Discrete Mathematics with Graph Theory (3rd Edition) 84

# Discrete Mathematics with Graph Theory (3rd Edition) 84 -...

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82 Solutions to Exercises (d) (2,2) and (2,3) are minimal; (4,3) and (4,4) are maximal; there are no minimum nor maximum elements. (e) a b glb(a, b) lub(a, b) i. (2,2) (3,3) * * ii. (4,2) (3,4) (3,2) (4,4) iii. (3,2) (2,4) (2,2) (3,4) iv. (3,2) (3,4) (3,2) (3,4) (f) Let Al = A2 = Z and let ~I and ~2 be the usual (total) order relation on the integers. Then Z x Z is not totally ordered since, for example, (1,2) and (0,3) are not comparable. 35. (a) First we note that (A, I ) is a poset, in the same way that (N, I ) is a poset. Now we show that any two elements a, b E A have a glb and a lub. Certainly lcm( a, b) has the properties required of a least upper bound and gcd( a, b) of a greatest lower bound. The key point is to show that these elements lie in A, and this follows from the definitions of gcd and lcm. If a, b E A, then a I n and bin, hence, lcm(a, b) I n by Exercise 29, so lcm(a, b) E A. Furthermore, since gcd(a, b) I a and a I n, we have gcd(a, b) I n and so gcd(a, b) E A too. (b) [BB]
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