1 - 10 1 P= 1 = 1 - io"~ n < 1. [9-244] [App. B] Hence, for each n, s < 1. Consequently, the set S of those real num- bers which occur as terms of the sequence s is a [nonempty] set of real numbers which has an upper bound. It follows from the iubp that S has a least upper bound- -say, i. Since s e S and s > 0, S. > 0. Since 1 is an upper bound of S, I < 1. This number S. which is uniquely determined by the decimal (*) is, by definition, the number represented by (*). As we have shown, it belongs to the segment 0, 1. [This, and the work in Unit 8 on base-m approximations completes the proof of Theorem 199a.] You are accustomed to getting rational approximations for real numbers by '* rounding -off" their decimal representations. For example, the decimal representation of v 2 /2 begins: 0.7071 . . . So, you say: y/J/Z = 0.7071 --that is, you take the number s as an approximation to i. Let's see how this works in general. If m > n then s m m ^_j p •10~ P p=l n ■ I* <—> P p = l • 10~ P + m p = n + 1 = s + n P m Iv = n + l • 10~ P 10 Now, using this result, we can show [just as we showed that, for each n, s < l] that, for each m > n, (1) s - s < 10~ n . [Use Theorems 143 and 167c] m n L J Since for each p, a > 0, it follows, also, that P (2) s - s > 0. m n — Consequently [(2)], for each m < n and [(1)] for each m > n, s < s + 10 .It follows that s + 10 is an upper bound of the set S n n — rt ^ [App. B] [9-245] whose least upper bound is the number S. which is represented by (*). Hence, for any n, _ I < s n + 10 n . So, since s € S and i is an upper bound of S, we have, for each n, (3) s < 1 < s + 10~ n . n — ~ n Applying this to the case of v2/2 we see [n = 4] that: I — 0. 7071 < ^j-< 0. 7071 + 10" 4 [=0.7072] [Since v2/2 is not rational the *<'s can be replaced by '<'s. ] So far, we have shown that each decimal of the form {*) represents a real number in the segment 0, 1. [And, in the optional section of Unit 8 previously referred to, we proved that each real number in 0, 1 is represented by a decimal of the form (*).] The next question is: Do different decimals represent different numbers? As you know, the answer is *No'--the repeating decimals '0. 10' and '0.09' both represent the number 1/10. We shall see that this is the only kind of situation in which two decimals do represent the same number. More explicitly, we shall see that if (*) represents 1, and a different decimal: (*') 0.a;a 2 'a 3 ' represents Sf t then, I = SL' if and only if, when the first digit in which (*) and (*') differ is the nth (a) a = a' + 1 [or a ' = a +1] n n L n n J and (b) for p > n, a =0 and a ' = 9 [or a = 9 and a ' = 01. P P P P [incidentally, it follows from this that two decimals each of whose digits is either a *0" or a 'l 1 must represent different numbers. ] To prove that this is the case, suppose that (*) and (#') are different decimals which represent the numbers I and 1' , respectively, and that the first digit in which (*) and (#') differ is the nth. Since a ^ a ', we & n ' n may assume "without loss of generality" that a > a ' . Since, for p < n, a = a', it follows [Theorem 135] that, for each m > n, P P ~ m m s - s ' = Ya • 10~ p - Ya' • 10 _p m m 6_i p Z_ , p p = n p = n [9-246] [App. B] and, so [Theorem 136, etc.] m (4) s - s ' = (a - a')10~ n + V ( a - a')l(T p . * ' m m n n' / . * p p 7 p=n + l There are two cases to consider: a - a' > 1 and: a - a' = 1 n n n n In the first case, a - a ' > 2 and, by (4) [m = ml, n n ~ J l J s - s' > 2* 10~ n . n n ~ Now, by (3), since (*) represents the number I, s < i n and, since (*') represents the number 1' , 1' < s' + 10" n . "~ n From these it follows that s - s' - 10~ n < 1 - f n n — and, since s - s ' - 10~ n > 2 • 10~ n - 10~ n , that SL - t' > 10~ n . Hence n n — — [in this first case], i £ I' . Now, con si for each m > n, m (5) s w -s' = 10" n + y (a - a')10"P. m m £__, p p p =n + 1 We shall consider two subcases: for each p > n, a =0 and a ' = 9, and: P P for some integer --say q- -q > n, and (a > or a ' < 9). q q In the first subcase, for each p > n, a - a ' = -9. Consequently, P P for each m > n, m m Now, consider the second case, that in which a - a' = 1. By (4), n n 3 = n+ 1 -9'10" (n+1) (l - 10 n " m ) P P p=n+l p=n+l 1 - 10" 1 = -10" n (l - io n ' m ) = -10" n + 10" m . [App. B] [9-247] Hence [from (5)], for each m > n, s - s ' = 10" n - 10" n + 10" m = 10" m . m m Now, by (3), since (*) represents 8. and (#') represents i\ s < I < s + 10~ m m — — m and s ' < j?'< s ' + 10 ~ m . m ~ — m From these it follows that s - s ' - 10~ m <£ - £' < s +10"" m -s'. mm — — m m Since s - s ' = 10 , it follows that, for each m > n, mm ~ (6) < JL - I' < 2' 10~ m . From this last result it follows that I - 1' . For, if not, i - £' > and, by Theorem 165, (1/I0) m < ^-y^ 1 20 if m > —, 1 — = -j-j- 77T . Now, there is a positive integer m 2 10 20 which is greater than or equal to both n and tt: -pr . Hence, if i - S. > 0, it follows that, for such an m, both Jt - £' 2*10" m . This being impossible, 2 - SL 1 ? and, by (6), i = I'. Finally, consider the second subcase- -that in which, for a certain integer q, q > n and (a > or a ' < 9). Since this is a subcase of the second case we have, by (5), that, for m > n, m s - s ' = 10 _n + V (a - a')10 _p . mm ,/_, p p p=n + 1 Since, in any case, < a < 9 and < aJ < 9, it follows that, for p > n, a - a' > -9. Moreover, q >n and since, by hypothesis, a - a' > —9, p p - n ' q q a - a ' > —9 + 1. This being the case, it follows that, for m > q, m m Y (a -a')10~ p > y -9- 10 _p + 1 • 10~ q p=n+l p=n+l = -io" n + io~ m + io" q [9-248] [App. B] Hence, for m > q, s - s ' > 10" n - 10~ n + 10" m + 10~ q m m — = 10~ m + 10" q . Now, just as in the first case, it follows from (3) that s - s ' - 10" m < 1 - £' mm — and, since s - s ' - 10~ m > 10 ~ q , it follows that I - JL' > 10 ~ q . Hence, mm - — ' Combining our results we see, from the first subcase of the second case, that if (*) and (*') satisfy (a) and (b) then £ = Jt r , while if (a) is not satisfied [first case], or (a) is satisfied but (b) is not [second subcase of second case], then S. £ £' . Consequently, SL = 1' if and only if (*) and (*') satisfy both (a) and (b). [This completes the proof of Theorem 199b.] EXERCISES 1. Prove that there are exactly as many real numbers between any two given real numbers as there are between and 1. [ Hint . Suppose that a and b are real numbers, with a < b. Establish a one-to-one mapping between 0, 1 and a, b by defining the simplest increasing function f you can think of for which f(0) = a and f(l) = b. ] 2. Prove that there are exactly as many (a) rational numbers (b) irrational numbers between any two given rational numbers as there are between and 1 [ Hint . Show that, if a and b are rational, the function f of Exercise 1 matches rational numbers with rational numbers and irrational num- bers with irrational numbers. ] [A similar result holds if the given numbers are not both rational. But the proof, at least of (b), is more difficult. ] 3. Prove that between each two rational numbers there is an irrational number. [App. B] [9-249] THE DENSITY OF THE RATIONALS IN THE REALS In section 9.05, you proved that the real numbers are dense — be- tween any two real numbers there is a real number. In addition to this, you proved that between any two rational numbers there is a rational number. Also, in solving Exercise 3, above, you have proved that be- tween any two rational numbers there is an irrational number — in fact, as follows from Exercise 2, between any two rational numbers there are uncountably many irrational numbers, and a countable infinity of ration- al numbers. By now, these results should not be too surprising. How- ever, there is still one more result which may surprise you- -in spite of the fact that there is a much larger number of irrational numbers than of rational numbers, there is a rational number between any two real numbers: w w -, ^ ^ [rr , ->ni i V V ^ 3 x x r ' J --the rational numbers are dense in the reals. To see how to prove that this is the case, suppose that you are given real numbers a and b with b > a. A little thought suggests that if we could find a positive integer q such ^ I that there is an integer --say, k-- between qa and qb then k/q would be a rational number between a and b. Now, for any q, [[qa]] + 1 is an inte- ger greater than qa and, since {[qa|] < qa, [[qa]] + 1 < qa + 1. So, we are qa qb all set if we can find an integer q such that qa + 1 < qb. But, since b - a > 0, this means, merely, that q > l/(b - a), and the cofinality principle tells us that there is such a q. Summing up--for b > a, ( ([qa]] + l)/q is a rational number between a and b, for any q > l/(b - a) [and, by the co- finality principle, there is such an integer q], EXERCISES 1. Prove: V V . 3 3 x x r s 2. Prove that between any two real numbers there is (a) a countable infinity of rational numbers, and (b) an uncountable infinity of irrational numbers. [9-250] [App. C] APPENDIX C [The purpose of this Appendix is to provide justification for clause (4 .) of the definition, on page 9-92, of the exponential functions, and to furnish proofs of Theorems 202-211 and of Theorem 220. The reader should previously have studied Appendix A.] The exponential functions . - -For any a > 0, the exponential function with base a and rational arguments has been defined in section 9.06; (3) VV <:T a r = ( lX vT) rm r m, rm el Using this definition we have proved Theorems 203-211, as stated on page 9-75, .under the restriction that the exponents are rational num- bers, [in this Appendix, we shall refer to these restricted theorems as Theorems 203 -211 .] r r J In section 9.07 the definition of the exponential functions was com- pleted by adopting: (4 X ) V x> j VjX U = the least upper bound of (y: 3 r < u y = * r } (4 2 > V 0~ U (4 ) V 1 U = 1 and V ^ ° U = [and, as previously, 0=1] 3 U U ■> r J As remarked on page 9-92, these definitions, particularly (4^, need some justification. As a basis for this justification, and for the proof of Theorem 202, we need to establish some further properties of the exponential functions with rational arguments. [As we shall see, Theo- rems 203-211 follow easily from Theorem 202 and Theorems 203 -211 .] ' r r Theorem 202 is: For each x > 0, the exponential function with base x is a continuous function whose domain is the set of all real numbers; if x / 1, its range is the set of all positive numbers; it is decreasing if < x < 1 and increasing if x > 1. The additional properties we need of the exponential functions with ra- tional arguments are like those which are asserted in Theorem 202 to [App. C] [9-251] hold of the complete exponential functions: For < a / 1, the exponential function with base a and rational arguments is a continuous monotonic function- -decreasing if < a < 1 and increasing if a > 1. For < a < 1, given M > 0, there is an N such that, for each r > N, a~ T > M and < a r < l/M. For a > 1, given M > 0, there is an N such that, for each r > N, a r > M and < a~ r < l/M. [The restriction 'a / 1' is introduced, above, to simplify the statement which needs to be proved. No such restriction is included in Theorem 202 because such a restriction would complicate the proofs of Theorems 203-211. However, the assertions made about the exponential function with base 1 in Theorem 202 are obviously correct. By (4 ), this func- tion is a constant function and is defined for all real numbers- -and each constant function is continuous. So, in proving Theorem 202 we need treat only two cases--the case in which the base is between and 1 and that in which the base is greater than 1. Incidentally, as far as the con- sistency of (4 ) is concerned, it is enough to check that it is consistent with (3). This is the case. By (3), for each r, 1 = ( VT) , for any m such that rm e I and, by previous theorems [Theorems 190c and 150a], ( m vT) rm = i.] We proceed, first, to establish the properties listed above of the exponential functions with rational arguments. Next, we shall use these results to justify (4^ and (4 2 ). Third, we shall prove Theorem 202. Then, we shall prove Theorems 203-211. Finally we shall prove a theorem- - Theorem 220--on power functions- -the functions {(x, y) , x >0: y = x }, for some u. MORE ON RATIONAL EXPONENTS We begin by proving that, for < a / 1, the exponential function with base a and rational arguments is monotonic --decreasing if < a < 1 and increasing if a > 1 . The proof is similar to that outlined for a = 2, in Part E on pages 9-87 and 9-88. [9-252] [App. C] For < a / 1, suppose that r > r x and consider a 2 _ a 1, Since R is closed with respect to subtraction, it follows by Theorem 205, that, , , r r - s s c for any r and s, a = a • a . bo, (*) a r 2 - a r i = (a r 2 " r i - l) a r l. r r Since, by Theorem 203 , a i > 0, it follows that a 2 is less than or r r - r greater than a i according as a 2 l is less than or greater than 1. We wish to show [assuming r 2 > r ] that this is the case according as 1. More conveniently stated, we wish to show, for a > 0, that, if r > 0, r r r r a < 1 or a > 1 according as a < 1 or a > 1. Now, by (3) [ and (R)], there is an m such that rm e I and r ,m/ — .rm , ,r ,mr-.rm a = ( Va ) and 1 = ( VI ) Moreover, if r > then rm > and, so rm£l + . Consequently, the (rm)th power function restricted to positive arguments is increasing [Theorem 185] and, so, a is less than or greater than 1 according as va is less than or greater than VI. But, since the mth principal root function is increasing [Theorem 189], this is the case according as a is less than or greater than 1. Our next job is to show that, for 1. What we wish to prove is that, for a > 1 and any r Q , given c > 0, there is a d > such that (**) V r 1 |r - r Q | < d => |a r - a r o | < c]. Taking a tip from (*)[in the proof of monotonicity], we begin by consid- ering the case in which r = 0, in which case (**) reduces to: V r [|r| |a r - 1| < c] Since, for a > 1, the exponential function with base a and rational arguments is increasing, we know that ■r 1 x . 1 t , - l/n . r l/n if < r < — then a ' < a < a . n n So, if, given c > 0, we can find an n such that both a ' and a dif- fer from 1 by at most c, it will follow that, for this n, [App. C] [9-253] if I r I < — then a differs from 1 by less than c i i n --and, for this c, we can choose d = l/n. We proceed, then, to esti- mate a --that is, va--and its reciprocal. To do so, we shall use Bernoulli's Inequality [Theorem 162]. Since the principal nth root function is increasing, it follows that, for a > 1, n VT > VT = 1. So, if h = n vT - 1 then h > 0. Also, a = (1-f h) n a - 1 and, by Bernoulli's Inequality, a > 1 + nh and, so, h < . Conse- quently, for a > 1, (i) o< r VT- i< a n Furthermore, x i n vr - i and, since, for a > 1, l/va < 1, Va" -l^.va-1 n/ — n vT 1 va - i . Consequently, for a > 1 , (2) < 1 - 1 n/ — Va n From (1) and (2) we see that, for a > 1, both a and a differ a - 1 a - 1 from 1 by at most . Consequently, if we choose n so thai < c 7 n n _ --that is, for c > 0, so that n > (a - l)/c--it will follow that both a and a differ from 1 by at most c. In fact, for such an n, -c (a - l)/c, if |r I < - then la 1 " - 1 I < c. 1 ■ n ■ ' Since, by the cofinality principle, there is an n such that n > (a - l)/c, it follows that, given c > 0, there is a d > such that V [ |r| < d ^> |a r - 1 | < c] [9-254] [App. C] --for a > 1, the exponential function with base a and rational arguments is continuous at 0. It is now easy to translate this result from to any argument r . For, as we know, a 1 " - a r o = (a 1 "" r o - l) a r ° and, so, since a ° > [Theorem 203], r r a - a o = la r - r o l|a r o. Hence, if we wish, for some c > 0, to restrict r so that |a - a ° | < c, all we need do is restrict r - r Q so that |a ° - 1 | < c/a °. Since c/a ° > 0, this can, as we have seen above, be done by requiring that r - r, 1 <— where n is some positive integer not less than [The least such positive integer is — (1 - a)a o c/a r o • 3 Consequently, for a > 1, the exponential function with base a and rational arguments is continuous at each of its arguments. To treat the case in which < a < 1 we note that, in this case, l/a > 1 [Theorem 164] and that, by Theorem 210 , for each r, a = (l/a) Consequently, for each r and r Ql |a - a °| = |( l/a) - ( l/a) ° | and, since l/a > 1 [and R is closed with respect to oppositing] it follows from the result just proved that, given r and c > 0, there is a d > such that if |-r - -r Q | < d then |a r - a X °| < c. Since -r - -r = -(r -r ), this is equivalent to: if |r - r | < d then |a - a °| < c. Hence, for < a < 1, the exponential function with base a and rational arguments is continuous at each of its arguments. Remark . When we come to apply the results just obtained to prove Theo- rem 202, we shall need to take note of a slight refinement which can be made in the preceding proof of continuity in the case a > 1. A sketch of a graph of an exponential function with base greater than 1 will con- vince you that the graph is less steep for small arguments than for large ones. One consequence of this is that once we have found, for [App. C] [9-255] some argument r, and some c > 0, a d > such that V [ |r - rj < d => |a r - a r i | < c], we can be sure that, for each r Q < r , V r [ |r - r Q | | a r - a r o| < c]. In short, given c > 0, a d > which "works" for some argument will also work for each smaller argument. In particular, for a > 1, given an r , there is a d > which will work for each argument r Q < r -- that is, a d > such that (***) v r |a r - a r o | < c j. o— 1 To prove this last [which is what we need for the proof of Theorem 202] it is sufficient to exhibit such a number d. In view of the preceding / i \ r« proof, this is easy. Let N. be some integer greater than and let d = l/N . From the proof we know that this d works for r . Suppose r r that r < r . Then since a>l, a ° and each r < r . [In technical terms, what we have just shown is that, for a >1 and any r , the exponential function with base a and rational argu- ments is uniformly continuous on { r : r < r } . ] Finally, concerning exponential functions with rational arguments, we need to know that for < a < 1, given M > 0, there is an N such that V r >N (a" r >M and < a r < l/M) and that for a > 1, given M > 0, there is an N such that V r >N (a r > M and < a~ r < l/M). The essential content of these statements is that, for < a / 1, the ex- ponential function with base a and rational arguments has, among its values, both arbitrarily large numbers and arbitrarily small positive numbers. (9-256] [App. C] It will be sufficient to prove the second statement. [The first fol- lows from the second somewhat as continuity for base < a < 1 follows from continuity for base a > 1. ] For a > 1, a = 1 + h, with h > 0. So, by Bernoulli's Inequality, for each k > 0, a > 1 + kh. Now, given any number M > [ no matter how large], 1+kh J> M for each k > (M - l)/h. If N is such an integer k-- say, the least one- -then, since, for a > 1, the exponential function with base a and rational arguments is increasing, it follows that if r > N then a > a > M. Moreover, since a > 0, M > 0, and l/a = a , it fol- lows that if r > N then a < l/M. And, of course, a > 0. JUSTIFICATION OF (4 X ) AND (4 2 ) To justify: (4^) V .V x = the least upper bound of{y:3 < y = x} we must show, first, that for a > 1 and any b, (y :3 r 1 and any r , the least upper bound of{y:3 < y = a } is, if r lP €l, (^T) r l P . r r " By the least upper bound principle, we shall have done the first job once we have shown that, for a > 1 and any b, (y: 3 1, the exponential function with base a and rational exponents is increasing, it follows that, r r r if r < b then a < a !