MATH 150 HOMEWORK 4 SOLUTIONS


 Alison Holland
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1 MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers , , nd is nonnegtive. Is your proof constructive or nonconstructive? For ny three rel numbers x, y, nd z, it must be the cse tht one of xy, xz, or yz is nonnegtive. There re severl wys to show this; one is csebycse rgument bsed on whether the numbers re positive, negtive, or zero. Insted we use proof by contrdiction. Assume tht xy, xz, nd yz re ll negtive. (This is the negtion of the sttement we re trying to prove.) So xy < 0. The only wy for product of numbers to be negtive is if they re of opposite signs. So either x > 0 nd y < 0, or x < 0 nd y > 0. Without loss of generlity, ssume tht x > 0 nd y < 0. (Otherwise, we could just switch the nmes of x nd y.) Now we know tht x > 0 nd xz < 0. So z < 0. Then yz > 0, s both y nd z re negtive. But we ssumed tht yz < 0. This is contrdiction, so t lest one of xy, xz, or yz must be nonnegtive. This proof is nonconstructive becuse it only shows tht one of the products is nonnegtive, but it does not identify which one. 14. Prove or disprove tht if nd b re rtionl numbers, then b is rtionl. This is flse. let = 2 nd b = 1/2. Then b = 2, which is n irrtionl number. (This ws shown in clss.) 16. Show tht if, b, nd c re rel numbers nd 0, then there is unique solution of the eqution x + b = c. First we show tht there is solution. Let x = c b. Then x + b = c b + b = c b + b = c. Now we show tht the solution is unique. Suppose tht there is some number y such tht y + b = c. Therefore y = c b, nd y = c b. So 1
2 2 MATH 150 HOMEWORK 4 SOLUTIONS c b is the only solution. 30. Prove tht there re no solutions in integers to the eqution 2x 2 + 5y 2 = 14. Both 2x 2 nd 5y 2 re nonnegtive, so if they sum to 14, then ech is less thn or equl to 14. So 2x 2 14, so x 2 7. The only integers x cn be re 2, 1, 0, 1, or 2. Likewise, 5y 2 14, so y 2 14, nd the 5 only integers tht solve this inequlity for y re 1, 0, or 1. As both x nd y re squred in this eqution, we my ssume without loss of generlity tht both re nonnegtive. (If negtive vlue of x gives solution to the eqution, then nonnegtive vlue of x would lso solve the eqution.) So x {0, 1, 2} nd y {0, 1}. Now we compute: 2(0) 2 + 5(0) 2 = 0 2(1) 2 + 5(0) 2 = 2 2(2) 2 + 5(0) 2 = 8 2(0) 2 + 5(1) 2 = 5 2(1) 2 + 5(1) 2 = 7 2(2) 2 + 5(1) 2 = 13 None of these give 14, so the eqution hs no solution in integers. 34. Prove tht 3 2 is irrtionl. Assume tht there exists rtionl number x such tht x 3 = 2. So x = p for some integers p nd q with q 0, nd we my ssume tht q ( ) 3 p p nd q hve no common fctors. We hve tht q = 2, so p 3 = 2q 3. Therefore p 3 is even. This implies tht p is even, s follows: if p is odd, then p = 2k + 1 for some integer k. Then p 3 = (2k + 1) 3 = 8k k 2 + 6k + 1 = 2(4k 3 + 6k 2 + 3k) + 1 nd so p 3 is odd. So if p 3 is not odd (i.e. even), then p is even. Therefore p = 2k for some integer k. Then nd dividing by 2, (2k) 3 = 2q 3 4k 3 = q 3.
3 MATH 150 HOMEWORK 4 SOLUTIONS 3 So q 3 is divisible by 4, nd in prticulr is even. So q is even by the rgument bove. But then p nd q hve common fctor of 2, which is contrdiction. Section Show tht if A nd B re sets with A B, then ) A B = B. Let x A B. So x A x B. If x A, then x B becuse A B. If x / A, then x B by disjunctive syllogism. Either wy, x B, so A B B. Now insted, let x B. By the ddition rule, x A x B. So x A B. So B A B. b) A B = A. Let x A B. So x A x B. By simplifiction, x A. So A B A. Insted, let x A. As we know A B, we hve x B. So x A x B. So x A B. 12. Determine whether ech of these functions from Z to Z is onetoone. ) f(n) = n 1 is onetoone; if n 1 = m 1 then n = m. b) f(n) = n is not onetoone; f(1) = 2 nd f( 1) = 2. c) f(n) = n 3 is onetoone; rel numbers (nd hence integers) hve unique cube root. d) f(n) = n/2 is not onetoone; f(1) = 1 nd f(2) = Let f : R R nd let f(x) > 0 for ll x R. Show tht f(x) is strictly incresing if nd only if g(x) = 1/f(x) is strictly decresing. Assume tht f(x) is incresing. Suppose tht x < y. Then f(x) < f(y). Tking reciprocls of both sides gives 1/f(x) > 1/f(y). So g(x) > g(y), nd g is decresing. Now ssume tht g(x) = 1/f(x) is decresing. So if x < y, then g(x) > g(y). So 1/g(x) < 1/g(y). So f(x) < f(y). So f is incresing. 28. Show tht f(x) = e x from the set of rel numbers to the set of rel numbers is not invertible, but if the codomin is restricted to the set
4 4 MATH 150 HOMEWORK 4 SOLUTIONS of positive rel numbers, then the resulting function is invertible. If f : R R is given by f(x) = e x, then f is not onto, becuse e x is lwys positive, so for exmple e x = 1 hs no solution. Let g : R R + be given by g(x) = e x, where here the codomin is the set of positve rel numbers. For ny y R +, the eqution e x = y is solved by x = ln(y), so g is onto. Also, g is incresing, so g is onetoone. Therefore g is invertible.
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