# Lampv

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• Publicado : 12 de septiembre de 2012

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MA 225 p.26: #8cﬁj, 9acg, 10cfgh. p.37: #5ci, 8.

Homework 2 Solutions

p. 26: #8: (c) F (f) T (i) T (j) F 9: (a)Every natural number is greater than or equal to one. (c) Every prime natural number except two is an odd integer. (g)For every natural number, if it is odd, then its square is also odd. 10. (c) F (f) F (g) T (h) F p. 37: # 5 (c) Pf. Let xand y be even integers. Then x = 2k for some integer k. and y = 2j for some integer j. Therefore, xy = 4kj, which isdivisible by 4. # (i) Pf. Let x be an even integer, and let y be an odd integer. Then x = 2k for some integer k and y = 2j+ 1 for some integer j. Therefore, xy = 2k(2j + 1) is a multiple of 2, and hence is even. # 8. (a) Pf. Let n be an oddinteger. Then n2 is odd. Therefore, n2 + n + 3 is the sum of three odd integers, and is consequently odd itself.Similarly, if n is even, then n2 is even, so n2 + n is even. Thus, n2 + n + 3 is the sum of an even and an odd integer, andhence is odd. # (b) Pf. Let n be an integer. Then n2 + n = n(n + 1), which is even (if n is even or if n is odd, making (n +1) even). Thus, n2 + n + 3 = n(n + 1) + 3 is odd, since it is the sum of an even integer and an odd integer.#