--so, the set in question has a i as an upper bound. Now, we tackle the second job. Much as in the preceding paragraph --for a > 1 — if r < r then a r < a*"*. So, a r l- - that is (^a ) lP - - is an ( r - * r upper bound of(y: 3 < y = a ). To prove that a * is, in fact, the [App. C] [9-257] least upper bound, we make use of the continuity of the exponential func- tion with base a and rational arguments and prove that if b < a 1 then b is not an upper bound of the set in question. In fact, suppose that r r b < a i. Then, a i- bis a positive number--say c--and, by continuity, there is a d > such that V r [|r - v x | |a r - a r i | < c]. Now, | r - r | < d <^=> -d < r - r < d i r r i . - - „, r r . and Ja - a l J < c < > -c < a - a l < c. In particular, if -d < r - r < then |r - r | < d and r r r r if | a - a i J < c then -c < a - a l . Consequently, by continuity, if -d < r - r < then -c < a* - a r i --that is, r r if r - d < r < r then a > a 1 - c = b. r c r^ Now, for each r < r , a belongs to(y:3 < y=a) and, so, for each r ^ r such that r -d 0, from Theorem 201. Consequently, since (-Va ) x ^is an upper bound of {y: 3 < y = a } r r i and since no smaller number is an upper bound of this set, it follows that the least upper bound of (y: 3 < y = a j is ("va ) x . So, (4 1 ) is r r i consistent with (3). The situation with respect to: ^ V 0 1 and, so, that, for any b, (l/a) is determined by (4 ). The problem of the consistency of (4 ) also reduces to that of (4^ via Theorem 210. [9-258] [App. C] PROOF OF THEOREM 202 What we need to prove is, mainly, that, for a > 1, the exponential function with base a is an increasing continuous function whose range is the set of all positive numbers. The proof of this will, of course, depend on: (4.) V ^ , V x = the least upper bound of { y : 3 ^ y = x ) i x > 1 u *■ r < u J Since (4^) defines the exponential functions with bases greater than 1 for all real numbers as arguments, the proof referred to will settle Theo- rem 202 for all such functions. Once this is done, it will follow, using: 1, the exponential function with base a is an increasing function. [To do so, we make use of the corres- ponding property of the exponential function with base a and rational arguments, and of Theorem 201.] Suppose that u > u,. By Theorem 201, there are rational numbers r and r such that u_ > r > r > u . Since the exponential function with base a > 1 has been shown to be in- creasing on the set of rational numbers, a 2 > a l. Now, from (4 Jl ), a 2 is an upper bound of {y: 3 < y = x). Consequently, since r 2 a 2. Hence, a 2 > a 1. Also, from (4 ), a * is the least upper bound of { y: 3 ^, y = x } . Since r. > u , , it follows that if r < u. then «-' r < -u 1 J 11 i r > r. Since the exponential function with base a is increasing on the r r set of rational numbers, it follows that if r < u then a * > a --that is, that a x is an upper bound of{y:3 < y = a). Consequently, a 1 ^a 1 . Since, as proved above, a 2 > a i, it. follows that a 2 > a 1. So, for a > 1, [App. CJ [9-259] V V [u 2 > u, =>a U 2 > a U i] Ul u 2 2 i --that is, for a > 1, the exponential function with base a is an increasing function. Next, we shall show that, for a > 1, the exponential function with base a is continuous--that is, that, for any u Q , given c > 0, there is a d > such that V u [|u- u | |a U - a u o| < C J. It would be natural to use the same procedure we used in proving that, for a > 1, the exponential function with base a, restricted to rational arguments, is continuous. In fact, the proof for continuity at will go through just as before- -all that is needed is to replace 4 r' by 'u' and make use of the result just proved that, for a > 1, the exponential func- tion with base a is increasing on the set of all real numbers. However, the remainder of the earlier proof makes use of Theorem 205 , and, since r r we have not yet proved Theorem 205 [and intend to use Theorem 202 in proving itl, this part of the proof cannot be carried over to the present situation. So, we must seek a different proof. As previously indicated [see Remark on page 9-254], we shall use the result that, for a > 1, given r and c > there is a d > such that (***) V r |a r - a r o | < c]. o— 1 To apply this result in proving that the exponential function with base a is continuous at u Q , we choose for r some rational number greater than u . [Given c > 0, a d > which works for all rational num- bers less than or equal to r should also work for any number u Q , rational or not, which is less than r .] Suppose, then, for a > 1 and some number u Q , we are given a num- ber c > 0, have chosen an r, > u Q , and that d is positive and satisfies (***). We proceed to show that, for any u, if |u - u Q | < d then |a U - a U o | < c. To do so, suppose that |u - u Q j < d. Ignoring the trivial case in which u = u Q --in which case, |a - a °| = 0< c--there are two cases to con- sider: [9-260] [App. C] Case 1. u - d < u < u Q Case 2. u Q < u < u Q + d r Q r r Q + d r-d r Q r 1 1 1 1 — | , — , f_ _| 1 1 1 1 , u - d u u Q u Q u u + d In Case 1, let r be a rational number such that u Q -d u 1 and any u Q , given c > 0, there is a d > such that if | u - u | < d then | a - a ° | < c --for a > 1, the exponential function with base a is continuous. The final step needed to complete the proof of Theorem 202 is the proof that, for a > 1, the range of the exponential function with base a is the set of all positive numbers. There are two things to be established, first, that each value of the function in question is positive, and second, that each positive number is a value of the function. To prove the first, we use (4^ and Theorem 203 . For any u, let r be some rational number less than u. By (4^, a 0. Consequently, a > 0. To prove the second, let y be any positive number, and choose a number M which is greater than both y and l/y. Then, M > and, for a given a > 1, there is an N such that, for each r > N, a r > M and a~ r < l/M. r Choose some such rational number r. Since M > y, a > y and since l/M < y, a < y. Now, consider the function {(u, y), |u| 1, the exponential function with base a is an increasing continuous function, the same is true of its restriction to -r, r. So, by Theorem 187, the range of this latter function is a , a • Since, by the choice of r, y belongs to this range, it also belongs to the range of the whole exponential function. This completes the proof of Theorem 202. PROOFS OF THEOREMS 20 3-211 The first and last of these theorem s- - Theorems 203 and 211--are almost immediate corollaries of Theorem 202. For, by Theorem 202, for < a / 1, the range of the exponential function with base a is (y: y > 0} . So, for each such base, V a >0. And, by (4 3 ), V 1 = 1 > 0. So, Theorem 203. As to Theorem 211, this merely says that, for < a / 1, the exponential function with base a has an inverse. Since, [9-262] [App. C] by Theorem 202, each such function is monotonic, Theorem 211 follows by Theorem 184. Of the remaining theorems, Theorems 206, 209, and 210 follow, as in the rational case, from preceding theorems. So, all we need prove here are Theorems 204, 205, 207, and 208. Each of these theorems can be derived from the corresponding theo- rem for rational exponents by using the fact that the exponential functions are continuous [Theorem 202], together with some standard theorems about continuous functions. The chief of these latter theorems is the following: If f and g are continuous functions whose domain is the set of all real numbers and are such that, for each rational number r, f(r) = g(r), then f = g --that is, V f(u) = g(u). In brief, there is aj. most one continuous function which has prescribed values for all rational numbers and which is defined for all real numbers. [Whether or not there ^s such a function depends on how its values are prescribed at the rational points of the number line. Since each subset of a continuous function is a continuous function, a necessary condition that it be possible to extend a function whose domain is the rationals to a continuous function whose domain is the real numbers is that the original function be continuous. It turns out that a necessary and sufficient con- dition is that the original function be uniformly continuous on each inter- val of rational numbers.] The proof of this uniqueness theorem is quite easy. Suppose that f and g are continuous functions defined for all real numbers and such that, for each r, f(r) = g(r). Given any real number u Q and any c > 0, there is, since f and g are continuous at u , a d > such that V u [|u - u Q | < d ^>(|f(u) - f(u )| and any b > 0, V (ab) = a b . u To do this, consider, for any given a > and b > 0, the functions f and g defined, for each real number u, by: f(u) = (ab) and: g(u) = a b We wish to prove that, for each u, f(u) = g(u)--that is, that f = g. Now, by Theorem 208 , we know that, for each rational number r, f(r) = g(r). So, we can conclude, by the uniqueness theorem, that f = g once we have shown that both f and g are continuous. Now, f is merely the exponential function with base ab [ since a > and b > 0, ab > 0]. So, by Theorem 202, f is continuous. And g is also continuous since it is the product of two functions each o£ which is, again by Theorem 202, continuous. Con- sequently, Theorem 208. The proof of Theorem 204 is just about as simple. Again, we need a standard theorem about continuous functions, whose proof we defer: The reciprocal of a continuous function is continuous. [9-264] [App. C] [Recall that an argument at which a function is does not belong to the domain of its reciprocal.] Aside from this theorem, we shall use only Theorem 204 , Theorem 202, and the uniqueness theorem. To prove Theorem 204 we need to show that, for any a > 0, ^, — u 1 V a = — — u u a To do this, consider, for any given a > 0, the functions f and g which are defined, for each real number u, by: f(u) = a and: g(u) = — a Note that, by Theorem 202, a ^ 0, for each u. ] We wish to prove that, for each u, f(u) = g(u)--that is, that f = g. Now, by Theorem 204 , we know that, for each rational number r, f(r) = g(r). So, we can conclude, by the uniqueness theorem, that f = g once we have shown that both f and g are continuous. Consider the function f. Since a > 0, it follows from Theorem 202 that, given any real number u Q , the exponential function with base a is continuous at -u Q . So, given c > 0, there is a d > such that if |_ u -_ Uo |0 r s [and Theorem 202 and the uniqueness theorem] to prove: w ww rv r+v V^ n VVxx=x (*) x >0 r v [App. C] [9-265] The second part consists in using (*) to prove Theorem 205: w ww uv u+v x >0 u v To prove (*), we must show that if, for any a > and any r, f(v) = a a and g(v) = a then f = g--that is, for each v, f(v) = g(v). As usual, a previous theo- rem--in this case. Theorem 205 --tells us that, for each s, f(s) = g(s). To complete the proof of (*) all we need show is that f and g are con- tinuous. Consider f. This function is the product of the constant function x whose only value is a and the exponential function with base a. Each constant function is continuous and, by Theorem 202, since a > 0, the exponential function with base a is continuous. So, by our standard theorem on the product of continuous functions, f is continuous. [Actu- ally, the proof that the product of a constant function by a continuous function is continuous is extremely easy. ] Consider g. Since a > 0, it follows that, given any real number v Q , the exponential function with base a is continuous at r + v Q . So, given c > 0, there is a d > such that if |(r + v) - (r + v Q )| < d then |a r + V - a*" + V ° | < c. But, j(r + v) - (r + v Q ) j = |v - v Q | . So, by the definition of g, this is equivalent to: if |v - v Q | < d then |g(v) - g(v Q ) | < c So, g is continuous at each of its arguments v . Consequently (*). Now, to prove Theorem 205, we proceed in just the same way, using (*) instead of Theorem 205 . For a > and any given v, let f / \ u v j / \ u + v f(u) = a a and g(u) = a By (*), f(r) = g(r), for each r, and, just as in the first part of the proof, it is easy to see that f and g are continuous. So, by the uniqueness theorem, for each u, f(u) = g(u). Consequently, Theorem 205. [9-266] [App. C] Finally, we need to prove Theorem 207. Once again we need a stan- dard theorem on continuous functions- -this time: The composition of two continuous functions is continuous. As in the case of Theorem Z05, the proof is in two parts. In the first part, we use Theorem 207 : r V x>0 V rV Xr ' S = xrS to prove: <*> V x>0 V uV xU > S=xUS In the second part, we use (*) to prove Theorem 207: w WW/ u \ v uv V ^ n V V (x ) = x x >0 u v To prove {*), consider any a > and any rational number s, and let f(u) = (a ) and g(u) = a By Theorem 207 , for each r, f(r) = g(r). So, as usual, we can show that, for each u, f(u) = g(u)--that is, we can prove (*)--merely by showing that f and g are continuous. Consider f. Since s€ R, there is a p€l + such that sp€l and, by defi- nition, f(u) = n/a u J . So, f is the composite of three functions- -the ex- ponential function with base a [a > 0], the pth principal root function, and the (sp)th power function. The first of these is continuous by Theorem 202, the second by Theorem 189. As to the third, it is continuous by Theorem 186 if sp > 0, is constant [and, so, continuous] if sp = 0, and is the reciprocal of a positive-integral power function [and, so, by Theorem 186 and a standard theorem, continuous] i£ sp < 0. Consequently, f is continuous. Consider g. If s = 0, g is constant and, so, continuous. Suppose that s / 0. Since a > 0, it follows that, given any real number v , the exponential function with base a is continuous at vs. So, given a c > 0, there is a d > such that if | vs - v s| < d then |a - a ° | < c. But, |vs - v Q s| = |v - v Q | • |s| and |s| / 0. So, by the definition of g, this is equivalent to: if | v - v | < d/ | s | then | g( v) - g(v Q ) | < c. [App. C] [9-267] So [ since, for d > 0, d/| s j > 0], g is continuous at each of its arguments v o« Consequently, by the uniqueness theorem (*). Now, to prove Theorem 207, we proceed in the usual fashion, using (*). For a > 0, and any given u, let f(v) = (a ) and g(v) = a . By (*), f(s) = g(s) for each s. f is the exponential function with base a and, so, by Theorem 202, is continuous. The proof that g is continuous is just like the corresponding portion of the first part of the proof. So by the uniqueness theorem, for each u, f(u) = g(u). Consequently, Theo- rem 207. This completes the proof of Theorems 203-211. THREE STANDARD THEOREMS ON CONTINUITY For completeness, we prove the three standard theorems we have used on continuous functions: (1) The product of two continuous functions is continuous. (2) The reciprocal of a continuous function is continuous. (3) The composition of two continuous functions is continuous. Proof of ( 1) . Since f(u)g(u) - f(u )g(u ) = [f(u) - f(u Q )]g(u) + f(u )[g(u) - g(u )]. it follows that |f(u)g(u)-f(u )g(u )|< |f(u)-f(u )| . |g(u) I + |f(u ).| . |g(u)-g(u )|. Now, suppose that g is continuous at u Q . Then, there is a d.>0 such that, if |u-u | 0, there are numbers d' > and d" > such that if | u - u Q | < d' then |f(u) - f(u Q ) | < c/(2M) [9-268] [App. C] and if |u - uj and if |u - u Q | < d then |f(u)g(u) - f(u )g(u ) | < ~ ' M+ m' M = C --that is, the product of f and g is continuous at u„. Proof of (2) . Suppose that u Q is an argument of the reciprocal of g. Then g(u Q ) / and, hence, j g(u ) |/2 > 0. So, assuming that g is con- tinuous at u Q , there is a d 1 > such that, for |u - u Q J |g(u)| > |g(u ) |/2- -and, in particular, g(u) i 0. Now, |g<«) - g(u >l g(u) g(u ) |g( u ) I ' |g( u > and, for |u - u Q | < d , it follows that g(u) g(u Q ) |g 0, there is a d' > such that if |u - u | < d' then |g(u) - g(u Q )| < c ( | g(u ) \/zf . Consequently, if d is the smaller of d and d' then d > and if | u - u Q | < d then < c g< u > g(u Q ) that is, the reciprocal of g is continuous at u Proof of (3) . Suppose that u Q € h„ % g(u )€^£, g is continuous at u , and f is continuous at g(u ). Then, given c > 0, there is a d' > such that if |v - g(u Q ) | < d' then |f(v) - f(g(u Q ) | < c and since d' > 0, there is a d > such that if |u - u Q | < d then |g(u) - g(u Q ) | < d' . Consequently, if |u - u Q | < d then |f(g(u)) - f(g(u Q ))| < c --that is, the composition of f with g is continuous at u Q . [App. Cj [9-269] CONTINUITY OF POWER FUNCTIONS As another application of (3) we shall use it and some theorems on ex- ponential and logarithm functions to prove: Theorem 220. For each u, the power function with exponent u and positive arguments is continuous. By Theorem 217, for any a > 0, log (a ) = u log a. So, by the defining principle (L), a u = 1Q u log a = (io u ) loga Theorem 207 .-u Consequently, if exp is the exponential function with base 10 , it follows that the power function with exponent u and positive arguments is exp olog. Since both exp and log are continuous [Theorems 202 and 213], so, by {3\ is the power function. [9-270] [App. D] APPENDIX D [This appendix contains an abbreviated development of the usual volume and surface area formulas for simple solids. ] Volume -measures. --In Unit 6 you derived formulas for the area-meas' ures of plane regions of certain kinds. For example: I—-- b — "A r b 2 -i K=fbh K = bh K = ~(b 1 + b 2 )h K = TTd' [The formulas in the first three examples which deal with polygonal re gions are derived by using two area-measure axioms, Axioms I and J. Axiom I. The area-measure of a triangular region is half the product of the measure of any of its sides by the measure of the altitude from the opposite vertex of the triangle. Axiom J. The area-measure of a polygonal reg ion is the sum of the area -measures of any set of component triangular regions into which it i :an be cut. The fourth formula, for the area-measure of a circular region, was not proved, but was made to appear reasonable by considering the area- measures of inscribed polygonal regions. In brief, using Axioms I and J, it was proved that the area-measure of a region bounded by a regular [App. D] [9-271] polygon is yap, where a is the apothem [the radius of the inscribed circle] of the polygon and p is its perimeter. Now, suppose that c is a circle of diameter d and, for each n, P is a regular n-gon, inscribed in c, whose apothem is a and whose perimeter is p . It seemed rea- sonable [in Unit 6] that, for sufficiently large values of n, a differs as little as one may wish from the radius, d/2, of c, and that p differs as little as one may wish from the circumference, 77d, of c. Consequently, for n sufficiently large, the area-measure, -r-a p , of the region bounded c n n by P differs as little as one may wish from ■=■ • y % fid- -that is, from — r— . Since any given point belonging to the circular region is, if n is suffic- iently large, inside R , it is natural to define the area-measure of the circular region to be ES— . Similar considerations suggested adopting the formula: K = -T. for the area-measure of a circular sector of radius r and arc-measure s.] In this Appendix we shall give informal justifications for some for- mulas for the volume-measures of some simple " solid" regions, includ- ing prisms, cylinders, pyramids, cones, and solid spheres, and for some formulas for the area-measures of the surfaces of such solids. We shall not attempt to base these justifications entirely on formally- stated axioms. Instead, we shall assume that you can recognize when two solids are congruent- -that is, when they have the same size and shape-- and that you agree that, given a solid, there is a solid congruent to it anywhere you please--in other words, that a solid can be "moved" any- where you like without changing its size and shape. We shall also as- sume that congruent solids have the same volume -measure, and that if a [9-272] [App. D] solid is cut up into other solids, the volume-measure of the given solid is the sum of the volume-measures of its parts. Finally, we shall state, later, two numerical axioms on volume-measures. Before we go into the matter of formulas for volumes and surface areas of solids, let's spend a little time in becoming acquainted with a few simple concepts about points, lines, and planes. These ideas will be useful to us in our discussions of the mensuration formulas. [As you read and work exercises, practice drawing three-dimensional illustra- tions. ] POINTS, LINES, AND PLANES Consider three noncollinear points A, B, and C. These points are the vertices of a triangle, A ABC. Pick any two points of A ABC. These points determine a line. Now, consider the union of all such lines. This union is called the plane ABC , or the plane determined by the three non- collinear points A, B, and C. [We shall use capital letters as variables whose domain is the set of points of space, the letters l i\ 'm', and l n' as variables whose do- main is the set of all lines, and the letters 'p\ l q\ and 4 r' as variables whose domain is the set of all planes. ] Given a line 1, there are many planes which contain I, But, given a line I and a point P such that P / i, there is exactly one plane p such that ICp and Pep. If Q and R are two points on 2 then p = plane PQR. (1) Given a circle c with diameter AB. How many planes contain c? How many planes contain AB ? (2) Suppose that A e p and B e p. Does it follow that AB C p? That ABCp? ThatABCp? ThatABCp? (3) Suppose that a line 1 is not contained in a plane p. How many points are there in I r\ p? [App. D] [9-273] (4) A line I is said to intersect a plane P if and only if I r> p con sists of exactly one point. A line 1 is said to be parallel to a plane p [i | | p] if and only if S. rs p = 0. Suppose that ^n p consists of more than one point, what can you say about i and p? tr^p = {P} i r\ p = (5) Planes p and q are said to be parallel if and only if p r^ q = 0. If p^f'q, what kind of geometric figure is pr\ q? 