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ROOF EVALUATION: (see instructions for Exercise b in Section 2.2): a) "THEOREM": If x is a non-zero integer, then x is both positive and negative. "Proof": By contradiction. Assume that the hypothesis is true and the conclusion is false; i.e., suppose that x is a nonzero integer, and that x is not positive and not nega- tive. The trichotomy property states that, for any two real numbers a and b, exactly one of the following holds: "a b". Applying this with a = x, b = 0, we have that either x < 0, x = 0, or x > 0. We are assuming that "x < 0" is false and that "x > 0" is false. Therefore "x 0" must be true. This contradicts our assumption that x is nonzero. = b) "THEOREM": For x, y E Z, if xy is even, then x and y are even. "Proof": Let x and y be integers, and assume that x and y are not even. Then there are integers m and n such that x = 2m + 1 and y 2n + 1. Therefore = xy (2m + 1) (2n + 1) = 2(2mn + m + n) + 1, - so xy is odd. By Theorem 2.16, this means xy cannot be even. Thus we have proved the contrapositive of this theorem, so the theorem is true. c) "THEOREM": For x, y = Z, if xy is even, then x and y are even. "Proof": Let x and y be integers. If x and y are even, then there are integers m and n such that x = 2m and y = 2n. Therefore xy = (2m) (2n) = 2(2mn) so xy is even, as desired

ROOF EVALUATION: (see instructions for Exercise b in Section 2.2): a) "THEOREM": If x is a non-zero integer, then x is both positive and negative. "Proof": By contradiction. Assume that the hypothesis is true and the conclusion is false; i.e., suppose that x is a nonzero integer, and that x is not positive and not nega- tive. The trichotomy property states that, for any two real numbers a and b, exactly one of the following holds: "a b". Applying this with a = x, b = 0, we have that either x < 0, x = 0, or x > 0. We are assuming that "x < 0" is false and that "x > 0" is false. Therefore "x 0" must be true. This contradicts our assumption that x is nonzero. = b) "THEOREM": For x, y E Z, if xy is even, then x and y are even. "Proof": Let x and y be integers, and assume that x and y are not even. Then there are integers m and n such that x = 2m + 1 and y 2n + 1. Therefore = xy (2m + 1) (2n + 1) = 2(2mn + m + n) + 1, - so xy is odd. By Theorem 2.16, this means xy cannot be even. Thus we have proved the contrapositive of this theorem, so the theorem is true. c) "THEOREM": For x, y = Z, if xy is even, then x and y are even. "Proof": Let x and y be integers. If x and y are even, then there are integers m and n such that x = 2m and y = 2n. Therefore xy = (2m) (2n) = 2(2mn) so xy is even, as desired

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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PROOF EVALUATION (This type of exercise will appear occasionally): Each of the follow- ing is a proposed "proof" of a "theorem". However the "theorem" may not be a true statement, and even if it is, the "proof" may not really be a proof. You should read each "theorem" and "proof" carefully and decide and state whether or not the "theorem" is true. Then: • 18 If the "theorem" is false, find where the "proof" fails. (There has to be some error.) If the "theorem" is true, decide and state whether or not the "proof" is correct. If it is not correct, find where the "proof" fails.
Transcribed Image Text:PROOF EVALUATION (This type of exercise will appear occasionally): Each of the follow- ing is a proposed "proof" of a "theorem". However the "theorem" may not be a true statement, and even if it is, the "proof" may not really be a proof. You should read each "theorem" and "proof" carefully and decide and state whether or not the "theorem" is true. Then: • 18 If the "theorem" is false, find where the "proof" fails. (There has to be some error.) If the "theorem" is true, decide and state whether or not the "proof" is correct. If it is not correct, find where the "proof" fails.
PROOF EVALUATION: (see instructions for Exercise 6 in Section 2.2): a) "THEOREM": If x is a non-zero integer, then x is both positive and negative. "Proof": By contradiction. Assume that the hypothesis is true and the conclusion is false; i.e., suppose that x is a nonzero integer, and that x is not positive and not nega- tive. The trichotomy property states that, for any two real numbers a and b, exactly one of the following holds: "a <b", "a = b", or "a > b". Applying this with a = x, b = 0, we have that either x < 0, x = 0, or x > 0. We are assuming that "x < 0" is false and that "x>0" is false. Therefore "x = 0" must be true. This contradicts our assumption that x is nonzero. b) "THEOREM": For x, y E Z, if xy is even, then x and y are even. "Proof": Let x and y be integers, and assume that x and y are not even. Then there are integers m and n such that x = 2m + 1 and y 2n + 1. Therefore xy (2m + 1) (2n + 1) = 2(2mn + m + n) + 1, so xy is odd. By Theorem 2.16, this means xy cannot be even. Thus we have proved the contrapositive of this theorem, so the theorem is true. c) "THEOREM": For x, y E Z, if xy is even, then x and y are even. "Proof": Let x and y be integers. If x and y are even, then there are integers m and n such that x = 2m and y - 2n. Therefore xy = (2m) (2n) = 2(2mn) so xy is even, as desired.
Transcribed Image Text:PROOF EVALUATION: (see instructions for Exercise 6 in Section 2.2): a) "THEOREM": If x is a non-zero integer, then x is both positive and negative. "Proof": By contradiction. Assume that the hypothesis is true and the conclusion is false; i.e., suppose that x is a nonzero integer, and that x is not positive and not nega- tive. The trichotomy property states that, for any two real numbers a and b, exactly one of the following holds: "a <b", "a = b", or "a > b". Applying this with a = x, b = 0, we have that either x < 0, x = 0, or x > 0. We are assuming that "x < 0" is false and that "x>0" is false. Therefore "x = 0" must be true. This contradicts our assumption that x is nonzero. b) "THEOREM": For x, y E Z, if xy is even, then x and y are even. "Proof": Let x and y be integers, and assume that x and y are not even. Then there are integers m and n such that x = 2m + 1 and y 2n + 1. Therefore xy (2m + 1) (2n + 1) = 2(2mn + m + n) + 1, so xy is odd. By Theorem 2.16, this means xy cannot be even. Thus we have proved the contrapositive of this theorem, so the theorem is true. c) "THEOREM": For x, y E Z, if xy is even, then x and y are even. "Proof": Let x and y be integers. If x and y are even, then there are integers m and n such that x = 2m and y - 2n. Therefore xy = (2m) (2n) = 2(2mn) so xy is even, as desired.
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