Discrete Mathematics with Graph Theory (3rd Edition) 98

# Discrete Mathematics with Graph Theory (3rd Edition) 98 -...

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96 Solutions to Exercises Transitivity: Suppose a, b, c E A, a j b and b j c; that is, ab == a 2 (mod p) and be == b 2 (mod p). We want to prove that a j c, that is, that ae == a 2 (mod p) or, equivalently, that p I (ae - a 2 ). Now we have that (ab)(be) == a 2 b 2 (mod p) and, hence, that p I b 2 (ae - a 2 ). Again, by Proposition 4.3.7, p I b 2 or p I (ae- a 2 ). The latter case is what we want. In the former case, we must have p I b, hence, that ab == 0 (mod p). Therefore, a 2 == ab == 0, so a == 0 and again ae - a 2 == 0, the desired result. (b) This is a partial order. Reflexivity: For any a E A, a j a because a(a) == a 2 (mod pq). Antisymmetry: If a, b E A, a j band b j a, then ab == a 2 (mod pq) and ba == b 2 (mod pq). Thus, ab == a 2 (mod p) and ba == b 2 (mod p) and, as in (a), we get a == b (mod p). Similarly, we obtain a == b (mod q). So we have a - b divisible by both p and q. By Exercise 14 of Section 4.3, pq I (a - b), giving a == b (mod pq) as desired. Transitivity: Suppose
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## This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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