Web Solutions for How to Read and Do ProofsWeb Solutions for How to Read and Do Proofs An Introduction to Mathematical Thought Processes Third Edition Daniel Solow Department of Operations Weatherhead School of Management Case Western Reserve University Cleveland, OH 44106 e-mail: [email protected] web: http://weatherhead.cwru.edu/solow/ A Wiley-Interscience Publication JOHN WILEY & SONS, INC. New York / Chichester / Weinheim / Brisbane / Singapore / Toronto Contents 1 2 3 4 5 6 7 8 9 Web Solutions to Exercises in Chapter 1 Web Solutions to Exercises in Chapter 2 Web Solutions to Exercises in Chapter 3 Web Solutions to Exercises in Chapter 4 Web Solutions to Exercises in Chapter 5 Web Solutions to Exercises in Chapter 6 Web Solutions to Exercises in Chapter 7 Web Solutions to Exercises in Chapter 8 Web Solutions to Exercises in Chapter 9 1 3 7 13 15 19 21 23 27 29 v 10 Web Solutions to Exercises in Chapter 10 . vi CONTENTS 11 Web Solutions to Exercises in Chapter 11 12 Web Solutions to Exercises in Chapter 12 13 Web Solutions to Exercises in Chapter 13 31 35 39 . F = false) C T F T F T F T F (A ⇒ B ) T T F F T T T T (A ⇒ B ) ⇒ C T F T T T F T F a. From part (a). r is a real and satisfies r2 = 2. you need a value for x so that the hypothesis A is true and the conclusion B is false. √ p and q are positive real numbers with pq = (p + q )/2.11 B T T F F T T F F (T = true. Hypothesis: Conclusion: c. Any value of x with 0 < x < 1 does this.1. The sum of the first n positive integers is n(n + 1)/2. p = q. Hypothesis: Conclusion: n is a positive integer. “If x > 0. 1. According to Table 1. you should try to show that A is true and B is false because this is the only time that “A implies B ” is false.10 A T T T T F F F F 1.4 a. 1 .1 Web Solutions to Exercises 1. b. r is irrational. then log7 (x) > 0” is false. Hypothesis: Conclusion: b. to show that the statement. . How can I show that an integer (namely. you start with the statement B that you are trying to conclude is true. With the backward process.1 The forward process is a process that makes specific use of the information contained in the hypothesis A. just as the last statement in the forward process helps you choose the right key question and answer. You then derive from A a sequence of new statements that are true as a result of A being true.2 Web Solutions to Exercises 2. 2n2 − 3n = −2)? How can I show that an integer (namely. Every new statement derived from A is directed toward linking up with the last statement obtained in the backward process. 2n2 − 3n + 2)? 3 . you derive a sequence of new statements with the property that if the sequence of new statements is true. n) is the root of a quadratic equation (namely. you begin with the statement A that you assume is true. With the forward process. The last statement of the backward process acts as the guiding light in the forward process. n) satisfies a given equation (namely.8 a. n2 ) is even? How can I show that the square of an integer (namely. The backward process is a process that tries to find a chain of statements leading to the fact that the conclusion B is true. How can I show that an integer (namely. then B is true. By asking and answering key questions. n) is even? b. The backward process continues until you obtain the statement A or until you can no longer ask and/or answer the key question. 2. the two lines are the legs of a right triangle. cn = a2 cn−2 + b2 cn−2 > a2 (an−2 ) + b2 (bn−2 ). from sentence 1. (1) How can I show that a triangle is equilateral? (2) Show that the three sides have equal length (or show that the three angles are equal). For sentence 1: The fact that cn = c2 cn−2 follows by algebra. Show Show Show b. cn−2 > an−2 and cn−2 > bn−2 and so.22 a. both sets are equal to a third set.20 For sentence 2: For sentence 3: 2. Because n > 2. (3) Show that RT = ST = SR (or show that R = S = T ). c > b. b. The author then substitutes c2 = a2 + b2 . Algebra from sentence 2. . 2. For a right triangle. the two lines intersect at a 90 degree angle. 