integers n, if n2 is odd, then n is odd. Proof: Assume that n2 is odd but n is not odd, so n must be even. By definition, there are some integers p and q such that n2 = 2p + 1 and n = %3D n? = (2q)² = 2q. Therefore 2p + 1 4q?. Rewriting above equation, we have 1 = 4q? - 2p = 2(2q? - p), which is even since 2q² - p is an integer. %3D Which proof method is used in above proof. Proof by counterexample Direct proof O Proof by contraposition Dreof b y Contradiction

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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Consider the following proof that for all
integers n, if n2 is odd, then n is odd.
Proof: Assume that n2 is odd but n is
not odd, so n must be even. By
definition, there are some integers p
and q such that n² = 2p + 1 and n =
2q. Therefore 2p + 1 = n² = (2q)² =
4q?. Rewriting above equation, we
have 1 = 4q? - 2p = 2(2q² - p), which
is even since 2q² - p is an integer.
%3|
Which proof method is used in above
proof.
Proof by counterexample
Direct proof
Proof by contraposition
Proof by contradiction
Transcribed Image Text:Consider the following proof that for all integers n, if n2 is odd, then n is odd. Proof: Assume that n2 is odd but n is not odd, so n must be even. By definition, there are some integers p and q such that n² = 2p + 1 and n = 2q. Therefore 2p + 1 = n² = (2q)² = 4q?. Rewriting above equation, we have 1 = 4q? - 2p = 2(2q² - p), which is even since 2q² - p is an integer. %3| Which proof method is used in above proof. Proof by counterexample Direct proof Proof by contraposition Proof by contradiction
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