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Proposition: Let a and b be integers. If a is even and a divides b. then b is also even Choose a match Proof: Assume p is true, so that a and b are integers, a is even, • Proof by Contradiction and a divides b.

Proposition: Let a and b be integers. If a is even and a divides b. then b is also even Choose a match Proof: Assume p is true, so that a and b are integers, a is even, • Proof by Contradiction and a divides b.

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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Proposition: Let a and b be integers. If a is even and a Choose a match divides b, then bis also even Proof: Assume p is true, so that a and b are integers, a is even, Proof by Contradiction and a divides b. By definition, there exists an integer k with a = 2k, and there exists an integer withb=a By substitution, we can writeb-a = (2k) = 2(k'). Since b = 2(k )b is even. O Proof by Contrapositive O Direct Proof Statement: There are no integers a. b for which 2a + 4b Choose a match = 1 Proof: Suppose the proposition is false, so that there are integers a, b for which 2a + 4b 1. Dividing both sides of this equation by 2. we conclude that a+ 2b = 1/2. Since a and b are integers, a + 2b is also an integer. But 1/2 is not an integer, so this is impossible. Therefore, the proposition cannot be false, so it must be true.
Transcribed Image Text:Proposition: Let a and b be integers. If a is even and a Choose a match divides b, then bis also even Proof: Assume p is true, so that a and b are integers, a is even, Proof by Contradiction and a divides b. By definition, there exists an integer k with a = 2k, and there exists an integer withb=a By substitution, we can writeb-a = (2k) = 2(k'). Since b = 2(k )b is even. O Proof by Contrapositive O Direct Proof Statement: There are no integers a. b for which 2a + 4b Choose a match = 1 Proof: Suppose the proposition is false, so that there are integers a, b for which 2a + 4b 1. Dividing both sides of this equation by 2. we conclude that a+ 2b = 1/2. Since a and b are integers, a + 2b is also an integer. But 1/2 is not an integer, so this is impossible. Therefore, the proposition cannot be false, so it must be true.
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