'q p^ q = (6) Given a plane p, how many planes q are there such that q j j p? Given a plane p and a line S. such that i | | p. How many planes q are there such that i. C q and q | | p? (7) Given a plane p and a line i such that Ajrfp. How many planes q are there such that Kq and q j | p? How many planes q are there such that Kq and q Jrf p ? (8) Suppose that i | | p, S. C q, and p r\ q / 0. Draw a sketch illus- trating this supposition. Suppose, further, that i' = p n q, Are i and i* subsets of the same plane? Are they parallel? [Parallel lines are nonintersecting lines which are contained in one plane. ] [9-274] [App. D] (9) Suppose that & r\ p = {p} and l£q, Draw a sketch illustrating this supposition. What can you say about p r\ q? What can you say about S. r\ (pn q)? (10) Given a line £ and a point P e i. Let p be a plane which contains &, and m be the line in p such that m is perpendicular to I at P. Let q be another plane which contains 1, and n be the line in q such that n is perpendicular to I at P. Does m = n? (11) How many lines are there which are perpendicular to a given line at a given point on the line ? (12) How many lines are there which are perpendicular to a given line and which contain a given point not on the given line? (13) Given a line I and a point Pel. Consider the set of all lines m such that m J- I at P. What kind of figure is the union of all such lines m? (14) Given a line 1. Consider the set of all lines m such that m ± I. What is the union of all such lines m? (15) Suppose that S. r\ p = {F} and, for each line m C p for which F e m, m -L i. Draw a sketch illustrating this supposition. For each line S. and for each plane p such that SL intersects p, S. is said to be perpendicular to p if and only if i is perpendicular to each line of p which contains the point of intersection of H and p. The point of intersection is said to be the foot of the perpendicular. xU •'i* (16) a. Given a plane p and a point P <-> EB, GD EB, GB EB, FC EA, GC EA, GB Intersection empty? Parallel lines ? (26) Suppose that a and b are sets of points such that, for some line JL K a C 9. and b£i. Then, by definition, a and b are collinear . Sup- pose that c and d are sets of points such that, for some plane p, c C p and d C p. Then, by definition, c and d are coplanar . Now, return to the tables, above, and in the next-to- the -last row, for each pair of lines, answer the question: Are the lines coplanar? (27) In Unit 6 we defined 'parallel lines' as lines whose intersection is empty. Such a definition is adequate as long as we restrict the do- main of the relation of parallelism of lines to lines in some one plane. But, if we wish to extend the domain of this relation to include lines in [App. D] [9-277] space, we need a new definition: JL | | m if and only if SL r~\ m = and SL and m are coplanar Pairs of lines which are not coplanar are called skew lines . Complete the tables by answering the question: Are the lines skew? (28) Suppose that SLj^m. Does it follow that SL and m are skew? (29) Suppose that SL, m, and n are three lines such that SL ± n and m X n. Under that condition would if follow that SL | | m? That 1 and m are skew? That SL and m are neither skew nor parallel? (30) Suppose that SL, m, and n are three lines such that SL j j n and m | | n. Does it follow that SL \ | m? (31) Suppose that p, q, and r are three planes such that p j [ r and q | | r. Does it follow that p | j q? (32) Given that p, q, and r are planes such that p j | r and q j | r. Under what condition would it follow that p^H^q? (33) Suppose that q | | r. (a) If p r\ q ^ 0, under what condition would it follow that p | | r? (b) If p^H^q and p ^ q, does it follow that p r\ r ^ 0? (c) If p intersects q in line SL and p intersects r in line m, what can you say about SL and m? [9-278] [App. D] Suppose that Pisa point and p is a plane, and let I be the line through P perpendicular to p. The foot of the perpendicular is called the projection of P on p. <— > Suppose that c is a set of points and c' is the set of points which are the pro- jections of the points of c on p. The set c' is called the projection of c on p. (34) Given a line AB and a plane p. Suppose that CD is the projec- tion of AB on p. Under what condition is CD = AB ? Under what condi tionisCD = 0? CD < AB? CD > AB? (35) Given circle c and plane p. Suppose that c' is the projection of c on p. Under what condition is c' = c? c' = c? c'a segment? (36) Suppose that SL and m are two lines, p is a plane, and SL' and m' are the projections of 1 and m on p. (a) If i | | p and m | | p, does it follow that I'nm' =0? (b) If SL and m are skew and m | | p, does it follow that f r\ m' (37) Suppose that I and m are skew lines. (a) Is there a plane through SL parallel to m? (b) If SL C p and m | | p and m' is the projection of m on p, do SL and m' intersect? (c) Is there a line which intersects both SL and m and is perpen- dicular to each of them? (d) How many common perpendiculars do two skew lines have? (38) Given an isosceles right angle A ABC, with ZC the right angle and AC 26 inches long. Suppose that p is a plane such that C £ p and A and B are each 10 inches from p. If A' and B' are the projections of A and B on p, find the measure of Z.A'CB'. [App. D] [9-279] Recall from Unit 6 that a line separates a plane into two half-planes Suppose that two planes p and q intersect in the line Si. Let i separate p into the half-planes p x and p 2 and I separate q into the half-planes q x and q 2 . Then, p x r\ p 2 = and p x o- I w p 2 = p, and q x r\ q 2 = and q w i u q = q. The union of two noncoplanar half-planes and their common edge is called a dihedral angle. So, for example, p^lu q 2 and q u I u p are dihedral angles. Consider the dihedral angle p x w 2. w q Let A be a point on i and — > — 7* let AP and AQ be half-lines contained in p x and q x , respectively, such <-> <-> that AP -Li and AQ ± i. Notice that A, P, and Q are noncollinear. The angle, ZPAQ, is called a plane angle of the dihedral angle and its mea- sure is said to be the measure of the dihedral angle. A dihedral angle is called a right dihedral angle if and only if its measure is 90. What do you think an acute dihedral angle is? An obtuse dihedral angle? The union of two intersecting planes containa four dihedral angles. Two intersecting planes are said to be pe rpendicular if and only if their union contains a right dihedral angle. (39) Consider the dihedral angle p^lw q Let ZPAQ be a plane angle of this dihedral angle. If PA = AQ = QP, what is the measure of the dihedral angle? (40) Suppose that the angle ZPAQ is a plane angle of the dihedral angle p 1 wluq r If B € I and B ^ A, compare m(ZPAQ) and m(ZPBQ). (41) Let L ABC be a plane angle of the right dihedral angle a.uiuc, such that AB = 10 = BC. Suppose that D is a point of I such that D ^ B and BD = 10. What is the area-measure of AACD? [9-280] [App. D] (42) The bisector of a dihedral angle is the closed half-plane whose edge is the edge of the dihedral angle and which contains a bisector of a plane angle of the dihedral angle. If a point in the bisector is 10 inches from the edge of a dihedral angle of 60°, how far is the point from each of the planes whose union contains the dihedral angle? (43) Suppose that line I intersects plane p and that plane q contains i. (a) If i J. p, does it follow that q lp? (b) If i is oblique to p [that is, intersecting but not perpendicu- lar], does it follow that q is not perpendicular to p? (44) Given the three planes p, q, and r. (a) If p | | q and p ± r, does it follow that q -L r? (b) If p ± r and q lr, does it follow that p | | q? (c) If p || q and q | | r, does it follow that p j [ r? (d) If p J- q and q ± r, does it follow that p lr? (45) Consider the two dihedral angles p w S. w q 1 and r x w mu s x . Let p and r be parallel and q and s be parallel. (a) What can you say about Si and m? (b) What can you say about the measures of the given dihedral angles ? (46) Given three parallel planes p , p 2 , and p , line S. which inter- sects the planes in the points A^ A 2 , and A , respectively, and line m which intersects the planes in the points B x , B , and B , respectively. (a) If j? | | m and A X A 2 = A 2 A 3 , does it follow that B i B 2 = B 2 B 3 ? (b) If l | | m, A 1 A 2 /A 1 A 3 = 0.4, and B 2 B 3 = 12, compute B X B 2 . (c) If A x = B i# A 1 A 2 = 3, A 2 A 3 = 8, and B X B 2 = 6, compute B X B 3 . (d) If A 2 = B 2 , A i B 1 = 4, A 3 B 3 = 12, and A X A 3 =8, compute A X A 2 . (47) In (46) let n be a third line which intersects the planes in the points C x , C 2 , and C , respectively. Suppose that A^^ = B x = C i , Ag, B 2 , and C 2 , are three points, and A X A 2 = l/2»A 1 A 3 . If the area-mea- sure of AA 2 B 2 C 2 = 7, compute the area-measure of AA 3 B 3 C 3< [App. D] [9-281] SOME SIMPLE SOLIDS Here are pictures of solids of some of the kinds we shall consider. A / i / / / / / r y Four of the pictured solids are prisms and two are pyramids. A prism may be thought of as a solid which is the union of parallel segments, all of the same length, connecting corresponding points of two congruent polygonal regions. The two congruent polygonal regions are the bases of the prism, and the union of those parallel segments which join points of any two corresponding sides of the bases is a lateral face of the prism. The union of the lateral faces of a prism is its lateral surface . Each segment which joins corresponding vertices of the bases is a lateral edge of the prism. The bases of a prism may be bounded by polygons of any kind, but the boundaries of the lateral faces are always parallelograms. A triangular prism is one whose bases are triangles, a pentagonal prism is one whose bases are pentagons, etc. A prism all of whose faces- -bases, as well as lateral faces- -are bounded by parallelograms is called a parallelepiped. [Any two opposite faces of a parallelpiped may be considered to be its bases. ] A prism which is the union of segments perpendicular to its bases is called a right prism. A rectangular parallelepiped is a right paral- lelepiped whose bases are rectangles. A cube is a rectangular paral- lelepiped whose faces are square regions. The altitude of a prism is the [perpendicular] distance between its bases. [9-282] [App. Dj A pyramid is also a union of segments- -the segments which join its vertex to the points of the polygonal region which is its base. The lateral faces of any pyramid are triangular regions, but its base may be a polygonal region of any kind. A pyramid with a 4- sided base is called a quadrangular pyramid. One whose base is a triangular region is called a triangular pyramid, [sometimes: a tetrahedron ]. [Any face of a tetrahedron may be con- sidered to be its base. ] The altitude of a pyramid is the [perpendicular] distance from its vertex to its base. A regular pyramid is one whose base is a regular polygonal region whose center is the foot of the perpendicular from its vertex to its base. The lateral faces of a regular pyramid are bounded by congruent isos- celes triangles, and its slant height is the common altitude of these triangles. Cylinders and cones are analogous to prisms and pyramids, respec- tively. We shall consider only circular cylinders and cones--those whose bases are circular regions. Among these we single out the right circular cylinder s- -those which are unions of segments perpendicular to their bases--and the right circular cones--those for which the feet of the perpendiculars from their vertices to their bases are the centers of their bases. [These right solids can be pictured as being generated by revolving a rectangular region about one of its sides or a right triangu- lar region about one of its legs. ] If a plane intersects a solid then their intersection is called a cross - section of the solid. Any solid is the union of its cross -section by all those planes parallel to any given plane which intersect the solid. When we speak of a cross-section of a prism, pyramid, cylinder or cone, we shall mean a cross- section by a plane parallel to the bases [or to the base]. We shall assume, as obvious, that all such cross-sections of a [App. D] [9-283] prism or cylinder are congruent and, so, have the same area-measure. Since it is less obvious, we shall indicate a proof that all such cross- sections of a pyramid or cone are similar, and shall compute the ratio of similitude of such a cross- section to the base of the solid. A Consider any pyramid with vertex A and base bounded by a polygon BCD..., and a cross- section of this pyramid by some plane p' parallel to the plane p which contains its base. The cross-section is bounded by a polygon B'C'D'. . . , Let A" be the foot of the perpendicular from A to p, and let A' be the point of intersection of this perpendicular with p'. Then AA" is the altitude of the given pyramid and AA' is the distance from A to p'. We shall show that the polygon B'C'D'. . . is similar to the polygon BCD . . . , and that the ratio of similitude is AA'/AA". Consider any two adjacent vertices of the base of the pyramid- -say B and C--and the plane which contains these points and the vertex A of the pyramid. Both the plane p and the plane containing A, B, and C con- tain B and C. So, the intersection of these two planes is the line BC. < — > Similarly, the line B'C is the intersection of the plane p' and the plane < — => <-> containing A, B, and C. Since B'C C p' and BC C p and since [because < . > . <-> , < . >. p and p' are parallel] p' r\ p = 0, B'C r^ BC = 0. So, since B'C and BC are coplanar [because both are subsets of the plane containing A, B, < — > <^-> and C], it follows by definition that B'C | | BC. [9-284] [App. D] It follows from this that [in the plane containing A, B, and C] the matching AB'C -— - ABC is a similarity and, in particular, that B'C _ AC' BC " AC ' Now, consider the plane which contains A, A", and C. [For the moment, we assume that, as in the figure, A" / C. ] Since A'C and < — > A"C are the intersections of this plane with the parallel planes p' and p, < — > < — > respectively, it follows [as above] that A'C' j A"C. So [as above], the matching AA'C— AA"C is a similarity and, in particular, AC' AA' . AC AA' 7 [Of course, if, as we assumed was not the case, A" = C, then A' = C' and, trivially, AC'/AC = AA'/AA".] Hence [in any case], B'C A A' t BC ~ AA" " Since B and C were any adjacent vertices of the base of the given pyramid, we have now shown that the sides of the boundary of the cross- section are proportional to the corresponding sides of the boundary of the base, and that the factor of proportionality is AA'/AA". To complete the proof that the polygons B'C'D'. . . and BCD . . . are similar [and that the ratio of similitude is AA'/AA"], it remains to be shown that corresponding angles of thu two polygons are congruent. To do this, consider any corresponding vertice s- - say, C and C. It is easy • — > • — > to see that the rays C'B' and CB are similarly directed [they are, as we have proved, parallel, and B' and B are on the same side of C'C], Like- wise, CD' and CD are similarly directed. Now, if ZBCD and Z B'C'D' were [as they aren't] in the same plane, it would follow from a theorem you proved in Unit 6 [Theorem 5-13 on page 6-154] that ZBCD and ZB'C'D' are congruent. Fortunately, Theorem 5-13 still holds for angles in space --if the sides of two angles can be matched in such a way that corre- sponding sides are similarly directed then the angles are congruent. So, ZB'C'D' ^ ZBCD and, since C and C were any corresponding vertices of the two polygons, we have proved that corresponding angles of the two polygons are congruent. Taken together with the fact already estab- lished that corresponding sides are proportional in the ratio AA'/AA", [App. D] [9-285] this shows that the polygons are similar and that the ratio of similitude is AA'/AA". [The "space-form" of Theorem 5-13 which we have ap- pealed to follows from the "plane -form" and the fact that the polygons bounding the two bases of a right prism are congruent. If you wish, you can show how it follows, by considering the right prism one of whose bases is the cross section we have been considering and whose other base is in the plane p. ] The result we have established concerning cross-sections of pyra- mids and the fact that the ratio of the area-measures of two similar polygonal regions is the square of the ratio of similitude, yield the fol- lowing result: The ratio of the area-measure of a cross- section at a dis- tance d from the vertex of a pyramid of altitude h to the area- 2 2 measure of the base of the pyramid is d /h . In particular: Given any two pyramids which have the same altitude and whose bases have the same area-measure, it follows that the area-measure of any two cross-sections which are at the same distance from the bases are the same. Since all cross-sections of a prism [because they are congruent] have the same area-measure, a statement similar to the last holds for prisms. The corresponding statements about cones and cylinders should now seem obvious enough to require no further comment. EXERCISES A. Makes sketches of the solids which fit these descriptions. 1. The three lateral faces are triangles and the base is a triangle. 2. The lateral faces are rectangles and a base is a triangle. 3. The edges are congruent and both bases are parallelograms. 4. A square pyramid whose lateral faces are equilateral triangles. [9-286] [App. D] B. 1. Compute the area-measure of the total surface of a triangular pyramid whose lateral faces are equilateral triangles, given that one of the edges has measure e. 2. Repeat Exercise 1 given that the slant height is s. 3. A lateral edge of a right pentagonal prism has measure x and the perimeter of a base is y. Compute the total area-measure of its lateral faces. 4. The slant height of a pyramid is s and its base has perimeter p and area-measure K. Compute the area-measure of the total surface of the pyramid. 5. Suppose that a plane perpendicular to one of the lateral edges of a hexagonal prism intersects each lateral edge of the prism, and that the perimeter of the resulting cross- section is p. If S is the total of the area-measures of the lateral faces, compute the measure of a lateral edge. 6. Let d be the diameter of a base of a right circular cylinder. If the altitude of the cylinder is h, what is the area-measure of the total surface of the cylinder? 7. If you double the dimensions of a cube, what change takes place in the area-measure of the total surface of the cube? 8. Suppose that the altitude of a regular quadrangular pyramid is x and each edge of the base has measure y. Compute the total area-measure of the lateral faces of the pyramid. [What change takes place in the total if the dimensions are tripled?] 9. Suppose that the outside diameter of a 10 inch length of pipe is 3 inches. If the pipe is a quarter of an inch thick, what is the area of the total surface of the pipe [inside, outside, top, and bottom]? 