510.7 I^63hl 1956 V.3 pt.l ILLINOIS. UNIVERSITY- COMMITTEE ON SCHOOL MATHEMATICS HIGH SCHOOL MATHEMATICS LIBRARY OF THE UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN 510.7 Ii63hl 1956 pt.l Digitized by the Internet Archive in 2011 with funding from University of Illinois Urbana-Champaign http://www.archive.org/details/highschoolmathem31univ HIGH SCHOOL MATHEMATICS THIRD COURSE TEACHERS' EDITION UNIT ONE 1956 ^fATHEMATICAL INDUCTION University OF Illinois Committee on School Mathematics Copyright 1957 ^by the University of Illinois This material may be quoted at a length of not over 100 words for each quotation in professional journals or reviews without further written permission. Requests for permission to quote under other conditions should be addressed to the University of Illinois Committee on School Mathematics, University High School, 1208 West Springfield, Urbana, Illinois. ^ lu' 1 ii^c.-^\\ )9H V.3 / Unit 1 MATHEMATICAL INDUCTION 1.01 If for one then for the next Followers Properties which are inherited 1.02 Generalizations about counting niimbers The problem of proving a generalization Figurate numbers Recursive definitions 1.03 Mathematical induction The principle of mathematical induction for counting numbers The principle of nnathematical induction for other sets of numbers 1.04 Sums of progressions S-notation More about 2-notation; recursive definition Still more about 2-notation; two theorems Arithmetic progressions Sums of arithmetic progressions Review Exercises Powers with counting niimber exponents UICSM University High School Urbana, Illinois UICSM-2-56, Third Course ,,«^.«;.,..*- :t .^to■:^li■:('Jlj'■ "-v LJ. .;;;; •iXi!,?: ERRATA Unit One THIRD COURSE Page Line 1-5 7 T.C. IIA 7 1-20 16 T.C. 24B 15 has a given property. no algebraic expression of degree n > -whose. expressed by (fd • • • delete the line 3 2 1-29 lb Z 4^ = S <4^) + [^(3)1 x=l x=l 1+1 1-30 5b \.^ -p.^) Ma i] k=p k=p ^ 1-33 6b and for every real integer x > 0. i!--iii'l : r. [1-1] irg ^ irsbi ;J lo Oi- I : '...-.ifijciSii}.','- y- dk ■■ji'.i ri-i .ii-.-J'-^-Tiu :..:! iy .'.' .•■.Mi .-s-uj-.;..> .oi/ii • J.T ■^ V/hen Mr. Jones hears no response to his request "Call out 'absent' if your follower is absent", he knows precisely that the property expressed by (1) is hereditary. From this he knows that if any student is present then so is every student whose name follows this one's name alphabetically. (It may still be the case that Mr. Jones is alone in the assembly hall'.) Hence, in order to know that (2) is true it is sufficient that he check on Bill Aaron. If someone calls out 'absent' then Mr. Jones knows that the property expressed by (1) is not hereditary. He knows then that some- one is present whose follower is absent. If there is only one call, he knows that the absentees form a group of one or more "consecu- tive" students. In fact if there are calls of 'absent' then the number of calls is the number of such groups of absentees. Exercises 1. -2. See discussion above. 3. Bill Aaron is the only absentee. 4. (a) The absentees form a block of consecutive students (starting with Bill Aaron, of course). [In an extreme case there may be no students present. ] (b) Yes. In this case the first ten students would necessarily be among the absentees. (c) Yes. In this case every student would be absent. T. C. IB Unit One, Third Course [1-1] »JCi\ TJ ';,,: ^urg '■i i^To'l; lI^' T. !:-ttc ..■i>. rjo-trli n rn I'i: ijlbi :)drrxy ' ■)ia Oils ■ sjoY ■ ' i :'j ' .1. ifoy t. i' ( I A IK rlyiH UJv.' -a .-fo ■ji, ..('.I. . r . , . , ■ . 'i , ■. ,!■','■ .-^r; .1.; ' •r,d.' ■Vtj Before reading further in the Com:.Tient ^.ry you should read rapidly through Sections 1. 01, 1. 02, and 1. 03. The illustration on page 1-1 embodies many of the ideac v/hich sre introduced later in the unit. You should become fb.rail-'.ar with the new terms in those sections as you read the follov/ing discussion. vi^ ^1.. 'i"> '1^ Mr. Jon3s v/islies to hnow whether every student at Zabranchburg High School has the property expressed by: (1) ... is present today or, in other words, v/hether the generalization: (2) Every student is present today. is true. He does this by introducing the notion of follower which permits him to ask the pertinent question: Is the property expressed by (1) hereditary? That is, is it the ci£.-e, for each student S who has a follower, that if S is present tod?.y then so is his follower? He knows that if the answer to this question is 'yes', and Bill Aaron is present, then (2) is true. One v/ay he might explain his certainty is by saying that if anyone were absent, there would be an alphabetically first absentee; that if Bill Aaron were present then this first absentee would have to be someone's follower: ?.nd that in this case this "someone" would then be present but his follower E.bsent--so the property of being present would not be hereditary. Hence, if Bill Aaron has the property expressed by (1), and this property is hereditary, then every student has the property in question. [As you have juct seen, v/hat is essential for this argximent is that every non-empty subset of the set of students have a member who is not the follower of any member of this subset. As long as 'follower' is defined in such a way that this is the case, every property which is hereditary with respect to this notion of follower, and which holds for every student who is no one's follower, holds for every student.] , ^. , ^ _ ,_, ' •• (continued on T. C. IB) T. C. lA TTnit One. TV-irH r.ovrse [1.01] [1-1] 1.01 If for one then for the next.- -Mr. Jones, principal of Zabranchburg High School, likes to keep track of how many assemblies have perfect attendance. He has an interesting way of finding out whether everyone is present at an assembly. At the beginning of the school year all of the students in school line up in "alphabetical order". Each student in the school (except Dick Zilch who is last in line) memorizes the name of his FOLLOWER (the student directly behind him) and is responsible for reporting the absence of his FOLLOWER throughout the year. Each time the students are seated in the auditorium, Mr. Jones first checks to see that Bill Aaron (who was first in the original "line-up')* is present. Then Mr. Jones instructs the audience: "Call out 'absent' if your FOLLOWER is absent. " No one calls out 'absent'. What can Mr. Jones conclude? EXERCISES A. Answer the following questions about Mr. Jones' way of checking for perfect attendance. 1. Why does Mr. Jones check to see if Bill Aaron is present? Could Bill Aaron be absent and no one call out 'absent'? 2. If after Mr. Jones makes his request, he hears exactly one call, would he be correct in cbnclu^ding that exactly one student is absent? 3. On one assembly day Bill Aaron is absent but his FOLLOWER is present. Mr. Jones notes this fact and then asks each student to call out if his FOLLOV/ER is absent. No one calls out. What can he conclude ? 4. On anoth'er assembly day Bill Aaron is again absent. Mr. Jones notes this fact and then asks each student to call out if his FOLLOWER is absent. (a) No one calls out. What can Mr. Jones conclude? (b) Could ten people be absent and no one call out? (c) Could Dick Zilch be absent and no one call out? UICSM-2-56, Third Course h'.il':>C!Ci-.> 'Sv 1/. A [■•ns 1 aJn'shiJ.Sa ;?••,• :.-jsci£; vino c ; ii i inobii'^- : fqrnBx;) b 11, ■ :i i;J:jj'0..j -Ltj-^bsov/ sxli v"B*ii"' ^livJJjl-: .;:>J5it-. -jci '..iJ bsxij'.'fji O! Hir-jrijii ttio jr- ■ Uim.'i.- :) : LT i • ' ■ ■ ■ , hj.a; :.ii!;;r-£i .bn:: I » -Ifll .:-j .1 ^xercises 1. Mr. Jones' conclusion is colnreci 2. In this case Mr. Jones is vrong. There are 299 counter- examples: student 1 is the only absentee; students 1 and 2 are the only absentees; . . . ; students 1 through 299 are the only absentees. The strongest conclusion that Mr. Jones can draw is that either one of these 299 possibilities is the case or that all students are present. He might, for example, say that if any students are absent then those who are present are precisely those whose numbers follow that of the "last" absentee, and that student 300 is among these. Or he might draw the weaker conclusion that all students with numbers > 300 are present. 3. Correct. In this case, for every student S / 1, S's follower may be defined to be student S - 1, and, if so, the conditions mentioned in the bracket on T. C. lA are satisfied. 4. Mr. Jones is again in error. The situation now is essentially the same as that in Exercise 2. There are now precisely 627 counter- examples. 5. Similar to Exercise 2. 6. Mr. Jones' instructions are not quite complete, but he obviously means they should apply to all students except student 1 and student 628. With this understanding, students 250 and 251 are not "followers" of any students and, to justify his conclusion, it would be necessary to deter- mine that these two are present. In the actual situation there are 250 X 378 counter-examples, and the most that Mr. Jones can conclude is that the absentees, if any , form a block of consecutive students to which one or both of students 250 and 251 belong. T. C. 2A Unit One, Third Course [1.01] [1-2] B. There are other ways in which Mr. Jones could check for perfect attendance. He could line the students up in any way (without bothering to get therri into alphabetical order) and then assign numbers in succession to the students. The first student in line is student 1, the second student is student 2, and so on with, say, student 628 being the last student in line. Each student remennbers his number . Below are described various methods Mr, Jones could use when the students have been numbered as described above. Some of the methods work and some do not. For the methods which do not work, find a counter-example, an example showing that students could be absent when Mr. Jones concludes that they are all present. In such cases, if there is a conclusion which Mr. Jones could make, tell it. 1. Mr. Jones cl.^cks that student 1 is present. Then he gives the instructions: "Ever/ student S except 628 check if student S + 1 is present. If he isn't, call out 'absent'. " Mr. Jones hears no calls so he concludes that every student is present. 2. Mr. Jones chrcks that student 300 is present and gives the same instructions a.T above. There are no calls; he concludes that every student is present. 3. Mr. Jones checks that student 628 is present and then gives the instructions; ''Every student S except 1 check if student S - 1 is present. " There are no calls and he concludes that everyone is present. 4. Mr. Jones gives the came instructions as in Exercise 3 and checks that scudent 1 is present. No calls. He concludes that everyone is present, 5. Mr. Jones gives the sanie instructions as in Exercise 3 and checks that student 475 is present. No calls. He concludes that everyone is present. 6. Mr. Jones gives the instructions: "Every student S, for S > 250, check if student S + 1 is present; every student S, for S < 250, check if student S - 1 is present. " No calls. He concludes that every student is present, (continued on next page) UICSM-2-56, Third Course •j:>j'.abjc .■ 'di rao- -.Ui i;-o;) 'v.rvfi-.-: .•xl.'.'. .T ;•l^^wui■ •b'-tc'. ..V iDL'J-.^ : 1 ;>■//:.:( i ■ i ! . ■.bjjlirJ 1. ..•jsxt : ■•> 1 .'.■irl ;; :'i J ciST i! i : i-. ,Al .0 .T n-.. boiK-; = i>ni -:A . ^I I .im-ifrr /•! ■? /9 -loii shfo/i yj[:'..:?!1c .;■' .ij...- o« - rrtj "to T-nJairj-L • . J. :\.J rsich X'- /d bi r!-v.cxlv, ■1:5 ■!(.. X>f 7. Mr. Jones' conclusion follows from the evidence. vl, vl, -'^ 'i- 'f 'I- Part C 1. Student 628 has no follower; student 1 is no one's follower. 2. Student 1 has no follower; student 628 is no one's follower. 3. Student 252 is the follower of student 251; student 249 is the follower of student 250; student 1 has no follower; student 1 is the follower of student 2; student 628 is the follower of student 627; student 628 has no follower; students 250 and 251 are not followers. 4. Student 3 is the follower of student 1; student 4 is the follower of student 2; student 627 has no follower; the follower of student 626 is student 628, 5. No one. xl, nI^ -j^ 'r 'r 'I- Part D The three preceding parts have explored the method of estab- lishing universal generalizations which was discussed on T. C. lA and IB. As indicated on T. C. lA, such generalizations assert that some property holds for every member of a certain set , i. e. , that every member of this set has the property in question. An instance of a generalization of this type is a sentence which asserts that some named member of the set has the property. For example, 'Bill Aaron is present today' is an instance of 'Every student is present today'. The property in question is that of being present today. We shall also say that the property in question is that which is expressed by the expression "... is present today'. [The circumlocutions 'that of being . . . ' and 'expressed by '. . . ' 'are necessitated by the fact that most of the properties which can be referred to in English (or in any other "natural" language) have no English names. Perhaps the closest one can come to an English name for the property in our example is 'present-today-ness' , If this is accepted then one can say that the property in question is_ present-today-ness. ] T.C. 3A Unit One, Third Course [1.01] [1-3] 7. Mr. Jones chides that students 1 and 2 are present. Then he gives the instructions: "Every student S except 627 and 628 check if student S + Z is present, " No calls. Mr. Jones con- cludes that every student is present. C. In Exercise 1 of Part B each scudent is told to check on his FOLLOWER by looking for the student who has been assigned the next higher number. Student I's FOLLOWER is student 2, student 2's FOLLOWER is student 3, etc. In Exercise 3 the student makes a different interpretation of •FOLLOWER'. In that case student 628's FOLLOWER is student 627, student 627*3 FOLLOWER is student 626, etc. Notice that no student has more than one FOLLOWER. 1. Which student does not have a FOLLOWER in Exercise 1? ■",'..■--■. .1 student is not a FOLLOWER in Exercise 1? 2. Which student does not have a FOLLOV/ER in Exercise 3? Which stndern: is not a FOLLOWER in Exercise 3? 3. Who is the FOLLOWER of student 251 in Exercise 6? Who is the FOLLOWER of student 250? Does student 1 have a FOLLOWER? Does student 2 have a FOLLOWER? Does student 627 have a FOLLOWER? Does studsnt 628 have a FOLLOWER? Which student is not a FOLLOWER? 4. In Exercise 7 who is the FOLLOWER of student 1? Of student 2? Qf student 627? Of student 626 ? 5. For wtioni is it the case in Exercise 6 that he FOLLOWS his FOLLOV/ER? D. Many things can be said ol each of the students at Zabranchburg High School- Take Bob Floogle, for example: Bob is 6 feet tall, has blue eyes, is present v/hen Mr. Jones checks for perfect attendance, does well in mathematics, is co-captain of the football team, lias a hi-fi set, etc. UICSM-2-50, Thiid Course ■■r>di in 4] '•■ v'-'if -;-• - '■■ '■-• ^- -i-u, ;-;j '. ti . . f. i .-. -J 1 LUX : ■ :(H;>Jji il.SVJti ^i>ul. ^i^ :j-ii.: : ^-^ (e) This sentence can be analyzed in several ways. If taken as asserting a property of a thing, the more usual interpretation would be that it asserts that John has the property of loving Joan. An equally correct interpretation would be that it asserts that Joan has the prop- erty of being loved by John. A reason for preferring the first to the second is that if, as has been done here, an English sentence is analyzed as being of subject-predicate form then it is customary to take as subject the name which first occurs in the sentence. With this point of view, the first analysis would be more natural for (e) while the second would be more natural for (f). [A third way of analyzing the sentence in question is that it asserts that John and Joan (in this order) have the relation expressed by "... loves '. You should not go out of your way to head off this analysis but, since we are interested in properties rather than in relations, it should not be stressed. ] (f) See discussion for (e). (g) [This is a quotation from a poenm by Edward Lear, See LAUGHABLE LYRICS (1888). ] The thing is the Pobble who has no toes; the property is that expressed by '. . . swam across the Bristol Channel' . (h) (The operation of) addition of integers; comnnutativity, (i) Truth; that of being beautiful, or beauty. [An alternative rendering of the same sentiment is: Truth has beauty. ] (j) The last dodo; that expressed by "... died before 1628'. (k) (3, 4, 5); that of being a Pythagorean triple or, in jest, Pythagorean triplicity. [You nnay need to review the concept of Pythagorean triples. ] T. C. 4B Unit One, Third Course :fji /.■ J f; ■',');). K to- •a I fVi n..,' ■>'' ••! '^^ ' . ;.i;f.>:jri . ■., ... Exercises 1 and 2 are intended to familiarize the student with the notion of a property. In Exercise 3 we return to the concept of "follower' and the question as to whether a given property is heredi- tary [See page 1-8] with respect to a given definition of 'follower'. 1. One such statement is 'Bob Floogle goes steady with Mary Jones'. The property in question is that of going steady with Mary Jones and is expressed by '. . . goes steady with Mary Jones'. It has no systematically formed name but it might be worthwhile bringing out the point that it, like anything else, might be assigned a name if it were expedient to do so. For example, we might decide to call it 'Bill'. This would of course require a definition, for example: 'Bill' is a name for the property of going steady with Mary Jones, or: 'Bill' is a nanae for the property expressed by '. . . goes steady with Mary Jones'. [But 'Bill' is not an abbreviation for '. . . goes steady with Mary Jones' since the latter is not a name for the property in question. ] So, a student who gives this example might indicate the property in question by saying the following: (a) The property is that of going steady with Mary Jones. (b) The property is expressed by: . . . goes steady with Mary Jones. (c) The property is Bill. [If he gives (c), and this is highly unlikely, then he would have to supply a definition of 'Bill' such as we have given above. ] 2. (a) The thing is George Washington; the property is that of being a president of the United States. t (b) 2; evenness. (c) Donald Duck; that of quacking, or that expressed by '. . . quacks' (d) 2; primeness, or that of being prime, or that expressed by '. . . is prime'. [You may need to review here the concept of prime number. We use this concept later in the unit. A prime number is a counting number other than 1 whose only divisors are itself and 1. ] (continued on T. C. 4B) T. C. 4A Unit One, Third Course [1.01] [1-4] Each of these facts can be stated in a single sentence. For example : (1) Bob Floogle has blue eyes. (2) Bob Floogle is present when Mr. Jones checks for perfect attendance. (3) Bob Floogle is co-captain of the football team. Each of these sentences states that Bob Floogle has a certain property. Sentence (1) states that he has the property BLUE- EYEDNESS. Sentence (Z) states that he has the property expressed by: ... is present when Mr. Jones checks for perfect attendance. Sentence (3) states that he has the property of being co-captain of the football teann. 1. Make up five other statements which might be true for Bob Floogle and indicate in two or three ways what property of Bob Floogle is referred to in each of your statements. 