2.13 a. the hypotenuse c is longer than either of the two legs a and b so. the elements of the two sets are identical. Show Show Show that that that that that that the product of the slopes of the two lines is −1. which is true from the Pythagorean theorem applied to the right triangle. c > a.11 a. each set is a subset of the other.4 WEB SOLUTIONS TO EXERCISES IN CHAPTER 2 2. (3) Show that the solution −b/2a is positive. The number to the left of each line in the following figure indicates which rule is used. (1) How can I show that the solution to a quadratic equation is positive? (2) Show that the quadratic formula gives a positive solution. A key question associated with the conclusion is.WEB SOLUTIONS TO EXERCISES IN CHAPTER 2 5 b. A3 : SU = SU . The number to the left of each line in the following figure indicates which rule is used. work forward from the hypothesis to establish that B 2 : triangle RSU is congruent to triangle SU T . so A1 : RU = U T . c. SU is a perpendicular bisector of RT . “How can I show that a triangle is equilateral?” One answer is to show that all three sides have equal length. A : A1 : A2 : A3 : A4 : B1 : B: s given ss rule 1 ssss rule 1 sssst rule 4 tst rule 3 s given tst from part (c) tsttst rule 1 tsst rule 2 tssttsst rule 1 tsssst rule 2 ttst rule 3 2. B 1 : RS = ST = RT . Working forward from the hypothesis you can obtain this because . It remains (from B 1) to show that B 3 : RS = RT . Specifically. from the hypothesis. specifically. Thus the side-angle-side theorem states that the two triangles are congruent and so B2 has been established.25 Analysis of Proof. To see that RS = ST . In addition. A2 : RU S = SU T = 90o . A: A1 : A2 : B1 : B: d. Proof. by the hypothesis. it will be shown that RS = ST = RT . . Hence. To see that triangle RST is equilateral. To see that RS = RT .6 WEB SOLUTIONS TO EXERCISES IN CHAPTER 2 A4 : RS = 2RU = RU + U T = RT . RS = ST . the hypothesis that SU is a perpendicular bisector of RT ensures (by the side-angle-side theorem) that triangle RSU is congruent to triangle SU T . To that end. one can conclude that RS = 2RU = RU + U T = RT . ” A B “A OR B ” T T F F T F T F T T T F 3. A1 : n = 2k + 1. d. A : a. A : Triangle RST is equilateral. where p and q are two integers. where p and q are integers with q = 0. F = false) a. e.3 Web Solutions to Exercises 3. A1 : s = p/q .3 (A is the hypothesis and A1 is obtained by working forward one step. and R = S = T . where a = 0 and b = 0 are integers. Also. A1 : RS = ST = RT . (T = true. b. A : n is an odd integer. c. A1 : b = pa and c = qb. A : s and t are rational numbers with t = 0. where k is an integer. A : sin(X ) = cos(X ). c are integers for which a|b and b|c.6 7 . A1 : x/z = y/z (or x = y ). b. Truth Table for “A OR B .) a. t = a/b. n2 ) is odd?” The definition for an odd integer is used to answer the key question. for some integer m. “How can I show an integer (namely. so A4 : n2 = 2(2m2 + 2m) + 1. . so A3 : n2 = 4m2 + 4m + 1. Both are false when A is true and B is false and true otherwise.” A B “A AN D B ” T T F F T F T F T F F F c. Truth Table for “A AN D B . Truth Table for “(N OT A) OR B . 3. A1 : n = 2m + 1. by definition. for some integer k .” A N OT A B “(N OT A) OR B ” “A Implies B ” T T F F F F T T T F T F T F T T T F T T “A implies B ” and “(N OT A) OR B ” are equivalent. Therefore. Truth Table for “A AN D N OT B ” A B N OT B “A AN D N OT B ” T T F F T F T F F T F T F T F F d. The forward process is now used to reach the desired conclusion. which means you have to show that B 1 : n2 = 2k + 1. A2 : n2 = (2m + 1)2 .8 WEB SOLUTIONS TO EXERCISES IN CHAPTER 3 b. Because n is an odd integer. The forward-backward method gives rise to the key question.9 Analysis of Proof. and D ⇒ A) required to show that A is equivalent to each of the three alternatives. however. The hypothesis states that “A implies B ” so it remains to show that B 2 : “B implies A”.” “B implies C . and because m is an integer. by the result in Exercise 3. follows from the hypothesis that “B implies C ” and “C implies A” so. C ⇒ D. The hypothesis states that “C implies A” so it remains to show that B 2 : “A implies C ”.” you have that “A implies D. If the four statements in part (a) are true. it must be shown that B 1 : “A implies C ” and “C implies A”. The proof is now complete. C ⇒ A. “A implies C ” and hence B 2 is true. 3. A ⇒ D.12.13. “How can I show that two statements are equivalent?” By definition. one is led to the key question.” and “C implies D.WEB SOLUTIONS TO EXERCISES IN CHAPTER 3 9 Thus the desired value for k in B 1 is 2m2 + 2m.15 a. “How can I show that two statements are equivalent?” By definition.13 To show that A is equivalent to B .” b. A ⇒ C . however. B ⇒ A. “B implies A” and hence B 2 is true. 3. and therefore. then you can show that A is equivalent to any of the alternatives by using Exercise 3. n2 = = = (2m + 1)2 4 m2 + 4 m + 1 2(2m2 + 2m) + 1. Proof. follows from the hypothesis that “A implies B ” and “B implies C ” so. 2m2 + 2m is an integer. Hence it has been shown that n2 can be expressed as 2 times some integer plus 1.13. Likewise. B ⇒ C . there is an integer m for which n = 2m+1. which completes the proof. to show that A is equivalent to C .” By Exercise 3. This. by the result in Exercise 3. you already know that “D implies A. The advantage of the approach in part (a) is that only four proofs are required (A ⇒ B . . and D ⇒ A) as opposed to the six proofs (A ⇒ B . to show that A is equivalent to D. For instance. one is led to the key question. Because n is an odd integer. because “A implies B . This. Hence n2 is an odd integer.12. it must be shown that B 1 : “A implies B ” and “B implies A”. 