10. A point-source of light is 10 feet from a vertical wall. If a cir- cular piece of cardboard is held parallel to the wall and between the light and the wall, it casts a circular shadow on the wall. If the cardboard is 1 foot in diameter and held 2 feet from the point- source, what is the area of the resulting shadow? [App. D] [9-287] AN AXIOM ON VOLUME- MEASURES As you may know, one way to make a model of a solid- -say, a tri- angular pyramid- -is to begin by drawing pictures, on a thick sheet of cardboard, of cross-sections of the solid by planes which are parallel to a given plane and which are distant from one another [consecutively] by the thickness of the cardboard. Next, one cuts out these pictures and pastes them together, one on top of the other. [The figures show a tri- angular pyramid and a model constructed in this way. ] Finally, one fills in the "grooves" with plaster of Paris. Notice that the volume of the stack of pieces of cardboard is very nearly the volume of the fin- ished model. In fact, no matter how short you are of plaster of Paris [as long as you have some ] you can get by just by choosing thin enough cardboard. One difficulty with making such a model is in properly centering each piece of cardboard over the preceding. And, if someone bumps into your stack of cardboard before the glue dries, you may end up with a model of a solid like the one pictured below. However, whatever happens, the volume of cardboard in your model will stay the same. Such considerations suggest that two solids- -for example, the two pictured above — have the same volume if there is a plane p such that [9-288] [App. D] each plane parallel to p which intersects either of the solids also inter- sects the other and that the cross- sections of the solids by any such plane are congruent. This is, in fact, the case. Although we shall want a more comprehensive result, we can use this one to prove that two prisms which have the same altitude and whose bases are congruent have the same volume-measure, and that two pyramids which have the same altitude and whose bases are congruent have the same volume-measure. The first is easy to prove. Since we have agreed that we can "move" solids around without changing their volume-measures, we may assume that the two prisms have bases in the same plane, p [and, except for their bases, are on the same side of p]. Since the prisms have the same altitude, it follows that each plane parallel to p which intersects either prism also intersects the other. Since all cross sections of a prism are congruent to the base, and since the bases of the two prisms are congruent, it follows that the cros s- sections of the two prisms by any such plane [or by p itself] are congruent. Consequently, the two prisms have the same volume-measure. The corresponding statement about pyramids is not much harder to prove. Again, we may assume that the two pyramids [with the same altitude, h, and congruent bases] have their bases in the same plane, p, and are otherwise "above" p. A plane parallel to p and at a distance d < h above it cuts out cross-sections of the pyramids which are similar to the respective bases, the ratio of similitude being the same in both cases. So, since the bases are congruent, so are the "corresponding" cross-sections. Hence, the two pyramids have the same volume-measure, [App. D] [9-289] In the same way one can argue that two cylinders [or two cones] which have the same altitude and congruent bases have the same vol- ume-measure. To obtain more complete results, let's return to the consideration of cardboard models and guess at a more comprehensive principle than the one we have been using about congruent cross-sections. This time, suppose that we have two solids such that each plane which is parallel to a given plane p and intersects either of the solids also intersects the other and that the two cross- sections of the solids by any such plane have the same area-measure. Suppose now, we picture, on cardboard, cross-sections of both solids by planes parallel to p and use them, as before, to build models of both solids. Each model will contain the same number of pieces of cardboard and corresponding pieces will have the same area and, of course, the same thickness--the thickness of the cardboard we choose to use. Consequently, it is natural to assume that corresponding pieces in the two models have the same volume--from which it follows that the two models have the same volume. Such con- siderations suggest: Cavalieri's Principle. Two solids have the same volume-measure if there is a plane p such that each plane which is parallel to p and intersects either of the solids also intersects the other, and the cross-sections of the solids by any such plane have the same area-measure. Using this principle [as an axiom] together with the fact that cross- sections of prisms and cylinders have the same area-measures as their bases, we can prove that two solids, each of which is either a prism or a cylinder, which have the same height and whose bases have the same area-measure, have the same volume -measure. The proof is much like that of the theorem concerning prisms with the same height and congruent bases. A similar proof establishes the state- ment with 'prism or cylinder' replaced by 'pyramid or cone*. [9-290] [App. D] VOLUME FORMULAS Cavalieri's principle gives us a sufficient condition in order that two solids have the same volume -measure. To obtain formulas for the volume -measures of solids we still need some place to start from. We shall choose as a starting place [as a second axiom] the formula for the volume-measure of a particular kind of prism- -a rectangular parallele- piped. V = = Bh Since the base of this prism is a rectangle with side-measures a and b, the area-measure of its base is ab. Since the altitude of the prism is c, the formula tells us that the volume -me a sure of a rectangu- lar parallelepiped is the product of the area-measure of its base by its altitude. We can now find a formula for the volume-measure of any prism or cylinder. For, consider any such solid whose base has area-measure B and whose altitude is h, and consider any rectangular parallelpiped whose base also has area-measure B and whose altitude is h. Using Cavalieri's principle, we have just proved that these two solids have the same volume-measure. Since that of the second is, by definition, Bh, that of the first is also Bh. [App. D] [9-291] In particular, the volume-measure of a circular cylinder whose base has radius r and whose altitude is h is 7Tr h [in terms of the dia- 2 meter of the base, V = ird h/4]. Having obtained pretty satisfactory results as to the volume-meas- ures of prisms and cylinders, we now turn our attention to pyramids and cones. Evidently, we can parallel our work on volume formulas of prisms and cylinders once we find a formula for the volume-measure of one kind of pyramid. Triangular pyramids turn out to be easiest. We shall find that the volume-measure of a triangular pyramid is one third that of a prism which has the same base and altitude. Fig. 1 Fig. 2 Any triangular pyramid with vertex A and base ABCD can be "enlarged" [as shown in Fig. 1] to a triangular prism which has the same altitude and base [DE and CF are parallel to BA, and DE = BA = CF], As shown in Fig. 2, this triangular prism is the union of three tri- angular pyramids. One way of seeing this is to notice, in Fig. 1, that the prism is the union of the given triangular pyramid and a pyramid whose vertex is A and whose base is #%'■$ CDEF. This latter quadran- gular pyramid is cut into two triangular ones by the plane which contains A, D, and F. Notice that, since ACDF ~ AEFD, these two triangular prisms have congruent bases. Since the altitude of each is the distance between A and the plane containing Mffl CDEF, these two triangular prisms have the same volume-measure. Consider, now, the first two pyramids pictured in Fig,2--the origi- nal pyramid with vertex A and base ^,BCD and the pyramid with vertex D and base AAEF. The bases of these pyramids are the two bases of the prism and, hence, are congruent. The altitude of each is merely the distance between the parallel planes which contain the bases of the prism. So, these two triangular pyramids have the same volume-measure. [9-292] [App. D] Consequently, all three triangular pyramids have the same volume measure- -which must, then, be one third that of the prism. B-^ A V V^Bh As in the case of prisms and cylinders, it now follows that the vol' ume-measure of any pyramid or cone whose base has area-measure B and whose altitude is h is (l/3)Bh. V = -jBh In particular, the volume-measure of a circular cone whose base has radius r and whose altitude is h is (l/3)77r h [in terms of the dia- meter of the base, V = nd h/12]. EXERCISES A. Solve these problems. 1. What is the volume of a 14-foot high right circular cylinder with a 6-foot diameter? 2. Compute the volume of a parallelpiped whose altitude is 10 and which has a rectangular base with side-measures 3 and 7. [App. D] [9-293] 3. [The axis of a circular cylinder is the line determined by the cen- ters of the bases. If the cylinder is not a right cylinder, the projec- tion of the axis on the plane of either base is a line. The union of the axis and its projection contains four angles, two acute and two obtuse. The common measure of the acute angles is called the inclination of the axis . ] The axis of a circular cylinder is inclined 30°. If the radius is 3 and the distance between centers of the bases is 10, what is the volume? 4. Suppose that the total area-measure of the six faces of a cube is the same number as the volume-measure. What is the measure of an edge of the cube? 5. If the volume of a circular cylinder is 18077 cubic inches and the altitude (in inches) is 5, what is the diameter (in inches)? 6. Suppose that the volume of a right trapezoidal prism is 26 cubic inches and its altitude is 16 inches long. Find the distance (in inches) between the two parallel lateral faces if their areas are 168 square inches and 192 square inches, respectively. 7. The common length of the four diagonals of a rectangular solid [that is, a right rectangular prism] is 13 inches and two of the edges are 7 inches and 8 inches, respectively. Compute the vol- ume of the solid. 8. The diagonal of a cube [that is, the measure of any of the four seg- ments which join vertices but are not contained in any of the faces] is 4v 3 . Compute the volume-measure of this cube. 9. The dimensions of a rectangular solid are in the ratio 2: 3:4 and the diagonal is 58. Compute the volume-measure of this solid. 10. If you double the dimensions of a rectangular solid, what change takes place in the volume? [9-294] [App. D] 11. If the area-measure of the total surface of a cube is t, what is the volume-measure? 12. If the diagonal of a face of a cube is d, what is the volume - measure ? 13. Find the volume-measure of the largest right circular cylinder which is contained in a cube of side -measure e. 14. Find the volume -measure of the largest right prism with a right- triangular base which is contained in a right circular cylinder whose radius is 3 and whose altitude is 10. 15. A water tank in the shape of a right circular cylinder has a dia- meter of d feet. How long will it take to fill the tank to a depth of x feet if the water flows into the tank through a pipe of inside diameter y inches at the rate of r feet per minute? 13. Solve these problems. 1. A regular square pyramid is 3 inches tall and the perimeter of its base is 16 inches. What is its volume? 2. A regular triangular pyramid has equilateral-triangular faces. If each edge of the pyramid is 10 inches long, what is its volume? 3. How tall is a pyramid whose volume is 21 cubic inches if its base is bounded by a rhombus with diagonals 6 inches and 7 inches, respectively? 4. What is the volume-measure of the largest circular cone con- tained in a circular cylinder whose volume -measure is V? 5. A cross-section of a pyramid 10 inches from its vertex has an area of 100 square inches. If the base of the pyramid has an area of 200 square inches, compute the volume of the pyramid. [App. D] H c 6. /' F V E ' i i A' A B [9-295] The diagram at the left ia a picture of a cube with edge-measure e. Compute the volume - measure of each of the pyramids described in the following exercises. (a) vertex F and base QABCD (b) vertex F and base AHGD (c) vertex P and base 0ABCD where {P> = EG r^ HF (d) vertex Q and base E2ABCD where {Q} = AG r\ EC (e) vertex A and base 1II1HFBD (f) vertex E and base tIEIDHABG (g) vertex F and base AHGA PRISMA TOIDS A prismatoid is a solid which, like a prism, has for its bases two polygonal regions in parallel planes. The prismatoid is the union of all segments which have an end point in each base. Its lateral faces are triangular or quadrangular regions whose vertices are vertices of the bases. The quadrangular faces are either trapezoidal or, in special cases, bounded by parallelograms. Each prism is a prismatoid, and pyramids can be considered limiting cases of prismatoids- -when one base is "shrunk" to a point. The midsection of a prismatoid is the section made by a plane mid- way between the planes which contain its bases. The altitude of a prismatoid is the distance between the planes which contain its bases. Despite the wide variety of prismatoids, there is a simple formula [9-296] [App. D] for the volume -measure of any of them. If the area-measures of the bases of a prismatoid are B. and B , that of its midsection is M, and its altitude is h, then its volume -meas- ure V is given by: V = ~(B X + B 2 + 4M)h This formula is called the prismoidal [or: prismatoidal] formula. Notice that, as it should, it gives the correct result for a prism [B = B = M = B] and for a pyramid [B x = B, B 2 = 0, M = B/4], The prismoidal formula can be derived by investigating the result of cutting a prismoid up into triangular pyramids. Here's how. Choose a point P in the midsection of a prismatoid. The prismatoid is, then, the union of the pyramids which have P as vertex and, for bases, the faces of the prismatoid. Two of these, those whose bases are the bases of the prismatoid, have easily computed volume -measures, These measures are respectively- -that is, K(! t B i h and and iMi 6 B 2 h - We shall see that the volume-measures of the other pyramids- - those whose bases are lateral faces of the prismatoid--add up to |Mh, thus completing the proof of the prismoidal formula. Since some lateral faces of a prismatoid may be trapezoidal and others triangular, some of the pyramids we have left to consider may [App. D] [9-297] have trapezoidal bases and others may have triangular bases. To sim- plify the argument we note that each pyramid with a trapezoidal base is split by a plane through its vertex and a diagonal of its base into two triangular pyramids. So, the pyramids whose volume-measures we have left to compute may be taken to be triangular pyramids with vertex P whose bases are either lateral faces of the prismatoid or "halves" of lateral faces. Consider such a pyramid, with base AABC. The mid- section of the prismatoid intersects AABC in a segment DE parallel to * • 1 BC, andK(AADE)= -r • K(AABC). Consequently, the volume-measure of the pyramid with vertex P and base A ABC is 4 times that of the pyramid with vertex P and base AADE. Since this latter pyramid may also be described as having vertex A and base APDE, its volume-measure is ^'K(APDE)|. So, the volume-measure of the pyramid witL vertex P and base A ABC is | • K(APDE)h. In other words, the volume-measure of each of the pyramids we are considering — those with vertex P and a lateral face, or "half" a lateral face, for base--is the product of two thirds the area-measures of its intersection with the midsection of the prismatoid by the altitude h of the prismatoid. Consequently, as we set out to show, the sum of the volume-measures of those pyramids is |Mh. 112 1 Since rB : h + T Q 2 h + T Mh = 6* B i + B 2 + 4M ) h > we have estab- lished the prismoidal formula. Note that we have assumed that the midsection is of such a shape that it is the union of nonoverlapping triangular regions with vertex P. This is the case for any choice of P in the midsection if the latter is convex- -the case usually treated. The prismoidal formula also works for solids which are like pris- matoids but whose bases, in parallel planes, are not polygonal regions --that is for solids which are unions of segments which have one end point in each of two parallel plane regions. [9-298] [App. Dj FRUSTUMS A frustum of a pyramid [or cone] is a solid consisting of the points of the pyramid [or cone] between or on the base of the pyramid [or cone] and a cross-section parallel to the base. A frustum of a pyramid is a prismatoid, and the prismoidal formula applies to frustums of both pyra mids and cones. In the case of a frustum one can express the area-measure M of the midsection rather simply in terms of the area-measures of the bases, and, so, obtain a useful special case of the prismoidal formula. To do so, suppose that the area-measure of the base of the pyramid [or cone], whose frustum we are considering, is B x> the area-measure of the other base of the frustum is B and that that of the midsection of the frustum is M. Suppose, also, that the altitude of the frustum is h and that that of the pyramid [or cone] is h.. It follows that the "upper" base of the frus- tum is at a distance h x - h from the vertex of the pyramid [or cone], and that the midsection is at a distance h x - — from the vertex. Conse- quently, and M = h x - h/2 \2 B \2 = [1 - 2h. B It follows that 1 - -^— = n/B 2 /B x and, so, that ~ = 1 - 'v/B 2 /B 1 . Consequently, h 2h 2h 1 - 4b. J*l 2 1 - • = 1 >/B 2 /B 1 2 1 + 4b. > a [App. D] [9-299] and, so, Hence, 1 - 2h 2 1 + 2^B 2 /B 1 + B 2 /B 1 M = ijLilEZL.i^i . B B x + 2^B 1 B 2 + B 2 Substituting into the prismoidal formula, one obtains, for the vol' ume-measure V of a frustum of a pyramid [or cone], the formula: V = ^(B x + B 2 +^B x B 2 )h Notice that the formula checks with earlier formulas in the case where the frustum is the entire pyramid [or cone], EXERCISES Solve these problems. 1. The altitude of a frustum of a regular square pyramid is 12 and the area-measures of the bases, 36 and 144, respectively. Compute the volume-measure of this frustum. 2. TT „ The diagram at the left pictures a rec- tangular solid such that AB = 14, BC = 10, and FB = 8. Let P and Q be points of EA and of HD, respectively, such that AP = DQ. Consider the plane containing the A B ^ points F, G, Q, and p. Determine AP so that this plane separates the rectangular solid into two solids whose volume-measures are in the ratio 1 : 2. 3. Use the same rectangular solid described in Exercise 2, but this time let AP = 7, DQ = 5, and let R be the point of CG such that CR = 3. Find the volume-measure of the portion of the rectangu- lar solid which is "under" the union of the planes FPQ and FRQ. [9-300] [App. D] SOLID SPHERES We shall now find a formula for the volume-measure of a spherical region. [A spherical region is one which is bounded by a sphere. Such a solid is sometimes confusingly called 4 a sphere'. Better short names are 'solid sphere' and 'ball'. ] As you might expect, we shall use Cava- lieri's principle. Consider a sphere of a radius r and a cross-section [of the solid sphere which it bounds] made by a plane at a distance e < r from the center of the sphere. This cross-section is a circular region whose radius, s, is \/r - e and whose area-measure is, consequently, 2 2 7T(r - e ). At this point, one needs an inspiration. What solid whose volume-measure we already know how to find has cross-sections whose area-measures are the same as those of these circular cross-sections? Struggling a bit, we note that .22. 2 2 77 {r - e ) = 77 r - 77e , and that all the cross- sections of a cylinder of radius r have area-meas- 2 2 ure 77r . Moreover, < e < r and, for each such e > 0, 77e is the area- measure of a circular region of radius e. Thinking of the cylinder and Cavalieri's principle may suggest that we consider a right circular cylinder of radius r and of altitude 2r--the "altitude" of the solid sphere. If we choose a point on the sphere and let p be the plane through this point and perpendicular to the corresponding diameter of the sphere [that is, the plane tangent to the sphere at the chosen point], and if we choose for the right circular cylinder one whose base is in the plane p, we have a set-up for exploiting Cavalieri's principle. Each plane paral- lel to p and at a distance e < r above or below the centers of the sphere and cylinder cuts out cross- sections of the two solids. For the solid [App. D] [9-301] sphere the cross-section is a circular region whose area-measure is TTr - 7Te and, for the cylinder, it is a circular region whose area- measure is TTr . Now, it is easy to find a circular region in the latter cross-section whose area-measure is TTe . Just take the one with the same center and radius e. Since the diameter and altitude of the cylin- der are equal, this circular region is merely a cross- section of one of the two right circular cones whose common vertex is the center of the cylinder and whose bases are the bases of the cylinder. By Cavalieri's principle, the volume-measure of the spherical region is the same as that of the solid which remains when these two cones are "bored out" of the cylinder. But, we know the volume -measure of this solid. It is 2 the difference between that of the cylinder, TTr • 2r, and that of the two 1 2 cones, 2(^-TTr *r). So, the volume-measure of the solid sphere is , 3 2 3 t , 4 3 277r - —TTr --that is, — 7Tr . V = 4 3 3* r rrd 3 It is interesting to notice that the volume -measure of the spherical region is 2/3 that of the smallest right circular cylinder which could contain it. [9-302] [App. D] SURFACE AREA FORMULAS The area-measures of the lateral surfaces or the total surfaces [including bases] of prisms and pyramids are usually fairly easy to come by, but, except in quite simple cases, the formulas are difficult to state. In general, one computes the area-measures of the faces in question and adds the results. All the faces are polygonal regions, and most of them are bounded by either parallelograms or triangles. As an example which leads to a simple formula for the area-measure of the total surface, we cite the case of a rectangular parallelepiped. —*/¥ S = 2(ab + be + ca) The only other example [among prisms and pyramids] worth bother- ing about is that of a regular pyramid--one whose base is bounded by a regular polygon whose center is the foot of the perpendicular from the vertex of the pyramid. Since all the lateral faces are congruent [isosceles] triangles whose bases are the sides of the regular polygon and whose common altitude is the slant height of the pyramid, the area-measure L of the lateral surface is very easy to compute. L, = yp-2, where p is the perimeter of the base and St is the slant height To obtain the area-measure S of the total surface, one adds to L. the area-measure of the base. [App. D] [9-303] This last result suggests the formula for the area-measure of the lateral surface of a right circular cone of radius r and slant height i. V 2 J . 