2. Each of the following sentences states that a certain thing has a certain property. What is the thing? What is the property? (a) George Washington was one of the presidents of the U.S. (b) 2 is even. (c) Donald Duck quacks. (d) 2 is prime . (e) John loves Joan. (f) Joan is loved by John. (g) "The Pobble who has no toes swam across the Bristol Channel. " (h) Addition of integers is commutative. (i) Truth is beautiful. (j) The last dodo died before 1628. (k) (3, 4, 5) is a Pythagorean triple. (continued on next page) no- •iiiTOi iJ(j :5] irtT'«/t( In Exercise 3 we bring together in a formal way the notions of set, follower, and property, and , without using the word, the notion of hereditariness. The principal purpose of these exercises is to get the student to state the question which he is told "to ask and to answer". [Another purpose of these exercises is to give the student opportunities to review mathematical ideas from earlier courses.] The student's justification for his answer may be informal. In the case of Sample 1 we have given an informal justification. A more formal justification is that, for every counting niimber n and for every counting number k, if n = 2k + 1 then n + 2 = 2(k + 1) + 1. Hence, if n is odd then so is n + 2. T. C. 5A Unit One, Third Course [1.01] [1-5] 3. In checking for perfect attendance Mr. Jones found that it was helpful to line the students up and if, say, Norman is directly behind James, then to declare that Norman is James" FOLLOWER. You will learn in this unit that such a procedure of "lining things up" and introducing the notion of "FOLLOWER" is often helpful in proving that each of the things in question has a certain property. (In Mr. Jones' case this was the property of being present at that assembly.) In each of the following exercises you are given a set, a definition of 'follower', and some properties. For each of the properties you are to ask and to answer a question of the form : Is it the case that whenever a nnember of the set has this property, then so does its follower ? Sample 1. Set: All of the counting numbers : 3, 2, 3, ... Follower : For every counting number n, n's follower is n + 2. Property: ODDNESS Solution . First, state the question in terms of the given information : Is it the case that whenever a counting number is odd, then the result of adding 2 to it is also odd? You can answer this question from your knowledge of counting numbers. The answer is 'yes'. In other words, if any member of the set has this property, then so does its follower . UICSM-2-56, Third Course 'Vrjtov/ \ 1 + r Suppose — = — were rational. Then 2 X — ^ — would be rational, and then (2 X 2 1 + r )-l would be rational also. But, this is r. So, r would be rational. But, v/e said that r was irrational, and we know that an irrational real number can't be a rational real number. T.C. 6E Third Course, Unit 1 1 '. .'IJ.i f.CK [-■■.-,' J:' • '.:,:-oy.fr-.-::-zi hs — ■ '.i.- ;:n.>- :;; '1 T1'..V.:> i^ ■i.) 1 J,-; No. Each number r such that -lw counter-examples the stu- dents should be able to describe the set of all counter- examples . ] (7) "Is it the case that, for every real number r, if r is irrational (i.e., if r is an irrational real number) then 1 + r — = — is irrational?" This is a good example of a situation where it is simpler to replace a statement by its contrapositive . The statement in question is : 1 + r (i) if r is irrational then — = — is irrational and this is equivalent to its contrapositive: i + r (ii) if — = — is rational then r is rational (since 'not irrational' means rational). Hence, we can reformulate the question as case that, for every real number r, if — = — then r is rational?" "Is it the is rational Now, the answer is obviously *yes ' since, for every real number r, if — =— is rational then so is 2( — = — )-l which is r. (cf, (4).) [It may be desirable to stress the difference between (ii), the contrapositive of (i), and: 1 + r (iii) if — = — is irrational then r is irrational which is the> converse of (i).] Before dealing formally with the concept of contrapositive, you nnight encourage the following informal justification: T.C. 6D (continued on T.C. 6E) Third Course, Unit 1 6] ■■'" {< (r :f.7 i J (3) **Is it the case that, for every real number r, if r = 1 th en Yes, 1 + = 1? 1 +(1) _ = 1, (4) "Is it the case that for every real number r, if r is a rational real number, then 1 + is a rational real num- ber ? " [Here you should review briefly the fact that a subset of the real numbers is isomorphic to the set of rational num- bers. This subset of the real numbers is called the set of rational real numbers. For example, the real number 1 3 •=- corresponds to the rational nunaber •=■ and is a rational real number. The sum or product of two rational real numbers is a rational real number.] Yes. For every rational real niomber r, 1 + r is a ration- 1 + r al real number, and — = — is a rational real number. (5) "Is it the case that, for every real number r, if r < 1 1 + r then i-1^ < 1?" Yes. Informal justification: The midpoint of the interval r, 1 belongs to the interval r, 1. Formal justification: For every real number r, if r < 1 then r + 1 < 2, and — = — < 1 . (6) "Is it the case that, for every real number r, if r < « 1 + r then -4-i < ? ' ' T.C. 6C (continued on T . C. 6D) Third Course, Unit 1 "„.,butb 1''^ .•voUo' :\-' \^: ,dr;nui Ib' x -fii ^" J ,iol"i ■'■ ■'■ '-.rft'i '■J-^ . , .■.■■■' T xr' \^' ..Sfci^l i.itoqb; .,ri: :'''^' vl^ ^1^ vl^ '1^ 'f w (b) Students should discover that for every real number r, the graph (on the number line) of 1 + r (i.e., the graph of r's follower) is the midpoint of the segment l,r. (1) "Is it the case that, for every real number r, if 1 < r < 3 then 1 < ^ ^ ^ < 3?" Yes. An informal justification: If r belongs to the seg- ment 1, 3 then the segment 1, r is a subset of the segment 1, 3, and the midpoint of l,r belongs to 1, 3. More formally: For every real number r, if 1 r 3 1 < r < 3 then •=■<•=•<•=■ and l=i4i) Note that in Sample 2 the counter-example is a number, the num- ber Z. The demonstration that it is a counter-example amounts to proving that 2's follower (the next larger prime) is not even. [Since 2 is the only even prime, there are no other counter- examples.] »l^ 0> vl^ "i* "1^ ■'«* (a) (1) "Is it the case that whenever a counting number is even, the result of adding 12 to it is even?" Yes. See discussion for Scimple 1; 2k + 12 = 2(k + 6). (2) Similar to (1). (3) "Is it the case that whenever a counting number is prime, the result of adding 12 to it is prinne ? " No. Counter-examples: 2, 3, 13, and others. (4) Similar to (1). (5) "Is it the case that whenever a counting number is greater than 5, the result of adding 12 to it is greater than 5?" Yes. For every n, n + 12 > n; and > is a transitive re- lation. (6) "Is it the case that whenever a counting nxomber is a mul- tiple of 5, the result of adding 12 to it is a multiple of 5?" An alternative fornnulation of the question is: "Is it the case that for every counting number n, if there exists a counting niimber k such that n = 5k, then there exists a counting number k ' such that || + 1 2 = 5k ' ? " No. Counter —example : 5. (continued on T.C. 6B) T.C. 6A Third Course, Unit 1 [1.01] [1-6] Sample 2. Set: All of the prime numbers. Follower : For every prime number n, n's follower is the next larger prime number. Property: EVENNESS Solution . Again, you state the question in terms of the given information: Is it the case that whenever a prime number is even, then the next larger prime is also even? You can answer 'no' to this question if you can find a counter-example. That is, you can answer 'no' if you can find an even prime number such that the next larger prime number is not even. The prime number 2 is one such. ( I s it the only one?) (a) Set : All of the counting numbers. Follower : For every n, n's follower is n + 12. Property: (1) EVENNESS (2) of being divisible by 3 (3) PRIMENESS (4) ODDNESS (5) of being greater than 5 (6) of being a multiple of 5 (b) Set : All of the real numbers. 1 + r Follower: For every r, r's follower is- 2 • Property: (1) of being between 1 and 3 (2) of being closer to 1 than 3 is (3) of being 1 (4) of being a rational real number (i.e., ,r. f, . 1 ^, T RATIONALITY) (5) of being less than 1 ' (6) of being a negative number (7) IRRATIONALITY UICSM-2-56, Third bourse »5i^/'7 . rfj'. •iLii ;:-. '-. ,'ji»i.!;i :»^ -: i>'^ 'u: i .1r,r\i ^'ZS'J ^.d! ti al' noiejjiDnoD arrtjsa srfJ nisJdo sw soniS '; . u ,1 £io-?>tt30qYfi is -noi ci>>I-s #n&^3 a i*i j ' tti' Hfptcl So niJ(s:'kt} Since we obtain the same conclusion: 2{y -vS) - 3(x - 2) =7 or 3(x' - 2) - 2(y' - 3) = 9 under either hypothesis 2y -'3x = 7' ) i.r:ai. ' or : a:-^ ' 3x - 2y = 9 -iion ct mv; ixvj:; iir.f-- . "• '-' this- CdrtGlusiOn is a consequence' of the alteirnation (disjunction) df the hypotheses. [Note that since ^we t^'' are considering the uhion of the Ibci, 'we are concern- i •. .. jg'■■ J ru |,C 1 li- ) I - ! i.n -..J-i :>:, '-o No. The set of all counter-examples is the set of all points (x, y) in the first quadrant with x < 2 or y < 3. (2) No. The set of all counter-examples is the set of all points (x, y) in the second quadrant with y < 3. (3) Yes. (4) No. The set of all counter-examples is the set of all points (x, y) in the fourth quadrant with x < 2. (5) Yes. For all real numbers x and y, if 2y - 3x = 7 then 2(y - 3) - 3(x - 2) = 7. [Here is a place to stress again the fact that a point is in the locus of an equation if and only if its coordinates satisfy the equation. ] (6) ''Is it the case that, for all real numbers x and y, if 2y - 3x = 7 or 3x - 2y = 9, then 2(y - 3) - 3(x - 2) = 7 or 3{x - 2) - 2(y - 3) = 9?" Yes. Infornnal justification : The union of the loci in question is a pair of parallel 3 lines each with slope •=- . If the point (x, y) belongs to one of these lines, then the point (x - 2, y - 3) also belongs to that line since the slope of the seg- . . 3 ment (x, y) (x - 2, y - 3) is ■=• . More formal justification: For every x and y, if 2y - 3x = 7 then 2(y - 3) - 3(x - 2) = 7; so, certainly, 2(y - 3) - 3(x - 2) = 7 or 3(x - 2) - 2(y - 3) = 9; if 3x - 2y = 9 then 3(x - 2) - 2(y - 3) = 9; so certainly, 2(y - 3) - 3(x - 2) = 7 or 3(x - 2) - 2(y - 3) = 9 T.C. 7C (continued to T.C. 7D) Third Course, Unit 1 'fr^xh vV O- vl. '■t- "i-' It would be instructive if at sometime during the discussion of Exercise 3 (c) the students discovered the point of view that (c) deals with a transformation, (x, y) -* (x, y +2) , of the coor- dinate plane onto itself, and that each part of (c) is concerned with the question of whether a certain subset of the coordiante plane is mapped ("nnoved") onto itself (or onto a part of it- self) by this transformation. For example, the transforma- tion in question clearly nmaps : the first quadrant onto part of itself; the second quadrant onto part of itself; the third quadrant onto the union of three sets : the third quadrant itself, the "negative half" of the first coordinate axis, and part of the second quadrant; the locus of 'x = 7' onto (the whole of) itself; 2 the curve whose equation is 'y = x ' onto a se- cond curve which does not intersect the first; the set of all points with real integral coordi- nates onto itself. A similar point of view can be adopted in the case of Exer- cise 3(d). (d) (1) "Is it the case that, for all real numbers x and y, if x > , and y > 0. then x - 2 > and y - 3 > 0? " T.C. 7B (continued on T.C. 7C) Third Course, Unit 1 'ir Hi biuoru " (r-, ■■:>Y li • J- J ; . J " ^ -: ■ i (c) [Coordinate plane paper should be used for (c) and (d).] (1) "Is it the case that, for all real numbers x and y, if x > and y > then x > and y + 2 > ? " Yes. For every real number y, if y > then y + 2 > 0. [See (a) (5).] [Compare these problems with the coordinate planes "games" in FIRST COURSE.] (2) Yes. Similar to (1). (3) No. Counter-example; (-3,-1). [Students should be able to describe the set of all counter-examples for this Exer- cise and for (4). ] (4) No. Counter-example: (5,-1). (5) "Is it the case that, for all real numbers x and y, if {x, y) belongs to the locus of 'x = 7' then (x, y + 2) belongs to the locus of 'x = 7'?" Yes. For every real number x, if x = 7 then x = 7. 2 (6) No. Counter-example: (3,9). (3,9) satisfies 'y = x ' but (3, 11) does not. In fact, for all real numbers x and 2 y, if y = X then (x, y) is a counter-example. (7) Yes. For every real nunaber y, if y is a real integer then y + 2 is a real integer. [As in the case of (b) (4), you should review the fact that that subset of the real numbers which is isomorphic to the set of integers is called the set of real integers.] (continued on T. C. 7B) T.C. 7A Third Course, Unit 1 [1.01] [1-7] (c) Set : All the points in the coordinate plane. Follower : For every x and y, the follower of the point (x, y) is the point (x, y + 2). Property : (1) of being in the first quadrant (2) of being in the second quadrant (3) of being in the third quadrant (4) of being in the fourth quadrant (5) of being in the locus of 'x = 7' 2 (6) of being in the locus of 'y = x ' (7) of being a point with real integral coordinates (d) Set : All of the points in the coordinate plane. Follower : For every x and y, (x, y) 's follower is (x - 2, y - 3). Property : ( 1) of being in the first quadrant (2) of being in the second quadrant (3) of being in the third quadrant (4) of being in the fourth quadrant (5) of being in the locus of '2y - 3x = 7' (6) of being in the union of the locus of '2y - 3x = 7' and the locus of '3x - 2y = 9 ' (7) of being in the union of the loci of '4x + 5y = 7' and of '4x + 5y = 9 ' (8) of being in the intersection of the loci of *4x + 5y = 7' and of '4x + 5y = 9' (9) of being in the intersection of the loci of '5(x - 4) + 7(3 - y) = 8x' and of '6x + 14y = 13' (10) of being in the intersection of the loci of '3x - 2y = 9' and of '2y - 3x = 7 ' ( 1 1) of being in the intersection of the loci of 'Bx - 4y = 2' and '2y - 4x + 1 =0" (12) of being in the intersection of the loci of '2y - 3x = 4' and of '3x - 5y = 1' UICSM-2-56, Third Course J.'. ■<;■ d "tB iri ,i!-'in /V :<> ii-.>: -8] !;njj :J.gj.''j;iJ[A ■/.'•■■jti Jii-jfjij;-. ;>n? Line 21: The phrase 'a property which is meaningful for the members of the set' should be interpreted as follows: If 'P' is a name of a property, then P is meaningful for the members of the set if the result of replacing ' . . , ' in: . . . has P by a name for a member of the set is a sentence which "makes sense" i. e. , is well-formed. For example, the property beauty is not meaningful for the counting numbers but is meaningful for physical objects. On the other hand, primeness is meaningful for the counting nvimbers but is not meaningful for people. ^1^ O^ -K '1^ '1^ 'p Exercises Although the exercises here are similar to those on page 1-1, the student now has a framework of ideas on which to base his answers, 1. No, but the absentees, if any, must be consecutive and include Bill Aaron. T.C. 8A Third Course, Unit 1 [1.0.1] [1-8] PROPERTIES WHICH ARE INHERITED Let us return to Mr. Jones and the Zabranchburg High assembly as described in Exercise 3 on page 1-5. The set Mr. Jones deals with is the set of all students enrolled at the school. ' follower ' is defined by lining up the students and declaring that the follower of each student (except the last in line) is the student directly behind him. The property he is interested in is that of being present. When no student calls 'absent', Mr. Jones knows that, in the case of each student, if that student is present then so is his follower. For short, we say in this case that the property of being present is hereditary . [You can guess the meaning of the word 'hereditary' if you know the nneaning of the word 'inherit'. If a property is hereditary, and if Ezra is Milton's follower and Milton has the property in question, then Ezra "inherits" the property from Milton. If you answered the questions in Exercise 3 correctly, then each yes-answer indicated an hereditary property. ] In general, given a set and a definition of 'follower', then a property which is meaningful for the members of the set is hereditary over the set if it is the case that whenever a member has the property in question, then so does its follower (if it has a follower). EXERCISES Recall the situation referred to in the first paragraph of Exercise 3 on page 1-5. Two bits of additional information are that Bill Aaron is first in line and Dick Zilch is last in line. Now answer the following questions assuming that Mr. Jones has just given his familiar instructions: "Call out 'absent' if your follower is absent. " 1. If Mr. Jones hears no response, can he conclude that every student is present? (continued on next page) UICSM-2-56, Third Course X- [1-9] .•.niiCi;i. ■ -li- :■.! ._-!■ Dor.'?. jur'' •■:.>r,3' .'l-e-'ii' bsx-.. . • jIm ; ;i ,s.nxd* ""no -.tc il fU' .D.J Sometimes a convention is made that the "domain of the index" is the desired set and, in this case it is not necessary to refer [< 1 to the set in the quantifie: on page 1-10. With respect to Exercise 3 it is to be understood that the domain of 'n' is the set of all counting numbers. ] vl^ -J^ vU ^,N '1- '|» A generalization such as the boxed one has many other instances besides those we have called 'first', 'second', etc. For example, 'the (5 + 2)nd odd number is 2 • (5 + 2) - 1' is an instance of the boxed statement and is a different sentence from any of those listed. [For one thing, it contains a ' + ' while this symbol does not occur in any of the enxxmerated instances. ] However, the instance in question is equivalent to what we would call the seventh instance of the boxed statement. T. C. 9B Third Course, Unit 1 [1-9] Ir. !i)yn i^'V' '.■tIJJX! M--. '.r^i.-tViC'.r ./•.•/ • ..ua.jffT j;xi'.i. .I-' 2. Yes. 3. All are present. 4. The absentees, if any, form a consecutive group of students, beginning with Bill Aaron, and ending before James Monk. 5. See solution for Exercise 4. 6. Sonne student is present whose follower is absent; the property of being present is not hereditary. xl^ xl, ^1^ We are using the word 'generalization' as short for 'universal generalization'. Logicians call statements like: There is a counting number n such that the nth odd nvimber is 2n - 1. 'existential generalizations'. Note the "tie" between 'existential' and 'There is'. This is the first formal use of the word 'generalization' which the student has encountered in the UICSM program. We intend to use the word in a technical sense. A generalization is a statement. A universal generalization such as we are concerned with here asserts that every member of some set has a given property. A generalization consists of a quantifier [for example, 'for every counting number',] a phrase which expresses the property in question [for exanmple, 'the . . . . th odd number is 2-. . . -1'], and an "index", [for exainple, 'n'] which "links" the quantifier with the blanks in the phrase. The quantifier usually contains reference to a set. [The assertion made by the generaliza- tion is that the property in question is "generalized over the set".] T.C. 9A ( continued on T. C. 9B) Third Course, Unit 1 [1.02] [1-9] 2. If Mr. Jones hears no response, can he conclude that the property of being present is hereditary? 3. If Mr. Jones hears no response and knows that Bill Aaron has the property, what can he conclude? 4. If Mr. Jones hears no response and knows that James Monk has the property, what can he conclude? 5. If Mr. Jones hears no response and knows that Dick Zilch has the property, what can he conclude? 6. If Mr. Jones hears a response, what can he conclude? Can he conclude that the property is hereditary? 1.02 Generalizations about counting numbers. --Given a property which is meaningful for each counting number, a problem which often arises is that of proving that each counting number has this property. A statement which asserts that every counting nunnber has a certain property is a generalization about counting numbers. For example: For every counting number n, the nth odd number is 2n - 1 . To begin with, let us make sure we know what this generalization states. What it states is most easily understood by considering some of its instances . An instance of it is obtained by filling the two blanks in: the . . . th odd number is 2- ... - 1 with a name for a counting number. For convenience, we shall say that its first instance is : the 1st odd nvimber is 2- 1 - 1 that its second instance is: the 2nd odd nvimber is 2 • 2 - 1 that its third instance is: the 3rd odd number is 2*3-1 etc. UICSM-2-56, Third Course qO": ■xrh ■-■ : i~ , ..II. i;- nt'i-iiiJii ii^ixirfi lV.; •J"j: :h -H .5 cio ! .;j .' 