18. w2 = u2 +v 2 . u/2v = u/w. and by algebraic manipulations. “How can I show that a triangle is isosceles?” Using the definition of an isosceles triangle. Furthermore. Thus it must be shown that √ B 1 : w = 2uv. or. Specifically. thus completing the proof. you have that A5 : u2 + v 2 = 2uv . sin(U ) = u/w. A3 : w2 = 2uv . w2 = 2uv . The forward-backward method gives rise to the key question. as in the proof in Exercise 3. u2 − 2uv + v 2 = 0. one has u = v . or. A1 : u/2v = u/w.17 Analysis of Proof. By the definition of sine. (Observe also that triangle U V W is a right triangle.) . Proof. To verify the hypothesis of Proposition 3 for the current triangle U V W . On factoring A5 and then taking the square root of both sides of the equality. On substituting 2uv for w2 and then performing algebraic manipulations. in this case. and t = w. 3.10 WEB SOLUTIONS TO EXERCISES IN CHAPTER 3 3. or. you obtain A3 : w2 = 2uv . you know that A1 : sin(U ) = u/2v. so A2 : u/2v = u/w. it is necessary to match up the notation. Now from the Pythagorean theorem. it follows that A6 : u − v = 0 and so u = v . A4 : u2 + v 2 = w2 . by the Pythagorean theorem. completing the proof. Working forward from the hypothesis. or. you must show that two of its sides are equal which. one obtains precisely B 1. s = v .19 Analysis of Proof. Substituting for w2 from A3 in A4. On taking the positive square root of both sides of the equality in A3. But. A2 : u/2v = u2 /w2 . means you must show that B 1 : u = v. r = u. Because sin(U ) = u/2v and also sin(U ) = u/w. By the hypothesis. thus one has u/2v = u/w. sin(U ) = u/w √. Hence the hypothesis of Proposition 3 holds for the current right triangle U V W . . By algebraic manipulations one obtains w = uv. sin(U ) = u/2v and from the definition of sine. and consequently the triangle is isosceles.WEB SOLUTIONS TO EXERCISES IN CHAPTER 3 11 Proof. . The forward-backward method gives rise to the key question. A triangle XY Z is isosceles if ∃ two sides − ⊃ their lengths are equal. 13 . The appearance of the quantifier “there are” suggests turning to the forward process to construct the desired p and q . Factoring x2 − 5x/2 + 3/2 = 0 yields (x − 3/2)(x − 1) = 0. when substituted in x2 − 5x/2 + 3/2 is 0. so x = 1 or x = 3/2. one answer is to show that B 1 : there are integers p and q with q = 0 such that s/t = p/q . ⊃ p(r1 ) = · · · = p(rn ) = 0. For a polynomial of degree n. b. .7 Analysis of Proof.4 Web Solutions to Exercises 4. From the hypothesis that s and t are rational numbers. Factoring this quadratic means you want to find a real number x such that (x − 3/2)(x − 1) = 0. there exists a real number. ∃ exactly n complex numbers.11 Analysis of Proof. “How can I show that a real number (namely. . Thus. x = 1 or x = 3/2. rn − 4. r1 . The appearance of the key words “there is” in the conclusion suggests using the construction method to find a real number x such that x2 − 5x/2 + 3/2 = 0. namely. such that x2 − 5x/2 + 3/2 = 0. So the desired real number is either x = 1 or x = 3/2 which. . Proof. s/t) is rational?”” By the definition. say p(x). The real number is not unique. 4. and by the definition.3 a. The real number is not unique as either x = 1 or x = 3/2 works. . 14 WEB SOLUTIONS TO EXERCISES IN CHAPTER 4 A1 : there are integers a. and hence s/t is rational. thus completing the proof.14 The proof is not correct. c. Proof. c. q = 0. c = 0. 4. thus completing the proof. b. . in fact. A2 : s/t = (a/b)/(c/d) = (ad)/(bc). Because s and t are rational. and noting that q = 0. Hence. and thus bc = 0. and d with b = 0 and d = 0 such that s = a/b and t = c/d. Because t = 0. s/t = p/q . there are integers a. Constructing p = ad and q = bc. b. also. Observe that because b = 0 and c = 0. The mistake occurs because the author uses the same symbol x for the element that is in R ∩ S and in S ∩ T where. the element in R ∩ S need not be the same as the element in S ∩ T . one has s/t = (a/b)/(c/d) = (ad)/(bc) = p/q . Because t = 0. c = 0. and d with b = 0 and d = 0 such that s = a/b and t = c/d. So the desired integers p and q are p = ad and q = bc. 5. for which it must be shown that B 1 : t is an upper bound for the set S . A key