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Mathematics 220 Homework for Week 7 Due March 6 If you are using the 2nd edition, be careful — question numbers may not agree. 5.4 Disprove the statement: Let n ∈ N. If Solution: Let n = 2, then n = 2 is a counterexample. (n + 1)(n + 2) n(n + 1) is odd, then is odd. 2 2 n(n + 1) (n + 1)(n + 2) = 3 is odd but = 6 is even. So 2 2 5.16 Prove that the product of an irrational number and a nonzero rational number is rational. Solution: Assume, to the contrary, that there exists a nonzero rational number r and an irrational number x, so that rx is rational. Since r and rx are rational, c a there are integers a, b, c, d with b and d nonzero, so that r = and rx = . Then b d a c x= b d If we multiply both sides by ab we have a b c b x= a b a d ⇒ x= bc ad But bc and ad are products of integers and are integers; so x is rational, which contradicts the assumption. 5.24 Prove that there exists no positive integer x such that 2x < x2 < 3x. Solution: Assume, to the contrary, that x is a positive integer with 2x < x2 < 3x. Since x is positive. 1/x is positive. Hence 2x(1/x) < x2 (1/x) < 3x(1/x) ⇒ 2 < x < 3. But there is no integer between 2 and 3 and this contradicts the assumption that x is an integer. 5.32 Prove that there exist no positive integers m and n for which m2 + m + 1 = n2 . Solution: Assume, to the contrary, that m and n are positive integers with m2 + m + 1 = n2 . Since m is positive m2 < m2 + m + 1 = n2 < m2 + 2m + 1 = (m + 1)2 . Mathematics 220 Homework for Week 7 Due March 6 Because m, n and m + 1 are positive, from the above inequality we conclude that m < n < m + 1. But there is no integer which is strictly between m and m + 1. This contradicts the assumption that n is an integer and proves the statement. 5.36 Let a, b ∈ R. Prove that if ab 6= 0, then a 6= 0 by using as many of the three proof techniques as possible. Solution: Direct Proof: Suppose a, b ∈ R are given and ab 6= 0. We cannot have ab b = 0 otherwise ab = 0 which is impossible, so b 6= 0. Because b 6= 0 then a = is b nonzero. Proof by contrapositive: We prove the contrapositive: if a = 0 then ab = 0. The product of every real number with zero is zero. So if a = 0 then ab = 0. Proof by contradiction: Assume, to the contrary, that ab 6= 0 but a = 0 but the product of every real number with zero is zero; this is a contradiction and proves the statement. 5.40 Show that there exist a rational number a and an irrational number b such that ab is rational. Solution: Let a = 1 and b = √ 2; then ab = 1 is rational. 6.2 Prove that if A is any well-ordered set of real numbers and B is a nonempty subset of A, then B is also well-ordered. Solution: Assume A ⊂ R is well-ordered and B is a nonempty subset. Given a nonempty subset C of B, C is also a nonempty subset of A. Since A is well-ordered, C has a least element. So we have showed that every nonempty subset of B has a least element. Hence B is well-ordered. 6.8 Find a formula for 1 + 4 + 7 + · · · + (3n − 2) for positive integer n, and then verify your formula by mathematical induction. Solution: We use induction to prove 1 + 4 + 7 + · · · + (3n − 2) = Page 2 n(3n − 1) . 2 Mathematics 220 Homework for Week 7 Due March 6 When n = 1, obviously both hand sides of the above identity are equal to 1; so the statement holds for n = 1. Assume the identity holds for k a positive integer: 1 + 4 + 7 + · · · + (3k − 2) = n(3k − 1) , 2 we use that to show the statement holds for k + 1: 1 + 4 + 7 + · · · + (3k − 2) + [3(k + 1) − 2] = [1 + 4 + 7 + · · · + (3k − 2)] + (3k + 1) n(3k − 1) = + (3k + 1) 2 3k 2 − k + 2(3k + 1) = 2 3k 2 + 5k + 2 = 2 (k + 1)(3k + 2) = 2 (k + 1)[(3(k + 1) − 1] = . 2 This shows that the identity holds for k + 1. By the principle of mathematical induction, this proves the statement for every positive integer n. 6.12 Consider the open sentence P (n) : 9 + 13 + · · · + (4n + 5) = n ∈ N. 4n2 + 14n + 1 , where 2 (a) Verify the implication P (k) ⇒ P (k + 1) for an arbitrary positive integer k. (b) Is ∀n ∈ N, P (n) true? Solution: (a) Assume P (k) holds, namely: 9 + 13 + · · · + (4k + 5) = 4k 2 + 14k + 1 , 2 we want to prove P (k + 1): 9 + 13 + · · · + (4k + 5) + [4(k + 1) + 5] = 4(k + 1)2 + 14(k + 1) + 1 . 2 We start from the left hand side and use P (k): 4k 2 + 14k + 1 + (4k + 9) 2 4k 2 + 14k + 1 + 8k + 18 = 2 4k 2 + 22k + 19 = 2 9 + 13 + · · · + (4k + 5) + [4(k + 1) + 5] = Page 3 Mathematics 220 Homework for Week 7 Due March 6 On the other hand for the right hand side of P (k + 1) we have: 4(k + 1)2 + 14(k + 1) + 1 4(k 2 + 2k + 1) + 14k + 14 + 1 4k 2 + 22k + 19 = = . 2 2 2 This shows the two sides of the identity in P (k + 1) are equal and therefore we have proved P (k + 1). (b) No! Let n = 1, the left hand side of the equality in P (1) has only one term and is equal to 9. But for the right hand side, we have 4 × 1 + 14 × 1 + 1 19 = 2 2 which is not equal to 9. Page 4