4. True 5. False 6. False 7. True 8. True. This generalization is itself the first instance of the generalization: For every m (and) for every n, (n + m) = n + 2nm + m . 9. True. This generalization is itself the 112th instance of the generalization which states that addition of counting numbers is commutative: For every m (and) for every n, m I- n = n + m. 10. True, 11. True. [The first instance is 'the (1 + 4)th even number is 2 • 1 + 8' which is a true sentence. ] 12. True. T.C. lOB Third Course, Unit 1 f ■ - > » > 1 . 1 c '<■■■ "iLVtl. 'itfr: I'liii: -.js! i . i :i!i.i^ ,\ *5i) ■vv/i.f.V .uuiJ hf ai ■: ili; i oija V, ■ :: • . y c: U; j . lofcr Students should become convinced that whether a sentence is a consequence of another (i, e. , can be inferred from it) is independent of the truth or falsity of the latter. Of course, if a sentence is true then so is each of its consequences, and a sentence from which a false sentence can be inferred is itself false. But whether a sentence is a consequence of another is a question of grammar alone (or, as a logician would be nnore likely to say, of syntax ) . The answer to it depends only on the forms of the two sentences. Point out to students that when they were dealing with the True-False exercises in the geometry unit, they established the falsity of some generalizations by discovering counter-examples. A counter-example to a generalization shows that some instance of the generalization is false. Since any instance is a consequence of the generalization, the falsity of any instance shows that the generalization is false. On the other hand, we hope that they recognize that they can not establish any generalization by establishing any niimber of its instances. K»^ ^'y, v'^ '!"• 'f^ 'l'' Exercises, 3. T.C. lOA Examples of instances: 1 is even, 2 is even, 3 is even, etc. These are alternately false and true. Since the generalization has at least one (in fact, infinitely many) false instances, it, itself, is false. 2 • 1 is even, 2 • 2 is even, 2 • 3 is even, 2 • 5285 is even, etc. Each of these instances is true and in fact from the definition of 'even number' we know that every instance of the generalization is true. So the generalization is true, by definition of 'even number'. The generalization is true. Third Course, Unit 1 (continued on T. C. lOB) [1.02] [1-10] Each instance of the generalization can be inferred from the generalization. In particular, if we accept the generalization, we must, logically, accept each of its instances; moreover, if the generalization is true, then each of its instances is true. Of course, there are generalizations some or all of whose instances are false. <' [Examples: (1) Every horse is white. (That is, for every horse h, h is white. ) (2) Every dog has seven legs. (That is, for every dog d, d has seven legs. )] Such generalizations are false. Even so, each instance of a false generalization can be inferred from it. EXERCISES A. Give three instances of the following generalizations. State for each instance whether the instance is true or false. Make a guess about the truth of each generalization. 1. For every counting number n, n is even. 2. For every counting n\imber n, 2n is even. Note: Since it is understood that we are talking about no other numbers than counting numbers in these exercises, we shall abbreviate 'For every counting niamber n' to 'For every n'. 3. For every n, 2n + 1 is odd. 4. For every n, the nth even niimber is 2n. 5. For every n, n > 2. 6. For every n, n > 7. 7. For every n, n ^ 1. 2 2 8. For every n, (n + 1) = n + 2n + 1 . 9. For every n, 112 + n = n + 112. 2 2 10. For every n, (n + 1) - n = 2n + 1. 11. For every n, the (n + 4)theven number is 2n + 8. 12. For every n, the (n + l)th odd number is 2n + 1 . (continued on next page) UICSM-2-56, Third Course i'lcv: . *.>£t ■<■ lu :,ic- >rt^L3.->/ivr: 'i'->..iiJ . a/rj. [1-11] ■^JOi/J ; . I 13. True. 14. True. 15. This is a classic problem frequently used to illustrate the fact that even a large niimber of true instances of a generalization do not constitute evidence that the geueralization is a theorem. Each of the first 40 instances of this generalization is true but the 41st is false. [As a matter of fact there is no algebraic expression whose value for each counting nximber is a prime number. (But don't bother trying to prove thisl )] B. 'i" 'f '(^ 1. For every n, 2n + 3 is an odd number. 2. For every n, 3(n + 2) is the nth multiple of 3. There are other possible answers: For every n, 3(n + 4) is the (n + 2) th multiple of 3. For every n, 3(n + 5) is the (n + 3)th multiple of 3. [Notice that although (a) and (c) are instances of: For every n, 3(n + 8) is the (n + 6)th multiple of 3. (b) is not. Note also that the three correct answers are not equivalent generalizations since they do not have the same instances.] 3. For every n, the (n + l)th^ square number is n + 2n + 1. 4. For every n, n is even. T.C. IIA Third Course, Unit 1 [1.02] [1-11] B. 13. For every n, the nth multiple of 5 is 5n. 14. For every n, the nth multiple of 6 is 6n. 15. For every n, n - n + 41 is a prime number. For each of the sets of statements given below state a generalization of which each statement in the set is an instance. 1, 2 ■ 9 + 3 is an odd number 2 • 1 + 3 is an odd number 2 • S + 3 is a.n odd number 3 • 9 is the 7th multiple of 3 3 • 7 is the 5th multiple of 3 3 • 100 is the 98th multiple of 3 the 3rd square number is 4 + 4 + 1 the 7th square number is 36 + 12 + 1 the 2nd square nujnber is 1 + 2+1 the 100th square number is 9801 + 198 + 1 28 is ever. 1 3 is even 97 is even 104 is even THE PROBLEM OF PROVING A GENERALIZATION In order to prove £. generalization about counting numbers such as the boxed statement of page 1- 9 For every counting number n, the nth odd number is 2n - 1. it is siofficient to show for each n how one can prove its nth instance. Since, for example, the 4th instance of this generalization: the 4th odd number is 2 • 4 - 1 asserts that 4 has the property expressed by: the ... th odd number is 2 1 we need to show for oach a how to prove that n has this property. For any particular counting number, say 4, we could give such a proof as follov/s : UICSM-2-56, Third Coui-se j:;J:.^c: '^txri:-! 'Ji'ri^j ^;r-i It. I !■ T:-fi.-;:f-,. T;:f;;; >■■::> -ti.; ■u • ri:i flru tll'.!'!l 'ihlW [1-12] ij'i Ola i .jiij Line 6: ". . . . we have no way. . . •' This is not quite exact. As in some of the preceding exercises, a generalization may be a rather immediate consequence of a definition. [For example. Exercise 2 on page 1-10. Each instance is a consequence of the definition of 'even number' and is, therefore, a theorem. ] v»^ O- v'^ ''4- 'I"* "i- Lines 18 and 19 : Every counting number has an immediate successor. 1 is the only co\inting number which is not the immediate successor of any counting number. T.C. 12A Third Course, Unit 1 [1.02] [1-12] The first 4 odd numbers are 1, 3, 5, 7, So, 7 is the 4th odd number. Since 7 = 2 • 4 - 1 ; it is the case that the 4th odd nunaber is 2 • 4 - 1, Of course, the same procedure could be used to test any instance of the generalization, but up to now we have no way of being certain that each such test would result in showing that the instance under test was true. Our earlier work suggests that it might be of use to see whether the property in question is hereditary. Before investigating this we must, of course, decide upon a definition of 'follower'. The most cona- monly useful definition when the set in question is the set of all count- ing numbers is, as we shall see: I For every counting nunaber n, n's follower is n + 1. [The follower, according to this definition, of any counting num- ber is often called its immediate successor. Thus 7 is the immediate successor of 6. Is there any counting number which does not have an immediate succeorci ? ] With this definition of 'fonower ' we are led to the definition: A property is hereditary over the set of all count- ing numbers if it is meaningful for each counting number and if, for every counting number k, if k has the property in question then so does k + 1. [Instead of saying that a property is hereditary over the set of all counting numbers ws shall usually say that it is an hereditary pro- perty of counting numbors. ] Now it is easy to prove that the property expressed by: the . . . th odd number is 2 • ... -1 is an hereditary property of counting numbers. To do so we must show that for every counting nuixiber k, if the kth odd number is 2k - 1 then the (k + l;'^i cdd number is 2{k +1) - 1 . UTCSM-2-56, Third Course . \ ilii\^ ■'■ n ! i^iiyh' : : J 1 • ■.-. . .r^^ •?.... :}■><. V.•l■>/■i^: 'V^n/ .\- !..«- ■iii''j-i-::>d iJj X.C; ^^' • li ( M 'J ■ 'Tiuesji T;/ h;j . ■■■rq ■l\ .Aysfji-r-x-. xj.i :.' ■ iO-iSiiliiri-b •• . jiti-.-'.k)';,q -f j;i- ai'cf;-: i'C!-OC;i: OlU ^ ■ f.h-.KClj A^ 1 j. I. Notice that in order to prove that a property is hereditary we have to prove a generalization, in this case: For every k, if the kth odd nvunber is 2k - 1 then the (k + l)th. odd number is (Zk + 1) - 1. We do so by deriving the needed generalization frona other generalizations. Ideally, these other generalizations should be previously proved theorems (or postulates) but in practice we are likely to use as premises generalizations which we may not actually have proved in this course but are sure that we could prove. In the derivation on page 1-13 we have referred (in lines 1 and Z) to the generalization: For every n and k, if n is the kth odd number then n + Z is the (k + 1) th_ odd number. We have not previously proved this theorem but it is a consequence of theorems proved or assumed in earlier courses. You can take the theorem as an assuinption at this point but you should make this fact explicit to your students. [Incidentally, the theorem is a simple consequence of the definitions of 'odd number' and 'even number', and elementary properties of the relation < . ] fj^ -j^ -'^ Vp ^^~. ^,^ Exercises, A. Sample 1. Here the generalization which asserts that the property in question is hereditary is derived from the sentence '> is transitive'. [This sentence is actually an abbreviation of the generalization: For every m, n, and p, if m > n and n > p then m > p. ] T.C. 13A Third Course, Unit 1 [1.02] [1-13] We know, however, that by adding 2 to the kth odd number we shall obtain the (k + l)th odd number. For every counting number k, if the kth odd number is 2k - 1 then for this k, the (k + i)th odd number is (2k - 1) + 2. But, (2k - 1) + 2 = (2k + 2) - 1 = 2(k + 1) - 1. Hence, it is the case that for every counting number k, if the kth odd nunnber is 2k - 1, then the (k + l)th odd number is 2(k + 1) - 1. We have shown that the property in question is an hereditary property of counting numbers. After the following exercises we shall show how this fact is used in proving the generalization which states that every counting number has this property. EXERCISES A. Consider each of the properties expressed below. If it is an hereditary property of counting numbers, prove that it is; if it is not, give a counter-example. Sample 1 . . . . > 5 Solution. If we are to prove that this property is an hereditary property of counting numbers, we must prove: For every counting number k, if k > 5 then k + 1 > 5. Now, for every counting number k, if k > 5 then, since k + 1 > k and since > is transitive, it follows that k + 1 > 5. [Note: It is clear that not all counting numbers have this property. Nonetheless, as we have proved, the property is hereditary. Compare this situation with that described in Exercise 2 on page 1-2. ] Sample 2. . . . < 5 Solution . In order to decide whether this is an hereditary pro- perty of counting numbers, we must consider the statement : UICSM-2-56, Third Course [1-14] Ti %r't .'jlu!/. ;:.ua'-. u: '^. o ri I ■ji>(.:b;,: ■oiJr J::i ■ -Ji' aJ ■■.V t^:i.! 'Ai! Note that the property in question, though hereditary, does not hold for any counting number. [Compare this situation with one in which Mr. Jones hears no calls of 'absent' and correctly concludes that the property (expressed by '. . . is present today') is hereditary even though no student is present . ] Contrast the theorem just established in Exercise 9: The property expressed by: the . . . _th odd number is 2 • . . is hereditary over the set of all counting numbers. with the generalization: For every k, the kth odd number is 2k + 5. which is not a theorem. 10. Hereditary. Proof like that for Exercise 4. + 5 T.C. 14B Third Course, Unit 1 [1-14] W;: '■>,r.i'.>b. a.\ •SS^xiii- I'JiS i BAs >ii.', i.-.-'iii ■•' 'i VsiiJ: ,., t.,:-i ■J - M ... Jn s iii:- i-.qrn-. ; iOij [In doing the exercises of Part A students should state the generalization they are trying to prove. Their work in formulating questions for exercises on pages 1-5 through 1-7 should have prepared them for stating the appropriate generalizations.] 1. Not hereditary. 1 is a counter-example, and so is every odd counting number. 2. Hereditary. For every counting number k, if the kth even number is 2k then the (k + l)th even number is 2k + 2, i. e. 2(k +1). [ See discussion on T. C. 1 3A. ] 3. Not hereditary. Any prime number other than 2 is a counter- example. 4. Hereditary, proof like that on pages 1-12 and 1-13. 5. Hereditary. For every k, if k is a counting number, then k + 1 is a counting number. [Justification: The operation of addition of counting numbers is closed. See Units 1 and 2 of SECOND COURSE, 1955-56.] 6. Not hereditary. 7 is the sole counter-example. 7. Hereditary. Conapare with Sample 1. 8. Not hereditary. Compare with Sample 2. 9. Hereditary. Generalization to be proved: For every k, if kth odd number is 2k + 5 then the (k + l)th odd nunaber is 2{k + 1) + 5. Proof: For every k, if the kth_ odd number is 2k + 5 then, for this k, the (k + l)tji odd number is {2k + 5) + 2, i. o. , 2{k + 1) + 5. (continued on T. C. 14B) T. C. 14A Third Course, Unit 1 [1.02] [1-14] f For every k, if k < 5 then k + 1 < 5. This statement is certainly false since its 4th instance: if 4 < 5 then 4 + 1 < 5 if false. Therefore, in view of the counter-example 4, we know that the property in question is not an hereditary property of counting numbers. 1 . ... is odd 2. the . . . th even number is 2 • ... 3. ... is a prime number 4. the {. . . + l)th odd number is 2 • ... +1 5. ... is a counting number 6. ... =7 7. ... > 1000 8. ... < 1000000 9. the . . . th odd number is 2 • . . . + 5 10. the (. . . + 10)th even number is 2 • ... +20 B. Figurate numbers One of the mathematical pastimes of the ancient Greeks was the discovery of interesting generalizations concerning counting num- bers. For example, consider the following diagram: • 1 • I • • • If we count the dots in each of these "triangles" and in those we would get by continuing the obvious construction, we obtain a sequence of numbers which begins: 1. 3, 6, 10. 15, 21, 28, ... UICSM-2-56, Third Course nuti yjn.; ■LOVfj ■' •(f A) [1-15] ibers , s. A 'r-3'JrriUfi .\>ji.t'rrfjn>u> v"^»''^ '-''^•'i (brus r'.jiiii : JHi/. ./;.; J .■;;: : ;.;.; : jfi.' •J i-i ..tfl'..' ;o rhod I ]ir. other examples of recursive definitions are: |i For evei •y re 1 a al niimber a a and. for every counting number k k a + 1 = k a • a. and: 1 t = 1 and. for every counting nvimber k. {k + 1)1 = k! • (k + 1). Recursive definition is a way of avoiding the use of ' . . . ' and 'etc. '. For example, compare the above recursive definition's with the iJtatcmonts: a • a and: k factors kl = 1 • 2 • . . .-k, both of which leave a good deal to the imagination. Recursive definitions form a natural basis for proof that properties are hereditary (See below ). T.C. 15A Third Course, Unit 1 [1.02] [1-15] The Greeks called the numbers in tlfis sequence triangular numbers, Having obtained these numbers in sequence it is clear what we shall mean by 'the first triangular number', 'the second triangular number', etc. As you can see from the picture the second triangu- lar number is 2 more than the first, the third triangular number is 3 more than the second, the fourth is 4 more than the third, etc. In general, we see that For every counting number k, the (k + l)th triangular number = the kth triangular number + (k + 1). [If, for every n, T is the nth_ triangular nvimber, then the above can be written more simply as: For every counting number k, Tk+l=Tk + (k+l). ] Notice that the boxed statement: ^1 = 1 and. for ever y k. T k+1 = Tk + (k + 1). provides us with a way of computing successive triangular num- bers. For example, T^ = l, T^ = Tj + (2) = 1 + (2) = 3, T3 = T^ + (3) = 3 + (3) = 6. T^ = T3 + (4) = 6 + (4) = 10, etc. [Such a statennent as the boxed one which specifies the first term of a sequence and tells you how, for every k, to compute the (k + l)th term once you know the kth ternn is called a recursive definition . V/e shall make considerable use of recursive defi- nitions in this and the following unit.] UICSM-2-56, Third Course ■f ?>c.h I ri-16] >6ni -T^l.l.f.■ i -t 1. - -..jy: 3. For every k, if Sq, = k then, by the recursive 2 2 definition, Sq , = k + (2k + 1) = (k + 1) . K, T 1 4. (b) For every k, if P, = — — r ' then, k(3k - 1) bv the recursive definition, P, , = — ^ — :: + (3k + 1) ■' k + 1 2 (k + i)(3rk+ n - 1) 2 T. C. 18A Third Course, Unit 1 [1.02] [1-18] You can see that the following is a recursive definition of the sequence : Sqi = 1 and, for every k. S^k+l =Sqk + (Zk + 1). 