2 L, = 77ri = 77r Vr + h [and, S = tit (I + r). ] Further evidence for the correctness of this formula [but not a proof] can be obtained by thinking of the lateral surface as slit along one of the 277r segments joining the vertex to the base [like an Indian tepee] and flat- tened out. The result is a circular sector of radius i and arc-measure 27Tr. And, as we know, the area-measure of such a sector is — i*27Tr. A similar procedure suggests formulas for right circular cylinders. L, = 27lrh i h S = 27Tr(h + r) Somewhat similar formulas [L, = ph and S = ph + 2b] hold for all right prisms and cylinders. [9-304] [App. D] Finally, we can arrive, in a somewhat backhanded way, at the for' mula for the area-measure of a sphere of radius r. Since the volume- 4 3 measure of the region bounded by such a sphere is — 77r , that of a spherical shell of outer radius r and thickness t must be 4 3 3 771 * 4 3 J7r(r - t)\ On the other hand, if t is small and S is the area-measure of the outer surface of the shell, then the same volume-measure should be approxi- mately St. So, S = |77[r 3 - (r - t) 3 ]/t, 2 2 Since, for sufficiently small values of t, 3r - 3rt + t is very nearly 2 4 2 3r , we can guess that S is exactly t^* 3r . This guess is correct [but, the preceding argument is not a proof]. S = 4flT = 7Td So, the area-measure of the surface of a sphere is 4 times that of its largest cross-section. Recalling our comparison of the volume- measures of a spherical region and of the smallest right circular cylin- der which contains it [the former was 2/3 of the latter], we see a rather [App. D] [9-305] surprising similarity. The ratio of the area-measures of the total sur- faces of the solids is 47Tr /[277r(Zr + r)]- -again 2/3. This so pleased the Greek geometer, Archimedes, who first gave adequate proofs of these results, that he asked that a picture of a sphere and circumscribed cylinder be engraved on his tomb. There is another way to make the formula for the area-measure of a sphere appear reasonable [and to help you to remember both this for- mula and that for the volume-measure of a solid sphere]. This is to think of a solid sphere as the union of a lot of narrow pyramid-like solids whose common vertex is the center of the sphere and whose "bases" are portions of the surface of the sphere. Since the bases are curved rather than plane, these solids will not actually be pyramids. However, the volume-measure of each should be pretty close to one third the product of the area-measure of its "base" by its "altitude", the radius of the sphere. Consequently, the area-measure S of the sphere and the volume -measure V of the region it bounds may be ex- pected to satisfy the equation: V = |sr 3 If you recall this and either one of the formulas: ... 4 3 S = 47Tr' it is easy to derive the other, [9-306] [App. D] SUMMARY OF MENSURATION FORMULAS Triangular regions K = -a-hb K = K = |ab sin x°, [0 < x < 90] ^-ab sin (180 - x)°, [90 < x < 180] K = Vs(s - a)(s - b)(s - c) [s = | (a +b + c)] K= ^ Quadrilateral regions /' w K = hb = 7 ab sin x°, [0 < x < 90] K ab sin(180 -x)°, [90 < x < 180] x; ''.■'.-■•} -4 • • • .• N . i x K = jd^ K = wi K = jh(b x +b 2 ) K = jd x d 2 sin x° [0 < x < 90] K = jd^ sin (180 -x)° [90 < x < 180] [App. D] [9-307] Regular polygonal regions n-gon K = rans = 2 a P 1 2„ /x = 4ns tan k 1 2„ / fa - 2)180 = _ ns tan _J Circular regions c = iTtx ~ ffd 2 1 ,2 K = TTr s — ;rd t^s t = sfo t° < * < 18 °J K = jfo^r 2 ) [0 -Uhd 2 4 Tt^j L » 2ffhr = ffhd S * 27rr(h + r) Pyramidal solids V=|hB V a ~7Thr 2 L= yip S = |-ip+B i t * V 2 . . 2 V L = 7rir = TirV r + h S = 7Tr(i + r) regular base, congruent faces [9-308] [App. D] Prismatoidal solids V = ^h(B 1 + B 2 + 4M) V = |h(B x + B 2 + v'B^) Spherical solids 4 3 1 ,3 V = -nr = z nd S = 47Tr 2 = TTd 2 EXERCISES A. Solve these problems. 1. Compute the altitude of a right prism whose total surface has area-measure 320 and one of whose bases is a rhombus whose diagonal -measures are 6 and 10, respectively. 2. Suppose that the radius of the regular base of a right hexagonal prism is 6. If the altitude is 12, compute the area-measure of the total surface and compute the volume -measure. 3. Repeat Exercise 2, but triple the radius and double the altitude. [Compare the new area- and volume -measures with the ones ob- tained in Exercise 2. ] 4. Compute the lateral area-measure of a right prism whose alti- tude is x and whose base is a right triangle with legs of measures 3x and 4x, respectively. [App. D] [9-309] 5. The edge-measures of two cubes are in the ratio p:q. Compute the ratio of the area-measures of their total surfaces and the ratio of their volume-measures. 6. Suppose that the radii of two circular cylinders are r 1 and r 2 , re- spectively, and their altitudes are h 1 and h 2 , respectively. Com- pute the ratios of their lateral area-measures and the ratio of their volume -measures. 7. If the ratio of the radii of two solid spheres is p to q, what are ra- tios of their surface area-measures and their volume-measures? 8. Compute the total area-measure of a regular hexagonal pyramid if the perimeter of the base is 24 and its altitude is 10. 9. Consider a right circular cone whose "apex angle" is an angle of 120°. If the radius of the base is 10, what are the lateral area- measure and the volume-measure? 10. Visualize the solid generated by rotating AABC around AC. Such a solid is called a solid of revolution . Of course, in this case, the solid is a right circular cone. Compute its volume-measure and its lateral area-measure. 11. Repeat Exercise 10 for the solids of revolution obtained by rotating AABC around BC and around AB. 12. Compute the total area-measure and the volume -measure of the solid of revolution obtained by rotating a rectangular region of side- measures 24 and 32 around the line containing one of the shorter sides. ^13. Repeat Exercise 12 for the solid of revolution obtained by rotating the region around the line containing a diagonal. [9-310] [App. D] B. Solve these problems. 1. Consider a solid sphere with radius 10. Compute the area-meas- ure of the intersection of the solid sphere and a plane whose dis- tance from the center is 3. 2. Given the circular regions with centers A and B such that AC = 3, CB = 4, and BA = 5. Consider the solid of revolution obtained by ro- tating both regions around AB. "What is the area-measure of the region generated by CD? 3. If 6 is the radius of the smallest solid sphere which contains a right circular cylinder of altitude 10, what is the volume-meas- ure of the cylinder? 4. The intersection of a sphere and a plane containing its center is called a great circle of the sphere. If the area-measure of the region [interior to the sphere] bounded by a great circle is x, what are the area- measure of the surface of the solid sphere and the volume - me asure ? 5. Write a formula for the surface area-measure S of a sphere in terms of its volume-measure V. 6. Compute the radius of the smallest solid sphere which contains the smallest cube which contains a solid sphere of radius r. 7. Two spherical balls of radii 7 and 10, respectively, are placed on a table so that they touch each other. How far apart are their points of contact with the table? 8. What is the volume-measure of the largest sphere which is con- tained in the frustum of a right circular cone whose bases have radii 4 and 16? [App. D] [9-311] 9. Compute the diameter of the smallest spherical solid which con- tains a rectangular solid with edge-measures x, y, and z. 10. Find a formula for the volume measure of the holed frustum, as pictured. **11. Consider the set of all rectangular regions of perimeter p. Which of these regions produces the solid of revolution of maximum vol- ume when it is rotated around the line containing one of its sides? [ Hint . Let w be the measure of the side contained in the axis of revolution. Find the function that V is of w. Graph the function and guess its maximum value. ] C. Each of the following exercises gives certain conditions. Describe the geometric figure which satisfies the given conditions. Example 1 . The set of points each of which belongs to a plane p and each of which is the distance d from the point Pep, Solution . The circle in p with center P and radius d. Example 2 . The set of points each of which is equidistant from the two points A and B. <-» Solution . The plane perpendicular to AB and which contains the midpoint of AB. 1. The set of points each of which is the distance d from P. 2. The set of points each of which belongs to a plane p and is equi- distant from the two points A € p and B e p. 3. The set of points each of which is the distance d from a line i. 4. The set of points each of which belongs to a plane p and is the distance d from a line £ C p. [9-312] [App. D] 5. The locus of points [that is, the set of points] each of which is the distance d from a plane p. 6. The locus of points each of which is equidistant from two parallel planes p and q. 7. The locus of points each of which is equidistant from two intersect- ing planes p and q. 8. The locus of points each of which is twice as far from plane p as from plane q where p | J q. 9. The locus of points P such that, for some point X of the circle c of radius r, PX = — r. 10. Repeat Exercise 9 but add the condition that P be in the plane of c. 11. The locus of points each of which belongs to plane ABC and is equi- distant from A, B, and C. 12. The locus of points each of which is equidistant from the vertices of AABC. 13. The locus of points each of which is equidistant from the parallel lines SL and m. 14. The locus of points each of which is equidistant from the nonco- planar parallel lines i, m, and n. 15. The locus of points P such that AP = d x , BP = d 2 , and A / B. [Consider all cases. ] 16. The locus of points P such that AP = d and P is equidistant from the parallel planes p and q. 17. The locus of points P such that PA = d and PB = PC. [App. E] [9-313] APPENDIX E [This Appendix contains proofs of Theorems 221-224. These theorems are the basis of the applications discussed in section 9. 10, and the Appen- dix ends with a summary of the results needed for such applications.] Some functional equations . --Many problems you have met in your study of mathematics require, for their solution, that one find numbers which satisfy given conditions. [Most of the "word problems" you have solved are of this kind. ] Many such problems boil down to finding the solutions of an algebraic equation. Other problems- -there are several examples in section 9. 10- -require one to find functions which satisfy given con- ditions. In many cases such problems boil down to solving a so-called functional equation. Here are two examples of functional equations: (1) v x f(x) = 2f(x) (2) V x V y f(x) = f(y) Both (1) and (2) are open sentences--neither true nor false. From either we can generate a statement by replacing *f by a name for some function. Some of the statements obtained in this way will be true and some will be false. [Some replacements will lead to nonsense--for example, 'V Vx = 2vx '--because the function in question is not defined for all real num- bers. ] Those functions for which (2), say, becomes a true statement are said to satisfy (2), or to be solutions of (2). To solve a functional equation is to determine which functions are its solutions. The functional equations (1) and (2) are easy to solve. For example, (1) is equivalent to: V f(x) = [Explain. ] So, the only solution of (1) is the constant function 0. What are the solu- tions of (2)? In this Appendix we shall study some functional equations which oc- cur frequently in mathematics. As we shall see, it is often difficult to solve a functional equation completely --that is, to find all its solutions. However, it is often possible to find all the solutions of a given functional equation which have some "desirable" property, such as monotonicity or continuity, and to prove that all its remaining solutions have some "undesirable" property. [9-314] [App. Ej THE LDPMA AND THEOREMS 205, 215, AND 218 Among the functions we have studied, four kinds stand out: (a) the homogeneous linear functions [f = {(x, y): y = ex}, for some c ^ 0] (b) the exponential functions [f = {(x, y): y = a }, for some a > 0] (c) the logarithm functions [f = {(x, y), x > 0: y = log x}, for some < b / 1] (d) the power functions with positive arguments [f = {(x, y), x > 0: y = x }, for some u] These are very simple functions. All are continuous and, except for the exponential function with base 1 and the power function with exponent 0, all are monotonic. For each of the four kinds of functions, we have proved a number of useful theorems. For example [one theorem for each kind of function], by the idpma, if f is a homogeneous linear function then (1) V x V y f(x + y) = f(x) + f(y), by Theorem 205 , if f is an exponential function then (2) V u V v f(u + v) = f(u)f(v), by Theorem 215 , if f is a logarithm function then (3) V u > V v > f(uv) = f(u) + f(v) [and *f = {u: U > ° }] and by Theorem 218 , if f is a power function with positive arguments then (4) V u>Q V v>0 f(uv) = f(u)f(v) [and^ f ={u: u > }]. It is interesting --and, as shown in section 9. 10, proves to be use- ful—to discover that each function f which is, say, monotonic and which satisfies either (1), (2), (3), or (4) is, respectively, either a homogene- ous linear function, an exponential function, a logarithm function, or a power function with positive arguments. As a matter of fact, we shall [App.E] [9-315] prove much more than this. Here is a sample of what we shall prove: Each function f which satisfies (1) and is not a homogeneous linear function either is the constant function whose value is or has the property that each square region of the number plane [whose sides are parallel to the axes], no matter how small it is and no matter where it is in the plane, contains an ordered pair belonging to f. Functions which have the peculiar property described above must be quite complicated. You may even doubt that there can be functions of this kind, but, in a moment we will give an example of such a function. Before doing so, let*s see that such a function cannot be monotonic. For this, it is sufficient to note that a function which contains ordered pairs in both the second and fourth quadrant cannot be increasing [Explain. ], and that one which contains ordered pairs in both the first and third quadrants cannot be decreasing. So, given any monotonic function f, there is at least one whole quadrant which contains no ordered pair be- longing to f. Hence, no monotonic function can contain ordered pairs in each square region of the number plane. [Since the constant function is not monotonic, it follows from the theorem stated above that each monotonic function which satisfies (1) is a homogeneous linear function.] A QUEER FUNCTION Although there actually are very many functions which both satisfy (1) and have the peculiar property we have been discussing, such func- tions are very difficult to define. Fortunately, it is not difficult to de- fine a function which has the property in question and "partly satisfies (1)". [What this last means will appear when we have defined the func- tion. ] Here is a definition of such a function g: If there are rational numbers r and s such that x = r + sv2 then g(x) = r; if there are no such rational numbers r and s then g(x) = 0. For example, for each rational number r, g(r) = r, g(r - v2 ) = r, and 147 i — g(r + Q - - • V 2 ) = r. Since there are no rational numbers r and s such that [9-316] [App. E] r + sv2 = v3, g(v3 ) = 0. The figure is meant to suggest the ordered pairs of g just mentioned as examples. The lines are dotted to indicate that the only points on them which belong to the graph of g are those at rational distances from the x-axis. You can imagine the complete graph of g to be constructed as follows: First, plot all the ''rational points" on the graph of 'y = x\ and call this set 4 S*. Next, plot all the points on the x-axis whose ab- scissas are rational multiples of v2 , and call this set 'T'. Next, plot all the points that can be obtained by shifting S to the left or right [but not up or down] so that the shifted set contains one of the points of T. [Do this for each point of T. ]. Now, in your imagi- nation, you have plotted all the ordered pairs (x, g(x)) such that, for some r and s, x = r + sv2 . Complete the job by plotting the points (x, 0) for which there are no rational numbers r and s such that x = r + sv2 . [For example, plot (v3, 0). ] Of course, if you tried to carry out these instructions on paper, the resulting picture would be a mess, and you are probably wondering whether g really is a function. Is it really the case that no two ordered pairs of g have the same first component? To see that this is so, all we If there were rational numbers r and s such that v 3 = r + sv2, there would, then, be a rational number s / such that v3 - sv2 is a rational number. The square of this number- -3 - 2sv6 + 2s --would be rational and, hence, 2sv6 would be rational. Since s £ 0, it follows that v6 is rational. But, this is not the case. [App. E] [9-317] need do is make sure that if x is a number for which there are rational numbers r x and s x such that x = r + S.V2 then there are not any other rational numbers-- say r 2 and s 2 ~-such that x = r 2 + s 2 v2 . This will show that there is no number x for which the definition tells us that g(x) = r and g(x) = r 2 where r 2 / r lt So, what we need do is show that if r 2 + s 2 v2 = r x + s 1 yZ then r 2 = r 1# This is easy. For, suppose that r 2 + s 2 v2 = r x + s 1 v2. Then, (s 2 - s 1 )v2 = r x - r 2 and, unless s 2 - s x = 0, it follows that v2 = (r x - r )/(s - s ), a rational number. So, since v2 is irrational, it follows that s 2 = s^ Hence, since r 2 + s 2 v2 = r + Sj^vZ, it follows that r = r,. [So, we could have obtained a queer function of the same kind as g by replacing 4 v2 ' in the definition of g by a name for any ir- rational number. The new function would be different from our function g if the irrational number we chose was not of the form *r + sv2 '. ] The description of how g might be graphed probably convinced you that g does have the odd property we are interested in- -that is, that each square region of the number plane contains an ordered pair belong- ing to g. At any rate, it is not difficult to prove that this is the case. To do so, consider, for any given numbers d > 0, x Q , and y_, the square region whose center is (x Q , y ) and whose side-measure is 2d, [Draw a picture. ] What we wish to do is find a number x such that (x, g(x)) belongs to this region- -that is, such that x Q -d is also closed with respect to addition. So, f satisfies the first part of (1'), We end by noting that, if x = r x + S;,^^ and y = r 2 + s^fl then x + y = (r 1 +r 2 ) + (s x + s 2 )v2 and, so, by definition, f(x) = r^ f(y) = r 2 , and f(x + y) = r l + r 2 . Hence, for each x and y in &,, f(x + y) = f(x) + f(y)--that is, f satisfies the second part of (!'). [App.E] [9-319] A THEOREM ON HOMOGENEOUS LINEAR FUNCTIONS Having considered some examples, it's now time to get back to our program of proving some theorems like the one displayed on page 9-315. In applications of mathematics one often has to deal with functions which are not defined for all real numbers and which, just for this reason, can- not possibly satisfy: (1) V V f(x +y) = f(x) + f(y) x y However, many of them [like the particular function f of our example] satisfy: (*'> v v, v v ve v < x + y € ^ and f < x + y) = f < x > + f I To be more explicit, some problems in, for example, physics or chemis- try, can be solved by discovering, experimentally, that one quantity is a function of another and that the function in question satisfies (1'). If this is the case then, using the theorem we are going to prove, it is possible to conclude that the function in question is a homogeneous linear function restricted to £■,, and this furnishes a way of predicting the value of the first of the two quantities when one knows the corresponding value of the second. In brief, one has discovered a law of nature. An example of this sort of discovery is given in the discussion of Gay- JLussac's Law which begins on page 9-139. The functions which arise in applications of the kind described above, besides satisfying (1'), have the further property of being defined for at least all the points of some interval. Our theorem will tell us that each such function is either a homogeneous linear function restricted to £y , or the constant function whose value is and whose domain is ^- f , or a function which, for arguments in as short an interval as you wish, has values as close as you wish to each real number. This suffices for the applications, since physical [or chemical] considerations will, in each case, show that the experimentally determined function can neither be constant nor extremely "jumpy", [in fact, in most cases it will be clear that it is monotonic.] We are now ready for the statement of our basic theorem on homo- geneous linear functions. [From it, we will derive similar theorems about exponential, logarithmic, and power functions.] [9-320] [App. E] Theorem 221. If f is a function such that (n V xe£ V ye^ U + V € > f a *<* *<* + y) = f(x) + f(y)) then either f is a subset of a homogeneous linear func- tion or f is not monotonic. In the latter case, either f is a subset of the constant function or, for each x_ which belongs to an interval contained in & f and for each y Q and each d > 0, there is an x e ,5-, such that x Q - d < x < x + d and y - d < f(x) < y Q + d. (Note that replacing (!'), above, by: (1) V V f(x + y) = £(x) + f(y), x y amounts to adding the hypothesis that & f is the set of all real numbers. So, the theorem which results when (!') is replaced by (1) is a corollary of Theorem 221. In this case, since >, is the set of all real numbers, each number x Q automatically belongs to an interval contained in £-. So, the corollary is just a restatement of the theorem displayed on page 9-315. Note, also, that the second sentence in Theorem 221 tells you nothing unless ,5y contains an interval. But, whatever &, is, the first sentence tells you that if f is monotonic and satisfies (1') then f is a sub- set of a homogeneous linear function. ] In order to simplify the proof of Theorem 221, we begin by proving three lemmas concerning any function f which satisfies (1'). Lemma 1 . € £ f ^ f(0) = Lemma 2. V .V (nx e ,$> and f(nx) = nf(x)) x£X n i Lemma 3 . For each x x and x 2 in &r and each r x and r 2 such that r i X l + r 2 X 2 € ^f ' f ( r i X l + r 2 X 2^ = r i f ( X j.) + r 2 f ^ X 2^ [Only Lemmas 1 and 2 are needed for the proof of the first part of Theo- rem 221. Go over the proofs, given below, for Lemmas 1 and 2 now, and make sure you understand Lemma 3. Then you may find it simpler to skip to the proof of Theorem 221 and, afterwards, study the proof of Lemma 3. The latter is somewhat tedious.] [App. E] [9-321] Proof of Lemma 1 . Since f satisfies (1'), it follows that, if e /5y then f(0) = f(0 + 0) = f(0) + f(0). So, if e a f then f(0) = 0. Proof of Lemma 2 . [By induction. ] For afl, since 1 • a = a, it fol- lows that 1-ael and f(l • a) = f(a) = 1 • f(a). For a 6^r, suppose that pa e £ f and f(pa) = pf(a). Since pa and a belong to b f , it follows from (1') that pa + a € &r and f(pa + a) = f(pa) + f(a). Consequently, (p + l)a € <5y and, since f(pa) = pf(a), £((p + l)a) = f(pa+a) = f(pa) + f (a) = pf(a) + f(a) = (p+l)f(a). So, by mathematical induction, Lemma 2. Proof of Lemma 3 . Consider arguments a x and a 2 of f and rational numbers r x and r 2 such that r JL a 1 + r 2 a 2 € X. Choose a positive integer p such that p + r x > and p + r 2 > and choose a positive integer m such that r x m e I and r m € I. By Lemrqa 2, since a x e St and a 2 e ,5-f , it follows that pa x 6 &f and pa 2 e ^f , and, so, by (l')> that pa x + pa 2 e ,5-f . Hence, since r 1 a 1 + r 2 a 2 e £ f , it follows, by (1'), that (pa x + pa 2 ) + (r^j^ +r 2 a 2 ) belongs to £y . This being the case, it follows, by (1'), that f((pa 1 +pa 2 ) + (r 1 a 1 + r 2 a 2 )) = f(pa x +pa 2 ) + f(r 1 a 1 + r 2 a 2 ) = f(pa x ) + f(pa 2 ) +f(r iai +r 2 a 2 ). Since (pa^^ + pa 2 ) + (r 1 a 1 + r 2 a 2 ) = (p + r i )a 1 + (p + r 2 )a 2 and since, by Lemma 2, f(pa x ) = pf(a 1 ) and f(pa 2 ) = pf(a 2 ), it follows that (i) £((p+r 1 )a 1 + (p + r 2 )a 2 ) = pf(a x ) + pf(a 2 ) + f(r i a 1 + r 2 a 2 ). Also, by Lemma 2, since (p + r )a + (p + r 2 )a e £•£ , it follows that f(m[(p + r 1 )a 1 + (p + r 2 )a 2 ]) = mf((p+r jL )a 1 + (p+r 2 )a 2 ). Since m[(p + r 1 )a 1 + (p + r 2 )a 2 ] = m(p + r 1 )a 1 + m(p + r 2 )a 2 , [9-322] [App. E] it follows that (ii) f(m(p + r 1 )a 1 + m(p+r 2 )a 2 ) = mf((p + r 1 )a 1 + (p + r 2 )a 2 ). Now, since r x m € I, it follows that m(p + r x ) € I and, since p + r L > 0, it follows that m(p + r J € I + . So, by Lemma 2, since a x € ^£, it follows that m(p + r 1 )a 1 e >f . Similarly, m(p + r 2 ) € I* and m(p + r 2 )a 2 e £f . Hence, by (l')» f(m(p + r 1 )a 1 + m(p+r 2 )a 2 ) s f(m(p + r i )a 1 ) +f(m(p + r 2 )a 2 ). So, by Lemma 2 [since a x and a 2 are arguments of f and m(p + r x ) and m(p + r 2 ) are positive integers], it follows that (iii) f(m(p + r 1 )a 1 +m(p+r 2 )a 2 ) = m(p +r 1 )f(a 1 ) + m(p + r 2 )f(a 2 ). Comparing (ii) and (iii), and noting that m ^ 0, we see that f((p+r 1 )a 1 + (p + r 2 )a 2 ) = (p+r^f^) + (p + r 2 )f(a 2 ) = pf(a x ) + pf(a 2 ) + r 1 f(a 1 ) + r 2 f(a 2 ). Comparing this last with (i), we see that f(r 1 a 1 + r 2 a 2 ) = r 1 f(a 1 ) + r 2 f(a 2 ). Proof of Theorem 221 . Suppose that f satisfies (1') but is not a subset of a homogeneous linear function. For the first part of the theorem, we wish to show that f is not monotonic. For the second part, we shall assume, also, that f is not a subset of the constant 0, and show that, for each x Q which be- longs to an interval contained in &, and for each y and each d > 0, there is an x € /$> such that jx - x | < d and jf(x) -y | < d, Since f is not a subset of a homogeneous linear function, it follows that f has nonzero arguments. For, if £, = then, trivially, for each x e ^£» f(x) = x; and, if ^ - {0} then, by Lemma 1, for each x € -$>» f(x) = x--in either case, f is a subset of a homogeneous linear function. Since, as we have shown, f has a nonzero argument and since ^, is closed with respect to addition, it follows that f has many arguments. If, for all of its arguments, f has the value then [since f has at least two arguments] f is not monotonic. Since this is the conclusion we wish to reach in the first part of the proof, we may, for this part, assume that f has an argument x x such that f(x x ) ^ 0. And, for the second part [App. E] [9-323] of the proof, this follows from our assumption that f is not a subset of the constant 0. So, from now on, we shall assume that x x is an argu- ment of f such that f(x x ) ^ 0. Setting c = f(x 1 )/x i , it follows that c / and f(x x ) = cx 1# Since, by hypothesis, f is not a subset of a homogeneous linear function, it follows that f has an argument x 2 such that f(x 2 ) / cx 2 . By Lemma 1, x 2 / 0. Since f(x x ) = cx x , x 2 £(x x ) = cx 1 x 2 . Since f(x 2 ) / cx 2 and x x / 0, x x f(x 2 ) / cx 1 x„. So, there are nonzero arguments x x and x 2 of f such that f(x x ) / and U) x x f(x 2 ) ^ x 2 f(x x ). We shall now use (i) to show that f is not monotonic, thus complet- ing the proof of the first part of Theorem 221. After doing so, we shall use (i), in another way, to prove the second part of Theorem 221. Since f satisfies (l')» it follows that -f satisfies (1'). Also, if -f is not monotonic then neither is f. So, without loss of generality, we may assume that f(x x ) > 0. [If f(x x ) < 0, we merely transfer our atten- tion to — f. ] Also, we may, without loss of generality, assume that x x and x are both positive. For, to begin with, suppose that x 2 < < x JL . In this case there is a positive integer p such that x + px x > 0. By Lemma 2 and (1'), it follows that f(x 2 + px x ) = f( x 2 ) + P*( x i)» So, x x f(x 2 + px x ) = x 1 f(x 2 ) + px 1 f(x 1 ). Since (x 2 + px i )f(x 1 ) = x 2 f(x x ) + px^U^, it follows from (i) that x 1 f(x 2 + px x ) ^ (x 2 + px 1 )f(x 1 ). So, if x < < x,, we can replace x by x + px x and have two positive arguments of f which satisfy (i). Note that one of these arguments is k x and, as before, f{x 1 ) ^0. Next, suppose that x ± < < x . In this case we can choose p so that x— + px x < and end up with two negative arguments of f, one of which is x x , which satisfy (i). So, without loss of generality, we may assume that x x and x 2 are either both positive or both negative. Now, since f satisfies (1') so does the function g where g(x) = f(-x). And, if g is not monotonic then neither is f. So, without loss of generality, we may assume that both x. x and x are positive, [if they are both negative, we merely transfer our attention to g. ] [9-324] [App. E] There are now, two cases to consider: f2 < i ^l and . ^2 > £ < X 2 ) x x f(x x ) x x f^) Since x 2 /x x > 0, there are, in either case, positive integers m and n such that m/n is between x 2 /x x and f(x 2 )/f(x 1 ). In the first case (a) — < — < 77 r. x 1 n f( X;L ) In this case it follows [since x. > and n > 0] that nx„ < mxj and [since n > and f(x x ) > 0] that nf(x 2 ) > mf(x 1 ). From the last, by Lemma 2, f(nx 2 ) > ffrnx^. So, f is not an increasing function. To show that f is not decreasing, let p be a positive integer such that mx. 1 < px 2 . Since f(x 1 ) > and f(x 2 )/f(x 1 ) > m/n > 0, it follows that f(x 2 ) > 0. And, since nx 2 < mx 1 < px 2 and x g > 0, it follows that n < p. So, by Lemma 2, f(m Xl ) = mf(x 1 ) < nf(x 2 ) < pf(x 2 ) = f(px 2 ). Since mx. < px and f(mx 1 ) < f(px ), it follows that f is not decreasing. Since f is neither increasing nor decreasing, it follows that [in case (a)] f is not monotonic. The alternative to (a) is that f(x ) m x b TT-^T < — < — . Mx.^ n x x It follows, as before, that mx 1 < nx^ and that ffrnx^ > f(nx 2 ). So, f is certainly not an increasing function. To show that f is not decreasing, choose p so that nx 2 < px x . Then, since m < p and i(x. x ) > 0, f(nx 2 ) < f(mx x ) = mf(x x ) < pf(x x ) = i(px x ). Since nx < px x and f(nx 2 ) < f(px x ), f is not decreasing. Hence, in case (b), f is not monotonic. This completes the proof of the first part of Theorem 221. To prove the second part, we assume that f satisfies (1'), is not a subset of a homogeneous linear function, and, also, is not a subset of the con- stant 0. As previously shown, it follows that there are arguments x x and x 2 of f such that (i) x x f(x 2 ) ^ XgfU^. Suppose, now, that x is any number which belongs to an interval [App.E] [9-325] all of whose members are arguments of f and that y Q is any number at all. Suppose, also, that d is any positive number. We shall show that there are rational numbers r and r such that if x = r x 1 + r 2 x 2 then x e £ f and x Q - d < x < x + d and y Q - d < f(x) < y Q + d. Notice that, since x belongs to an interval all of whose members are arguments of f, the first of the pairs of inequations just displayed will ensure that x£^ if d is a sufficiently small positive number. Without loss of generality, we can suppose that d is small enough for this to be the case. Notice, also, that, since x. and x are arguments of f and since r. and r are to be chosen so that r 1 x 1 + r 2 x p e ^f» ** ^ onows from Lemma 3 that f(x)--that is f(r 1 x 1 + r 2 x 2 )--will be r f(x x ) + r 2 f(x 2 ). So, our job is just, given x Q , y_, and d > 0, to find r and r such that (ii) x - d < r 1 x 1 + r 2 x 2 < x Q + d and y Q - d < r 1 f(x 1 ) + r 2 f(x 2 ) < y Q + d, We note first that, because of (i), the simultaneous equations in (a, b): (iii) / *i a + X 2 b = X ^ f{ Xl )a +f(x 2 )b = y have a unique solution. If the components of this solution happened to be rational numbers r and r then things would be very simple --for then the value of f at x Q , itself, would be precisely y and (ii) would be satisfied. This is not likely to be the case, but (ii) leaves us enough leeway to find a pair (r , r ) of rational numbers which comes near enough to satisfying (iii) that it will also satisfy (ii). Briefly, if the solution of (iii) is (a, b) and r and r are chosen sufficiently near a and b, respectively, then x 1 r 1 + x p r 2 will be as close as we like to x and f(x 1 )r 1 + f(x 2 )r 2 will be as close as we wish to y Q . More explicitly, for any rational numbers r and r , by (iii), x i r i + x 2 r 2 = x o + x J r i " a ) + x 2^ r 2 " b * and i(x 1 )r 1 + f(x 2 )r 2 = y Q + f(x 1 )(r 1 - a) + f(x 2 )(r 2 - b). From this one sees that if | r - a | and Jr - b| are sufficiently small [how small they need be depends on x , x , f(x ) and f(x 2 )] then (ii) will be satisfied. This completes the proof of Theorem 221. [9-326] [App. E] THREE OTHER THEOREMS Using Theorem 221 and some theorems concerning logarithmic and exponential functions it is not difficult to derive theorems about func- tions which satisfy: (2) V u V v f(u + v) = f(u)f(v) or, the less restrictive: < 2 '> V n* s \c s (u + v e ^ f and f(u + v) = f(u)f(v)) As indicated earlier, one might expect the solutions of (2) to be expo- nential functions. The discussion leading up to Theorem 221, however, should prepare us to expect that (2) also has some queer solutions. This is the case. In Theorem 222 [see below] we assume, besides that f satisfies (2')» that each value of f is positive. As shown, following the proof of Theo- rem 222, this latter assumption is unnecessary in case f satisfies (2)-- that is, in case f satisfies (2') and ,5> is the set of all real numbers. Theorem 222. If f is a function whose values are positive and is such that {r) V ue£ V ve^ (u + v € *f and f(u + v) = f(u > f < v » then either f is a subset of an exponential function with positive base different from 1 or f is not mono- tonic. In the latter case either f is a subset of the constant function 1 or, for each u which belongs to an interval contained in &. and for each v Q > and each d > 0, there is a u e & f such that u. d < u < u_ + d and v. - d < f(u) < v. + d. Proof of Theorem 222 . Since all values of f are positive, the function F for which F(u) = in(f(u)) has the same domain as f. So, for u e ,5-p and v e -5p , u + v e bp and [App. E] [9-327] F(u + v) = in[f(u + v)] ~\ = in[f(u)f(v)] J = in(f(u)) + in(f(v)) = F(u) + F(v). Consequently, F satisfies (1'). Note, also, that since in is increasing, F is monotonic if and only if f is monotonic. It follows from these remarks and the first part of Theorem 221 that if f is monotonic then F is a subset of a homogeneous linear func- tion- -that is, there is ac /0 such that V . in(f(u)) = cu. Since £.-, = X , it follows that, for each u e X, CI \ cu u -..1- c f(u) = e = a , with a = e . Since c / 0, < a ^ 1. This completes the proof of the first part of Theorem 222. To prove the second part we use the second part of Theorem 221. This, together with the fact that £„ = &, and that F is monotonic if and only if f is, tells us that if f is not monotonic then either F is a subset of the constant function 0--in which case f is a subset of the constant function l--or, for each u Q which belongs to an interval contained in &* and for each v Q > and each d' > 0, there is a u £ £, such that u Q - d' < u < u Q + d' and in v Q - d' < in (f(u)) < in v Q + d\ Since the exponential function is continuous at inv , given any d > 0, there is a d' > such that (*) if | in (f(u) - inv Q | < d' then |f(u) - v Q | < d. So, if, given d > 0, we choose for the d' of the second case a number which is less than or equal to d and satisfies (*) then the second case tells us that there is a u e £y such that u Q - do. So, in this case, (2') [or (2)] ensures that the values of f are at least nonnegative. Now, suppose that for some v , f(v ) = 0. Since, for each u, u - v Q e ^£, it follows, again by (2'), that, for each u, f(u) = f((u - v ) + v Q ) = f(u - v )f(v Q ) = 0. So, if f satisfies (2) V u V v f(u + v) =f(u)f(v) then either f is the constant function or all values of f are positive. Since a function which satisfies (2) must also satisfy (2'), Theorem 222 tells us what happens in the second case--f is either an exponential func- tion with positive base, or a queer function. A similar remark applies to situations in which it is known that &, is the set of nonnegative numbers and that (**) V u>0 V v>0 f(u +V) = f < u > f < v >- As before, this implies that the values off are all nonnegative. Also, as before, if f(v Q ) = then, for each u > v , f(u) = 0. Now, suppose that f(v Q ) = and that there is a u > such that f(u ) / 0. [By the results just noted, u Q < v Q , and f(u Q ) > 0.] It is easy to prove, using (**), that, for each n, f(nu Q ) = [f(u Q )] n . [See the proof of Lemma 2 on page 9-321.] So, since f(u Q ) ^ 0, it follows that, for each n, f(nu Q ) ^ 0. However, for n = j[v /u ]j + 1, nu Q > v Q and, so, for this n, f(nu Q ) = 0. Conse- quently, (**) implies that either, for each u > 0, f(u) = or, for each u > 0, f(u) > 0. Also, it follows from (**) that f(0) = f(0 + 0) = [f(0)] 2 and, so, that either f(0) =0 or f(0) = 1. As we have seen [v = 0], if f(0) = then f(u) = for each u > 0--that is, f is the constant function whose domain is the set of nonnegative numbers and whose value is 0. On the other hand, if f(0) = 1 then, as we have seen, either, for each u > 0, f(u) = 0--that is, f is the exponential function with base 0--or, for each u > 0, f(u) > 0. In this last case, since, also, f(0) = 1 > 0, Theorem 222 tells us that f is either an exponential function with posi- tive base restricted to nonnegative arguments or is a queer function. [App. E] [9-329] Summarizing, if ,5y is the set of nonnegative numbers and f satisfies (**) then, either f is an exponential function with nonnegative base restricted to nonnegative arguments, or is the constant with nonnegative argu- ments, or is a queer function. Finally, a portion of the preceding argument suffices to show that if £> is the set of positive numbers and f satisfies (2') then f is one of the functions just mentioned [but restricted to positive arguments]. In a very similar manner, we can derive a theorem about functions which satisfy a generalization of: (3) V V f(uv) = f(u) + f(v). u V As we expect, the "nice" solutions of (3) are logarithm functions. Theorem 223 . If f is a function whose arguments are positive and is such that < 3 '> X V (uv e 3v and f(uv) = f(u) + f(v)) then either f is a subset of a logarithm function [to a positive base different from 1] or f is not monotonic. In the latter case either f is a subset of the constant function or, for each u which belongs to an interval contained in ,5> and for each v Q and each d > 0, there is a u 6 > f such that u. d < u < u + d and v d < f(u) < v Q + d. Proof of Theorem 223 . Since all arguments of f are positive and since, for each u > 0, u = e , the function F for which F(x) = f(e X ) has for its domain {x: 3 x = inu}. [Note that, for each u € &,, f(u) u t I = F(inu). ] So, for any arguments x and y of F there are arguments u [9-330] [App. E] and v of f such that x = inu and y = in v. It follows that x + y=inu+inv = in(uv) and since, by (3')» uve &., it follows that x + y e -3- F . More- over, F(x +y) = f(e X + y ) = f(e X «e y ) ^ = f(e x ) +f(e y ) = F(x) + F(y). (3') Consequently, F satisfies (1'). Note, also, that since the exponential function is increasing, F is monotonic if and only if f is monotonic. If follows from these remarks and the first part of Theorem 221 that if f is monotonic then F is a subset of a homogeneous linear func- tion- -that is, there is a c / such that Yi* v f ( e nu ) = c ^ nu . U t rJr Since c -f 0, it follows that, for each uel, 1/c f(u) = log, u, with b = e ' This completes the proof of the first part of Theorem 223. To prove the second part, we use the second part of Theorem 221. In view of the fact that F is monotonic if and only if f is, this tells us that if f is not monotonic then either F is a subset of the constant func- tion 0--in which case, the same holds for f--or, for each x_ which be- longs to an interval contained in ^^ and for each v and each d' > 0, there is an x e £_ such that F x Q - d' < x < x Q + d' and v Q - d' < F(x) < v Q + d\ To see what this means for f, we begin by noting that, since in is an in- creasing function, if a number u Q belongs to an interval contained in ,?)