# Hmk4sol - COT 3100 Homework # 4 Solutions Spring 2000 1)...

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COT 3100 Homework # 4 Solutions Spring 2000 1) Define a relation T N x N such that T = {(a,b)| a A b A a – b = 2c+1 for some integer c}. (N is the set of non-negative integers.) a) Prove that this relation is not reflexive. We have that a – a = 0 for all integers a. But 0 2c + 1 for any integer c. Thus, for all integers a we have (a,a) R. So, not only is the relation not reflexive, it is in fact irreflexive. b) Prove that this relation is symmetric. If we have that (a,b) R, that means that a – b = 2c+1, for some integer c. Now consider the quantity b – a. b – a = -(a – b) = -(2c+1) = -2c – 1 = -2c –2 + 1 =2(-c-1) + 1 Thus, we have that (b,a) R as well. c) Define the term anti-transitive as the following: Given a set A and a relation R, if for all a,b,c A, (aRb bRc cRa) (a = b b = c) Prove that the relation T is anti-transitive. The relation is vacuously anti-transitive. That is because it is impossible for (a,b) R, (b,c) R, (c,a) R all at the same time. To see this consider the following : a –b = 2d +1 for some integer d b – c = 2e +1 for some integer e, so a – c = 2d + 2e +2 = 2(d+e+1). But this makes a – c even, which means (a,c) R. Proving that (a,b) R, (b,c) R, (c,a) R can NOT all be true. This also proves that the relation is anti-transitive.

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2) Let g: A A be a bijection. For n 2, define g n = g ° g ° ... ° g, where g is composed with itself n times. Prove that for n 2, that g n is a bijection from A to A as well, and show that (g n ) -1 = (g -1 ) n . You can show this inductively. You know that composing a bijection with another one results in a bijection. Consider the base case of n=1. Clearly g is a bijection, since this is given. Now, assume that g n is also a bijection, for some arbitrary n. Now, we must show that g n+1 is also. We have that g n+1 = g ° g n . But by the inductive hypothesis, g n is a bijection. Thus, all we are doing is composing two bijections together which results in another bijection, proving that g
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Hmk4sol - COT 3100 Homework # 4 Solutions Spring 2000 1)...

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