4. Prove that the property expressed by: is an hereditary property of counting numbers. Still another kind of figurate number is the pentagonal number, The fourth pentagonal number is the number of dots in the following figure: • " X\' • ^ • \ . \ . • \ The recursive definition of the sequence of pentagonal numbers is : Pi = 1 and, for every k. Pk+l=Pk + (3k + 1). (a) Make diagrams like the above corresponding to p., p^, P3' P5' ^"^^ P6- (b) Prove that the property expressed by: , ... (3 • ... - 1) ... Z is an hereditary property of counting numbers. UlCSM-2-56, Third Course -19] . .IstUir- ': The logical status of the principle of mathematical induction depends on how the counting numbers were arrived at. In a pos- tulational treatment the principle might be taken as a postulate. This was done by Peano [See Stabler 's INTRODUCTION TO MATHEMATICAL THOUGHT (Cambridge, Mass.: Addison- W esley Publishing Company, 1953).] On the other hand, if one defines cardinal numbers as equivalence classes of equinumerous sets as did Frege and Russell [See SECOND COURSE, 1955- 1956.], the principle of mathematical induction for counting numbers follows from the definition: 'counting number' is an abbreviation for 'cardinal number which has all the hereditary properties of 1'. Thus, in Peano's system there are two postulates according to which 1 is a counting number and the follower of each count- ing number is a counting number, and the role of the principle of mathematical induction (a third postulate) is to assert that there are no counting numbers other than those whose existence is ensured by the first two postulates. On the other hand, in the Frege -Russell development the principle of mathematical induc- tion serves to single out the counting numbers from the set of all cardinals . In the Commentary for page 1-22, we describe a postula- tional basis for the counting numbers, different from Peano's, with respect to which the principle of mathematical induction is a theorem. T. C. 19C Third Course, Unit 1 ■19] 'For every n, the nth odd nvimber is 2n - 1', and the set of all its instances, for example, the set whose members are 'the 1st odd number is 2 • 1-1', 'the 2nd odd number is 2 • 2-1', etc. The nature of this gap can perhaps be seen more clearly in a simpler situation in which the generalization covers only a finite number of instances. Suppose, whether you know it or not, that there are only three boys, say John, Charles, and Henry, in a room. If you notice that John has red hair, Charles has red hair, and Henry has red hair and if you also observe that there are no other boys in the room , then you are entitled to as- sert that every boy in the room has red hair. If you don't make the last observation, then all that you are justified in inferring is that John, Charles, and Henry have red hair. Similarly, if you could somehow prove separately each instance of 'For every n, the nth odd number is 2n - 1', you would still only be justified in claiming to have proved 'the nth odd number is 2n - 1 when n = 1, 2, 3, . , . '. Clearly, it is impossible to write the infinitely long "sentence" suggested by the dots. But even if it could be written, it would still be the case that before you could assert 'For every n, the nt^i odd number is 2n - 1', you would also have to prove 'Every counting number is either 1 or 2 or 3 or . . . '. The principle of mathematical induction for counting nxim- bers, by implying that every counting number > 1 can be obtained from 1 by successive additions of 1, has the effect of ensuring that every counting number is either 1 or 1 + 1 or 1 + 1 + 1 or... . As in the case of the red-headed boys in the room, aside from establishing each instance of the generalization, you must also establish that these are all the instances. vl^ O, vl^ '1^ 'p '1^ (continued on T . C. 19C) T. C. 19B Third Course, Unit 1 -19] icvni stress the last paragraph on page 1-19. A proof by mathematical induction is not an infinite sequence of syllogisms: 1 has the property. If 1 has the property then 2 has the property . .'. 2 has the property. If Z has the property then 3 has the property . ." • 3 has the property. etc. A proof must be of finite length. However, even if we allowed "proofs" of infinite length and so admitted as a proof such an infinite sequence of syllogisms, we would still need a principle to the effect that each instance of the generalization: For every x, x has the property is equivalent to the conclusion of one of these syllogiams. The princi- ple of mathenaatical induction for counting numbers is, essentially, such a principle. In fact, the principle of mathematical induction for counting numbers allows us to infer the generalization fronn the pre- mises (i) the property is hereditary, and (ii) 1 has the property and, hence, obviates the necessity for infinite proofs. The principle of mathematical induction is not merely a handy gadget which allows one to by-pass the impossible task of stating a proof for each of infinitely many cases, or to abbreviate the practi- cally impossible task of, say, proving step-by-step that the lOOOtJi odd number is 2 • 1000 - 1. Rather, its importance is in bridging the gap between a generalization about counting numbers, for example, (continued on T. C. 19B) T. C. 19A Third Course, Unit 1 [1.03] [1-19] 1.03 Ivlathernatical Induction . -- Mr, Jones (see Exercises on page 1-1) knew that if the property of being present was hereditary (with respect to the notion of 'follower' he had defined) and if Bill Aaron was present, then every student had the property of being present. Perhaps -^e, having proved: (i) the property expressed by: '(-■/ the . . . th odd number is 2 • ... - 1 is an hereditary property of counting numbers can, if we also prove: (ii) 1 has the property expressed by {^) conclude tl -.it every coua;. -~ raur..- sr has /.lic property. That is, perhaps we can, by using these facts, establish the boxed statement on page 1-9: For every counting number n, the nth odd number is 2n - i . [Before continuing you should check that 1 does have the property ex- pressed by (■^). ] Now, it is easy to see that, becav.se of (i) and (ii) alone, 2, 3,4, etc. must have the property expressed by (t^). For, by (i), if 1 has this property, then so does its follower 1 + 1, But by (ii), 1 does have this property, so 1 + 1 also has it. Since 1 + 1 = 2, 2 has the property. Again by (i), if 2 has the property then so dees 2+1. But we have just shown that 2 has the property, so ?. -r 1 also has it. Since 2+1 = 3, 3 has the property. Again by (i), if 3 has the property then so does 3+1. But we have just shown that 3 has the property, so 3 + 1 also has it. Since 3 + 1=4, 4 has the property. Etc. Evidently (i) and (ii) nnaks it possible for us to show that every nxun- ber which we can obtain from 1 by a finite number of steps, each of which consists in adding 1 to the number obtained in the preceding step, has the property in question. Now, it is a fundamental property of the set of all counting numbers that each of its members can be obtained by such a step -by-step process. Consequently the following principle is a theorem : UICSM-2-56, Third Course ■20] The property in question is that expressed by: the (. • . + 10)th even number is 2 • ... +20. (i) The property is hereditary . For every k, if the (k + 10)th even number is 2k + 20 then the [(k + 1) + 10]th, or the (k + ll)th even number is (2k + 20) + 2, or 2(k + 1) + 20. (ii) 1 has the property . The (1 + I0)th, or the Uth even number is 2 • 1 + 20, or 22. Therefore, in view of (i) and (ii), by the principle of mathe- matical induction for counting numbers, the property holds for every counting number. T. C. 20C Third Course, Unit 1 •20] (i) The property is hereditary . For every k, if the (k + l)th odd number is 2k + 1 then the [(k + 1) + l]th or (k + 2)th odd number is (2k + 1) + 2, or 2(k + 1) + 1. (ii) 1 has the property . The (1 + l)th, or the 2 nd odd number is 2 • 1+1, or 3. Therefore, in view of (i) and (ii), by the principle of mathe- matical induction for counting numbers, the property holds for every counting number. 3. The property in question is that expressed by: ... is a counting nximber. (i) The property is hereditary . For every k, if k is a counting number then, since addition over the counting numbers is closed, k + 1 is a count- ing number . (ii) 1 has the property . 1 is a counting number. Therefore, in view of (i) and (ii), by the principle of mathe- matical induction for counting numbers, the property holds for every counting number, that is, every counting number is a counting nunnber! [This generalization is, of course, trivial and could be proved without the principle of mathe- matical induction. It is included for comic relief.] (continued on T. C. 20 C) T. C. 20B Third Course, Unit 1 •20] Note, in the resume, that >we conclude the proof by citing the principle of mathematical induction. Although it is not customary in proofs to cite the justification of each step, we believe that this should be done for any step whose justification is the principle of mathenaatical induction. Our reason for this insistence is the wide- spread misconception of inductive proofs as consisting of infinitely many syllogisms [See T. C. 19A ff. ]. We hope that by doing this consistently students will be reminded of the distinction. O^ vl, ~.l^ 'C- 'f 't* (i) The property in question in each of these exercises was shown to be hereditary in an exercise on page 1-14. We repeat the proofs here in order to exhibit the style of proofs by mathema- tical induction. 1. The property in question is that expressed by: the . . .t^i even number is 2 • . . . The property is hereditary . For every k, if the kth even number is 2k then the (k + l)th even number is 2k + 2 = 2(k + 1). 1 has the property . The first even number is 2 • 1, or 2. Therefore, in view of (i) and (ii), by [the definition of 'even number', and the distributive principle, and] the principle of mathematical induction for counting numbers, the proper- ty holds for every counting number. The property in question is that expressed by: the (. . . + l)th odd number is 2 • ... +1. (ii) 2. (continued on T. C. 20B) T. C. 20A Third Course, Unit 1 [1.03] [1-20] Every hereditary property of counting numbers which holds for 1 holds for every counting number. This principle is called the principle of mathematical induction for counting numbers . It is important! By virtue of this principle we can conclude from (i) and (ii) that every counting niomber has the property expressed by(ij^J , that is, that For every counting nximber n, the nth odd nvimber is 2n - 1 . We give nov/ a resume of the proof of this generalization: (i) The property expressed by: {H) the . . . th odd number is 2 • ... - 1 is hereditary. For, for every k, if the ktji odd number is 2k - 1 then the (k + l)th odd number is (2k - 1) + 2 = 2{k + 1) - 1. (ii) The number 1 has the property expressed by (1). For, the first odd number is 1 and 2 • 1-1 = 1. Hence, by the principle of mathematical induction for counting numbers, every counting number has the property expressed by {it), that is the generalization : For every counting number n, the nth odd number is 2n - 1. is a theorem. EXERCISES A. Use the principle of mathematical induction for counting numbers to prove each of the following generalizations. Whenever possible, make use of what you have already done in answer to the exercises on page 1 -14. 1. For every n, the nth even number is 2n. 2. For every n, the (n + Ijt^i odd number is 2n + 1. 3. For every n, n is a counting number. 4. For every n, the (n + 10)th even number is 2n + 20. UICSM-2-26, Third Course iu: . '-^ri^v ■ r'f'f.f-y-- > .'o't^ .*"'; -'iM ri-21] t . 10. Property is that expressed by: (i) The property is hereditary . Suppose that, for a given k, k + 1 Then, for this k, S 1 then k + 1 > 1 . (ii) 1 has the property . 1 > 1 because 1=1. Therefore, in view of (i) and (ii), by the principle of mathe- matical induction for counting numbers, the property holds for every counting number. Property is that expressed by: (...) + ... is even. (i) The property is hereditary . 2 Suppose that, for a given k, k + k is even. Then, for this k, (k + 1)^+ (k + 1) = k^ + 2k + 1 + k + 1 = (k^ + k) + 2{k + 1). 2 Since, by hypothesis, k + k is even, and since, by definition, 2(k + 1) is even, and since the sum of two 2 even numbers is even, we know that (k + k) + 2{k + 1), 2 or (k + 1) + (k + 1) is even. 2 Hence, for every k, if k + k is even then 2 (k + 1) + (k + 1) is even. (ii) 1 has the property . 1^+1 = 2 Therefore, in view of (i) and (ii) , by the principle of mathe- matical induction for counting numbers, the property holds for every counting number. (continued on T . C. 21E) T. C. 21D Third Course, Unit 1 ri-21] For example, in order to prove: if 1 + 2 = 2+1 then 1 + (2 + 1) = (2 + 1) + 1 one might proceed as follows: From the associative principle 1 + (2 + 1) = (1 + 2) + 1. If 1 + 2 = 2+1 then ( 1 + 2) + 1 = (2 + 1) + 1. Hence, if 1+2 = 2 + 1 then 1 + (2+ 1) = (2 + 1) + 1. The foregoing proof would be equally valid if, throughout it, the numeral '2' were replaced by any other numeral, for example, '7' or '27'. The variation essentially consists in exhibiting the form of all proofs so obtained. In the proof the symbol 'k' serves as a parameter; if one replaces 'k' by the appropriate numeral [and omits such explanatory phrases as 'for this k'] one obtains a' proof of any instance of the generalization. vl^ vl^ vU '1^ 'i- 'I" 3. The proof is similar to that given for Exercise 1. 4. Property is that expressed by: ... > 1. (i) The property is hereditary . Suppose that., for a given k, k > 1. Then, for this k, k + 1 > k Hence, since > is transitive: if k > 1 then k + 1 >_ 1, (continued on T . C. 2 ID) T. C. 21C Third Course, Unit 1 ri-21] Hence, for every k, if 1 + k = k + 1 then 1 + (k + 1) = (k + 1) + 1. (ii) 1 has the property . 1 + 1 = 1 + 1 (=is reflexive) Therefore, in view of (i) and (ii), by the principle of mathe- matical induction for counting numbers, the property holds for every counting number. vl^ -.1^ vU '1^ '1^ 'r The foregoing proof of (i) is a variation of the kind of proof we have illustrated in connection with Exercise 1. Let students com- pare this variation with the following: For every k, if 1 + k = k + 1 then since, by the associative principle, 1 + (k + 1) = (1 + k) + 1, 1 + (k + 1) = (k + 1) + 1. In connection with the variation it is important to note that the in- ductive hypothesis does not assert that, for every k, l+k = k+l. This is what we are trying to prove! The phrase 'Suppose that, for a given k', gives notice that we shall not justify a step in the proof by inferring from the inductive hypothesis one of its substitution instances [e.g., we shall not infer ' 1 + (k + 1) = (k + 1) + 1 ' from the inductive hypothesis by substituting 'k + 1 ' for 'k']. To get students to understand this variation, let thenn consider the problem of proving an instance of the generalization: For every k, if 1 + k = k + 1 then 1 + (k + 1) = (k + 1) + 1 . (continued on T . C. 21C) T. C. 21B Third Course, Unit 1 ri-21] 1. Property is that expressed by: the . . .th multiple of 5 is 5 • . . . (i) The property is hereditary . For every k, if the kth multiple of 5 is 5k then the (k + l)th multiple of 5 is, by the recursive definition, (the kth multiple of 5) + 5, or 5k + 5, or 5(k + 1). (ii) 1 has the property . By the recursive definition, the first multiple of 5 is 5. Hence, the first multiple of 5 is 5 • 1. Therefore, in view of (i) and (ii), by the principle of nnathe- matical induction for counting numbers, the property holds for every counting number. 2, Property is that expressed by: 1 + . . . = . . . + 1. (i) The property is hereditary . Suppose that, for a given k, 1 + k = k + 1. [The foregoing supposition '1 + k = k + 1' is often called the inductive hypothesis . Students should be taught this term.] Then, for this k, 1 + (k + 1) = ( 1 + k) + 1 [associative principle] = (k + 1) + 1 [inductive hypothesis] (continued on T. C. 21B) T. C. 21A Third Course, Unit 1 [1.03] [1-21] B. Use the principle of mathematical induction for counting numbers to prove the following generalizations. 1. For every n, the nth multiple of 5 is 5n. [Hint: Use the recursive definition: The 1st multiple of 5 is 5 and, for every k, the (k + l)th multiple of 5 is (the kth multiple of 5) + 5.] 2. For every n, l+n = n+l. [Of course, you are expected not to use the commutative prin- ciple for addition of counting numbers in your proof.] 3. For every n, the nth multiple of 6 is 6n. 4. For every n, n > 1. 5. For every n, n + n is even. 6. For every n, the nt^i triangular number is . [See the recursive definition on page 1-15.] 2 7. For every n, the nth square number is n . 8. For every n, the nth pentagonal n\imber is -^ — = ■ . 1S9. For every n, Sq^^^ = T^^^ + T^. ^10. For every n, P , , = Sq_ , , + T . •* n+i Ti+i n THE PRINCIPLE OF MATHEiMATICAL INDUCTION FOR OTHER SETS OF NUMBERS You have seen that the reason that the principle of mathematical induction for counting numbers holds is that each counting number can be reached by starting at the first counting number (1) and proceeding through a finite number of steps each of which consists in passing from the number reached by the preceding step to its follower (its immediate successor). There are many other sets of numbers for which one can define 'first niember' and 'follower' in such a way that each member can be reached from the first member in the set through such a step- by-step process. In fact, one can do this for each set which can be "ordered (by some relation) in the same way" in which the set of count- ing numbers is ordered by the relation < . UlCSM-2-56, Third Course i If : , n J '2] ' >J.^: ■!■ If A C K, and (a) 'k 6 A, and (b) for every x €K, if X / % and X € A then + X e A, th en A = K This principle of finite induction holds for every simply or- dered finite set. It can be proved, using (I ), (II ), (III ), the denial of (IV ), and (V ) in just the same way as the principle of mathematical induction is proved from (I )-(V ). It is sufficient to remark, at the appropriate point of the proof, that {i r) / gr^ j\ is, and so, as required in the proof, has an immediate successor. T. C. 22P Third Course, Unit 1 ly. 1 1 72] -'■■' rii :;[ ".' '■ ) Pictorially, with 'R' corresponding to 'is to the left of: 1, 3, 5, . . . ; 2, 4, 6 4. It is clear that the relation in question satisfies postulates (I )- {IV ) but does not satisfy postulate (V ), since 2 has no immedi- ate predecessor. Also, the usual form, of the principle of mathe- matical induction is not valid here, but that if A C K and (a) 1 e A and 2 e A, and (b) for every x e C, if x e A then x e A, then A = C. [Here, of course, for every xeC, x =x+2.] In the preceding, we have merely touched on the theory of simply ordered sets. A convenient source for further material is Huntington's THE CONTINUUM (Cambridge, Mass. : Harvard University Press, 1942). The examples used at the beginning of this unit to introduce the notion of mathematical induction dealt with finite ordered sets, and a few words concerning these are in order. If postulate (IV ) is not satisfied [but (I )-{IIl ) and (V ) are] then there is an x e K such that, for every y e K, y R x or y = x. [The proof of this, in particular the proof that if it is not the case that x R y then either y R X or y = X, depends on the connectedness of R which we have said follows from (i^) and (V V ] Thus K has a "greatest number" and, using (I ) it is easily shown that there can be only one such. If we denote the greatest member of K by 'gj^' then it is easy to prove, under the specified assunnptions concerning K and R, that (continued on T. C. 22P) T. C. 220 Third Course, Unit 1 •2] is necessarily asymmetric. Such a relation is called an irreflexive order relation . [A good example of an irreflexive order relation is that expressed by '. . . is supported by ... ' and defined over the parts (trunk, boughs, branches, twigs) of a tree.] A relation R defined over K is said to be connected if, for every x and y in K, if x / y then either x R y or y R x. [The example just referred to is not connected (except in the case of an extremely one-sided tree).] A connected irreflexive order relation is called a simple (or some- times a linear) order relation. It is not difficult to show that a relation R which satisfies pos- tulates (I ) and (II ) is necessarily a simple order relation. Con- sequently, every progression is a simply ordered set. Postulate {II ), however, says much more. It is equivalent, for a simply ordered set, to the statement that no subset of K is orde ed by R in the same way that the set of negative integers is ordered by < . [Other ways of saying this are (1) that no subset of K is ordered by R as a regression , or (Z) no subset of K is ordered by the con - verse of R as a progression.] Thus, for example, the set of inte- gers (negative, zero, and positive) is simply ordered by <, but is not so ordered as a progression. Another example of a simply ordered set which is not a progression is given by the peculiar ordering of the counting numbers suggested in Part B on page 1-25. A relation which satisfies postulates (I ) and (II ) is called a well-ordering relation. An example of a well -ordering of C which is not a progression is given by the relation R such that: X R y if (1) X and y are both odd or both even, and X < y, or (Z) X is odd and y is even. (continued on T . C. 2ZO) T. C. Z2N Third Course, Unit 1 '•2] These serve as a recursive definition of the relation < . [The word 'must' above should be qualified, but a discussion of how and why would take us too far into logical theory. ] Neither of the above systems of postulates is adequate for arithmetic. [This also should be qualified but, as in the case of 'must', we cannot do it