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Discrete Mathematics with Graph Theory (3rd Edition) 78

# Discrete Mathematics with Graph Theory (3rd Edition) 78 -...

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76 Solutions to Exercises (c) By the result of part (a), 319 566 has 1+ LloglO 319 566 J digits. Since 10glO 319 566 = 566 log 10 319 ~ 1417.15, the integer 319 566 has 1418 digits. Exercises 4.2 1. (a) [BB] Not totally ordered; for example, 6 and 21 are not comparable. (b) [BB] 1 is a minimum element since 1 1 n for all n E N, but there is no maximum element since given any n E N, n 1 2n, so n can't be maximum. 2. (a) The greatest lower bound of natural numbers is their greatest common divisor. Since the gcd of any two elements of this poset is, in each case, still an element of the poset, every pair of elements has a glb. (b) The least upper bound of natural numbers is their least common multiple. The lcm of 4 and 6 is 12. Since 12 is not in the poset, 4 and 6 have no lub. (c) This is not a lattice (because not all pairs of elements have a lub.) 3. (a) [BB] 4 6 V1 0 0 (6) 8 9 2 3 5 7 7 1 4. (a) 2, 3, 5 and 7 are minimal; 4,5, 6 and 7 are maximal. There are no minimum nor maximum elements. (b) 1 is minimal and minimum; 6, 7, 8, 9 and 10 are maximal. There is no maximum. 5. [BB] As in Problem 6, write a = nq + r, where 0 ::;
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