question associated with B 1 is. c. The appearance of the quantifier “for every” in B 2 suggests using the choose method. The appearance of the quantifier “for every” in the conclusion suggests using the choose method.7 a. sin(2t) = 2 sin(t) cos(t).5 Web Solutions to Exercises 5. ∀ real numbers x and y with x < y . ∀ angle t. this one is taller than the others.11 Analysis of Proof. whereby one chooses A2 : an element x ∈ S . one must show that B 2 : for every element x ∈ S . pq ≥ (p + q )/2. √ ∀ nonnegative real numbers p and q . ∃ a rational number r − ⊃ x < r < y. d. for which it must be shown that B 3 : x ≤ t. b. 15 . whereby one chooses A1 : an element t ∈ T . x ≤ t. “How can I show that a real number (t) is an upper bound for a set (S )?” By definition. ∃ a mountain − ⊃ ∀ other mountains. C ) is convex?” One answer is by the definition. and for all . so A4 : x ≥ 0 and x − 3 ≤ 0. Proof. Let t be a real number with 0 ≤ t ≤ 1. So. and adding yields a[tx + (1 − t)y ] ≤ b. But. for which it must be shown that B 2 : tx + (1 − t)y ∈ C . A2 : ax ≤ b and ay ≤ b. either x ≥ 0 and x − 3 ≤ 0. that is. or. tx + (1 − t)y ∈ C . But the latter cannot happen. Combining A5 and A6 yields B 3. Turning to the forward process. work forward from A2 and the definition of the set S in the hypothesis to obtain A3 : x(x − 3) ≤ 0. because x and y are in C (see A1). x ≤ 0 and x − 3 ≥ 0. 5. and for all real numbers t with 0 ≤ t ≤ 1. respectively. thus completing the proof. for every real number t with 0 ≤ t ≤ 1. Multiplying both sides of these inequalities by t ≥ 0 and 1 − t ≥ 0. Hence. thus completing the proof.17 Analysis of Proof. tx + (1 − t)y ∈ C . and a real number t with 0 ≤ t ≤ 1. The forward-backward method gives rise to the key question. Performing algebraic manipulations on A3 yields B 2. The appearance of the quantifiers “for all” in B 1 suggests using the choose method to choose A1 : elements x and y in C . a[tx + (1 − t)y ] ≤ b. respectively. from A1 and the definition of the set T in the hypothesis A6 : t ≥ 3. From A3.16 WEB SOLUTIONS TO EXERCISES IN CHAPTER 5 To do so. Multiplying both sides of the two inequalities in A2. and let x and y be in C . Then ax ≤ b and ay ≤ b. whereby it must be shown that B 1 : for all elements x and y in C . From A4. by the nonnegative numbers t and 1 − t (see A1) and adding the inequalities yields: A3 : tax + (1 − t)ay ≤ tb + (1 − t)b. “How can I show that a set (namely. A5 : x ≤ 3. WEB SOLUTIONS TO EXERCISES IN CHAPTER 5 17 elements x and y in C . . tx + (1 − t)y ∈ C . C is a convex set and the proof is complete. Therefore. . and 0 ≤ t ≤ 1.11 Analysis of Proof. 6. and for all t with 0 ≤ t ≤ 1. The appearance of the quantifier “for all” in the backward process suggests using the choose method to choose A2 : real numbers x and y .6 Web Solutions to Exercises 6. “How can I show that a function (namely. You do not know that the something happens for objects that do not satisfy the certain property. one answer is to show that B 2 : for all real numbers x and y . Therefore. s f ) is convex?” Using the definition in Exercise 5. 19 . The appearance of the quantifier “for all” in the conclusion indicates that you should use the choose method to choose A1 : a real number s ≥ 0.1 The reason you need to show that Y has the certain property is because you only know that the something happens for objects with the certain property. for which it must be shown that B 1 : the function s f is convex.1(e). An associated key question is. if you want to use specialization to claim that the something happens for this particular object Y . you must be sure that Y has the certain property. s f (tx + (1 − t)y ) ≤ ts f (x) + (1 − t)s f (y ). Let s ≥ 0. it follows from the definition that f (t x + (1 − t )y ) ≤ t f (x ) + (1 − t )f (y ). Proof. Specializing the statement in A3 to x = x . let x and y be real numbers. The desired statement B 3 is obtained by multiplying both sides of the inequality in A4 by the nonnegative number s . It will be shown that s f (t x + (1 − t )y ) ≤ t s f (x ) + (1 − t )s f (y ). and for all t with 0 ≤ t ≤ 1. thus completing the proof. Because f is a convex function by hypothesis. you have that A3 : for all real numbers x and y . f (tx + (1 − t)y ) ≤ tf (x) + (1 − t)f (y ). The desired result is obtained by multiplying both sides of this inequality by the nonnegative number s . The desired result is obtained by working forward from the hypothesis that f is a convex function. To show that s f is convex. By the definition in Exercise 5. . and t = t (noting that 0 ≤ t ≤ 1) yields A4 : f (t x + (1 − t )y ) ≤ t f (x ) + (1 − t )f (y ). and let t with 0 ≤ t ≤ 1. y = y .1(e).20 WEB SOLUTIONS TO EXERCISES IN CHAPTER 6 for which it must be shown that B 3 : s f (t x + (1 − t )y ) ≤ t s f (x ) + (1 − t )s f (y ). whereby one must show that B 1 : there is a real number y such that for every real number x. that is. that is: B 2 : for every real number x. x2 − 2x + 1 ≥ 0.9 Analysis of Proof. But because x2 − 2x + 1 = (x − 1)2 . Thus B 3 is true. for which it must be shown that B 3 : −x2 + 2x ≤ 1. −x2 + 2x ≤ 1. completing the proof. −x2 + 2x ≤ y . 