here.] Either of them becomes so if we adjoin postulates which define recursively the operations of addi- tion and multiplication. In the case of the Peano postulates these would be: (VII ) For every x, if x e K then x + 1 = x ''. (VII , ) For every x and y, if x € K and y € K then x;+ 7^= (x + y/. (VIII ) For every x, if x e K then x • 1 = x. (VIII , ) For every x and y, if x e K and y € K then x • y' = X • y + X. These recursive definitions form bases for inductive proofs of the usual theorems of arithraetic. ■j^ o^ o* 'C 'l- 'I- A better understanding of the nature of a progression can be obtained from a more detailed investigation of possible kinds of relations. A relation R defined over a set K is said to be transi - tive if, for every x, y and z in K, if x R y and y R z then x R z. A relation which satisfies postulate (I ) is said to be asymnmetric. A relation which is asymmetric has the further property of being irreflexive : for every x in K, it is not the case that x R x. It is readily seen that a relation which is both irreflexive and transitive (continued on T. C. 22N) T. C. 22M Third Course, Unit 1 zz] for every counting number n, the set of all counting numbers '. [*< ' and '> ' are, of course, ambiguous since they are used in various contexts to refer to a relation defined for counting numbers, another defined for real numbers, etc., and these are, strictly, different relations.] Non-numerical interpretations are a little difficult to find since, for exannple, it is doubtful that there are infinitely many physical objects. Availing oneself of some degree of poetic license, one can, however, imagine an infinite row of people for which the relation expressed by *. . . is to the right of ..." satisfies the above postulates. Another interesting mathe- matical interpretation consists of the set whose members are, (continued on T . C. 22Li) T. C. 22K Third Course, Unit 1 -■.'jij-i ,,)■:. r ! r i2] them is that the former does so by speaking of properties of counting numbers, while the latter refers to sets of counting numbers. The preceding developnnent shows that the validity of the principle of mathematical induction for counting numbers is a consequence solely of the manner in which the relation < orders these numbers. Since we have used only those properties of the counting numbers which are expressed by (I), (II), (III), (IV), and (V) it is evident that the following is true: Let K be any set for which there is a relation R such that (I ) For every x and y, if x € K and y e K then it is not the case that both x R y and y R X, ("ll^ If A C K and A /^ (i.e. A is non- empty) then there is an x such that X e A and, for every z, if z £ A then either x R z or x = z, (IIlS K ^ 0. (IV ) For every x, if xe K then there is a y such that y e K and x R y, (V ) For every x, if x e K and there exists a z e K such that z R x (i. e. x j^ i„), then there exists a y e K such that y R X and, for every z 6 K, if z R x then either z R y or z = y. (continued on T . C. 22K) T. C. 22J Third Course, Unit 1 ZZ] >' ..-.rl (1) ^\ e A,- and that (2) for every zeC, if ze A then i ~ < z. A ~ From (1) and (a) it follows that i~ /^ i„. Hence, by (V). i~ A ^ A has an immediate predecessor, (i~ ") . We know that A (3) (i~ )~ < i~ , and that A A (4) for every z, if z < i~ then z < (i~ ) A J\. Evidently (i~ ) eA for, if not, it would follow from (2) that i^r < (l~)~ and this, by (1), contradicts (3). Hence, by (b), ^ ~ A ((i~) ) eA. But, by the previous theorem, ({Jl-r) ) = £ ~ . Consequently, i~ eA, in contradiction to (1). Since our assumption that A is non-empty has led to a con- tradiction, A = C. In order to relate the theorem we have just proved to the principle of mathematical induction, we notice that to say that a property holds for every counting number is to say that the set A of those numbers for which the property holds is C. And to say that the property holds for 1 is to say that leA (or, alter- natively, that i„ eA); while to say that the property is hereditary is to say that, for every xeC, if xeA then so does its follower (or, equivalently, then x eA). Consequently, the principle of mathematical induction and the theorem which we have just proved express the same characteristic of the set of all count- ing numbers, as it is ordered by < . The only difference between (continued on T . C. 22J) T. C. 221 Third Course, Unit 1 7.2] !- 1 A) L (1) k' < k, and (2) for every zeC if z < k then z < k . From, the similar theorem concerning immediate successors, whose statement precedes that of postulate (V), we know that (3) k" < (k")"*", and (4) for every zeC, if k < z then (k ) £ z. Now from (1) and (4) it follows that (k ) < k; that is, that either (k")"^ = k or (k")"^ < k. But, if (k")"*" < k then, by (2), (k")"*" < k". However, by (3), k < (k ) . But, by (I), it is impossible that (k ) < k and k < (k ) . Hence it is inapossible that (k ) < k, and we are left with the alternative that (k ) = k. We shall now prove a theorem which is merely a restate- ment of the principle of mathematical induction for counting numbers : If A C C and (a) i_eA, and (b) for every xeC, if xe A then X eA, then A = C. Proof. Suppose that there is a counting number which does not belong to A. Then A, the complement of A, is non- empty and, by (II), has a least member, Jl~ . We know that (continued on T. C. 221) T. C. 22H Third Course, Unit 1 '^2] Our last postulate is : (V) Every counting number except i„ has an immediate predecessor. [In other words: For every xeC, if x ^ i then there is a yeC such that y < x and, for every z eC, if z < x then z < y. ] From (I ) it follows that each number has at most one imme- diate predecessor. For suppose that some number x had two innmediate predecessors, say y, and y^. Then, for every z < x, z < y, and z < y . Since y^ < x, it follows that y-, < y,; since y, < x, that y, < y^ . Hence, by (I), y, = y^. Every number, then, has at most one immediate predecessor. Hence, from (V) it follows that every number other than i„ has a unique immediate predecessor. Thus we are justified in speaking of the immediate predecessor of any number other than i„. Using a self explanatory abbreviation for 'immediate predecessor of ' we have the theorem; For every xeC, if x / i„ then x < x and, for every zeC, if z < x then z < x . In deriving the principle of mathematical induction from our five postulate's we shall make use of the following theorem: For every xeC if x / i„ then (x ) = x . In order to prove this let us suppose that keC such that k ^ £ . Then, from the theorem noted at the end of the preceding para- graph, it follows that (continued on T. C. 22 H) T. C. 22G Third Course, Unit 1 '^2] for the number 1. However, before we can introduce the sym- bol 'i ' we need an additional postulate: (III) The set C of all counting numbers is non-empty (i.e. there are counting numbers). As a consequence of the definition of '^^ ' we know that I eC and, for every zeC, ^„ < z. (IV) There is no greatest counting number. [This means that, for every xeC, there exists a yeC such that x< y.] From (IV) it follows that, for every xeC, the set G(x) of all numbers y such that y > x is non-empty. Hence i„, .eG(x), and, for every z, if zeG(x) then i„, . < z. In other words: for every ■' G(x) — — xeC, X < i_, , and, for every zeC, if x < z then i„, . < z. It y UjX) - ^mx) -^ y^ is convenient to abbreviate 'i^, . ' to '( ) ', to omit the parentheses G( ) when doing so will not lead to confusion, and to read the resulting symbol as 'the immediate successor of '. For example, 'i_ ' denotes the immediate successor of the least member of C, which is, of course, the number two. We can now restate the theorem underlined above : For every xeC, x < x and, for every zeC, if X < z then x < z. [As a matter of fact, for every xeC, x = x + 1, and we shall want to recall this later. But, at present, since we wish to use only properties of the relation <, it is better not to think of the operation of addition.] (continued on T . C. 22G) T. C. 22F Third Course, Unit 1 IZ] ':•:.',', I ,'■ .1- : iS y [:"'' immediately that for every x and y, if x < y and y < X then x = y; and that for no x and y do we have both X < y and y < x. ] (II) Every non-empty set of counting numbers has a least member . [This means that if A is any non-empty set of counting numbers then there is a number x such that xeA (i.e. X is a member of A) and such that, for every number z, if zeA then x < z. If, for example, A is the set of even numbers, then 2eA and, for every even number z, 2 < z. 2 is the least even number.] Fronn (I) it follows that there is at most one least member of any given set A. For suppose that there were two least members, x, andx^. Then, for every zeA, x, < z and x^ < z. Since x^eA, it follows that x. < x . Since x,eA it follows that X- < X- . Hence, by (I), x, = x^. So, every set of counting numbers has at most one least member. But, by (II), every non - empty set of counting numbers has at least one least num- ber. Therefore, by (I) and (II), every non-empty set of count- ing numbers has a unique least mennber. This last assertion justifies our use of the definite article in speaking of the least number of a given non-empty set of counting numbers. It will be convenient to name the least member of a non- empty set by a symbol obtained by attaching a name for the set as a subscript to the letter 'i'. Thus, if we use 'C as a nanme for the set of all counting nunnbers, then 'St' is a name (continued on T . C. 22F) T. C. 22E Third Course, Unit 1 >Z] JO.' .•..r| <.. • .: With this definition the principle of mathematical induction is again a theorem. ] [B.] We shall now take up a postulational basis for the arithmetic of the counting numbers v/hich is quite different from Peano's. The latter emphasizes the operation of obtaining the innmediate successor of a counting number; the one which we are about to take up emphasizes, instead, the order relation < . What we shall do is to show how, from five postulates con- cerning <, one can establish the existence and uniqueness of an immediate successor of each counting nuniber, and derive the principle of mathematical induction. The procedure to be followed is implicit in the last two paragraphs of T . C. lA (q. V. ). Our postulates will be chosen in such a way as to jus- tify the following argument: If there exists a counting number which does not have a given property P then there is a least such number. If 1 has P then this least number which does not have P is not 1, and so has an immedi- ate predecessor. This latter number nnust then have P and so, if P is hereditary, its immedi- ate successor has P. But its immediate suc- cessor is the least number which does not have P. Consequently, if P is an hereditary property of 1 then there is no counting number which does not have P; i.e., every counting number has P. Our first two postulates are: (I) For no counting numbers x and y do we ever have both X < y and y < x. [From (I) and the obvious definition of ' < ' it follows (continued on T. C. 22E) T. C. 22D Third Course, Unit 1 ?2] We shall now consider two alternative postulational treat- ments of the principle of mathematical induction. [a. ] In explaining the principle of mathematical induction for counting numbers we have said that it is valid because each counting number o ther than 1 can be reached by starting at 1 and nnaking _a finite sequence of steps each of which consists in passing to the follower (or, as we shall now say: the imme- diate successor) of the number reached in the preceding step (or, in the first step, to the immediate successor of 1). It might seem natural to take this as a postulate, in place of the principle of mathematical induction, and it is perfectly feasible to do this provided that one has first defined the notion of finite sequence. One cannot use the most natural definition according to which a finite sequence is a function whose domain is the set of all counting numbers < some counting number, for then the underlined statement above would assert no more than that, for every counting number n, there is a 1-1 correspondence between the counting numbers < n and the counting numbers < n. But it is possible to define 'finite sequence' without referring to counting nuinbers, and on this basis to carry out the program suggested above. However, this procedure is somewhat com- plicated and while the underlined statement may in itself seem more intuitively simple than does the principle of mathematical induction, it beconnes much less so when backed up by an appro- priate definition of 'finite sequence'. [a sinnilar modification can be made in the Frege - Russell procedure: first define 'finite set' without reference to count- ing nunnbers and then define 'counting number' as an abbrevia- tion for 'cardinal nunnber whose members are finite sets'. (continued on T . C. 22D) T. C. 22C Third Course, Unit 1 ^2] and are true and '1 + 2 = 3' ' for all counting numbers a, b and c, a(b + c) = ab + ac*, if 'counting number' means even counting number, if '1' stands for the number two, if, for every even counting number a, a's follower is the next even counting number, i^ for all even counting numbers a and b, a + b is the ordinary sum of a and b, and a • b is half the ordinary product of a and b. With this interpretation of the symbolism, '1 + 2' denotes the svun of two and four, or six, whose name, in the present interpreta- tion, is '3', and '3 • 5' denotes half the product of six and ten, or thirty, whose name is now '15'. Hence '1 + 2 = 3' now states that the sum of two and four is six, while '3 • 5 = 15' states that half the product of six and ten is thirty. On the other hand, the Frege -Russell definition, also referred to on T . C. 19C, not only gives a basis from which to develop the arithmetic of the counting numbers, but also tells us, in terms of the notion of set, what the counting numbers are.] (continued on T . C. 22C) T. C. 22B Third Course, Unit 1 -= ■;;i:jii< iv • = .;': ■ ti; '-2] (3) The first member is 2 and, for every counting number n, n's follower is n + 2. (4) 2; for every prime number n, n's follower is the least prime number > n. (9) 1; for every unit fraction n, n's follower is t — ; — . X ; » 7 1 + n The following comments have relevance for the text material on pages 1-21 through 1-23 as well as for Part B of the Exercises on page 1-25. As was pointed out on T . C. 19C, the principle of mathemati- cal induction may be taken as one of a set of postulates chosen to characterize the arithmetic of the counting numbers. [We say ' the arithmetic of the counting numbers' because a postulational treatment such as Peano's cannot characterize the counting numibers themselves (i.e., cannot tell what they are), but only their behavior, with respect to operations such as addi- tion and multiplication. For example, all the theorems of arith- metic, such as T. C. 22A (continued on T. C. 22B) Third Course, Unit 1 [1.03] [1-22] Here are some examples : (1) 1. 2. 3, 4, 5, 6, 7, 8, (2) 4, 5. 6, (3) 2, 4. 6, 7, 8. 9, 10. 11, 8, 10, 12, 14, 16, (4) 2, 3, 5, 7, 11, 13, 17, 19, (5) 0, 1, 2, 3, 4, 5, 6, 7, (6) +1, +2, +3, +4, +5, +6, +7, +8, (7) -5, -4, -3. -2, -1, 0, +1. +2, (8) -1, -2, -3, -4, -5, -6, -7, -8, (9) 1, 1 2' 1 3' 1 4' 1 5' 1 1 7' 1 8' Set (2) is the set of all counting numbers > 4. The first member in that set is 4, and for every x, x's follower is x + 1 . Set (8) is the set of all negative real integers. The first member in that set is -1, and for every x, x's follower is x - 1. The student should define in a simi- lar way 'first member' and 'follower' for (3), the set of even counting nximbers; (4), the set of prime counting numbers; (5), the set of finite cardinal numbers; (6), the set of positive real integers; (7), the set of real integers > -5; (9), the set of unit fractional nxunbers. [Note: Actually 'first member' and 'follower' could be defined for each of the above sets in many different ways. We have indicated the definitions which we shall find most useful in practice. ] Once 'follower' has been defined with respect to the members of a set, one can introduce a notion of "hereditariness ". For example, in case (3) above we would make the following definition: UICSM-2-56, Third Course ^^ft^ nf TO ;; - -i J rf.. . . .:''. .■.j.f'.i ,j.r-| p.". te) r.rf. [1.03] [1-23] A property is hereditary over the set of all even counting numbers if it is meaningful for each even counting n\amber and if, for every even counting number k, if k has the property in question then so has k + 2. The student should make corresponding definitions for each of the sets (2). (4), (5), (6), (7), (8), and (9). Since for each of the given sets each of its members can be reached from its first member through a finite number of "steps", we have in each case a "principle of mathennatical induction". For example, in case (8), the principle of mathematical induction for negative real integers is: Every hereditary property of negative real integers which holds for -1 holds for every negative real integer. The student should state principles of mathematical induction for the other sets. We give now an example of the use of one such principle. Let us attempt to prove the following generalization: For every real integer x < 0, 1 - 2x > 0. Note that this is a generalization about the negative real integers, set (8). As remarked before, the first member in that set is -1 and for every X, x's follower is x - 1. A property is hereditary over the set of all negative real integers if it is meaningful for each negative real integer, and if, for every negative integer y, if y has the property in question then so does y - 1. The principle of mathematical induction for negative real integers has been stated above. The property in which we are interested is that expressed by: 1 - 2 ■ . . . ■> The proof of the generalization runs as follows: UICSM-2-56, Third Course biife ■24] \ (i) The property is hereditary . Suppose that, for a given real integer y < 3, the (3 - y)th even integer is 6 - 2y. Then, for this integer y, the [3 - (y - l)]th, or [3 - y + l]th even integer is 6 - 2y + 2, or 6 - 2(y - 1). (ii) 2 has the property . The (3 - 2)th or first even integer is 6 - 2 • 2, or 2. Therefore^ in view of (i) and (ii), by (d), the property holds for every real integer < 3. (a) All counting numbers which are multiples of 5: (5, 10, 15. ...}. (b) 5; k's follower is k + 5. (c) A property is hereditary over the set of all counting num- bers which are multiples of 5 if it is meaningful for each multiple of 5 and if, for every k which is a multiple of 5, if k has the property in question then so has k + 5. (d) Every hereditary property of counting numbers which are multiples of 5 which holds for 5 holds for every counting number which is a multiple of 5. (e) The property in question is that expressed by; ... + 1 is even. The generalization is false; 10 is a counter-example. T. C. 24C Third Course, Unit 1 1-24] (ii) 6 has the property. _ (6 - 4)(6 - 5) = l; T^.3 = T^ = 1. 6-5 2 Therefore, ;n view of (i) and (ii), by (d), the property holds for every counting number > 6. 2. (a) The set of all counting numbers > 3. (b) First number is 3; k's follower is k + 1. (c) A property is hereditary over the set of all counting numbers ^ 3 if it is meaningful for each counting num- ber > 3 d.nd if, for every counting number k > 3, if k has the property in question then so has k + 1. (d) Every h&Teditary property of counting numbers > 3 which hoMs for 3 holds for every counting number > 3. (e) The property in question is that expressed by: (...) - 1 is a multiple of 8. The generalization is false; 4 is a counter-example. 