Discrete Mathematics with Graph Theory (3rd Edition) 54

Discrete Mathematics with Graph Theory (3rd Edition) 54 -...

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52 Solutions to Review Exercises (b) We have 0'R7 and 7'R0, yet 0 =f 7. The relation is not antisymmetric, so it is not a partial order. 17. (-----+) Assume x f'V a. Let t E x. Then t f'V x so t rv a by transitivity. Thus tEa. This proves x ~ a. Similarly, a ~ x, so X = a. (f--) Assume x = a. Since x E x, by symmetry, x E a. Thus x f'V a. 18. Since a E b, a f'V b and hence a = b by Proposition 2.4.3. Similarly dEb implies d = b, hence a = d. Now d ¢: c implies d =f c, so end = 0 by Proposition 2.4.4. Since a = d, so also a n c = 0. 19. (a) We must show that ~ is reflexive, anti symmetric and transitive on A. The relation is reflexive: For any (a, b) E A, (a, b) f'V (a, b) because a::; a and b ::; b. It is antisymmetric: If (a, b), (c, d) E A, with (a, b) f'V (c, d) and (c, d) f'V (a, b), then a ::; c, d::; b, c ::; a and b ::; d. So a = c, b = d, hence, (a, b) = (c, d). It is transitive: If (a, b), (c, d), (e, f) E A with (a, b) f'V (c, d) and (c, d) f'V (e, f), then a ::; c, d ::; b,
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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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