21 . The forward-backward method gives rise to the key question. Now it must be shown that this value of y = 1 is correct. Trial and error might lead you to construct y = 1 (any value of y ≥ 1 will also work). The appearance of the quantifier “there is” in B 1 suggests using the construction method to produce the desired value for y . The appearance of the quantifier “for all” in B 2 suggests using the choose method to choose A1 : a real number x. this number is always ≥ 0. “How can I show that a function (namely.7 Web Solutions to Exercises 7. f ) is bounded above?” One answer is by the definition. for which it must be shown that B 1 : there is an element x ∈ S such that x > 1 − . you can write S as follows: S = {real numbers x : there is an integer n ≥ 2 with x = 1 − 1/n}. It then follows from the defining property that x = 1 − 1/n ∈ S and by the choice of n that x = 1 − 1/n > 1 − 1/ . Proof. namely. noting that > 0 from A1. the construction method is used to produce the desired element in S . that x = 1 − 1/n > 1 − . Then x2 − 2x + 1 = (x − 1)2 ≥ 0. To see that there is an element x ∈ S such that x > 1 − . if you let n > 2 be any integer > 1/ . from the hint. let n > 2 be an integer for which n > 1/ .” so the choose method is used to choose A1 : a real number > 0. thus completing the proof. To that end. n > 1/ . you want x = 1 − 1/n > 1 − . from B 1. it will be shown that for all real numbers x. Thus. 1/n < . To see that the function f (x) = −x2 + 2x is bounded above.11 Analysis of Proof. . 7. there is an element x ∈ S such that x > 1 − . −x2 + 2x ≤ 1. you can construct x = 1 − 1/n.22 WEB SOLUTIONS TO EXERCISES IN CHAPTER 7 Proof. then x = 1 − 1/n ∈ S satisfies the desired property in B 1. To that end. for an appropriate choice of the integer n ≥ 2. The first key words in the conclusion from the left are “for all. It has thus been shown that for every real number > 0. that is. The proof is now complete. let x be any real number. To find the value for n. that is. In summary. Recognizing the key words “there is” in B 1. Let > 0. thus completing the proof. From A4. rewriting A1 by algebra.11 is: Analysis of Proof. or A3 : n3 − 6n2 = 2. and n2 ≥ 49.8 Web Solutions to Exercises 8. A contradiction is reached by showing that B 1 : n2 (n − 6) = 2 and n2 (n − 6) ≥ 49. By contradiction. because n2 > 0. To that end. This contradiction completes the proof. so A6 : n2 (n − 6) ≥ n2 ≥ 49. n − 6 ≥ 1. or A4 : n2 (n − 6) = 2. and n + 1 are consecutive positive integers. that A : n − 1. you have: A2 : n3 + 3n2 + 3n + 1 = n3 + n3 − 3n2 + 3n − 1. But when n ≥ 7. it must be that A5 : n − 6 > 0. 23 . assume A and N OT B . Now A6 contradicts A4 because A6 states that n2 (n − 6) ≥ 49 while A4 states that n2 (n − 6) = 2. n ≥ 7. that is. n. A1 (N OT B ): (n + 1)3 = n3 + (n − 1)3 . Because n is the .15 that Analysis of Proof. . and n + 1 satisfy (n + 1)3 = n3 + (n − 1)3 . This contradicts the fact that n2 (n − 6) = 2. A contradiction is reached by showing that B 1 : p ≤ n and p > n. Assume. . Because n is the largest prime and p is prime. observe that when n! + 1 is divided by 2.24 WEB SOLUTIONS TO EXERCISES IN CHAPTER 8 Proof. there is a remainder of 1. one has that n! 1 n! + 1 = + . there will be a prime number that is larger than all the other prime numbers. Assume. where 1 < r ≤ n. Expanding these expressions and rewriting yields n2 (n − 6) = 2. Let p be any prime divisor of n! + 1. n ≥ 7. that there are a finite number of primes. that the three consecutive integers n − 1. which contradicts the assumption that n is the largest prime number. But then n2 (n − 6) ≥ n2 ≥ 49. From A1. Therefore the prime factor p is greater than n. So. n − 6 > 0. and let A3 : p be any prime number that divides n! + 1. it can be assumed A1 : the number of primes is finite. r r r Hence n! + 1 has no prime factor between 1 and n. Consider the number n! + 1. A2 : let n be the largest prime number... To see that B 2 is true. thus completing the proof. there is a remainder of 1 because. By the contradiction method. to the contrary. to the contrary. p = 3.. thus completing the proof.