3. (a) All real integers < 3. (b) 2; y's follower is y - 1. (c) A property is hereditary over the set of all real integers < 3 if it is meaningful for each real integer < 3 and if, for every real integer y < 3, if y has the property in question then so has y - 1. (d) Every hereditary property of real integers < 3 which holds for Z holds for every real integer < 3. (e) The property in question is that expressed by: the (3 - . . . )th even integer is 6 - 2 • . . • (continued on T. C. 24C) T. C. 24B Third Course, Unit 1 1-24] 1. (a) The set' of alTcountirig^numbers >'6V (b) First number is 6; k's follower is k + 1 , (c) A property is hereditary over the set of all counting numbers > 6 if it is meaningful for each counting num- ber > 6 and if, for every counting number k > 6, if k has the property in question, then so has k + 1. (d) Every hereditary property of counting numbers > 6 which holds for 6 holds for every counting number > 6. (e) The property in question is that expressed by: _ (... - 4)( 5) (i) The property is hereditary . Suppose that, for a given k > 6, ■^k - 5 " 2 Then, for this k , = <^ - ^f - ^) + [(k - 5) + 1] ^ (k - 4)[(k - 5) + 2] 2 _ (k - 3)(k - 4) 2 ^ [(k + 1) - 4][(k + 1) - 5] 2 (continued on T. C. 24B) T. C. 24A Third Course, Unit 1 [1.03] [1-24] ( i) The property in question i_s hereditary: For every y, if 1 - Zy ^ then 1 - 2(y - 1) = (1 - 2y) + 2 > + 2 ^ 0. (ii) -1 has the property in question : 1 - 2(-l) - 3 > 0. Having established (i) and (ii) the principle of mathematical induction for negative real integers now tells us that every negative real integer has the property in question. That is, for every real integer x < 0, 1 - 2x > 0. EXERCISES A. For each of the follov/ing generalizations (a) identify the set with which the generalization is concerned, (b) decide on appropriate definitions of 'first member' and 'follower', (c) define 'hereditary property' appropriately, (d) state the appropriate principle of mathematical induction, and (e) either use the answer to (d) to prove the generalization, or give a counter-example. 1. For every counting number n •> 6, T , = s^ • — n - 5 2 2 2. For every odd counting number n j^ 3, n - 1 is a multiple of 8. 3. For every real integer x < 3, the (3 - x)th_ even integer is 6 - 2x. 4. For every counting number m which is a multiple of 5, m + 1 is even. UICSM-2-56, Third Course u\ 1-25] 1. 3, 11, 10, 98. 2. Yes, 2. 3. Yes, 1. 4. 1, 3, 5, ...,..., 6, 4, 2. 5. [See Commentary to page 1-22.] With this definition of 'follower' for the counting numbers it is not the case that each counting number which is a follower can be reached by a finite number of steps each of which consists in passing from the num- ber reached in the preceding step to its follower. In fact, no even counting number can be so reached. Hence, (although there is just one counting number which is not the follower of any other) there is no principle of mathematical induction corresponding to this notion of follower. The given notion of follower does correspond to an order- ing relation (see answer to Exercise 4, above) but, with this ordering there are non-empty sets of counting numbers which have no least members. In fact, the set of all even counting numbers is of this kind. T. C. 25A Third Course, Unit 1 [1.03] [1-25] B. Suppose you make the following definition of 'follower' for counting numbers : For every odd counting number n, n's follower is n + 2; for every even counting number n ^ 2, n's follower is n - 2. 1. What is the follower of 1? Of 9 ? Of 12? Of 100? 2. Is there a counting number which does not have a follower? 3. Is there a 'first member"? 4. Indicate how to list all of the counting numbers in a horizontal line such that the first member is "first in line" and the follower of each number except 2 is listed immediately to the right of it. 5. We are going to 'prove ' now that every counting nximber is odd. (i) The propert y is hereditary : For every counting number k, if k is odd, then it has a follower, k + 2, and its follower is odd. (ii) 1 has the property: 1 is odd. Therefore, every counting number is odd. Criticize the above "proof". C. V/e are going to "prove" that for every counting number n, every n boys have the same weight. (i) The property is hereditary : Suppose that every k boys have the same weight. (continued on next page) UICSM-2-56, Third Course O i •■.! 1-26] :i.lqf o-A .:■!;: The fallacy is in the proof of ( i), and illustrates the dangers of using ' . . . ' (as in line 2). The error can be discovered by attempting to specialize the given proof of (i) to a proof that if all the boys in every set of 1 boys have the same weight then so do all the boys in every set of 1 + 1 boys. T. C. 26A Third Course, Unit 1 [1.04] [1-26] Consider a group of k + 1 boys, say, B^, B^. By .... Bj^, Bj^ ^ y The boys Bj. B^, B3, ..., B^ form a set of k boys. Therefore, they have the same weight. Also, the group B^, B3, .... B^, B^^ J form a set of k boys; so they too have the same weight. But, since B^ belongs to both sets, each boy weighs the same as B . Hence, for every counting number k, if all the boys in every set of k boys have the same weight then so do all the boys in every set of k + 1 boys. (ii) 1 has the property : Clearly, all the boys in every set which consists of just 1 boy have the same weight. Therefore, by the principle of mathematical induction for counting numbers, for every counting number n, all the boys in every set of n boys have the same weight. Note that the second instance of this generalization states that any 2 boys have the same weightl Criticize the above proof", 1. 04 Sums of progressions . -- Sequences of numbers such as those given on page 1-22 are often called progressions . We are sometimes interested in computing the sum of the members of a set of successive terms in a progression. For example, you have learned earlier in the unit that, for every counting number n, the nth triangular number is the sum of the first n terms of the progression which begins: 1, 2, 3, 4, 5. ... UICSM-2-56, Third Course [1-27] \ in a 1-1 way on the set (O, 1, 2, 3, 4}, and /_, (k + k) = 2[(j + 1)^ + j]. Also, £ (k^ + k) = Yj [(-^)^ + (-"^)]; i^e^e j=0 k=l m = -5 the transformation is expressed by 'm - -k'. In general, allow- able transformations of 2j"expressions will be of the form 'j = ak + b' with 'a' replaced by ' 1 ' and 'b' by a counting numeral, if the interval of summation is a set of counting numbers; or 'a' replaced by '1' or by '-1' and 'b' by and integer numeral, if the interval of summation is a set of integers. The restriction on replacennents for 'a' is necessitated by the requirement that the transformation be 1-1. Other examples of such transforma- tions will be found on T.C. 29 A. T. C. 27C Third Course, Unit 1 [1-27] <• ■»■)-■ hv. in the range of the variables. For example, if 'u' is a variable whose range is the set of real numbers then \j\i{l - u) has a value for each value of 'u' between and 1, inclusive. Sinnilarly, if the range of 'p' is the set of integers, ^r has a value for r=p each value of 'p' < 5, i. e. for . . . , -3, -2, . . . , 4, 5, and these values are the same as the "corresponding" values of the expres- . 5 . . 5 . sion 2j ^ • 2j"^^P-'^^^^^°'^^ like ^j ^ in which no variables x=p k=l occur are themselves numerals. [If one replaces 'c', in the ex- ample above in which it occurs, by a name for a real nunnber then the resulting expression names a set of real numbers. The ori- ginal expression, then, has sets as values.] Returning to the analogy between 2j" expressions and J -expres- sions, you will recall that in the case of the latter one can "trans- form variables" as illustrated by: ( N/t dt = I y • 2y • dy. •^1 ^1 Here the transformation is that expressed by 'y = 'vt ', and for every real number x > 0, transforms the segment l,x (or x, 1) • — » ■ • in a 1-1 manner on the segment 1, nTx (or \rx, 1). In a similar 5 -expression, say 2j (^ + ^) can be transformed by k=l using any transformation which maps the "interval of summation" in a 1-1 way on a similar set of numbers. For example, the transformation expressed by ' j = k - 1 ' maps the set (1. 2, 3, 4, 5} (continued on T . C. 27C) T. C. 27B Third Course, Unit 1 [1-27] There is an analogy between the notations ^j ^W ^^'^ k=p ' rb ' J f(x)dx . [The concepts expressed by the notations are, of course, quite different.] The symbols 'k' and 'x' are apparent variables (or "dummy" variables) and can be replaced by any ap- propriate symbols without changing the meaning of the expressions q q for every p, q and f, ^^ f (k.) = ^i(''f); for every a, b and f, k=p *=p J f(x)dx = J f (z)dz. Other examples of apparent variables are the symbols 't' and 'y' in 'for every counting number t, t + 1 = 1 + t' and 'the set of all real numbers y such that y + 2 < 3c'. The symbols 'p', 'q', 'a', 'b', and 'c' in these examples are variables (i.e. pronumerals)(and 'f is also a variable, a "pro- functional"), 'p' and 'q' are called 'limits of summation' and 'a' and 'b' are called 'limits of integration'. [The same termino- logy is also applied to any expressions which occupy the same positions with respect to a /^ as do 'p' and 'q', or with respect . x+5 , to an J as do 'a' and 'b'. For example, in ^j j '1' ^^nd n=l 'x + 5' are the limits of summation. ] 5 In interpreting a 2j~expi"ession such as ^r the variables r=p which occur in the limits of summation (in this case, the variable 'p') must have specified ranges, for example, the set of counting numbers, or the set of non-negative integers, or the set of in- tegers, etc. The expression, then, like any other algebraic expression has a value corresponding to each of a set of values (continued on T . C. 27B) T. C. 27A Third Course, Unit 1 [1.04] [1-27] It is convenient to have a simple notation to use when referring to such Slims. Such is the so-called "2-notation". As an example of the use of this notation consider the 5th triangular number, T_, which, you remember, is 1 + 2+3 + 4+5. Using the 2 -notation we would write: , I'' k=l as a name for the 5t_h_ triangular number. Similarly, we could use as a name for the 4th square number: 4 ^(2k - 1) k=l [This symbol is read "sigma from k = 1 to 4 of 2k - 1". 'S' is the Greek letter corresponding to our 'S' and is meant to remind you of 'sum' . ] The expression: 4 Y^(zk - 1) k=l stands for the siim of the four values of ' 2k - 1' which corresponds to the values 1, 2, 3, and 4 of 'k'. In other words: 4 Y^ (2k - 1) = [2(1) - 1] + [2(2) - 1] + [2(3) - 1] + [2(4) - 1] = [1] + [3] + [5] + [7] = 16 EXERCISES A. Study each of the first six statements xintil you understand them, and then complete the last seven in the same manner. 3 1. Yj^Z - 3x) = [2 - 3(1)] + [2 - 3(2)] + [2 - 3(3)] = [-1] + [-4] + [-7] x=l 3 2^ ^(2 - 3y) = [2 - 3(1)] + [2 - 3(2)] + [2 - 3(3)] = [-1] + [-4] + [-7] y=l (continued on next page) UICSM-2-56, Third Course ■28] rs ^. oi oi 01 ■^ ••. - .■ > ! T' 8. 9. 10. 11. 12. j^5_132j.29^ 65 22222 2 9 + 4+1+0 + 1 + 4+9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 399 26 + 33 + 40 + 47 = 146 -10-5 + + 5+10 + 15 + 20 = 35 22 2 6 12 20 ^ 30 42 56 8 Remind students of the following procedure for finding the low- est common multiple for the denominator numbers: 13. L.C.M. = 2" 2860 T. C. 28A Third Course, Unit 1 [1.04] 13 3 y 4x = 40 + 44 + 48 + 52 x=10 3 4, J(3k + 7) = 1 + 4 + 7 + 10 + 13 + 16 k= -2 3 5^ ^(a + nd) = [a] + [a + d] + [a + 2d] + [a + 3d] n=0 5 6. ^ Mikl-i) = 1 + 5+12+22+35 k=l ^ [1-28] 7. I 4 8t - 3 t=0 2 10 8. I - = s= -3 6 i=3 7 10. Z'^"' ^> ~- x=l 3 11. J (5P+ 7) = p=3 7 12. Z n{n + i) n=l 5 13. 2j 11^^+711-8 n=2 UICSM-2-56, Third Course ^9] \ • i<-' 2j ^ [Correction: insert ' + ' between the first two terms m Exercise 6 . J By the recursive definition on page 1-18, P3 =P^ + (3 • 4+ 1) = P^ + (3 • 3 + 1) + (3 • 4+1) = P^ + (3 • 2 + 1) + (3 • 3 + 1) + (3 • 4 + 1) = P^ + (3 • 1 + 1) + (3 • 2 + 1) + (3 • 3 + 1) + (3 • 4+1) = [3- 0+1] + (3- l + l)+(3- 2 + l) + (3- 3 + 1) + (3 • 4 + 1), Therefore, Pr = Tj (3k + 1). or £ (3k - 2), (a) T = £ (k + 1), or £ k . k=0 k=i n-1 n (b) Sq^ = 2 (2k + 1), or fj (2k - 1) . k=0 k=l (c) P^ = L (3k + 1). or I (3k - 2) . k=0 k=l vl, vt. ^1, 'r '1^ '1^ The exercises in Part C are extremely easy. Note that they lead to the recursive definition given on page 1-30. T. C. 29B Third Course, Unit 1 ^.9] B. Here is an illustration of a procedure for obtaining an unlimited number of answers to problem posed in the Sample [See T. C. 27A ff.]: 3 k-l=3 4 X (3k + 4) = I [3(k - 1) + 4] = 2^ (3k + 1); k=0 k-l=0 k=l 3 k-2=3 5 2 (3k + 4) = I [3(k - 2) + 4] = 2 (3k - 2); k=0 k-2=0 k=2 3 ^i^ = ^ -J- 2 (3k + 4) = I [3(k + 5) + 4] = I { k=0 k+5=0 k=-5 A second procedure : k=3 -k=3 k=0 2 (3k + 4) = 2 [3(-k)+4] - Yj (4 - 3k); k=0 -k=0 k=-3 k= 8 -k=8 k=-5 2 (3k - 11) = 2 [3(-k) - 11] = 2j (-3k - 11). k=5 -k::-5 k=-8 6 ° 2 1. £ (5k + 3) 2. X (k - 1) k=2 k=-3 11 2, 7 3. y. 10k 4. 2^1^' E(io + 5^) k=5 k=3 k=l k=2 k=2 (continued on T . C. 29B) T. C. 29A Third Course, Unit 1 [1.04] [1-29] B. Express each of the following in Z)-notation. Sample. [3(0) + 4] + [3(1) + 4] + [3(2) + 4] + [3(3) + 4] 3 Solution . Here is_an answer: 2j ^^^ "*" ^^' k=0 Other possible answers are: 8 ^(3k-ll) and: li^ - ^^) . k=5 k= -3 1. [5(2) + 3] + [5(3) + 3] + [5(4) + 3] + [5(5) + 3] + [5(6) + 3] 2. [(-3)^ - 1] + [(-Z)^ - 1] + [(-1)^ - 1] + [(0)^ - 1] 3. 10(5) + 10(6) + 10(7)+ 10(8) + 10(9) + 10(10) + 10(11) 4. 15 + 20 + 25 + 30 + 35 + 40 + 45 5. [0(2) + 5] + [0(3) + 5] + [0(4) + 5] + [0(5) + 5] 6. [ 4 ] [ 4 ] + [ 4 ] + [ 4 ] 7. the 5 th pentagonal nvimber ijS. (a) the nth triangular number (b) the nth square number (c) the n th pentagonal number C. Verify each of the following. 5 I ^ (9 + 3k) =9 2. j; (3 - 2k) = -7 k=0 k=5 3 2 3. X 4x = 24x+ [4(3)] x=l x=l (continued on next page) UICSM-2-56, Third Course 30] •iiiJ le recursive definition in the box on page 1-30 is annbigubus to the extent that we have not specified the range of the variables 'p' and 'q'. Their range may be specified as any progression but in practice will be either the counting numbers, or the real inte- gers. T. C. 30 A Third Course, Unit 1 [1.04] [1-30] -1 y=-2 1000 y=-2 999 5. Yj(^^ + 2x - 3) = J (Sx"^ + 2x - 3) + [5(1000)'^ + 2(1000) x=6 x=6 9 8 6. Z ^k = Z^k + ^^9^ 3] k=3 k=3 [ Note: Think of 'a, ' as an abbreviation of an expression which is meaningful when 'k' is replaced by '3', '4', '5', '6', '7', '8', or '9'. For excunple, the expression '12 + 7k', in which case, 'ag' is an abbreviation for '12 + 7(9)'. ] MORE ABOUT S -NOTATION The exercises in Part C suggest the following recursive definition; 1\- % and for every q ^ Pi q+ 1 q y a, = y a, + [a , ] Zjk /ijk '■q+1-' k=p k=p As in the case of any recursive definition [See page 1-16. ] this provides us with a way of computing sums of successive values of 'a, ' . For example, if we replace 'a, ' by '5 - 2k' and 'p' by '3', we get the following recursive definition: UICSM-2-56, Third Course vh .;i- \i [1.04] [1-31] J (5 - ^k) = 5 - 2(3) k=3 and for every q ^ 3, q+1 q 2 (5 - 2k) = y (5 - 2k) + [5 - 2(q+ 1)] k=3 k=3 Hence, 3 2 (5 - 2k) . -1 k=3 j;{5 - 2k) = y (5 - 2k) + [5 - 2(4)] k=3 k=3 = -1 + [-3] = -4 5 4 2; (5 - 2k) = j(5 -2k) +[5- 2(5)] k=3 k=3 ■4 + [-5] = -9 etc. Another important use of recursive definitions is to prove generalizations like the following: For every counting number n, the siiin of the first n even numbers is n(n + 1). Introducing 2 -notation this can be written: For every counting number n, n y 2k = n(n + 1). k=l UICSM-2-56, Third Course rs, N Therefore, in view of (a) and (b), by the principle of mathema- tical induction for counting numbers, the property in question holds for every counting number. Be sure to call attention to the similarity between this proof and the proof for Exercise 6 on page 1-21. T. C. 32C Third Course, Unit 1 *^^'< .»» , ri-32] ^rs, The proof that yn(n + 1) is the sum of the first n counting num- bers makes use of the principle of mathematical induction for counting numbers. xl, -I, o^ '1^ ',V ^,N Property is that expressed by: k=l (... + 1) (a) 1 has the property . By the recursive definition on page 1-30, 1 E k = 1; k=l and 1(1 + 1) _ , 2 (b) The property is hereditary . For every q, q if ^k _ q(q + 1) k=l q+1 then ^k = Yj ]<. + (q+1) k=l k=l [recursive definition] q(q + 1) + {q+ 1) [inductive hypothesis] T. C. 32B ^ (q-f l)(q + Z) 2 - (q+ l)[(q+ 1) + 11 2 (continued on T . C. 32 C) Third Course, Unit 1 ri-32] ^rs, Be sure that students understand the inductive proof given on page 1-32. The answer to the first '[Why?]' is: the recursive definition and the answer to the second '[Why?]' is : the inductive hypothesis. .1.- ^i^ xi, '<"■ 'I- '1- 1. Point out to students that in many college algebra texts (in which 2j"n.otation is not discussed) exercises of this type are often stated: Prove : 1 +2 + 3+ ... +n = J}ilL^. The use of 2j~i^otation avoid the vagueness of the three dots. Students should be able to state what the generalization . in this exercise asserts, viz . that the sunn of the first n counting numbers is = . It is natural for students to wonder about the source of the form — — ^ ~. They nnay guess at the derivation of this form if you lead them through the following : Sum = l + 2 + 3+...+n Sum = n + (n -1) + (n - 2) + . . ■ + 1 2 X Sum = (n + 1) + (n + 1) + (n + 1) + , . . + (n + 1) = n(n + 1) Sunn = y-n (n + 1) . (continued on T. C 32B) T. C. 32A Third Course, Unit 1 [1.04] [1-32] According to the principle of mathematical induction for counting numbers , it is sufficient to show that 1 (a) I 2k = 1(1 + 1) k=l and that (b) for every counting number q, q q + 1 if J 2k = q(q + 1) then ^^ 2k = (q + l)[(q + 1) + 1]. k=l k=l Proof of £a) : 1 y 2k = 2(1) = 1(2) = 1(1 + 1). k=l Proof of (b): For every q, q k=l q(q+ 1) q+l q then 2 2k= 2; 2k + [2(q+ 1)] [Why?] k=l k:.l = q(q+ 1) + [2(q+ 1)] [Why?] = (q+ l)(q+ 2) = (q+ l)[(q+ 1) + 1]. Hence, the boxed statement follows by the principle of mathematical induction for counting numbers, EXERCISES Prove each of the following. Be sure to state the principle of mathematical induction you use for each proof. n 1. For every counting nxomber n, )^ ^ ~ 2 k=l (continued on next pag^ UICSM-2-56, Third Course ri-33] bf.c (.'■,! w V T*-; ; Hence, for every real number y different from and -1, 3 - - + ^ (y+1)' So, for every real integer y ^ 1, if X , 2 < 3 - - k=l^ y < 3 y + r y+1 1 1 then y -7- < 3 Therefore, in view of (a) and (b), by the principle of miathematical induction for positive real integers, the property in question holds for every positive real integer. T. C. 33N Third Course, Unit 1 ri-33] (b) The property is hereditary . Suppose that, for a given y > 0, Tj 71 < 3 - - . k = 1 ^ y Then, for this y. Now, y+1 y 1 k2 %4^, k^ ^ (y + 1) Z dz = Z. w2 + '- . -2 k= 1 " k= 1 Since, by the inductive hypothesis. lb k=r <^-\- Z k^ k = l ^ < 3 - i + y 1 1 (y+ i)^-v y^(y+ 1)^ '- ^ ^ y(y + 1)^ {y+ 1)^ = 3 . y .+ y + ^ y(y + 1) 2 1 < 3 - ^ i-^ ["add 3 "] y(y+i) y(y+i) = 3- ' y+l (continued on T. C. 33N) T. C. 33M Third Course, Unit 1 'in: i-»' .•■ i "i ■ j-^iin ri-33] •xhl"' Then, for this z, z + 1 z z <= = z <= + ^ k= y k = y = c(z-y+l) + c = c[(z+ 1) - y + 1]. Hence, for every real integer z > y, z if 2y c = c{z - y + 1) k = y z+1 *^-" Z c = c[(z+l)-y+l]. k = y Therefore, for every real integer y, by the principle of mathematical induction for real integers > y, it follows in view of (i) and (ii), that the property in question holds for every real integer _> y. 11. Property is that expressed by: z ,,2 k= 1 (a) J has the property 1 z ^ k= 1 = ~J - 1; and 1^ 1 < 3 - Y = 2 . (continued on T. C. 33M) T. C. 33L Third Course, Unit 1 ri-33] ■; f Therefore, in view of (i) and (ii), by the principle of mathematical induction for positive real integers, the property in question holds for every positive real integer. Note that the generalization in this exercise asserts that, for every real number c and for every real integer x > 0, X 2j c = ^c -f c + c + . . . + c . k= 1 r^^^- X terms Connpare with Exercises 5 and 6 on page 1-29. 