•, then in u belongs to an interval contained in ,5-p . So, taking x = in u Q and x = in u, we see, in the second case, that, for each u_ which be- longs to an interval contained in &r and for each v Q and each d' > 0, there is a u 6 1 such that in u Q - d' < in u < in u Q + d' and v Q - d' < f (u) < v Q + d' . Now, since the exponential function is continuous at in u Q , given any [App. E] [9-331] d > 0, there is ad' > such that (*) if |inu - inu | < d' then | u - u Q | < d. So, if, given d, we choose for d' a number which is less than or equal to d and satisfies (*) then the second case tells us that there is a u £ y such that u Q -d and each d > 0, there is a u € &t such that u Q - d < u < u Q + d and v Q - d < f(u) < v Q + d. Proof of Theorem 224 . Since both the arguments and the values of f are positive, it follows, as in the proofs of Theorems 222 and 223, that the function F for which F(x) = in(f(e X )) has for its domain {x: 3 x = inu}. [Note that, for each u e & ( , in (f(u)) = F(inu). ] So, as in the proof of Theorem 223, if x and y are arguments of F, so is x + y. Moreover, (4') [9-332] [App. E] F(x + y) = in(f(e X + y )) = in(f(e X 'e y )) = in(f(e X )'f(e y )) = in(f(e X )) +in(f(e y )) = F(x) + F(y). Consequently, F satisfies (1'). Note, also, that since both the expo- nential function and in are increasing, F is monotonic if and only if f is monotonic. It follows from these remarks and the first part of Theorem 221 that if f is monotonic then F is a subset of a homogeneous linear func- tion- -that is, that there isa c /0 such that V in (fle^ U )) = c in u = in (u C ), whence, for each u e ^ f > f(<-0 = u . This completes the proof of the first part of Theorem 224. To prove the second part we use the second part of Theorem 221. In view of the fact that F is monotonic if and only if f is, this tells us that either F is a subset of the constant 0--in which case f is a subset of the constant function l--or, for each x_ which belongs to an interval contained in ,$p and for each y and each d' > 0, there is an x e ,5p such that x - d' < x < x + d' and y - d' < F(x) < y Q + d' . As in the proof of Theorem 223, if u belongs to an interval con- tained in ,5>£ then fnu Q belongs to an interval contained in ,5p . So, if f is not a power function, we know that, for each u_ which belongs to an interval contained in ^ and for each v Q > and each d' > 0, there is a u e &r such that inu Q - d' 0, there is a d x >0 and a d > such that if |in u - in u Q | < d x then | u - u Q | < d and if |in(f(u)) - invj < d a then |f(u) - v Q | < d. [App. E] [9-333] So, if, given d > 0, we take for d' the smaller of d x and d 2 , we know that u Q - d0 V v>0 £(uv) = £(u)f(v) ' one can dispense with the assumption that the values of f are positive. In this case, for each u > 0, Vu e ^ and, by (4), f(u) = f(Vu~Vu~) = [f(Vu~) 2 > 0. So, at any rate, the values of f are nonnegative. Also, if, for some v Q > 0, f(v Q ) = 0, it follows from (4) that, for all u > 0--that is, for all u e <5y- - u \ , / u f(u) = f-f--vl =fhf-£(v o ) = 0. V u \v ,0 / \ o So, if f satisfies (4) then either f is the constant whose domain is the set of positive numbers or each value of f is positive. AN APPLICATION OF THE FOUR THEOREMS One normally compares two numbers by considering either their difference or, in the case of positive numbers, their ratio. So, for a function g, one's standards of comparison for two values g(x) and g(y) are g(x) - g(y) and, if (ft consists of positive numbers, g(x)/g(y). Often when such a function g crops up in applications, it turns out that one of the quantities g(x) - g(y) or g(x)/g(y) depends only on x - y or, if & con- sists of positive numbers, only on x/y. That is, often there will be a function f such that, either **^ V xe£ V ye£ * x " y € ^f and g * x * " g * y) = f * x " y ^' or g g ( *2 ) V xe^ V ya (g(y) > and x - y € ^ f and g(x)/g(y) = f(x-y)), or g g ( *3 ) V xe^ V ve^ (V > and x/y € ^- f and g(x) - g(y) = f(x/y)), or g g (* 4 ) V V (y >0 and g(y) > and x/y e > f and g(x)/g(y) = f (x/y)). g g [9-334] [App. E] The functions g which are encountered in such applications are usu- ally monotonic. In cases (* 1 ) and (* 2 ) their domains are closed with respect to addition and in cases (* ) and (# ) their domains are closed with respect to multiplication. If these supplementary conditions are met then, as is readily shown [see below], the function f is monotonic and satisfies (1'), (2'), (3'), or (4'), respectively. Consequently, in the four cases, respectively, the function g is a subset of (t,) a linear function, or of (t 2 ) the product of a positive constant and an exponential function with positive base different from 1, or of (t 3 ) the sum of a logarithm function [to a positive base different from 1] and a constant function, or of (t 4 ) the product of a positive constant and a power function with nonzero exponent. [For example, in the case of (*), once it has been shown that f is mono- tonic and satisfies (1'), it follows from the first part of Theorem 221 that there is a number c ^ such that [by (*)] for each x € £- and y e^-i g(x) - g(y) = c(x - y). Choosing some argument x_ of g, we see that g Since c ^ 0, it follows that g is the linear function with slope c which con- tains (x Q , g(x Q )) restricted to ^ . ] To prove what has been asserted above, all that remains is to show, fen each of the four cases, that if g is monotonic then so is f, and to show, for each case, under the appropriate hypothesis on £ g , that f satisfies the appropriate one of (l')» (2'), (3'), and (4'). We begin by considering the first two cases, {* x ) and (# 2 ), in which we assume that ,5- is closed with respect to addition. As in all four g ^ cases, we assume that g is monotonic. Without loss of generality we may- -and shall- -assume that fr r = {u: 3 * 3 • _ u = x - y}. With 3 f *• x € £_ y € £~ this assumption it follows that if u and u 2 are arguments of f then u x = x 1 - y 1 and u 2 = x 2 - y 2 , where x x , x 2 , y v and y 2 are arguments [App. E] [9-335] of g. And, since u 1 + u 2 = (x^^ + x 2 ) - (y. +y 2 ), it follows that, since ^ is closed with respect to addition, so is &,. Moreover [with the same g f notation], y x + y 2 e d , \i 1 + y x + y 2 = x x + y 2 e £ g , and u x + u 2 + y x + y 2 = x x + x 2 e i5- . So if, in the case of (*,), * ^" denotes subtraction and, in the case of (* 2 ), '' "& ' denotes division, we have, in either case, f(u 1 + u 2 ) = g^ + u 2 + x x + x 2 ) if g{x x + x 2 ) and f(u 2 ) = g(u 2 + x. x + x^) ix g(x x + x 2 ). Hence, in either case, i{n 1 + u 2 ) * f(u 2 ) = g(u 1 + u 2 + x x + x 2 ) u --that is, suppose that x 2 + y > x + y . We are assuming that g is monotonic. Let's assume, more explicitly, that g is increasing. It follows that g(x 2 + y x ) > g{x. x + y 2 ) and, so, that in the case of (*,) that f(u 2 - u x ) > 0, while in the case of (* 2 ) [in which the values of g are positive] that f(u - u ) > 1, In the first case, as we have proved, f(u 2 ) =f(u 2 - a x ) +f(u x ) So, since f(u 2 - u 1 ) > 0, it follows [for u 2 > u ] that f(u 2 ) > f(u ). Hence, f is increasing and, in particular, is monotonic. In the second case, as we have proved, f(u 2 ) = f(u 2 - uj-fluj. So, since f(u 2 - u x ) > 1 [and since, by (* 2 ), ffu^ > 0], it follows [for u > u ] that f(u ) > f(u ). In this case, also, f is increasing and, hence, is monotonic. The alternative assumption- -that g is decreasing- -leads, [9-336] [App. E] in the same way, to the conclusion that f is monotonic. Consequently, if g is monotonic and > is closed with respect to addition then, in cases (* 1 ) or (* 2 ), f is monotonic and satisfies (1') or (2'), respectively. This, as far as these cases are concerned, is what we set out to prove. The cases (* ) and (* ) are handled by a minor reinterpretation of the preceding argument. In these cases the assumption about £ is that it is closed with respect to multiplication. If, everywhere in the pre- ceding paragraph, we replace '-' by */' an< i '+' by ' * '» the resulting argument shows, first, that in case (* 3 ), f satisfies (3') and, in case (* ), f satisfies (4')» and, second, that, in either case, if g is monotonic then so is f. This is what we needed to prove concerning cases (* 3 ) and (* 4 ). [9-337] BASIC PRINCIPLES AND THEOREMS Commutative principles for addition and multiplication VVx + y = y + x V V xy = yx x y 7 7 x y 7 7 Associative principles for addition and multiplication VVVx+y + z = x + (y+z) VVV xyz = x(yz) xyz xyz 7 7 Distributive principle [for multiplication over addition] VVV (x + y)z = xz + yz xyz 7 7 Principles for and 1 V x + = x V xl = x 1^0 X X Principle of Opposites Principle for Subtraction V x + -x = VVx-y = x+-y x x y 7 7 Principle of Quotients V V x v j. n — y = x x y f y 7 «.<^ «.t„ o* rg* ^,n if. 1. VVV x(y + z) = xy + xz [page 2-60] xyz 2. V lx = x [2-611 x L J 3. V V V, V ax + bx + ex = (a + b + c)x [2-611 x a b c ' L J 4. V x V y V a V b (ax)(by) = (ab)(xy) [2-61] 5. V x V y V a V b (a + x) + (b + y) = (a + b) + (x + y) [2-61] 6 - v x v y v z [x = y=> x + z = y + z 3 £ 2 - 64 3 7 * V x V y V z [x + Z = y+ Z= * X= ^ [2-65] 8 * V x V y V z [x = y=> z +x = z + y] [2-66] [9-338] 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. VVV fz+x = z + v ' == > x = yl x y z L ' ' J V V fx = V == ^> -x = -yl x y l ' ' J VVV fx = y == > xz = yz] x y z l 7 7 J [page VVV [x = y x y z L 7 zx = z y] V V V V [(u = v and x = y) => u + x = v + y] u v x y L 7 7J V V V V [(u = v and u + x = v + y) ^> x = yl u v x y L ' } J V xO = x V V [x + y = => -x = y] x y L ' ' J V x = x x V V -(x + y) = -x + -y x y ' } V V -(x + -y) = y + -x x y ' J V V — (xy) = x • — y x y 3 } V V -(xy) = -xy x y ' ' V V [x = — y => —x = yl X y L / /J V V —x • — y = xy x y ' J V V — xy = x • -y x y ' 7 VVV -x(y + z) = -(xy) + -(xz) x y z ' VVV -x(-y + -z) = xy + xz x y z 7 7 V x • - 1 = -x x V -x = -lx X V V (x + y) + -y = x x y 7 7 VV (x + y)-y = x x y 7 7 2-66] [2-66] [2-66] [2-66] [2-66] [2-66] [2-66] [2-68] [2-69] [2-69] [2-69] [2-69] [2-69] [2-69] [2-70] [2-70] [2-70] [2-70] [2-70] [2-70] [2-71] [2-71] 31. VVV x-yz = x+ -yz x y z 7 J 32. VV x-y + y = x x y * 7 33. V V -(x - y) = y - x x y * 7 34. VVV x + (y-z)=x + y-z x y z 7 J 35. VVV x-(y + z)=x-y-z x y z ' 7 36. VVV x-(y-z)=x-y+z x y z J ' 37. VVV x+(y-z)=x-z + y x y z w / j 38. VVV x(y - z) = xy - xz x y z 7 7 39. VVV (x - y)z = xz - yz x y z 7 7 40. VVV x - (-y -z)=x + y + z x y z 7 41. VVVV x-(y-z-u)=x-y+z + u x y z u 7 7 42. V - x = -x x [9 [page 43. V x - = x x 44. VVV x+z-(y+z)=x-y x y z 7 45. VVV x-z-(y-z)=x-y x y z 7 7 4S * V a V b V c V d < a ~ b ) + ( c ~ d ) = (a + c) - (b + d ) 47. VVV [z + y = x => z = x - yl x y z L 7 7 J 48. V V V /_ [xz = yz ^^ x = yl x y z/= L 7 7J 49. V V y n V [zy = x=> z = -] x y f z L 7 y J 50. V ~ = x x 1 51. V , A - = 1 xfi x 52. V -^V = -x x - 1 -339] 2-71] [2-72] [2-72] [2-73] [2-73] [2-73] [2-73] [2-74] [2-74] [2-74] [2-74] [2-75] [2-75] [2-75] [2-75] [2-75] [2-89] [2-90] [2-91] [2-91] [2-91] [2-91] [9-340] 53. V /^0 x 54. V V , [- = 0=> x = 0] x y p L y J 55. V V [(x ^ and y £ 0) => xy / 0] 56. V V [xy = => (x = or y = 0)] x y l ' j >j •t-fww ww x ± u xv + uy 57. V V / V V # — + — = *- x y^O u v/0 y v yv 58. V V ,„ V V ^ xv ' u y ' v In* v 1 r\ — - ~ x y /= u v/t y v yv X u 59 V V / V V , x y^O u V7=0 y v xu yv" 60. V V xz V / V / — x y/0 z^O y; x 61. V V ,_ V , - x y £0 z/£ y y X -r Z y~Tz 62. V V V ,- SL : ly x y z/t z x z 63. V V ,_ - = x- x ypO y 64. V V ,. 2L = x x y/0 y 65. V ,. V V Xy + XZ = y + z x/ y z x 7 66. V V ,_ V V ,. V ,_ ™ = (x t Z { U xy/^0 uv/tO z^Oyv (y - z)v 67. v v v , n x + x. = *JLy xyz^Oz z z 68. V V ,_ V , V V , JL + JL x y/=0 Z/=0 u v/=0 yz vz 69. V V V ,. - . X x y z/= z z xv + uy yvz 70. V V V V x xv - V / V / V V / - — — ' x y/0 z /= u v^O yz vz 71. V V V , n x+ X = *£+JL x y z ^ z z uy_ yvz [page 2-91] [2-91] [2-91] [2-91] [2-92] [2-92] [2-93] [2-94] [2-95] [2-96] [2-97] [2-97] [2-97] [2-98] [2-99] [2-99] [2-100] [2-100] [2-100] 72. VV ,„ V ,_ x* £ S* x y / z /* z_ y 73. V V x . u x y/^0 u^O v^O y v 74 V / V / -i— = y - x^O y/O x/y x XV yu 75. V V /_ V ,. - -r z = x y f- z/ y x yz 76. V V /. -- x y^O y 77. V V , x y/O x "y 78. V V , n ^± x y/O -y — x y X X y 79. V [x t => -x/ 0] 80. -0=0 k»^ -ju *-,X ,f. ?,v (P L ) V [x / => either x e P or -x € P] (P 2 ) V not both x e P and -x e P (P 3 ) V x V [(xe P and y € P) => x+ y € P] (P 4 ) V x V y [(x€ P and ye P) => xy e P] 81. i P 82. 1 e P (G) V V [y > x <= x y t7 83. V [x > <=> x e P] 84. V V [y > x <=> y - x > 0] x y " 7 J €P] o^ o. »u 'i'* *"j x "i v [9-341 [page 2-101 [2-101 [2-101 [2-101 [2-103 [2-103 [2-103 [7-18 [7-18 [7-22 [7-22 [7-23 [7-24 O* *l^ o* "1" "4^ *V 1 x e P ' J [7-23 [1 > 0] [7-23 OU «X O- [7-30 [7-30 [7-31 [9-342] 85. V [x < <=> -x > 0] x L J [page 7-32] 86 a. V V [x ^ y => (x > y or y > x)] x y b. V V not both x > y and y > x — x y ' c. VV V [(x>y and y > z)=> x > z] — x y z d. VVV [x>y=>x+z>y+z] — x y z L ' ' i e. VVV [(z > and x > y) ==> xz > yz] — x y z u "*\ [7-32] 87. V x f x 88. V V [y > x <=> x t y] x y " — [V V (x = y L x y ' x/y)] [7-33] [7-33] 89. VVV [x + z > y + z x y z v ' => x > y] [7-33] 90. V x + 1 > x x [7-35] 91. V V V V [(x > y and u > v) => x + u > y + v] xyuv 7 7J 92. VVV [(x > y and y > z) => x > z] x y z ~ [7-35] [7-35] 93. V V [(x > y and y > x) => x = y] [7-35] 94. V V [-x > -y <=> y > x] x y *• ' J J [7-35] 95 a. V V V _ n [xz > yz <=> x > y] — x y z > l 7 ,J b. V V V _, _ [xz < yz <=£> x > y] — x y z < l } 7J [7-36] 96 a. V V [xy > <=> ( [x > and y > 0] or [x < and y < 0] ) ] b . V V [xy < <=> ([x > and y < 0] or [x < and y > 0] ) ] [7-36; 97 »• V x/0 X2> ° b. V x V y [x/y=> x* + y*> 2xy] c. V^ A x+i>2 — x> x — [7-38] [7-39] [7-40] [9-343] Jl - ,,2 = y] 98 £■ V x>0 V y>0 [X y b - V V . - [y 2 > x 2 => y > x > -y] — x y > ' ' c V,.V[y>x=>y 2 >x 2 ] — x> y ' ' i 99 a. VVV,„[->^ <=> xz > yz] — xyz^O t z z 7J "\ ) [page 7-38] J b. V / n \ — x/£ l x ( [ 1 > <=> x > 0] and [ - < <=> x < 0] ) [7-41] 100. V,_V,_V_[y> x>0 y /= z > l7 * > *] z y J *1- v»* o^ '4" "V *"l % (V) (I 2 + ) (V) [domain of *m', 'n\ 4 p', and 4 q' is I + ] 1 €l + V n n + 1 e I + V g [( 1 € S and V n [n e S => n + 1 e S] ) = V neS] n J o^ «.i„ +}* *"4 X "l x 'I s [7-41] [7-49] ioi. r c p 102. V V m + n e I* m n 103. V V mneT m n 104. V n > 1 n ~ 105. V V [n > m => n - me T] m n L J 106. V V [n > m + 1 <=> n > ml m n L — J [V n e P] L n J [7-49] [7-56] [7-56] [7-84] [7-84] [7-86] 107 a. V n {■ 1 n b. V V [n < m + 1 — m n l n < m] [7-86] [9-344] 108. Each nonempty set of positive integers has a least member. [V s [0/SC I + =* 3 m £ s V n £ s m < n]] [page 7-88] «.», Jy *t, "V *"l > * "l V (C) V 3 n > x x n ( I ) V [x e I <=> (x e 1* or x = or -x e 1* ) ] ».U O- V»^ V" "l"* 'l x [7-89] [7-94] 109. V [x e I <=> 3 3 x = m - n ] x l m n [domain of V, 'j\ and 'k' is I] 110 a. V. -j e I b. V. V k k + jel £ . V. V k k - j e I d. V. V k kj e I [7-94] > > [7-95] J 111. V. V k [k> j k-jel + ] 112. V. V, [k+ 1 > j <=> k > j] 113. Each nonempty set of integers which has a lower bound has a least member. [7-96] [7-96] [7-98] 114. V. Vg [(j € S and V k> . [keS=>k+ leS])=» V R> . ke S] [7-99] 115. Each nonempty set of integers which has an upper bound [7-100] has a greatest member. 116. V. V s [(je S and V k < . [ke S=> k- 1 €S])=> V R< . k e S] [7-100] [9-345] 117. \/ s [(0 € S and V, [k e S=> (k+ 1 e S and k - 1 e S)])=> V R ke S] [page 7-100) „,„ „,.» ^p V [[x]] = the greatest integer k such that k£x [7-102) V x «xj = x-|[x]| ».i> «a. +t+ -,x *,v *-,* 118 a. V V, [k < ffxj <=£> k < x] mm X K b. V V, [k > ffx]) <==> k > x) c . V V, [k>Ixl <=> k + 1 > x) d. V x V k [k < [xj <=^ k + 1 < x) e. VV, [k = f x] <==> k < x < k + 1] ^~ X K. "\ [7-107) [7-103) [7-104) > [7-105) J 119. v v. |[x + jfl = I[x]) + j x J 120. V V . ft 3, 3 [x = ky + z and < z < y] x y > k z u 7 — /J 121. V V . A 3 ny>x x v> n ' 122. V - j[ ~x ]) = the least integer k such that k > x A. 123. V V x m x m ffxB m (T/ and m 124. V V < x m ~~ ffxl m im = [x] - m m ¥) m ^ m J, O- si/ '•* *i* •v V V. [m I j <=> 3, j = mk] m j l IJ k J J >U o- ^ ^ ^. .*,«. 125. V (1 In and nln) n [7-105) [7-106J [7-106) [7-106) [7-111] [7-111] [7-115 and 7-129] [7-115] m [9-3463 126 a. V V [m|n= t> m < nl — m n L ' J b. V V V f(m|n and nlp)^ 1 ^ mlp] — m n p l • ' ■ J c . V V [(m I n and n | m) => m = n] — m n L ' ' J d. V V V [(mln and mlp)^^ m|n + pl — m n p l ' ,tr ' ri e. V V V [(mln and mln + p) =S > mlp] — m n p l ' ' r ltrj f. V V V [mln 11 ^ mplnp] — m n p L ' riri ) [page 7-115] [7-116] 127. V V 3. 3. HCF(m, n) = mi + nj m n i j J 128. V V V k [(HCF(m, n) = 1 and m|nk) m Ik] [7-122] [7-129] 129. V V [HCF(m, n) = 1 =5 m n l V. V. [mi + nj = <=> 3, (i = nk and j = -mk)]] [7-129] o^ o, ^, *>,x ~f. ^ For each j € I and for each function a -^ whose domain includes {k: k > j}, r j-i i = j k+ l k ^ k> ) a. = } a. + a, , j - i Z_j i Z_i i k+ i i=J i=J > [page 8-36] [An earlier form is on page 8-9] J J, «!L «J, 'J x "V" T 130. For any sequences a and b, n fb, = a, and V b =b+a ) =* > V V \ i i n n+i n n+i/ n / , a = b P n p=l [8-17] [9-347] n 131 p=l n P =i n - 2/ -^ l) 2 ^ V n ^ p2 = n(n+lH2n+l) fi> ^ £ p3 = n 2 ^ p=l p=l [page 8-24] n n 132 * V nZ 1 = * V n Z P = n(n + 1) "A p=l p=l n c. V n Yp(p+ 1) = p=l n n(n + l)(n + 2) 3 d. V n ]Tp(P + D(P+ 2) p=l n(n + l)(n + 2)(n + 3) > [8-24] J 133. V V. V. . Yxa. =xYa. x j k> j - l Z_i i Z_j i i=j x=j k k m - v i v k>j-iZ (a i +b i ) = 2 a i + Z b i 1=J 1=J X=j [8-39] [8-42] k Ji k 135. V. \' . V. Va. =Va. + V a. [8-44] 136. V. V, . . Ya. = a. + V a. J k>j ZL. i J Zli i i = J i = j+i [8-44] [9-348] 137 i=j i=J+J! [page 8-44] n 138. V V(a . -a) = a -a n £_j p+i P n+ l l p=l [8-53] 139 a. V y P = n t« V nl p = n & V nZ P = n ± V nZ P = p- 1) 2 — p-D(p-2) = Hi^iHn^ « iw« >u„ » n(n- l)(n- Z)(n- 3) p - l)(p - 2)(p - 3) - ^ [8-55] "i }[8-56] p- l)(p-Z)(p-3)(p-4) = n(n- l)(n-2)(n- 3)(n - 4) ■V "l N 'C V (Aa) = a , P P P+ l [8-57] O, x^ nU *r "» s *v n- 1 140 • V n a n = a i + y« Aa » f 'l x ~f "I s - [9-349] 141 If a is an AP with common difference d, and, for each > n, s is the sum of its first n terms, then b. V V / — n m^ n m - n a. V a = a, + (n - l)d, — n n x c. V s = %-[2aL,+(n- l)d], d. V a = 5-(a +a ). — nn Z l i 'J' _ nn 2 l n d = * m ~ * n . )fo a s e - 8 ;*7 J and 8-68] 142. V. V, . . . Va. = Ya. ,. . i=J i=J [8-72] 143. V n n n V . a < b ■■ m < n m m z_, a p Z P p=l p=l - [8-76] 144. V V x n n Lp=i 3 < a > * m < n m n J* xU o- 'C- "V 'I- [8-81] For each j € I and for each function a whose domain includes {k: k > j}, Ti a i = ' i = j k+ 1 k v k>j-irf a i = TTv a k +1 V 1=J 1=J K** *i» O* "I" «v "f ■"N / [8-94] J V k« k> * p=l [8-98] +f~ *", O, 'l v "» N T [9-350] i45 - Vk>i-ii i (a i b i ) = / ivTh » = j i=j i=j [page 8-99] 146 - V^j-i^Jh-llv /I a * i = j i = j i = j x +i [8-99] 147 - Vk>il l a i = Y I 1 a i i=j i = j+i [8-99] 148. V. V. V . . . 