(2)(1)] + 1. . Proof. p = n. It remains to show that p > n. n. Indeed. Let n be the largest prime. use the fact that p divides n! + 1 to show that B 2 : p = 2. Because n2 > 0. Similarly. 8. that is. when n! + 1 is divided by r. n! + 1 = [n(n − 1). To do so. p ≤ n. 19 Analysis of Proof. and d is even (see A7). is even. b is odd (±1) and d is even (±2) or vice versa. ac is even (from A9). and d are integers. A similar contradiction is reached in Case 2. it follows that A3 : A4 : A5 : A6 : a + c = 0. It must therefore be the case that A8 : bc is even. that A1 (N OT B ): the polynomial x4 + 2x2 + 2x + 2 can be expressed as the product of the two polynomials x2 + ax + b and x2 + cx + d in which a. This is because b is odd (see A7). A10 : the left side of A4 = odd + even + even = odd. But then. you have that A2 : x4 + 2x2 + 2x + 2 = (x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (b + ac + d)x2 + (bc + ad)x + bd. b is odd. so is ad. b + ac + d = 2. Assume. Suppose. to the contrary. which is a contradiction. This establishes a contradiction for Case 1 in A7. p must be less than or equal to n. Equating coefficients of like powers of x on both sides. bd = 2. c. From A6. namely. thus completing the proof. (Subsequently. the case when b is even and d is odd is considered. so it must be that A9 : c is even. However. Because d is even. from A7. because the right side of A5 is even.) It now follows that. 8. 2. when b is even and d is odd. bc + ad = 2. Working forward by multiplying the two polynomials.WEB SOLUTIONS TO EXERCISES IN CHAPTER 8 25 largest prime. A10 is a contradiction because the right side of A4. b. that A7 : Case 1: b is odd and d is even. first. However. the left side is also. Hence p > n. But n! + 1 cannot be divided by any number between 1 and n. . . It must be shown that B 1 (N OT A): there is an obtuse angle. say m.9 Analysis of Proof. 27 . But c = m + m2 = m(m + 1) is even because the product of two consecutive integers is even. With the contrapositive method. to the equation n2 + n − c = 0. Proof. But from A1. It must be shown that B 1 (N OT A): c is not odd. to the equation n2 + n − c = 0. It will be shown that c is even.9 Web Solutions to Exercises 9. 9. you assume that A1 (N OT B ): there is an integer solution. say m. you can assume A1 (N OT B ): the quadrilateral RST U is not a rectangle. Assume that there is an integer solution. you have that A2: c = m + m2 . thus establishing B 1 and completing the proof.11 that Analysis of Proof. By the contrapositive method. that is. c is even. Observe that m + m2 = m(m + 1) is the product of two consecutive integers and is therefore even. .28 WEB SOLUTIONS TO EXERCISES IN CHAPTER 9 The appearance of the quantifier “there is” in B 1 suggests turning to the forward process to produce the obtuse angle. because the sum of all the angles in RST U is 360 degrees. Assume that the quadrilateral RST U is not a rectangle. Otherwise the remaining three angles add up to more than 270 degrees. Proof. Thus one of the remaining three angles is obtuse. An obtuse angle will be found. The proof is now complete. one of them must be greater than 90 degrees. say. then it is the desired obtuse angle. A3 : angle R has less than 90 degrees. This in turn means that A4 : the remaining angles of the quadrilateral must add up to more than 270 degrees. and hence. Otherwise. one of its angles. angle R. Working forward from A1. Among these three angles that add up to more than 270 degrees. say. then it is the desired angle and the proof is complete. is not 90 degrees. you can conclude that A2: at least one angle of the quadrilateral is not 90 degrees. If angle R has more than 90 degrees. and that is the desired obtuse angle. R. If angle R has more than 90 degrees. > 0 and. you can assume the hypothesis that A : x ≥ 0. (Note that this construction is possible because −x > 0. that x < 0. thus completing the proof. Assume. and the proof is complete. When using the contradiction method. x + y = 0. because < −x. and also that A1 (N OT B ): either x = 0 or y = 0. 29 . because x < 0. you can assume A1 (N OT B ): x < 0. Recognizing the key words “there is” in B 1. By the contrapositive method. To that end.9 Analysis of Proof. 