10, (b) Property in question is that expressed by: for every real nvimber c and for every real inte ger y, Yj c = c(... -y + 1). k =y (i) For every real integer j/-, jr has the property. y 2 c = c; k = y and c(y - y + 1) = c. (ii) For every real intege r y^, the property is hereditary over the set of real integers > y. Suppose that, for a given z > y, T. C. 33K Yj c = c(z - y + 1). k = y (continued on T. C. 33L) Third Course, Unit 1 ri-33] 10. (a) Property is that expressed by: for every real number c V c = c k= 1 (i) 1^ has the property . For every real number c, 1 J c = c; k= 1 and c • 1 = c, [c = c + • k] (ii) The property is hereditary . Suppose that, for a given y, T. C. 33J 2y c = cy (for every real number k= 1 Then, for this y, y+1 y E c = Z c + c k= 1 k= 1 = cy + c = c(y + 1). Hence, for every y, if, for every real number c, Z, ^ ~ ^V k= 1 then, for every real number c, y + 1 Yj c = c(y + 1) . k= 1 (continued on T. C. 33K) Third Course, Unit 1 c). ri-33] Then, for this q, ^+^ i_ "^ L_ 1 2 k(k+l) Yj k(k+l) + {q+ l)(q+ 2) k= 1 k- 1 q+ 1 ' (q+ l)(q+ 2) 1 1 q + 1 1 q + q + q + 2 q + ^q + 1 q -1- 2 q+^ (q+ 1) + 1 * Hence, for every q, ?. 1 q if 2j k{k + 1) ~ q+ 1 k= 1 then 2j ^(k+ 1) (q + 1) + 1 k= 1 Therefore, in view of (a) and (b), by the principle of mathematical induction for positive real integers, the property in question holds for every positive real integer. «i^ «i^ «•- "I" 'I* 'I"* Correction: Insert 'every' between 'for' and 'real' in the sixth line from the bottom. ■^i, >u »i^ "i^ 'r '1^ T. C. 331 Third Course, Unit 1 /( . O ■ ; 1 1 1 ri-33] monr',' !0 -vv..,! .■ ; .jili\r. .}:■):, = (q+ l)(q+ 2){f + 1) _ (q+ l)[(q+ 1)+ l][(q+ 1) + 2] 3 Hence, for every q, q if 2k(k+ 1) = q 0, ^ 1__ _3_ 2; k(k+ 1) -q+l • k= 1 (continued on T. C. 331) T. C. 33H Third Course, Unit 1 ri-33] then 2 (5. + 10) . f5(x -H)1f(x + 1) + 5] ^ z = -4 "^ Therefore, in view of (a) and (b), by the principle of mathe- matical induction for the set of real integers >-4, the property in question holds for every real integer > -4. "7. Similar to foregoing proof except that here we use the principle of mathematical induction for real integers > -2. 8. Property is that expressed by: Z Mk + 1) k = 1 (a) 1 has the property . 1 (••• + 1)(. 2 k(k + 1) = 1(1 + 1) = 2; k = 1 1(1 + 1)(1 + 2) _ , (b) The property is hereditary . Suppose that, for a given q, q J_2) y;k(k+ 1) = q{•:', /■ Ui ■■c; >r: (a) -4 has the property . By the recursive definition, -4 2 (5z + 10) - 5{-4) + 10 z= -4 10: and ^(-^n^^+^) -_ -10. (b) The property is hereditary. Suppose that, for a given x > -4, X 1 (5Z+10) = 5x(x+_5)_ z= -4 '^ Then, for this x, x+ 1 X 2 (5z + 10) = J(5z + 10) + [5(x + 1) + 10] z = - 4 z = -4 = ^"^^"^^^ ^^ + [5(x+ 1) + 10] 5x + 25x + lOx + 30 5x + 35x + 30 2 5(x + l)(x H- 6) 2 [5(x+ l)][(x+ 1) + 5] Hence, for every x > -4, if I (5z + 10) = ^-^^i^ z= -4 T. C. 33F (continued on T. C. 33G) Third Course, Unit 1 ri-33] Hence, equation (2) becomes: (3) (x+ 1)^ - 1 = 3 2k^+ 3 k=l X Solving for ' 2j ■'^ ' ^^ obtain: x(x + 1) 2 + X k= 1 X k= 1 (x+ 1)" - 1 - 3x(x + 1) {x+ 1) {(x+ l)^ -^ - 1} i [(x + l)(2x^ + 4x + 2 - 3x - 2)] ^[(x+ l)(2x^ + x)] x(x + l)(2x + 1) 6 In a similar manner you can derive a form for 2j ^ ^Y considering the generalization: for every k, {k + 1)"^ - k^ = 4k^ + 6k^ + 4k + 1. K^ O^ %,•,. ♦"i^ 'C T 6. Property is that expressed by: E (5z + 10) = z= -4 (... +5) (continued on T. C. 33F) T. C. 33E Third Course, Unit 1 ri-33] Therefore, in view of (a) and (b), by the principle of mathematical induction for positive real integers, the property in question holds for every positive real integer. V^ *'^ o.* ?,- *,» r^p students may inquire about the derivation of the form ' — ^ 'y ' ' in Exercise 5. This form can be derived with the help of the theorenns stated in the box on page 1-34 and of the theorem in Exercise 10(a) on page 1-33, Since, for every k, (k + 1)^ = k^ + 3k^ + 3k + 1 then (k + 1)^ - k^ = 3k^ + 3k + 1. Now, X X ^^) Z [(k + 1)^ - k^] = 2 (3k^ + 3k + 1) k= 1 k= 1 XX XX 3 V , 3 ,V> , 2 or X (2) Yj (k+ 1)^ - E k^ = 3^ k^ + 3y; k +2 1 k= 1 k= 1 k:s 1 k= 1 k= 1 Since 2 (k+ 1)"* = 2^ + 3^ + 4-^ + . . . + X + (X + 1)- k=l V , 3 ,3 ^3 _3 ,3 and 2^k = 1 +2 +3 +...+X, k= 1 the left member of (2) simplifies to: (x+ 1)^ - 1^. T. C. 33D {continued on T. C. 33E) Third Course, Unit 1 ri-33] ,„d ^<^^^M^'^-^^) = 1. (b) The propert y is hereditary . Suppose that, for a given real integer y > 1, y r= 1 2 y(y 4- l)(2y + 1) 6 Then, for this y, y+1 y 2 r = 2 r + (y + 1)^ r= 1 r= 1 - y(y + Y^ ' ^) +(y+i)' _ y(y + l)(2y + 1) + 6(y + 1)^ 6 _ (y + l)ry(3y + i) + 6(y + 1)1 6 (y + l)(2y + y + 6y + 6) 6 _ (V+ l)(2y^ + 7y + 6) 6 ^ (y + 1) (y + 2) (2y + 3) 6 _ (y + l)[{y + 1) + l][2(y + 1) + 1] Hence, for every y, if V r^ - y(y + i)(2y + i) r.i ^ y+l then y r^ = (y + l)[(y + 1) + l][2(y + 1) + 1] r= 1 6 T.C. 33C Third Course, Unit I 1^' P! [1-33] Then, for this q, q+1 q 2 (5 + 4k) = 2(5 + 4k) + [5 + 4(q + 1)] k= 1 k= 1 = q{2q+ 7) + [4q + 9] = 2q + llq + 9 = (q+ l)(2q+ 9) = (q+1) [2(q+ 1) + 7]. Therefore, in view of (a) and (b), by the principle of mathe- matical induction for counting numbers, the property in question holds for every number. 3. Similar to foregoing proof. 4. Proof is similar to that for Exercise 1 except that here we use the principle of mathematical induction for non-negative real integers. The generalization to be proved asserts that the sum of the squares of the first x real positive integers is x(x + l)(2x + 1) 6 Property is that expressed by: I'' r= 1 (... + 1)(2 • ... + 1) 6 (a) 1 has the property . T. C. 33B By the recursive definition, 1 2 r = 1 , or 1; r= 1 (continued on T. C. 33C) Third Course, Unit 1 [1-33] i. Here the student is to find the sum of the first n terms of the progression 9, 13, 17, ... , The kth term of this progression is 5 + 4k. In traditional terminology, the student is asked to prove: 9 + 13 + 17 + . . . + (5 + 4n) = n(2n + 7). As in Exercise 1, the form 'n(2n + 7)' is suggested by: Sum = 9+13+17+.. . +{5 + 4n) Sum = (5 + 4n) + (1 +4n) + (-3 + 4n) + . . . +9 2x Sum = (14 + 4n) + (14 + 4n) + (14 + 4n) + . . . + (14 + 4n) = n (14 + 4n) Sum = -111(14 + 4n) = n(7 + 2n). Property is that expressed by: 2 (5 + 4k) = ... (2 • ... + 7). k=l (a) 1^ has the property . By the recursive definition, Yj (5 + 4k) = 5 + 4(1) - 9; k= 1 and 1(7 + 2-1) = 9. (b) The property is hereditary . Suppose that, for a given q, q ^ (5 + 4k) = q(2q+ 7). k=l (continued on T. C. 33B) T. C. 33A ' Third Course, Unit 1 [1.04] [1-33] n 2. For every counting number n, 2j ^^ ^ ^^^ ~ n(2n + 7). k=l n Z, ,, p \ _ n( 3n - 1) (:iK. - d) - 2 k=l Z, x(x + 1 ) 2 ' k=l y Z _ x(x + l)(2x + 1) 5. For every real integer x > 0, 2j ''^ ~ 6 r=l y 6. For every real integer y > -4, 2j '^^ ^ ^'^^ ~ 2 z=-4 w V ^>. + 7^ - (w+ 3)(w+ 12) ^ 7. For every real integer w _> -2, 2j 2 b= -2 n 8. For every counting number n, £^ * "•■ */ ^ k=l 1 _ X 9. For every real integer x > 0, 2_, k{k +1) x + 1 ' k=l 10. (a) For every real number c and for real integer x > 0, x J c = ex. k=l (b) For every real number c and for all real integers x and y such that X _> y, X Yj c = c(x - y +1). k=y x V 1 1 ■6^ 1 1 . For every real integer x>0, Li TTZ '^ -^ ~ ~' k=l ^ UICSM-2-56, Third Course •J- ••.if!." [1.04] [1-34] STILL MORE ABOUT 2-NOTATION The recursive defintion of a S-©xpression mcikes it possible to use mathematical induction to prove theorems about sums of any number of ternns of progression. We shall state, illustrate, and prove two such theorems: (I) k=p k=p k=p q q (11) For every real number c, 2j ^^^u^ • k=.p k=p Illustrations: 5 5 ^jj. 2(3k+ 1) + v(k - 2) k=2 k=2 = [7+10+13+16+19] + [0+1 + 2+3+4] = (7 + 0) + (10 + 1) + (13 + 2) + (16 + 3) + (19 + 4) 5 = 2([3k+ 1] + [k -2]) k=2 (11): 2 2m . 2(-3 )'^ + 2(-l)'^ + 2(0)^^ + 2(1)"^ + 2(2)"^ m= -3 = 2[(-3)^ + (-2)^ + (-1)^ + (0)^ + (1)^ + (2)^] m=-3 Proofs: Theorem (I) P P (a) Z"k + l\ k=p k=p a + b P P = Z <^k ^ \) . k=p UlCSM-2-56. Third Course [Why?] [Why?] :i..'Ai> [1.04] [1-35] (b) For every q ^ p, itl%* IK- Z'^^V k=p k=p k=p then E ^k +^Z ^k = t ^ -k + ^ + l] + [ E^k + \,l ] k=p k=p k=p k=p k=p k=p q k=p q+1 = E^^^V- k=p Hence, theorem (!) follows either by the principle of mathematical induction for the set of counting numbers > p (or by the principle of mathematical induction for the set of real integers > p, according to how the theorem is interpreted). Theorem II (a) For every resil number c, /^ ^(^i,) - '^(^r.) k=p ^ P k=p (b) For every real number c and for every q > p, q q if E^(V= ^ E^ k=p k=p q+1 q then E^^V = E ^(V + ^(^q+ l) k=p k=p (continued on next page) UICSM-2-56, Third Course sOrilw [1-36] q, q N 2k = S (k + k) k = p k = p 3. q q k = p k = p • q k = p .3)^ = y(k^. kfp 6k + c ) q k = p q ^6k k = p q k = p q k = p q k = p q k = p _q N(k + k= 1 3)^ . Vk^ + k'Tl q 6 N k kt^i q k= 1 q(q+ 1){2 6 iq+ 1) + 6 q(q + 1) 2 + 9q. T. C. 36A Third Course, Unit 1 [1.04] [1-36] q = cla^ + c(a j) k=p ^ q = cf ), a, + a , ] ■• , ^ k q + 1 -■ k=p ^ q+1 k=p Hence, theorem (II) follows either by the principle of mathematical induction for the set of counting numbers > p (or by the principle of mathe- matical induction for the set of real integers > p). EXERCISES rove each of the following. Sample. /^ (1 + 3k + k ) = £ 1 + q . l^' k=p k=p k=p k=p Solution. £ (1 + 3k + k ) k=p q q Zl 4. V/^k 4- k^\ [Associative principle for ^^p k^p addition and Theorem (I)] q q q E 1 + 2 3^ + Tj^^ [Theorem (I)] k=p k=p k=p q q q Jl + 3jk+2k^ [Theorem (II)] k=p k-p k=p q q q V^ V V 1 . l^ Zk = l^k + l^y: k=p k=p k=p 2. £ (k+ 3)^ = 2^k + 6 ^k + 2^ 9 k=p k=p k=p k=p 3. Use the result in Exercise 2 together with the results in Exercises 4, 5, and 10(a) on page 1-33 to show that I (k + 3)^ = qi£Ji^l2i+_li , 6[^^a_Lj)] -f 9q . k=l (continued on next page) UICSM-2-56, Third Course -37] s_ 2 £ 2 6s + 9s + 3 + 24s + 24 + 32 6s + 33s + 59 12 12 y (k - 12){k - 2) = y (k^ - 14k + 24) k=2 k=2 12 12 12 2 y .^ - 14 y . . ^ k = 2 k = 2 k = 2 12 12 = N k^ - 14 y k - (0 + 1) k=2 k=0 + 24(12 -2 + 1) 12 12 k^ - 14 \ k + 14 + 264 k = 2 k = 12 12 - N k^ - 14 S k + 278 . k=2 k=0 T.C. 37C Third Course, Unit 1 •37] 6(k^ + 3k^ + 3k + 1) + 33(k^ + 2k + 1) + 59(k + 1) 2 k + 1 6{k + 1) + 33(k + 1) + 59 Hence, for every k, k if if N (3t + 4)^ = 7 {6k^ + 33k + 59) the k+ 1 n S (3t + 4)' k + 1 6(k + 1) + 33(k + 1) + 59 Therefore, in view of (i) and (ii), by the principle of mathematical induction for counting numbers (or for real integers > 1), the property holds for every counting number (or for every real integer > 1). (b) r ,-,. .,2 >r ,„.2 N (3t + 4) = S (91" + 24t + 16) t= 1 t= 1 s i. S S t= 1 t=: Q 's(s + l)(2s + 1) " 7 6 16 + 24 s(s + 1) 2 + 16s 3s(s + l)(2s + 1) + 24s(s + 1) + 32s 2 T. C. 37B 3(s + l)(2s + 1) + 24(s + 1) + 32 (continued on T. C. 37C) Third Course, Unit 1 -37] , J r ! 4. (a) Property is that expressed by: I (3t + 4)^ = ^ [6(...)^ + 33- ... +59]. (i) 1 has the property. (3t + 4)^ = (3 + 4)^ = 49; and t= 1 ^ (6 + 33 + 59) = 49. (ii) The property i_s hereditary . Suppose that, for a given k, 1 (3t + 4)^ = |(6k^ + 33k + 59). Then, for this k, k+ 1 k t = {3t + 4)^ = S (3t + 4)^ + [3{k + 1) + 4]' t= 1 I (6k^ + 33k + 59) + (9k^ + 42k + 49) 6k "^ + 51k^ + 143k + 98 'T. C. 37A (continued on T. C. 37B) Third Course, Unit 1 [1.04] [1-37] 4. Prove that, for every s _> 1, E(3t + 4)' t=l |(6s^ + 33s + 59) (a) by mathematical induction, and (b) by the method suggested by Exercise 3 on page 1-36. 12 IZ 12 5. 2 (k - 12) (k - 2) = 2 k^ - 14 J k + 278 k=2 k=2 k=0 ARITHMETIC PROGRESSIONS Progressions such as (1) 5, 7, 9, 11, 13, 15. 17, 19, ... (2) -3, 1, 5, 9, 13, 17, 21, 25, ... 17 13 19 (3) 7 ' ^» 7 ' -*' "T > ^' "2 ' ^^' (4) 8, 1, -6, -13, -20, -27, -34, -41, ... (5) TT, 2tt, 3it, 4tt, Sit, 6tt, Ttt, 8it, 9tt, are called arithmetic progressions . Notice that a characteristic property of arithmetic progressions is this: The difference of any te rm of an arithmetic progression from its follower is the same as the difference of any othe r term from its follower. We frequently abbreviate the boxed statement by saying: The difference between successive terms of an A. P. is constant . This difference is the common difference of the A. P. . UICSM-2-56, Third Course ^-38] A. Let students discover their own methods for the exercises in this part. Formal methods are developed in subsequent parts . vl^ -^l- vl- 'I'" 'r '1^ Exercise 11 does not have a unique solution. T. C. 38A Third Course, Unit 1 [1.04] [1-38] For example, for the A. P. (1) the common difference is Z, and for the A. P. (4) the common difference is -7. For all real numbers a and d, the successive terms of the A. P. whose first term is a and whose common difference is d are the values of: a + {x - 1) d for the values 1, 2, 3, etc. of 'x'. For example, in A. P. (1) above, the first term is 5 and the common difference is 2. Identify the first term and the common difference in (2), (3), (4), and (5). EXERCISES A. Fill in the blanks in each of the following so that the result gives an arithmetic progression, ^» -5, , , DO, , , • • ■ 3 9 10 11 5 16 24 6. , , , , , -30, -35, 8 i i • O ' / » » 9 t t • • • 10. -3, , -3+i^l^, 11 7 B. In filling the blanks between '3' and '66' in Exercise 2 of Part A you "inserted two arithmetic means between 3 and 66". In Exercise 4 you inserted five arithmetic means between -8 and 17. UlCSM-2-56, Third Course ri-39] \ 4. X, X + d, and x + Zd are consecutive terms in an A. P . , and X + d is the arithmetic mean of x and x + 2d. Since y-x x + 2d = y, we have d and X + d = X + y-x _ X + y T. C. 39A Third Course, Unit 1 [1.04] [1-39] In each of the following exercises insert arithmetic means as indicated. 1. two arithmetic means between 4 and 10 2. three between 1 and 7 3. five between 8 and 3 4. ten between -6 and -14 5. two between 1 and -3 If three numbers are consecutive terms of an A. P. , then the second is called the arithmetic mean of the first and third . 1. Find the arithmetic mean of 6 and 10. 2. Find the arithmetic mean of 10 and 6. 3. Find the arithmetic mean of -3 and 3 4. Prove the following generalization: For every x and y, the arithmetic mean of x and y is X + y 2 ' p. We can develop a procedure for inserting any number of arithmetic means between any two real niimbers. Sample . Develop a procedure for inserting three arithmetic means between any two real numbers. Solution . For all real numbers x and y, there exists three arithmetic means between X and y if and only if there exists a real number d such that X, X + d, x + 2d, X + 3d, y UICSM-2-56, Third Course -40] i:; I 3. Suppose you want to insert p arithmetic means between X and y. This would give us p + 2 consecutive terms of an A. F. in which x is the first term and y is the (p + 2)th term. Hence, y = X + (p + 1) d, so P+ 1 Then the p + 2 terms of the A. P. are: V PX + y (p - 1) X + 2y (p - 2) X + 3y x -f py 'p+1' p+1 ' p+1 '•••'p+1' T. C. 40B Third Course, Unit 1 •40] D. 1. If X and y are two numbers such that x is the first term of an A. F. and y is the seventh term then y = X + 6d so For every x and y, the five arithmetic means between X and y are the values of: for the values 1, 2, 3, 4, and 5 of 'k'. Hence, the following are successive terms of an A. P.: 5x -I- y 2x + y x + y x + 2y x + 5y X, ^ , 3 ' 2 ' 3 ' 6 ' ^" If X is the first term of an A. p. and y is the 102nd term then y = X + lOld, so ^ - 101 • The 102 terms of the A. P. are: lOlx + y 99x + 2y 98x + 3y 2x + 99y x + lOOy ^' 101 ' 101 ' 101 ' • • •' 101 • 101 T. C. 40A Third Course, Unit 1 [1.04] [1-40] are successive terms in an A. P. This is the case if and only if the difference between successive terms is constant. Since each of the first three differences is d, all four differences will be the same if and only if y - (x + 3d) = d, that is, d =-i^^ 4 Hence, for all x and y, there exist three arithmetic means between x and y and, in fact, these are the values of the expression: 1 /V - X . x + k{--2f-) for the values 1, 2, and 3 of 'k'. [You could have applied this procedure in Exercise 2 of Part B. In that case the required arithmetic means are the values of: 1 + k(^-^) for the values 1, Z, and 3 of 'k': 7 - 1 ^ _ 5 ' " 2 ' 1 + 1( -'■ = - 4 7 - 1, 4 7 - 1 , 1 + 2( -~-^] = 4 , 1 + 3(^^)=^ .] Develop a procedure for inserting five arithmetic means between any two numbers. Tell how to insert one hundred arithmetic means between any two numbers. Tell how to insert any number of arithmetic means between any two numbers. UICSM-2-56, Third Course ;- r •- 1.0 [1.04] [1-41] SUMS OF ARITHMETIC PROGRESSIONS Consider each of the sums (I) 1 + 2+3 + 4 + 5 (11) 7+10+13+16+19 (in) (+4) + (-1) + (-6) + (-11) + (-16) Notice that each sum can be computed by adding the first 5 terms of an arithmetic progression. In (I) the terms of the progression are the values of '1 + (x - 1)(1)' for the values +1, +2, +3, ,.. of 'x'. The first five terms are obtained by using the first five values: +1, +2, +3, +4, +5. Therefore, (I) can be abbreviated as: + 5 Yj [+1 + (x - 1)(1)] x=+l Similarly, (II) + 5 ^ [7 + (X - 1)(3)] x=+l and + 5 (III) J[4 + {x- l)(-5)] x=+l In general, For every counting number n, the sum of the first n terms of the arithm etic progression whose first term is a and whose common difference is d is n x=l + (x - l)d] Let us use the boxed theorems on page 1-34 and compute (III). UICSM-2-56, Third Course .-'■r:-^,-i OO ii^' [1.04] [1-42] 5 5 J [4+ (X - l)(-5)] - Z t*^ ■ ^^J x= 1 x= 1 5 5 x= 1 x= 1 5 5 = Z 9 - 5 z - x=l x=l By Exercises 4 and 10a on page 1-33, j;9 -5 ^ X = 9(5) -5^ (5)( 5-f l) j,.3o x= 1 x= 1 In this particular example the result, -30, could have been obtained without using 2-notation. However, in cases involving a large number of tei'ins or when proving generalizations, the use of 2-notation results in a great saving of labor. Using S-notation we can prove a generalization covering all cases. Suppose we wish to find the sum of the first n terms of the A. P. whose first term is a and whose common difference is d. V/e have seen that this sum is n Z [a + (X - 1) d] . x=l Proceeding as in the example above n n Y^ [a+ (x - l)d] = Zna - d) + dx] x= 1 x= 1 n n = Z (a - d) + d Z X x= 1 x= 1 = (a-d)(n) + d[atnJ_lJ-] . (n) [(a - d) + -^lS_Ll) J = Y [ 2a - 2d + d(n + 1)] =: I [ 2a + (n- l)d]. UICSM-2-56, Third Course < • r ^i ^. '»"» '!"• •'4'' A. 20 [2(-2) + 19-7] = 10 (-4 + 133) = 1290 12 [2(-100) + 11 • 100] = 6{-200 + 1100) ^ 5400 3. |[2{1) + 6(-0.5)] = |(2 - 3) = -3.5 T. C. 43A Third Course, Unit 3 [1.04] [1-43] We have just proved the following theorem: For every countin g number n , and for all real numbers a and d, the sum of the first n terms of the A. P . whose first te rm is a and whose common difference is d is ft^ a + (n - l)d ]. The use of this theorem is illustrated by the following Example . Find the sum of the first 53 terms of the A. P. which begins: 8, 17, 26, ... Solution . The first term is 8 and the common difference is 9. Therefore, the sum of the first 53 terms is ^ [ 2(8) + {53 - 1)(9) ] - ^ [ 16 + 468 ] = 12826 The labor-saving value of the theorem is obvious. EXERCISES A. Find the sum as indicated. 1. First 20 terms of -2, 5, 12, . . . 2. First 12 terms of -100, 0, lOO, ... 3. First 7 terms of 1, 0.5, 0, ... {continued on next page) UICSM-2-56, Third Course .■1- ! -iXi •■1-44] Upon simplification, (1) and (2) becom( (lO 3a + Zld = 20 (2^) 2a + 19d = 13 or: (2''0 6a + 42d 6a + 57d = 40 = 39 a : -15d d = 1 1 15 ' 107 15 6. 630 = |[12 + (n - 1)3] 1260 = 12n + 3n{n - 1} 1260 = 12n + 3n^ - 3n 3n^ + 9n - 1260 = n^ + 3n - 420= -3 ± -v/ns? n = — 2 Since the given equation does not have a positive integral root, there is no A. P. v/hich meets the stated conditions. Modify the problem so that students are asked to find the number of terms of an A. P. (first term 6, difference 3) for which the sum is 627. T. C.44C Third Course, Unit 1 •■1-44] 2. d = n 2 S - 2na n (n - 1) [n(n - 1) ,t 0] 3. Since S,„ = 98, a = 10, n = 10, and 10 2 S„ - 2na n n (n - 1) then d = 2(98) - 2(10)(10) 10(10 - 1) 196 - 200 90 2 45 4. a = 2Sn - n(n - i)d 2n ^'i' - 30(29)(|) 2(30) 1 - 435 60 217 30 5. (1) 100 = ^[2a + (15 - l)d] 20 (2) 130 = ^[2a + (20 - l)d] (continued on T. C. 44C) T. C. 44B Third Course, Unit 1 '"1-44] ^ : u I ' ■ ; A. (continued) 4. 