7~|a. = / I a. . i = j ^j + jj [8-99] i49 - v i v k>j-iTi a i = / i a k +j -i i=j i=j vl^ o^ o, "l* *"l"» "V [8-99] r V x V k>0 X = / ! X p=l x/^0 k< X ~ [8-100] [8-114] J» xt- O, "I - * *V 'l" 150 a. V. 1 = 1 — k [8-103 and 8-115] b. Y k (-l) k+2 = (-l) k c. V k [(-l) 2k = 1 and(>l) 2k+1 = -1] d. 0° = 1 and V n = — n [8-102] [8-103] [8-103] [9-351] 151 a. V V, ._ m k €l + — m k> b. V % . V % n m >k — m> 1 k > [page 8-103] [8-107] 152 i- V x>0 V k xk>0 t- V x/0 V k xk ^° x> 1 k ) [8-103 and 8-115] 153 ' V x V k>0 (x ' 1) P* lsxk - p=l [8-103] 154 V / V x~ k = — 13 ** x/0 V k X k r x [8-114] 155. V , n V.V, x j x k = x j + k x/*0 j k lV x V j>0 V k>0 Ak = xJ + k l (8-117] 156. V , n V. V. ^ =x j " k xpO j k x k [8-118] 157 « V x/o V j V k( xJ ) k = xJk t V x V j>o V k>o( xJ ) k = xJk i t 8 - 118 ! k k k 158 - V x^O V y/0 V kH " X y ' V x V y V k> (**) = ") I 8 " 11 '] k k k k 159 ' V x/0 V y/ !0\(y) = T l V xVo V k>o(7) =Tcl t 8 " 119 ) -k k 16 °- VoVo v k(?) "(J) [8-119] [9-352] 161 . V . n V. V. [x j = x k <*=> (x = 1 or j = k)] x>Ojk t * J [page 8-122] 162 * V x>-l V k>0 (1 +X) - l +kx [8-125] 163 - V x>l V v V nl^ST^T = * xn> vl [8-125] 164. V Jn [- > 1 <==> < x< 1] x/e l x J [8-125] 165 - Vi^>o v n[( 0xno V y>of x ^^ i T I> ^ [8-131] 167. If a is a GP with common ratio r, and, for each n, s n is the sum of its first n terms, then w n - 1 a. v a = a, r , — n n i b. for r £ 0, V n+1 n a = r , n n c. for r js 1. V s = — r n n a^l-r") 1 -r a i- a n r d. for r / 1, V s = ~4 — ^ n n 1-r *\ > [8-130 and 8-133] J [9-353] 168 a. For any GP, a, with common ratio r such that |r| < 1, I p=l a = p 1 - r [pages 8-144 \ and 8-146] b. For each x such that Ox *U Ox ■t* t "r ^ -s n I x l = x and V ^ n l x l = ~ x x > * * x < ' ' [8-145] ».»x Ox Ox '»"» "t" T 169 a. v x v y |x|-|y| = l x vl b. V x V y [|x| < y <*= -y < x < y] c. V x V y |x| - |y| < |x + y| < |x| + |y ^ V [8-145] J 17 °- V k>0 V x V y x k -y k = (x-y)^x k -PyP- 1 p=l [8-160] v jl0 c(j,o) = 1 >Q C(j, k+1) = CU.k)-^ Ox Ox Ox 0! = 1 \>o (k+1)! " k! «(k + 1) Ox Ox Ox [8-168] [8-169] [9-354] m - V j>o v k>o c( J' k) k- 1 i = k! [page 8-169] 172 - V j>o V k>o c( J + k ' k) = (j +k)? j!k! [8-171] 173. V V C(m, n) m n = C(m - 1, n) + C(m - 1, n - 1) [8-172] Two sets have the same number of members if (C )( and only if the members of one set can be matched in a one-to-one way with those of the other. [8-173] (C 2 ) (~ If no two of a family of sets have a common member then the number of members in the union of the sets is the sum of the numbers of members in the individual sets. [8-174] (C.) f If a first event can occur in any of m ways, and, after it has occurred, a second event can occur in any of n ways, then the number of ways in which the two events can occur successively is mn. [8-177] o v k>o p( J- k)= | | { J- iJ i = [page 8-180] 175. The number of permutations of p things, of which p are of a first kind, p 2 of a second kind, . . . , p are of an nth kind, and the remainder are of different kinds, is P? n / rq q=l -\ [8-183] 176. V.. n C. = 2 J [8-185] 177. The number of odd-membered subsets of a nonempty set is the same as the number of its even-membered [8-185] subsets. k k 178. V x V y V.> (x+y) j = ^C(j, k)x*- k y k=0 [8-197] n 179 *• V ra V n T^P- 1, rn - 1 > = C < n » m ) p=l n > \ [8-205] b. V k>Q V n ^C(k + p-l, k) = C(n + k, k+1) p=l [9-356] 180. For any sequence a whose mth difference- sequence is a constant, m a. V n a n =a i + ^ C(n - 1, k)(A k a) x k=l and n m b. V n ^a p = nai + ^C(n, k+l)(A k a) ia p = l k= 1 [page 8-206] 181. Each composite number n has a prime divisor p 2 such that p < n. [8-207] 182. For any sequence n of positive integers, and any prime number p, m V [p| / In. => 3 ^ p|n ]. m r ' / / i q ■'I s Least upper bound principle [iubp]: Each nonempty set [of real numbers] which has an upper bound has a least upper bound. [8-213] [9-32] [ Note . With the adoption of this basic principle, basic principle (C) on page 9-344 becomes a theorem. See page 9-33. ] [9-357] A function f is increasing on a set E if and only if E C * £ and V XieE V x2£E fx s >*, =*£(x 2 ) > «« x )l An increasing function is one which is increasing on its domain. A function f is decreasing on a set E if and only if EC^ f andV Xi€E V x2eE [x 2 >x 1 => f<^) < flx,)]. A decreasing function is one which is decreasing on its domain. A function is monotonic on a set if and only if it is either increasing on the set or decreasing on the set. A function is monotonic if and only if it is either increasing or decreasing. A J [9-35, 9-36, 9-37, and 9-194, 9-195] vU %.** v«^ 'I" 'C 1" 184. Each monotonic function has a monotonic inverse of the same type. [9-37, 9-197] 185. VV . n V . n [x_ >x. =>x" > x. n ] nx>Ox„>0 2 1 2 i J l — [9-37. 9-198] 185'. V V V [x p >x, n x, x„ l 2 1 2n - 1 . 2n- 1 , x_ > x. J [9-54] Jx O* v»* *I N "l" '!"* A function f is continuous at x Q if and only if x Q £ ,5-r and f(x) differs arbitrarily little from f(x ) for each x e ^r which is sufficiently close to x . [ Note . For a more explicit definition, see page 9-211.] [9-42] A function is continuous if and only if it is con" tinuous at each of its arguments. [9-42] a, ..u o*. [9-358] 186. Each positive -integral power function is continuous 187. Each continuous monotonic function f whose domain is a segment a, b has a continuous monotonic in- verse of the same type whose domain is the segment f(a), f(b). ij, ^ «J, T" "J* "f> (PR) V n V x>Q (^^O and (*/x~) n = x) •A. «A- .A, '• v 'l x "l v 188. V V ^ n V [(y > and y n = x) => y = Tx"] n x>0 y lw — ' ' J 189. Each principal positive-integral root function is continuous and increasing on the set of nonnegative numbers. [9-45, 9-218] [9-45, 9-228] [9-49, 9-230] [9-49, 9-230] [9-49, 9-230] 190. a. V "Vo" = — m b. V V . ft n vx"> — m x>0 c v m vT = l — m 191. a. V m V x>0 V n Vx" m v7= n Vx7 nm. b. V V V ™Vx~ = Vx* — m n x>0 — m n x>0 V d. V.V V ^ n ( m Vx") j = YJ — j m x >0 \ ' «.!«. «A- vl^ 'r *v "i- (PR') V V n x 2n-l /x -)2n- ^ J - x A. vU O* ,,v ^,x .y. 188'. V V V [y 2n ~ 1 =x => y = 2n n x y L/ 7 VST] [9-51] [9-52] [9-51] [9-52] [9-55] [9-55] [9-359] 189'. Each principal odd positive- integral root function is continuous and increasing on the set of all real numbers. [9-55] 190'. V V 2n * \T^L = - 2n ~ n x Vx- [9-55] 191'. a. V _ V V n Vx" n v7= m V^ [O = set of odd posi- tive integers] *\ nm. -• V meO V neO 1U11 /— 'n 4 oV^ \i — mn/ — Vx = vx — meO d. V,V V^ rt ( n Vx")J= -\/x J m -' j m e O x ^ > [9-56] nm r- — m f [Note. For n even, V V Vx = ~\/lx| , *■ m x V nm. For n even and m odd, V v x = | vx | . ] J vl^ -J, «J* •v "i" "i* (R) V [x e R <=> 3 xn e I] x l n J [9-62] [domain of 4 r\ 4 s\ and 't' is R] V»- XU *•„ "l* '| N "I* 192. a. V -r € R b. VV r+seR — r s [9-65] [9-63] c. VV r-seR — r s d. V V rs e R — r s e. V V , n - eR — r s ?Q s N J 193. V V vm is irrational unless m is a perfect nth power, n m [9-65] [9-65, 9-231] [9-360] *.** *j~ «j* ■v *r> -v A set is finite if and only if it is or, for some n, its members can be matched [9-2341 in a one-to-one way with those of L J {m: m < n}. A set is infinite if and only if its members can be matched in a one-to-one way with those of one of its proper subsets. [B is a [9-235] proper subset of A if and only if B C A and B^ A. ] A set is countably infinite if and only if its members can be matched in a one-to-one [9-237] way with the positive integers. "I" A first set has a greater number of members than a second set if and only if members of ir . i the first set can be matched in a one-to-one r_ . ,.i way with all those of the second set and, no matter how this is done, there are members ^ of the first set left over. «b> «A» 'i x *T* T" 194. All sets which are countably infinite have the same number of members, and a set which has the same number of members as some countably infinite set is, also, countably infinite. 195. If each of two [disjoint] sets is countably infinite then so is their union. 196. Each subset of a countably infinite set is either finite or countably infinite. [9-237] [9-237] [9-237] [9-361] 197. Each Cartesian product of two countably infinite sets [as well as the Cartesian square of each countably infinite set] is countably infinite. [9-240] 198. The sets I + , I, and R are countably infinite. •J* sU «J«« f '« N 'l N A decimal of the form *0. a x a 2 a . . . represents the real number S. if and only if i is the least upper bound of n {: 4 r = ) a • p = l lo-P}. x r ' [9-67, 9-249] [9-362] x >0 r m, rm el rm [9-72) V' . V x = the least upper bound of {y: 3 ^ y = x } x>lu rr w r < u 7 J ^n ^ ^ i ^ x = ( 1 / x ) J.. O^ .j> 'i v "i-> "r J [9-92, 9-250] 20 2. For each x > 0, the exponential function with base x is a continuous function whose domain is the set of all real numbers; if x ^ 1, its range is the set of all positive numbers; it is decreasing if < x < 1 and increasing if x > 1. [9-93, 9-258] 203. V . n V x u > x >0 u 204 • V ^ n V x =l/x x >0 u ' [V . n V . n x U > 0] x > u > [9-75, 9-261] 205. V , n V V xV=x U+v x >0 u v [V V V x U x V =x U+V l 1 x > u > v > J U / v u - V 206. V _ ft V V x u /x = x x >0 u v ' u.v uv 207. V >n V V (x")= x x >0 u v tV x>0 V u>0 V v>0 (x > =X ] u u u u u u- 208. V Sft V , n V(xy) u =xy [V >n V ^ n V ,Jxy) u =xy] x>0 y >0 u ' 3 x>0 y>0 u>0 ' ' J 209 > V x>0 V y>0 V u(7J =7 21 °- V x>0 V y>o(f)" U =( X J" U ^ V x>0 V y>0 V u>0 \y) = ~u' 211. V s . V V [x U = x V => (x = 1 or u = v)] x >0 u v l J [9-363] v»^ vl, vO "I" 'l N "I" log X < L > V 0O a = * 19-115] ».!.» vl<. vl-» r,«. ^ r ^,v 212. V_^ ,.V..V [a y =x=>y = logx] [9-115] 00 y L 7 6 a J L J 213. The domain of each logarithm function is the set of positive numbers and its range is the set of real numbers. Each such function is continuous and [9-117] monotonic - -decreasing if its base is between and 1 and increasing if its base is greater than 1. 214 ' V 00 V y>0 lo ga (x y )=1 °ga X + lo gay [9-117] 216 ' V 00 V y>0 lo ga(f) =1 °ga X - lo gay [9-118] 211 ' V 00 V u lo ga (xU) =U>1 °ga X t 9 " 118 ] 218 « *• V 00 log a (1 / x)= " lo ga X ^ V 00 lo gb X = lo ga X / log a b [ 9 * 131 ^ 220. For each u, the power function with exponent u fo_ 2691 and positive arguments is continuous. L J [On pages 9-267 through 9-268 it is proved that products, reciprocals and composites of con- tinuous functions are continuous. ] [9-364] [Theorems 221-224 are used in section 9. 10 and are proved in Appendix B, pages 9-313 through 9-336. We give., below, abbreviated statements of these four theorems, ] 221. If f is a monotonic function such that V xe> V ye^ (x + y 6 *f and f{x + y) = f(x) + f(y)) [9-320] then f is a subset of a homogeneous linear function. 222. If f is a monotonic function whose values are posi- tive and is such that V , ic v Kcx < u + ve ^f and f < u + v > = f ( u ) f ( v )) [9-327] U fc njr V £ rJr I then f is a subset of an exponential function with positive base different from 1. 223. If f is a monotonic function whose arguments are positive and is such that V ,^v Kc*. (uv € a f and f(uv) = £(u) + £(v)) [9-330] U fc rtfr V fc rjc J- then f is a subset of a logarithm function. 224. If f is a monotonic function whose arguments and values are positive and is such that V a V \rc^ (UV £ *f aild f(UV) = f ( U ) f < V » [9-332] U fc ,J- , V fc &- I then f is a subset of a power function with nonzero exponent. [9-365] TABLE OF SQUARES AND SQUARE ROOTS n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 n 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 25 6 289 324 361 400 441 484 529 576 625 676 729 784 841 900 961 1024 1089 115 6 1225 1296 1369 1444 1521 1600 1681 1764 1849 1936 2025 2116 2209 2304 2401 1.000 1.414 1.732 2.000 2. 236 2.449 2. 646 2.828 3. 000 3. 162 3.317 3.464 3. 606 3. 742 3.873 4.000 4. 123 4.243 4.359 4.472 4.5 83 4. 690 4.796 4.899 5.000 5.099 5. 196 5.292 5. 385 5.477 5.5 68 5. 657 5.745 5.831 5.916 6.0 00 6.083 6. 164 6. 245 6.325 6.403 6.481 6.557 6. 633 6. 708 6. 782 6.85 6 6.928 7.000 3. 162 4. 472 5.477 6.325 7.071 7.746 8.367 8.944 9.487 10.000 10.488 10.954 1 1 . 40 2 11.832 12. 247 12. 649 13. 038 13.416 13.784 14. 142 14.491 14.832 15. 166 15.492 15.811 16. 125 16.432 16.733 17.029 17. 321 17.607 17.889 18. 166 18.439 18. 708 18.974 19.235 19.494 19.748 20.000 20. 248 20.494 20.73 6 20.976 21.213 21.448 21. 679 21.909 22. 136 2500 7.071 22.361 n 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 n 2601 2704 2809 2916 3025 3136 3249 3364 3481 3600 3721 3844 3969 4096 4225 435 6 4489 4624 4761 4900 5041 5184 5329 5476 5 625 5776 5929 6084 6241 6400 65 61 6724 6889 705 6 7225 7396 75 69 7744 7921 8100 8281 8464 8649 8836 9025 9216 9409 9604 9801 7.141 7. 211 7. 280 7.348 7.416 7.483 7.550 7.616 7.681 7.746 7.810 7.874 7.937 8.000 8.062 8. 124 8. 185 8. 246 8.307 8.367 8.426 8.485 8.544 8. 602 8.660 8.718 8.775 8. 832 8.888 8.944 9.000 9.055 9. HO 9. 165 9.220 9.274 9.327 9.381 9.434 9.487 9.539 9.592 9. 644 9. 695 9.747 9.798 9.849 9.899 9.950 VlOn 22.583 22.804 23.022 23.238 23.452 23.664 23.875 24.083 24.290 24.495 24.698 24.900 25. 100 25.298 25.495 25.690 25.884 26.077 26.268 26.458 26.646 26.833 27.019 27.203 27.386 27.568 27.749 27.928 28. 107 28.284 28.460 28.636 28.810 28.983 29.155 29.326 29.496 29.665 29.833 30.000 30. 166 30.332 30.496 30.659 30.822 30.984 31. 145 31.305 31.464 100 10000 10.000 31.623 [9-366] TABLE OF TRIGONOMETRIC RATIOS Angle sin cos tan Angle sin cos tan 1° .0175 .9998 .0175 46* .7193 .6947 1. 0355 2° .0349 .9994 .0349 47° .7314 .6820 1.0724 3° .05 23 .9986 .05 24 48° . 7431 .6691 1. 1106 4° .0698 .9976 .0699 49° .7547 .6561 1. 1504 5° .0872 .9962 .0875 50° . 7660 .6428 1. 1918 6° . 1045 .9945 . 1051 51° . 7771 .6293 1.2349 7° . 1219 .9925 . 1228 5 2° . 7880 .6157 1.2799 8° . 1392 .9903 . 1405 53° . 7986 .6018 1. 3270 9° . 1564 . 9877 . 1584 54° . 8090 .5878 1.3764 10* . 1736 . 9848 . 1763 55° .8192 .5736 1.4281 11° . 1908 .9816 . 1944 5 6° . 8290 .5592 1.4826 12° . 20 79 .9781 . 2126 5 7° . 8387 .5446 1.5399 13* .2250 .9744 . 2309 5 8 c . 8480 .5 299 1. 6003 14° .2419 .9703 . 2493 59° . 8572 .5150 1. 6643 15° .25 88 .9659 . 2679 60° .8660 .5000 1. 7321 16° . 2756 .9613 . 2867 61° . 8746 .4848 1.8040 17° . 2924 .9563 . 3057 62* . 8829 .4695 1. 8807 18* .3090 .9511 .3249 63° .8910 .4540 1.9626 19° .3256 .9455 . 3443 64° . 8988 .4384 2. 0503 20° . 3420 .9397 .3640 65° .9063 .4226 2. 1445 21° .35 84 .9336 . 3839 66° .9135 .4067 2. 2460 22° .3746 .9272 .4040 67° .9205 .3907 2.3559 23° .3907 . 9205 .4245 68° .9272 .3746 2.4751 24° .4067 .9135 .445 2 69 e .9336 .3584 2. 605 1 25° .4226 .90 63 .4663 70° .9397 .3420 2. 7475 26* .4384 . 8988 .4877 71° . 9455 .3256 2.9042 27° .4540 .8910 .5095 72° .9511 .3090 3.0777 28" .4695 .8829 .5317 73° .9563 . 2924 3.2709 29° .4848 .8746 .5543 74° .9613 .275 6 3.4874 30* .5000 .8660 .5774 75° .9659 .2588 3.7321 3 jo .5150 . 85 72 . 6009 76° . 9703 .2419 4. 0108 32° .5299 . 8480 . 6249 IT .9744 .2250 4.3315 33° .5446 .8387 . 6494 78° .9781 .2079 4. 7046 34° .5592 .8290 . 6745 79° .9816 .1908 5. 1446 35* .5736 .8192 . 7002 80- . 9848 . 1736 5. 6713 3 6" .5878 . 8090 . 7265 81° .9877 . 1564 6. 3138 37° . 6018 .7986 . 7536 82° .9903 . 1392 7. 1154 38° .6157 . 7880 . 7813 83° .9925 .1219 8. 1443 39° .6293 . 7771 . 8098 84° .9945 . 1045 9.5144 40° . 6428 . 7660 .8391 85* .9962 .0872 11.4301 41° . 65 61 . 7547 .8693 86° .9976 .0698 14. 3007 42° .6691 .7431 .9004 87* .9986 .0523 19.0811 43° . 6820 . 7314 . 93 25 88* . 9994 .0349 28.6363 44° . 6947 .7193 .9657 89° .9998 .0175 57. 2900 45* .7071 . 7071 1. 0000 [9-367] Table of pairs (x, y) belonging to the inverse of the exponential function with base 10: {( x, y): I0 y = x} X 1 2 3 4 5 6 7 8 9 TTo .0000 .0043 .0086 .0128 .0170 .0212 .0253 .0294 .0334 .0374 1.1 .0414 .0453 .0492 .0531 .0569 .0607 .0645 .0682 .0719 .0755 1.2 .0792 .0828 .0864 .0899 .0934 .0969 . 1004 . 1038 .1072 . 1106 1.3 .1139 . 1173 . 1206 .1239 . 1271 . 1303 . 1335 . 1367 . 1399 . 1430 1.4 . 1461 . 1492 . 1523 . 1553 . 1584 . 1614 . 1644 . 1673 . 1703 . 1732 1.5 . 1761 . 1790 . 1818 . 1847 . 1875 . 1903 .1931 .1959 . 1987 .2014 1.6 .2041 .2068 .2095 .2122 .2148 .2175 .2201 .2227 .2253 .2279 1.7 .2304 .2330 .2355 .2 380 .2405 .2430 .2455 .2480 .2504 .2529 1.8 .2553 .2577 .2601 .2625 .2648 .2672 .2695 .2718 .2742 .2765 1.9 .2788 .2810 .2833 .2856 .2878 .2900 .2923 .2945 .2967 .2989 2.0 .3010 .3032 . 3054 .3075 .3096 .3118 .3139 .3160 .3181 .3201 2.1 .3222 .3243 .3263 .3284 .3304 .3324 . 3345 .3365 .3385 .3404 2.2 .3424 .3444 .3464 .3483 .3502 .3522 .3541 . 3560 .3579 .3598 2.3 .3617 .3636 .3655 .3674 .3692 .3711 .3729 .3747 .3766 .3784 2.4 .3802 .3820 .3838 .3856 . 3874 .3892 .3909 .3927 .3945 .3962 2.5 .3979 .3997 .4014 .4031 .4048 .4065 .4082 .4099 .4116 .4133 2.6 .4150 .4166 .4183 .4200 .4216 .4232 .4249 .4265 .4281 .4298 2.7 .4314 .4330 .4346 .4362 .4378 .4393 .4409 .4425 .4440 .4456 2.8 .4472 .4487 .4502 .4518 .4533 .4548 .4564 .4579 .4594 .4609 2.9 .4624 .4639 .4654 .4669 .4683 .4698 .4713 .4728 .4742 .4757 3.0 .4771 .4786 .4800 .4814 .4829 .4843 .4857 .4871 .4886 .4900 3. 1 .4914 .4928 .4942 .4955 .4969 .4983 .4997 .5011 .5024 .5038 3.2 .5051 .5065 .5079 .5092 .5105 .5119 .5132 .5145 .5159 .5172 3.3 .5185 .5198 .5211 .5224 .5237 .5250 .5263 .5276 .5289 .5302 3.4 .5315 .5328 .5340 .5353 .5366 .5378 .5391 .5403 .5416 .5428 3.5 .5441 .5453 . 5465 .5478 .5490 .5502 .5514 .5527 .5539 .5551 3.6 .5563 .5575 .5587 .5599 .5611 .5623 .5635 .5647 .5658 .5670 3.7 .5682 .5694 .5705 .5717 .5729 .5740 .5752 . 5763 .5775 .5786 3.8 .5798 .5809 .5821 .5832 .5843 . 5855 .5866 .5877 .5888 .5899 3.9 .5911 .5922 .5933 .5944 .5955 .5966 .5977 .5988 . 5999 .6010 [9-368] 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0 5. 1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 6.0 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 12 3 4 .6021 .6031 .6042 .6053 .6064 .6128 .6138 .6149 .6160 .6170 .6232 .6243 .6253 .6263 .6274 .6335 .6345 .6355 .6365 .6375 .8129 .8195 .8261 .8325 .8388 5 6 7 8 9 6075 .6085 .6096 .6107 .6117 6180 .6191 ,6201 .6212 .6222 6284 .6294 .6304 .6314 .6325 6385 .6395 .6405 .6415 .6425 .6435 .6444 .6454 .6464 .6474 .6484 .6493 .6503 .6513 .6522 .6532 .6542 .6551 .6561 .6571 .6628 .6637 .6646 .6656 .6665 .6721 .6730 .6739 .6749 .6758 .6812 .6821 .6830 .6839 .6848 .6902 .6911 .6920 .6928 .6937 .6990 .6998 .7007 .7016 .7024 .7076 .7084 .7093 .7101 .7110 .7160 .7168 .7177 .7185 .7193 .7243 .7251 .7259 .7267 .7275 .7324 .7332 .7340 .7348 .7356 .7404 .7412 .7419 .7427 .7435 .7482 .7490 .7497 .7505 .7513 .7559 .7566 .7574 .7582 .7589 .7634 .7642 .7649 .7657 .7664 .7709 .7716 .7723 .7731 .7738 .7782 .7789 .7853 .7860 .7924 .7931 .7993 .8000 . 8062 . 8069 8136 8202 8267 8331 8395 .7796 .7803 .7810 .7868 .7875 .7882 .7938 .7945 .7952 .8007 .8014 .8021 .8075 .8082 .8089 .8142 .8149 .8156 .8209 .8215 .8222 .8274 .8280 .8287 .8338 .8344 .8351 .8401 .8407 .8414 .6580 .6590 .6599 .6609 .6618 .6675 .6684 .6693 .6702 .6712 .6767 .6776 .6785 .6794 .6803 .6857 .6866 .6875 .6884 .6893 .6946 .6955 .6964 .6972 .6981 .7033 .7042 .7050 .7059 .7067 .7118 .7126 .7135 .7143 .7152 .7202 .7210 .7218 .7226 .7235 .7284 .7292 .7300 .7308 .7316 .7364 .7372 .7380 .7388 .7396 .7443 .7451 .7459 .7466 .7474 .7520 .7528 .7536 .7543 .7551 .7597 .7604 .7612 .7619 .7627 .7672 .7679 .7686 .7694 .7701 .7745 .7752 .7760 .7767 .7774 .7818 .7825 .7832 .7839 .7846 .7889 .7896 .7903 .7910 .7917 .7959 .7966 .7973 .7980 .7987 .8028 .8035 .8041 .8048 .8055 .8096 .8102 .8109 .8116 .8122 .8162 .8169 .8176 .8182 .8189 .8228 .8235 .8241 .8248 .8254 8299 .8306 .8312 .8319 8370 .8376 .8382 .8293 .8357 .8363 .8420 .8426 .8432 .8439 .8445 1 [9-369] 7.0 .8451 .8457 .8463 .8470 .8476 .8482 .8488 .8494 .8500 . 8506 7. 1 .8513 .8519 .8525 .8531 .8537 .8543 .8549 .8555 .8561 .8567 7.2 .8573 .8579 .8585 .8591 .8597 .8603 .8609 .8615 .8621 . 8627 7.3 .8633 .8639 .8645 .8651 .8657 .8663 .8669 .8675 .8681 .8686 7.4 .8692 .8698 .8704 .8710 .8716 .8722 .8727 .8733 .8739 . 8745 7.5 .8751 . 8756 .8762 .8768 . 8774 .8779 .8785 .8791 .8797 .8802 7.6 .8808 .8814 .8820 . 8825 .8831 . 8837 .8842 .8848 .8854 .8859 7.7 .8865 .8871 .8876 .8882 .8887 .8893 .8899 . 8904 .8910 .8915 7.8 .8921 .8927 .8932 .89 38 .8943 .8949 .8954 .8960 .8965 . 8971 7.9 .8976 .8982 .8987 .8993 .8998 .9004 .9009 .9015 .9020 .9025 8.0 .9031 .9036 .9042 .9047 .9053 .9058 .9063 .9069 .9074 .9079 8.1 .9085 .9090 .9096 .9101 .9106 .9112 .9117 .9122 .9128 .9133 8.2 .9138 .9143 .9149 .9154 .9159 .9165 .9170 .9175 .9180 .9186 8.3 .9191 .9196 .9201 .9206 .9212 .9217 .9222 .9227 .9232 .9238 8.4 .9243 .9248 .9253 .9258 .9263 .9269 .9274 .9279 .9284 .9289 8.5 .9294 .9299 .9 304 .9 309 .9315 .9320 .9325 .9330 .9335 .9340 8.6 .9345 .9350 .9355 .9 360 .9365 .9370 .9375 .9 380 .9385 .9390 8.7 .9395 .9400 .9405 .9410 .9415 .9420 .9425 .9430 .9435 .9440 8.8 .9445 .9450 .9455 .9 460 .9465 .9469 .9474 .9479 .9484 .9489 8.9 .9494 .9499 .9504 .9509 .9513 .9518 .9523 .9528 .9533 .9538 9.0 .9542 .9547 .9552 .9557 .9562 .9566 .9571 .9576 .9581 .9586 9.1 .9590 .9595 .9 600 .9605 .9609 .9614 .9619 .9624 .9628 .9633 9.2 .9638 .9643 .9647 .9652 .9657 .9661 .9666 .9671 .9675 .9680 9.3 .9685 .9689 .9694 .9 699 .9703 .9708 .9713 .9717 .9722 .9727 9.4 .9731 .9736 .9741 .9745 .9750 .9754 .9759 .9763 .9768 .9773 9.5 .9777 .9782 .9786 .9791 .9795 .9800 .9805 .9 809 .9814 .9818 9.6 .9823 .9827 .9832 .9836 .9841 .9845 .9850 .9854 .9859 .9863 9.7 .9868 .9872 .9877 .9881 .9886 .9 890 .9894 .9899 .9903 .9908 9.8 .9912 .9917 .9921 .9926 .99 30 .9934 .9939 .9943 .9948 .9952 9.9 .9956 .9961 .9965 .9969 .9974 .9978 .9983 .9987 .9991 .9996 [H^Ywkiiii&liiluttuRKtll^s