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Discrete Mathematics with Graph Theory (3rd Edition) 55

# Discrete Mathematics with Graph Theory (3rd Edition) 55 -...

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Section 3.1 53 23. Two elements of a poset can have at most one least upper bound and here's why. Let £1,£2 each be least upper bounds for elements a and b. Then £1 is an upper bound for a and b, so £2 ~ £1 because £2 is least. Interchanging £1, £2 in the preceding statement gives £1 ~ £2. So £1 = £2 by antisymmetry. Exercises 3.1 1. (a) [BB] Not a function; f contains two different pairs of the form (3, -). (b) Not a function with domain {I, 2, 3, 4}. There's no pair of the form (3, -). (c) [BB] This is a function. (d) Certainly not a function. There is more than one pair of the form (1, -): In fact, there are four! (e) This is a function. 2. (a) [BB] This is not a function unless each student at the University of Calgary has just one professor, for if student a is taking courses from professors b1 and b2, the given set contains (a, bI) and (a,b2). (b) Assuming that each student currently registered at the University of Calgary is taking at least one course, then this is a function.
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