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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.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.1: Infinite Sequences And Summation Notation
Problem 65E
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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.
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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