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Discuss which of the following are valid proofs of the following statement: If ab is even, then a is even or b is even. Some of the proofs below are correct but some have errors and some are using the wrong logic. Proof1: Suppose a and b are odd. That is a = 2k+1 and b = 2k+1 for some integer k, then ab = (2k+1) (2k+1) = 4k2+4k+1 = 2(2k2 +2k) +1 Therefore, ab is odd. Proof2: Suppose a and b are odd. That is a = 2k+1 and b = 2m+1 for some integers k and m, then ab = (2k+1) (2m+1) = 4km+4k+4m+1 = 2(2km+k+m) +1 Therefore, ab is odd. Proof3: Suppose a or b is even. Say it is a (b even would be identical). That is a = 2k for some integer k. In that ca ab = (2k) b = 2 (kb) Therefore is even. Proof4: Suppose ab is even, but a and b are both odd. In that case there exist integers k, m, and n such that: ab = (2m+1) (2n+1) 2k = (2m+1) (2n+1) 2k = 4mn + 2m +2n +1 2k = 2 (2mn + m + n) +1 The left-hand side is an even integer, but the right-hand side is an odd integer, which is impossible. Therefore, a or b must be even.

Discuss which of the following are valid proofs of the following statement: If ab is even, then a is even or b is even. Some of the proofs below are correct but some have errors and some are using the wrong logic. Proof1: Suppose a and b are odd. That is a = 2k+1 and b = 2k+1 for some integer k, then ab = (2k+1) (2k+1) = 4k2+4k+1 = 2(2k2 +2k) +1 Therefore, ab is odd. Proof2: Suppose a and b are odd. That is a = 2k+1 and b = 2m+1 for some integers k and m, then ab = (2k+1) (2m+1) = 4km+4k+4m+1 = 2(2km+k+m) +1 Therefore, ab is odd. Proof3: Suppose a or b is even. Say it is a (b even would be identical). That is a = 2k for some integer k. In that ca ab = (2k) b = 2 (kb) Therefore is even. Proof4: Suppose ab is even, but a and b are both odd. In that case there exist integers k, m, and n such that: ab = (2m+1) (2n+1) 2k = (2m+1) (2n+1) 2k = 4mn + 2m +2n +1 2k = 2 (2mn + m + n) +1 The left-hand side is an even integer, but the right-hand side is an odd integer, which is impossible. Therefore, a or b must be even.

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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Discuss which of the following are valid proofs of the following statement: If ab is even, then a is even or b is even. Some of the proofs below are correct but some have errors and some are using the wrong logic. Proof1: Suppose a and b are odd. That is a = 2k+1 and b = 2k+1 for some integer k, then ab = (2k+1) (2k+1) = 4k2+4k+1 = 2(2k2 +2k) +1 Therefore, ab is odd. Proof2: Suppose a and b are odd. That is a = 2k+1 and b = 2m+1 for some integers k and m, then ab = (2k+1) (2m+1) = 4km+4k+4m+1 = 2(2km+k+m) +1 Therefore, ab is odd. Proof3: Suppose a or b is even. Say it is a (b even would be identical). That is a = 2k for some integer k. In that ca ab = (2k) b = 2 (kb) Therefore is even. Proof4: Suppose ab is even, but a and b are both odd. In that case there exist integers k, m, and n such that: ab = (2m+1) (2n+1) 2k = (2m+1) (2n+1) 2k = 4mn + 2m +2n +1 2k = 2 (2mn + m + n) +1 The left-hand side is an even integer, but the right-hand side is an odd integer, which is impossible. Therefore, a or b must be even.
Transcribed Image Text:Discuss which of the following are valid proofs of the following statement: If ab is even, then a is even or b is even. Some of the proofs below are correct but some have errors and some are using the wrong logic. Proof1: Suppose a and b are odd. That is a = 2k+1 and b = 2k+1 for some integer k, then ab = (2k+1) (2k+1) = 4k2+4k+1 = 2(2k2 +2k) +1 Therefore, ab is odd. Proof2: Suppose a and b are odd. That is a = 2k+1 and b = 2m+1 for some integers k and m, then ab = (2k+1) (2m+1) = 4km+4k+4m+1 = 2(2km+k+m) +1 Therefore, ab is odd. Proof3: Suppose a or b is even. Say it is a (b even would be identical). That is a = 2k for some integer k. In that ca ab = (2k) b = 2 (kb) Therefore is even. Proof4: Suppose ab is even, but a and b are both odd. In that case there exist integers k, m, and n such that: ab = (2m+1) (2n+1) 2k = (2m+1) (2n+1) 2k = 4mn + 2m +2n +1 2k = 2 (2mn + m + n) +1 The left-hand side is an even integer, but the right-hand side is an odd integer, which is impossible. Therefore, a or b must be even.
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