10. from A1. It will be shown that there is a real number > 0 such that x < − . It must be shown that B 1 (N OT A): there is a real number > 0 such that x < − . the construction method is used to produce the desired > 0. construct as any value with 0 < < −x (noting that this is possible because −x > 0).10 Web Solutions to Exercises 10. to the contrary.) By design. construct as any value with 0 < < −x. it follows that x < − .7 Analysis of Proof. Clearly > 0 and because < −x. y ≥ 0. Turning to the forward process to do so. Proof. x < − . Thus has all the needed properties in B 1. Because x ≥ 0 from A. y ≥ 0. but this contradicts the fact that x ≥ 0. A contradiction to the fact that y ≥ 0 is reached by showing that B 1 : y < 0. Assume that x ≥ 0. If x = 0. then y > 0. A4 : y = −x. . it must be that A3 : x > 0. x + y = 0. if y = 0. then x > 0 and y = −x < 0. and x = −y < 0. Specifically. Because −x < 0 from A3. because x + y = 0 from A. a contradiction has been reached. A similar argument applies for the case where y = 0 (see A1). suppose first that A2 : x = 0. Proof. but this contradicts the fact that y ≥ 0.30 WEB SOLUTIONS TO EXERCISES IN CHAPTER 10 From A1. Similarly. and that either x = 0 or y = 0. 31 . from A2 and A3. To establish the uniqueness by the indirect uniqueness method. suppose that A3 : y is a real number with y = x such that my + b = 0.11 Web Solutions to Exercises 11. According to the indirect uniqueness method. Specifically. A contradiction to the hypothesis that m = 0 is reached by showing that B 1 : m = 0. A4 : mx + b = my + b. you can construct A1 : x = −b/m. But because the hypothesis states the m = 0.3 Analysis of Proof. This value is correct because A2 : mx + b = m(−b/m) + b = −b + b = 0. one must first construct a real number x for which mx + b = 0. Subtracting the right side of the equality in A4 from the left side and rewriting yields A5 : m(x − y ) = 0. “for every integer n ≥ n0 . b. it must be that m = 0. First it is shown that P (n) is true for n = 5. P (n) is true” is when you can relate P (n + 1) to P (n). With the choose method. for which it must be shown that B 1 : P (n) is true. Hence P (n) is true for n = 5. Then mx+b = my +b. Starting with the left side of P (n + 1). 11. and so m(x − y ) = 0. Then mx + b = m(−b/m) + b = 0. This last statement is true because. you cannot use the assumption that P (n − 1) is true to do so. the statement will not have been proved for such values. To construct the number x for which mx + b = 0. . you must then prove that P (n + 1) is true. This contradiction establishes the uniqueness. If you were to use the choose method. So assume P (n) : 2n > n2 . 2n2 > (n + 1)2 = n2 + 2n + 1.32 WEB SOLUTIONS TO EXERCISES IN CHAPTER 11 On dividing both sides of the equality in A5 by the nonzero number x − y (see A3).12 Proof. so 25 > 52 . for then you can use the induction hypothesis that P (n) is true. To obtain P (n + 1). that (n − 1)2 > 2.18 Proof. you would choose A1 : an integer n ≥ n0 . But because x − y = 0. and this should help you establish that P (n + 1) is true. by subtracting n2 + 2n − 1 from both sides and factoring.8 a. It must be shown that P (n + 1) : 2n+1 > (n + 1)2 . you have: 2n+1 = 2(2n ) > 2(n2 ). let x = −b/m (because m = 0). it must still be shown that for n > 5. and this contradicts the hypothesis that m = 0. for n > 5. it follows that m = 0. As a result. (n − 1)2 ≥ 42 = 16 > 2. Proof. But 25 = 32 and 52 = 25. Assuming that P (n) is true. 11. or. 11. For n = 1 the statement becomes: P (1) : [cos(x) + i sin(x)]1 = cos(1x) + i sin(1x). It is not possible to use induction when the object is not an integer because showing that P (n) implies P (n + 1) may “skip over” many values of the object. and using the fact that P (n) is true. Now suppose that y = x and also satisfies my +b = 0. The time to use induction instead of the choose method to show that. A similar problem arises throughout the rest of the proof.23 The author relates P (n + 1) to P (n) by using the product rule of differentiation to express [x(xn )] = (x) (xn ) + x(xn ) .WEB SOLUTIONS TO EXERCISES IN CHAPTER 11 33 Now P (1) is true because cos(x) + i sin(x) = cos(x) + i sin(x). . Now assume the statement is true for n. The author then uses the induction hypothesis to replace (xn ) with nxn−1 . This establishes that P (n + 1) is true. Using the fact that P (n) is true and the facts that cos(a + b) = sin(a + b) = you have: [cos(x) + i sin(x)]n+1 = [cos(nx) + i sin(nx)][cos(x) + i sin(x)] = [cos(nx) cos(x) − sin(nx) sin(x)]+ i[sin(nx) cos(x) + cos(nx) sin(x)] = cos((n + 1)x) + i sin((n + 1)x). the right side of P (1) is undefined since you cannot divide by zero. 11. [cos(x) + i sin(x)]n+1 = [cos(x) + i sin(x)]n [cos(x) + i sin(x)]. that is: P (n + 1) : [cos(x) + i sin(x)]n+1 = cos((n + 1)x) + i sin((n + 1)x). thus completing the proof. cos(a) cos(b) − sin(a) sin(b). 