5. 50 2 12 2 ii 2 1000 ^[2{-3) + 49(0. 5)] = 25(-6 + 24. 5) = 462. 5 [2(9) + ll(-l)] = 6(18 - 11) = 42 [2(2) + 18(2)] = 19(2 + 18) = 380 [2(1) + 999(2)] = 1000(1000) = 1,000,000 8. Correction: Replace '6, 's/2 ' by '6+'v/2', —-[2 ^/2 + 27(3)] = 1134+28n/2 1001 [2(1) + 1000(-5)] = 1001(1 - 2500) = -2,501,499 10. 11. ifiw.,.iooo,-,-^,]= loom - i§5^ 1 = 1 In this case we let the first term of the A. P. be 7 + 4(4). ^[2(7 + 16) + 19(4)] = 10(46 + 76) = 1220 12. 4?- 4_8 2 2[64 + 47(-6)] + 47(-6 ) = 24[128 + 47(-18)] = -17,232 B. I. 2S a = n 2na + n(n - l)d 2S - n(n - l)d 2n , [n # 0] T. C. 44A Third Course, Unit 1 [1.04] [1-44] 4. First 50 terms of -3, -Z. 5, -2, . . . 5. First 12 terms of 9, 8, 7, ... 6. First 19 terms of 2, 4, 6, . . . 7. First 1000 terms of 1, 3, 5, ... 8. First 28 terms of \fl, 3 + 'v/I , 6, \fz , ... 9. First 1001 terms of 1, -4,-9, ... 999 997 10. First 1001 terms of 1, iqq, . iqqi • ••• 11. Twenty successive terms starting with the fifth term of 7, 11, 15, ... 12. Forty-eight successive terms starting with the forty-eighth term of 64, 58, 52, ... B. The boxed theorem on page 1-43 is often abbreviated to: S = f [ 2a + (n - l)d ] n 2 "• ■" 1. Solve this equation for 'a'. 2. Solve the equation for 'd'. 3. Find the common difference of an arithmetic progression whose first term is 10 and the sum of whose first 10 ternns is 98. 4. Find the first term of an arithmetic progression whose common difference is — and the sum of whose first 30 terms is —. 5. Find the first term and the common difference of an arithmetic progression the s\im of whose first 15 terms is 100 and the sum of whose first 20 terms is 130. 6. Given the arithmetic progression with first term 6 and comnnon difference 3. How many terms of this arithmetic progression must be added starting with the first to get the sum 630? UICSM-2-56, Third Course . ■ .Ar-. .,: O'-. ! ..Ju Sfff-r;!! f I.- ' ! -ui '^l Note that each of the sample problems can be "translated' into one involving arithmetic progressions. It is instructive to do this once or twice in order to increase the student's familiarity with S "notation and with arithmetic progressions. For example, in Sample 1. the problem can be viewed as finding the sum of the first 14 terms of the A. P. whose first term is 7 and whose common difference is 4. S,4 =T[14+13- 4] - 7(66) = 462 However, it is probably more innportant that a student develop dexterity with 2-notation by solving the exercises on 1-46 using the S -notation theorems. vl, ^u «.', '1^ '!>. 'r The bracketed note at the bottom of page 1-45 exemplifies the "change of apparent variable" procedure illustrated on T. C. 29A. Each occurrence of 'x' is replaced by 'x + 8' in the expression on the left of ' = '. Of course, the limits of summation are changed automatically. x= 20 X +8 = 20 12 Jx = J (x + 8) = 2(^< + 8) x=9 x+8=9 -^=1 T. C. 45A Third Course, Unit 1 Solution. [1.04] [1-45] C. Compute each of the given sums. Sample 1. ^^ 2 (4x + 3) x=l 14 14 14 2;{4x+3) = 4 y X + 2 3 x= 1 x= 1 x= 1 __ ,^ 14il£j_D ] , 3(14) 28(15) + 42 462 20 Sample Z. Vx x=9 Solution . Note that in this case you are asked to compute a sxim of 12 successive terms starting with the ninth term. We can compute the sum of these 12 terms by- subtracting the sum of the first 8 terms from the sum of the first 20 terms: 20 20 8 Z " = Z " ■ Z" x=9 x=l x=l 20(20 + 1) 8(8 + 1) 2 " 2 = 210 - 36 = 174 20 12 [Note: Do you see that Z ^ "" Z ^'^ ■*" ^^ ^^ x=9 x=l UlCSM-2-56, Third Course [1-46] (b) The property is hereditary . Suppose that, for a given k > 1, Then, for this k, ^"^k + 1 = ^\ + ^^^ + ^) > Tj^ + (2k + 1) = [T^ + (k + 1)] + k = T, , + k k + 1 k + 1. Hence, for every k > 1, [hypothesis] [k>0] if Sq, > T, ^k k thenSqj^^^ > T^^^. Therefore, in view of (a) and (b), by the principle of mathematical induction for counting numbers > 1, the property in question holds for every counting number > 1, [Ask students why the set in question is not the set of all counting numbers. ] Property is that expressed by: Sq . . . - T . . . > ... 2 is a counter-example . Sq^ - T^ = 4 - 3 = 1 < 2. [Note: The property in Exercise 2 is hereditary. Ask students to prove this and then to modify the Exercise so that it becomes a theorem. (Modification: replace '1' by '2'.) The proof of hereditariness is apparent if one analyzes the proof of (b) in Exercise 1 . ] T. C. 46F Third Course, Unit 1 [1-46] 'f> ia--. ■.:-■'.;■ if , 10. Since for every positive integer x, Zuk + i) = zf^^i^n+x. k=i L -^ J the equation in question is equivalent to: x(x + 1) + x = 15 and X + 2x - 15 = (x+5)(x -3) = 15 and x is a positive integer. 3; 3 is the root. *^'^ s.1^ v'^ »,•» ^,-. ^y\ Review Exercises A. 1. Property is that expressed by: Sq . . . > T . . . (a) Z has the property . Sq2 = Sqj + (2 • 1 + 1) = 1 + 3 = 4 T^ = Tj + {1 + 1) = 1 + 2 = 3; and 4 > 3. (continued on T. C. 46F) T. C. 46E Third Course, Unit 1 [1-46] Alternative method for Exercise 7: 18 22 22 2 (x+ 5) = J(x - 4+ 5) = 2{x+ 1) x=-3 x=l x=l 22(23) + 22 = 253 + 22 = 275 12 Yj (11 - 2x) = (11 - 2x)(12 + 5+1) [Ex. 10(b), page 1-33] k= -5 = 18(11 - 2x) The assertion to be proved states that the sum of the first 100 odd numbers and the first 100 even numbers is the sum of the first 200 counting numbers. Students who recognize this need do no more! Mechanical procedure: 100 100 100 2 (2w - 1) + 2]2w = 2(4w - 1) W=:l W=l W=l 200 I w= 1 ^iMioi) .100 100(202 - 1) 20100; 200(201) 2 T. C.46D = 20100. Third Course, Unit 1 [1-46] Alternative method for Exercise 4: 20 17 2(i X- 5) = X [^(x+ 3) 17 x = 4 ,^2 X = i 5] V 1 7 , Z (2 ^-2) x= 1 1 ■l7U8) 2 2 17 -] 2 9 - 7 - - 7(17) = 17 50 50 50 5. 2 (4 - 3z) + J (3z -4) = Z = 0(50) z=l z=l z=l = 6. Change the apparent variable 't' to 'u' in the second term and proceed as in Exercise 5, 18 18 18 7. 2 (x+ 5) = 2 x+ 25- x=-3 x=-3 x=-3 18 I Jx + 2^ + 5(18 + 3+1) Lx= - 3 X=: 1 - _3,, -2) + ,-1), 1-2(12) + no = 165 + 110 = 275 'J" "t" 'l" T. C. 46C Third Course, Unit 1 [1-46] Alternative method for Exercise 3: 19 X+7:=19 Z i^ + '^^) = 2;[2 + 4(x+7)] x = 8 x+ 7 = 8 12 = z (30 + 4x) x= 1 = 12 12 230 + 4^^ x= 1 x= 1 = 30(12) + 2(12)(13) = 360 + 312 = • 672 Note first that 15 , 20 20 E(i-5) x = 4 + x = l6 x=4 Then, 20 1 x = 4 , 20 20 x=4 x=4 20 3 1 " 2 -v- 1 x= 1 -■ - 5(20 -4+1) 1 ~ 2 r 20(21) 3(4] 2 2 1 -86 = 102 - 85 = r r , T. C. 46B ^{i^liili Third Course, Unit \ [1-46] I : 17 17 17 2(3-2x) -13 -21 x=l x= 1 x= 1 3(17) - 2 17(17 + 1) 51 - 306 •255 100 100 100 2. 2(2y+7) = 2 ^y + E^ y=i y=l y=l = 2 ^-^Opi , ,(100) = 10100 + 700 = 10800 19 2(2 + 4x) x= 8 = 19 7 2(2 + 4x) - 2 (2 + 4x) x= 1 x= 1 19 19 7 7 22 + 4 2^-^2 -4 I ^ x=l x=l x=l x= 1 2(19) + 2(19)(20) - 2(7) ■ ■ 2(7)(8) 2(19){21) - 2(7)(9) 2(399 - 63) 672 vl- vO ^ly "f "i- '1- T. C. 46a Third Course, Unit 1 [1.04] [1-46] 17 100 1. X (3 - 2x) 2. l(^y+^) x=l y=l 19 15 20 x=8 x=4 x=l6 50 50 z=l z=l 98 98 u= 1 t"^l 18 12 7.: Y^ {^+ ^) 8. E ^^^ ■ ^''^ x=-3 k=-5 v'»- x>^ >.'. 100 100 200 9. Prove: J (2w - 1) + ^ 2w = ^ ^ w=l w=l w=l X 10. Solve the equation: ^ (^^ + 1) ^ 1^ REVIEW EXERCISES A. Use mathematical induction to prove each of the following generalizations if it is true; give a counter-example if it is false. 1. For every counting number n > 1, the nth square number (Sq^) is greater than the nt_h triangular number (T^). 2. For every counting number n > 1, Sq - T > n. ' " ^n n - (continued on next page) UlCSM-2-56, Third Course ifi-cri U-47] s + t s + t 2 s + t 2 s_+ t 2a + (s + t - l)d (1 - t - s)d + (s + t 2 t«J l)d [Students may be puzzled over this theorem in that it is difficult to imagine arithmetic progressions which behave according to the conditions of the theorem. However, the equation 1 d' pr provides an unlimited number of illustrations.]' T. C. 47G Third Course, Unit 1 ^1-47] i)'- 1 S^Q = 5{2a + 9d) S^ = 3(2a + 5d) Since Sio = h' 10a + 45d = 6a + 15d 4a = -30d a = 15d " 2 • Sl6 = 8(2a + 15d) — 8 \zi-''/) + 15d L - = 8[-15d + 15d] _ 7. Exercise 6 is a special case of the theorem in Exercise 7. S^ = |[2a + (s - l)d] S^ = I [2a + (t - l)d] Since S = S^ , s t sa + s(s - 1) d = ta + t(t - 1) 2 2 t -t-S +Sj _ (t - s)(t + s) - (t - s) , 2 ( t - s)(t + s - 1) , (1 - t - s)d 2 T. C.47F [s - t -i 0] Third Course, Unit 1 ^1-47] <;■. ;;v -'Vw^iSj: Therefore, in view of (a) and (b), by the principle of mathennatical induction for real integers > "3, the property in question holds for every real integer > -3. vt^ sl^ sl^ 'r '•- T 1. ^[l4+15(-8)] 8[14 - 120] = -848 2. 8[8'v/2 + 15{-n/2)] = -56^i2 3. a + 4d = 9 a + 7d - 10 3d = 1 1 23 d = 3, a = 3 • ^16 = ' ''t^lH\) ] 488 ~ 3 ^- '! 2a + 9d = 5 Cj ; 12 2 2a + Ud = 13 -z.-..-f 7 17 d= j^, a = - g S16 = « '17 35 " [-4 + 4 J = 36 5. 50[2a + 99(5)] = 100 2a + 495 =2 2a = -493 ^16 =■ « -493 + 15(5) = 8(-^ T. C. 47E Third ■3344 U-47] !■; 1 (a) 2^ has the property . -3 Yj 2k = 2(-3) = -6; and (-3 + 4)(-3-3) k= -3 (b) The property is hereditary . Suppose that, for a given y^ - 3, y Yj 2k = (y + 4)(y - 3). k= -3 Then, for this y, y+1 y 2 2k = 2 2y + [2{y + 1)] k = - 3 k = -3 -6. = (y + 4)(y - 3) + 2(y + 1) = y +y-12+2y+2 = y + 3y - 10 = (y + 5)(y - 2) = [{y + 1) + 4][(y+ 1) - 3]. Hence, for every y > -3, y ^^ J 2k = (y + 4)(y - 3) k= - 3 then y+1 2 2k = [(y + 1) + 4][(y + 1) - 3]. k= - 3 (continued on T, C. 47E) T. C. 47D Third Course, Unit 1 U-47] I ! ■■I -It.; !'!:..■•":' •n[2(n T l)(2n + 1) - 6(n + 1) + 3] :Tn[4n + 6n + 2 6n - 6 + 3] ^n(4n^ - 1). '!-• 'l-~ '.^ 4. Property is that expressed by: k= 1 ^ 1 is a counter-example because k= 1 = 1" 1 and |(1)^ (1 + 2)^ 9 4 vl, nU *I^ 'J- 'P T The generalization that for every real integer x > 0, L k = -x (X + 1) k = 1 can be proved by mathematical induction. »«^ »u «i^ 5. Property is that expressed by: T. C. 47C Yj 2k = (... + 4)(... -3). k= - 3 (continued on T. C. 47D) Third Course, Unit 1 ^1-47] m)I - (2q+ l)(^q+ 3)(q+ 1) ■(q+ 1) [ {2(q+ 1) - 1} {2(q+ 1) + 1}] hq+ 1)[4 {q+ 1)^ - 1] . then Hence, for every q, q if Yj (2k - 1)^ =\q{4q- " 1) k- 1 q+1 2 (2k - 1)^ = i(q+ l)[4(q+ 1)^ - 1]. k= 1 Therefore, in view of (a) and (b), by the principle of mathematical induction for counting numbers, the property in question holds for every counting number. vl^ vl^ -.•■^ ""4^ 'I" 'I"- The generalization in Exercise 3 can be proved, without using mathematical induction directly, by using the results of earlier exercises and theorems. It is instructive to have students go through an alternative proof such as the following: n n 2 (2k - 1)^ . Yj (4^^ - 4k + 1) k= 1 k= 1 n n n 4^ k^ - 4^ k+ 2 1 k= 1 k= 1 k= 1 T. C. 47B = jn(n + l)(2n + 1) - 2n(n + 1) + n (continued on T. C. 47C) Third Course, Unit 1 ■"1-47] 3. Property is that expressed by: 2 (2k - 1)2 = U...) [4{...)2 - 1]. k= 1 ^ (a) J has the property. 1 J(2k - 1)2 = (Z ■ 1 - 1)2 = 1; and k= 1 |(1)(4- l2- 1) = 1. (b) The property is hereditary . Suppose that, for a given q. 2 (2k - 1)2 . iq(4q2 - 1) k= 1 ^ Then, for this q, q+ 1 q Yj (2k - 1)2 = 2 (2k - 1)2 + [2(q+ 1) - 1]2 k= 1 k= 1 iq(4q2 -1) + (2q + 1)2 iq(2q+ l)(2q- 1) + (2q + 1)2 = (2q+ 1) [§(2q- 1) + (2q + 1)] 2q + 1 ,^ 2 = -3 (2q -q+6q+3) 2q + 1 ,, 2 = —^^'-3 (2q + 5q + 3) (continued on T. C. 47B) T. C. 47A Third Course, Unit 1 [1.04] [1-47] 3. For every counting number n, n y (2k - 1)^ = \n (4n^ - 1). k=l 4. For every real integer x > 0, V^i,3 12, ,,2 k=l 5. For every real integer x > -3, X 2^ 2k = (x + 4)(x - 3) . k= -3 B. Find the sum of the first 16 terms of the arithmetic progression 1. whose first term is 7 and whose common difference is -8. 2. whose first term is 4^/2 and whose common differ?;nce is -n/T . 3. whose fifth term is 9 and whose eighth term is 10. 4. the sum of whose first 10 terms is 5 and the suin of whose first 12 terms is 13. 5. whose common difference is 5 and the sum of whose first 100 terms is 100. 6. the sum of whose first 10 terms is equal to the sum of whose first 6 terms. O^ vU o, '!•• '»* 'I'* 7. Prove : For all counting niimbers s and t, if s 7^ t and if the sum of the first s terms of an arithmetic progression is equal to the sum of the first t terms of that progresLjion then the sum of the first s + t terms is 0. UICSM-2-56, Third Course -48] In this section of the review, our aim is to get students to fornnulate for themselves the multiplication and addition rules for counting nximber exponents. These rules do not need to be stated; students will encounter no difficulty in discovering them. T. C. 48A Third Course, Unit 1 [1-04] 11-48] REVIEW OF V/ORK YvITH COUNTING NUMBER EXPONENTS In earlier courses you learned such facts as: 3 (1) ' 4 ' is an abbreviation for "4X4X4' 5 2 7 (2) 3X3=3 5^ 2 (3) ^ = 5^ ,A\ ^ 2 4 6 (4) For every x, x x = x . /t\ TT A I '^ 3,4 8 12 (5) For every x and y, {x y ) = x y 2 5 (6) For every x and y, if y / then — y- y 2 3 = X y . 3 Statement (1) indicates that you consider the '3' in ' 4 ' as a name for a counting number because it tells the number of factors in ' 4X4X4' 3 However, the '4' in ' 4 ' may be a name for a real number, or a name for a rational number, etc. In the exercises which follow we shall con- tinue to regard exponent symbols as names for counting numbers and regard the symbols to which they are attached as names for real num- bers or, as the case may be, as pronumerals whose domain is the set of all real numbers. In the next unit we shall deal with real number exponents . Consider the expression: 4^ 3 3 We know that ' 4 ' is a name for 64 because ' 4 ' is an abbreviation 3 for '4X4X4' . They symbol ' 4 ' is called an exponential ; the num- 3 3 ber 4 is called the third power of 4 . When 64 is named by ' 4 ' , we say that with respect to the base 4, the exponent of 64 is 3. [V^'ith respect to the base 2, the exponent of 64 is 6.] Note that exponentials are symbols and that powers, bases, and ex- ponents are numbers. UICSM-2-56. Third Course [1.04] fl.49] Study each of the following statements until you are sure you under- stand them. (a) 8 is the third power of 2 2 (b) ' 2' is the exponent symbol in the exponential ' 3 ' (c) 4 is the second power of 4 (d) 16 is the fourth power of 2 2 (e) 4 is the fourth power of 2 (f) 81 is the second power of 9 (g) 81 is the fourth power of 3 (h) 2 X 50 is the second power of 10 2 5 5 (i) (3 ) is the second power of 3 A. Write the simplest name which is not an exponential for the power given in each of the following exercises. 3 Sample. (3.5) Solution. (3. 5)- ^ = 3. 5 X 3.5 X 3.5 = 42.875 1. 2^ 2. 3^ 3. (2.5)^ 4. 2^ 5. 3^ 6. 10^ 7. 15^ 8. 2.3^ 9. 1.1^ 10. (-2)^ 11. (-2)^ 12. (-1)^ 13. (-1)^ 14. (-1)^^ 15. (-2)^ 16. 3^ 17. 3^ 18. 5^ 19. 3^ 20. 7^ 21. 0^^ 22. qIOOO 23. 1.414^ 24. (/T)2 UICSM-2-56, Third Course I' [1.04] [1-50] B. Simplify these exponentials. Leave answers in exponential form. 3 5 Sample 1 . 4X4 Solution. 4^ X 4^ = (4 X 4 X 4) X (4 X 4 X 4 X 4 X 4) = 4 Q 1 , 2 3 3 5 Sample 2 . x y x y Solution . For every x and y, 2335 2335 xyxy =xxyy 4x4x4x4x4x4x4x4 8 = XX X XXX X yyy X yyyyy = xxxxx X yyyyyyyy 5 8 = X y . Sample 3 . 5. 7(5. 7r)(5. 7r)^r^ Solution. For every r, 5.7(5.7r)(5.7r)^r^ = 5.7(5.7r){5.7r)(5, 7r)rrr = (5.7 X 5.7 X 5.7 X 5.7) (rrrrrr) = 5. 7 r . 3 3 2 3 2 1. 2-'2 2. 2-^2"^ 3. XX 4. 4^ • 5 • 4^ 5. x^yx^ 6. 6^6^ 4 9 4 9 5 10 7. 3 3"^ 8. d d^ 9. 19 19 in 5 9 ,, ,7_3,2^6 ,_ 7 3 2 5 10. yy 11.3232 12.xzxz 13. 3^'^3^'^ 14. 3^'^3^'^3^'' 15. 5^9^5'^9 w 2„3 9q ,7 ,106,207 ,„ 674 65 186 16. r 9 r 9 17, 3 3 18, x y x 19, 5^100^6^100^ 20, a^lOoS^lOO^ 21. O^O^^O^O^O^"^^ 22. (5a)^(5ab)^ab)^ 23, 8^63*^0^ ^S^'^ (continued on next page) UICSM-2-56, Third Course [1-0^] [1-bl] 24. (3. 14) (2.28) (3. 14) 25. xS^x-y zx 26. ab^c^r-^'b^c 27. (2, 9)t(2. 9t)" 28. (4.6p) (4.6p) (^.6p)^ 29. {6.7mn) (6 , 7mn)^(6. 7mn) (6.7mn)^ 30. (2ai(4c)(3a)^(Zc:(4a)^(2c'^)^ C. Partition the set of expressions which foj.low by using the relation STANDS FOP. THE SAME NUMBER AS 5 fourth power cf 5 (4 ) 4X4 4 3-*- 3 ^3 ?3 2 5X5 4' 5 X 5 50" v ..J (2 } 4X4 X 4 5 cubed (3^')^ fifth power of 3 4x4x4x4x4 5^ ~ 5^ 5"^' X S'^ 5X5^ the cube of 5 8^ 22 2? ?3 3 '^'- 5 4 X4 (5) (5 X 5 ) -=- 5 4 5' 4^4 5 -r 5 third power of 5 5" 3 (3 ) 4 2 (5 X 5) X 5 fifth power of 4 >. 2 ) 4 to the fourth power 8 5 4 8 3 3 4" 4- 4^ 125 2 X 2 X 2 2^ 8" v 4 fourth power of 4 3 "[{3 X 3) X 3] D. Partition the set of expressions v/hich follow by using the relation IS EQUIVALENT TO. [P.ecall that two ex-pressions which contain pronumerals are equivalent if they give a pair of expressions for the same nuinber upcn each reiilaceixient of the pronumerals by numerals . 1 UICSM-2-56, Third Course uTl^'^^'^ °'' laiNOIS LIBRARY AT URBANA-CHAMPAIGN 52] E. We introduce the student to a recursive definition of exponentials. This type of definition will be met repeatedly in the next unit. Note that we do not define an exponential with exponent symbol ' 1 ' at this point. The student is asked to do so in Exercise 7 on page 1-53. T. C. 52A Third Course, Unit 1 [1.04] [1-52] (xxx)" (x ) , 2.3 (x ) , 2.3 (x ) xxxyyy(y)' 3 8 X X 5, 2,4 X (x ) . 3,4 8 (x ) y 2 3 4 xxx . 2,4 (x ) 2 5 x y , 5,2 (x ) , 4 ,2 (y x) X 3 6 X y X X • x" 3 5 X y y 3, 2,4 X (y ) 3 3 3 XXX X X X X • X third power of x / 2,3 x(x ) X 4 2 X X third power of xy 3 3 2 X y y ninth power of x , 2 2,2 (x X ) 2.4 . 3 2,4 (x y ) x{x ) sixth power of x , 2,3 3 2 (y ) X y X X X I 2,3 (xy ) 2 3 2 x(xy) (y ) / 6 4,2 (x y ) , ,3 3 (xy) y 3 3 X X X to the ninth power 2 7 X X (xy) y 2 4 4 X y xy n, E. We could define ' a ' as an abbreviation for • the product of n factors a' . Instead of this we shall state the following recursive definition. F or every i 2 a -3. ■eal number • a a. and, for every counting number n > 2, n +1 a = n a • a Sample. Prove: 3 = 3 • 3 • 3 • 3 • 3 5 4 Solution. 3=3 "3 = (3-^ . 3) • 3 = [(3^ • 3) • 3] • 3 = [(3 • 3) • 3] • 3 • 3 = 3 • 3 • 3 • 3 • 3. [Why?] UICSM-2-56, Third Course ^3] Hence, for every k, if{-l)^^ = 1 ^, , ,,2(k+l) , then{-l) ^ =1. Therefore, in view of (a) and (b), by the principle of mathematical induction for counting numbers, the property in question holds for every counting number. 6. For every n> 1, by the recursive definition, (-1)2" = (-l)2n-l(.ij But, by the theorem in Exercise 5, for every n, {-!)'" = 1. Hence, for every n > 1, or or 1 = (■ .l)2n- \- •1 1 -1 = (- .l)2n- 1 -1 - (- .l)2n- - 1 • [The theorem in Exercise 6 can also be proven using mathematical induction. ] 7. (a) A reasonable definition: '(-1) ' means -1. (b) For every real number a, 1 a = a and, for every counting niimber n, n+ 1 n a = a • a. T. C. 53B Third Course, Unit 1 ^3] ■*.j Urge students to use the recursive definition in proving the statements 1-4 rather than use their conjectured addition rule. «»'-^ ^'^ «►'. 5. Property is that expressed by: (-1)^ ■ •••= 1 (a) _1_ has the property . By the recursive definition, (-1)^ = {-1){-1) -- (b) The property is hereditary. Suppose that, for a given k. (-1)^^ = 1. Then, for this k, ^_jj2{k+l) ^ ^_^j(2k+l) + 1 = (-1)^^+^ • (-1) = [(-1)^^{-1)][-1] = (-1)^^[(-1)(-1)] [recursive definition] = 1. [inductive hypothesis] T. C. 53A Third Course, Unit 1 [1.04] [1-53] As exercises in the use of this recursive definition prove each of the following: 5 2 3 5 1. (-2) = -32 2. 3^ • 3 = 3^^ 3. (-1)^ = 1 4. (VT)^ = 4;/y 5. Prove that for every counting number n, (-1) =1. [ Hint : Use mathematical induction.] 6. Prove that for every counting nunnber n>l, (-1) =-1. 7. (a) Suggest a definition for ' ("1) ' so that the theorem in Exercise 6 can be extended to every counting number. (b) In light of your ans\wer to (a) franne a reasonable recursive definition of exponentiation which allows every counting number to be an exponent (and every real number to be a base). UICSM-2-56, Third Course x^^ ^ a \^' .A^ 0S'i224622