19. Evaluation of Proofs See the instructions for Exercise (19) on page 100 from Section 3.1. (a) Proposition. If m is an odd integer, then (m + 6) is an odd integer. Proof. For m + 6 to be an odd integer, there must exist an integer n such that m +6=2n + 1, By subtracting 6 from both sides of this equation, we obtain m = 2n – 6 † I = 2 (n=3) +1. By the closure properties of the integers, (n÷3) is an integer, and hence, the last equation implies that m is an odd integer. This proves that if m is an odd integer, then m + 6 is an oddinteger. (b) Proposition. For all integers m and n, if mn is an even integer, then m is even orn is even. Proof. For either m or n to be even, there exists an integer k such that m = 2k or n = 2k. So if we multiply m and n, the product will contain a factor of 2 and, hence, nn will be even.
19. Evaluation of Proofs See the instructions for Exercise (19) on page 100 from Section 3.1. (a) Proposition. If m is an odd integer, then (m + 6) is an odd integer. Proof. For m + 6 to be an odd integer, there must exist an integer n such that m +6=2n + 1, By subtracting 6 from both sides of this equation, we obtain m = 2n – 6 † I = 2 (n=3) +1. By the closure properties of the integers, (n÷3) is an integer, and hence, the last equation implies that m is an odd integer. This proves that if m is an odd integer, then m + 6 is an oddinteger. (b) Proposition. For all integers m and n, if mn is an even integer, then m is even orn is even. Proof. For either m or n to be even, there exists an integer k such that m = 2k or n = 2k. So if we multiply m and n, the product will contain a factor of 2 and, hence, nn will be even.
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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19 please
Transcribed Image Text:19. Evaluation of Proofs
See the instructions for Exercise (19) on page 100 from Section 3.1.
(a) Proposition. If m is an odd integer, then (m + 6) is an odd integer.
Proof. For m + 6 to be an odd integer, there must exist an integer n
such that
m + 6 = 2n H 1,
By subtracting 6 from both sides of this equation, we obtain
m = 2n – 6 † I
= 2 (n=3) +1.
By the closure properties of the integers, (n+3) is an integer, and
hence, the last equation implies that m is an odd integer. This proves
that if m is an odd integer, then m + 6 is an odd integer.
(b) Proposition. For all integers m and n, if mn is an even integer, then m
1s even or n is even.
Proof. For either m or n to be even, there exists an integer k such that
m = 2k or n = 2k. So if we multiply m and n, the product will contain
a factor of 2 and, hence, nn will be even.
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