11. It must be shown that P (n + 1) is true.26 The proof is incorrect because when n = 1 and r = 1. Starting with the left side of P (n + 1). that is: P (n) : [cos(x) + i sin(x)]n = cos(nx) + i sin(nx). sin(a) cos(b) + cos(a) sin(b). . it follows that A2 : x3 + 3x2 − 9x − 27 = (x − 3)(x + 3)2 ≥ 0. without providing details. The author then works forward from this information to establish the contradiction that the left side of (2) is odd and yet is equal to the even number 2. 35 . or x + 3 ≤ 0. the author considers the following two cases: Case 1: The factor b is odd. 12.5 The author uses a proof by cases when the following statement containing the key words “either/or” is encountered in the forward process: A1 : either the factor b is odd or the factor d is odd. Case 2: The factor d is odd. It must be shown that B 1 (C ): x ≤ −3. According to this either/or method. By factoring A. that this case also leads to a contradiction. The author claims. you can assume that A : x3 + 3x2 − 9x − 27 ≥ 0.8 Analysis of Proof. Accordingly.12 Web Solutions to Exercises 12. and A1 (N OT D): x < 3. so.9 Analysis of Proof. so x + 3 = 0. 12. Proof. x = −3. it must be that A4 : (x + 3)2 = 0. If b = 0 then. or.36 WEB SOLUTIONS TO EXERCISES IN CHAPTER 12 From A1. x = −3. Then it follows that x3 + 3x2 − 9x − 27 = (x − 3)(x + 3)2 ≥ 0. by the definition it follows that: A2 : there is an integer k such that b = ka. from A3. Accordingly. Working forward from the hypotheses that a|b and b|a. x + 3 = 0. Thus. on dividing both sides of the equality in A4 by b you obtain: A5 : km = 1. Because x < 3. Assume that x3 + 3x2 − 9x − 27 ≥ 0 and x < 3. you can assume the hypothesis and A1 : a = b. The key words “either/or” in B now suggest proceeding with a proof by elimination. Observe that the conclusion can be written as: B : either a = b or a = −b. because x < 3. from A3. dividing both sides of A2 by x − 3 yields A3 : (x + 3)2 ≤ 0. A3 : there is an integer m such that a = mb. Thus. completing the proof. Therefore. From A5 and the fact that k and m are integers (see A2 and A3). Recognizing the key words “either/or” in A6. . you can assume that b = 0. Because (x + 3)2 is also ≥ 0. it must be that A6 : (k = 1 and m = 1) or (k = −1 and m = −1). a = 0 and so B 1 is clearly true and the proof is complete. Substituting a = mb in the equality in A2 yields: A4 : b = kmb. x − 3 < 0. It must be shown that B 1 : a = −b. Thus B 1 is true. (x + 3)2 must be 0. a proof by cases is used. or equivalently. t ≥ t∗ ensures that s ≥ t∗ . However. it follows that there are integers k and m such that b = ka and a = mb. from A2. To see that a = ±b. b = kmb. The appearance of the quantifier “for all” in the backward process suggests using the choose method to choose A1 : an element s ∈ S . it follows that A3 : s ∈ T . it follows that A5 : s ≥ t∗ . which is precisely B 2 and so the proof is complete. and so the proof is complete. it must be that k = m = 1 or k = m = −1. It then follows that km = 1. From this it follows that a = mb = −b. it follows that a = −b. t ≥ t∗ . Specifically. it follows that s ∈ T . In this case. 12. Proof. assume that b = 0. Specializing A2 to s = s . Case 2: k = −1 and m = −1. But then the hypothesis that for all t ∈ T . for which it must be shown that B 2 : s ≥ t∗ . the hypothesis states that A4 : for all t ∈ T . thus completing the proof. let s ∈ S . and a = b. To reach the conclusion. Because k and m are integers.WEB SOLUTIONS TO EXERCISES IN CHAPTER 12 37 Case 1: k = 1 and m = 1. It will be shown that a = −b. because a = b. It will be shown that s ≥ t∗ . b|a. which cannot happen according to A1. By definition. The max/min method is used to convert the conclusion to the equivalent statement B 1 : for all s ∈ S . Specializing A4 with t = s . . which is in S (see A1). s ≥ t∗ . A2 leads to a = b. it follows by definition that A2 : for all elements s ∈ S .16 Analysis of Proof. Also. then a = mb = 0 and so a = −b. Proof. which is precisely B 1. The desired conclusion is obtained by working forward. which is in T (see A3). because S is a subset of T . s ∈ T . assume that a|b. In this case. Consequently. By the hypothesis that S ⊆ T . it must be that k = m = −1. If b = 0. Thus. . then . . d. one would choose an integer n for which n ≥ 4. 39 .” One would therefore assume that n is an integer ≥ 4 and try to show that n! > n2 . . One would then try to show that (n )! > (n )2 . b. Using the induction method one would first have to show that 4! > 42 . Then one would assume that n! > n2 and n ≥ 4. . Converting the statement to the form “if . Using the contradiction method. Using the choose method.13 Web Solutions to Exercises 13. and try to show that (n + 1)! > (n + 1)2 . and try to reach a contradiction. then n! > n2 . .3 a. c. one would assume that there is an integer n ≥ 4 such that n! ≤ n2 .” one obtains “if n is an integer ≥ 4.