Introductory Discrete Mathematics Week 3 Homework Page 79 Determine whether each of the following relations is a function with domain {1, 2, 3}. For any relation that is not a function, explain why it isn’t. f = {(1, 1), (2, 1), (3, 1), (4, 1), (3,3)} Solution: This is not a function there are two different pairs of the form (3,-) in f f = {(1, 2), (2, 3), (4, 2)} Solution: With the given domain, this is not a function since a pair of the form (3,-) does not exist. f = {(1, 1), (1,2), (1,3), (1,4)} Solution: This is not a function since more than one pair of the form (1,-) exists. e. f = {(1, 4), (2, 3), (3,2), (4,1)} Solution: This is function since the pairs of the (1,-) and (3,-) exists. Suppose A is the set of students currently registered at the University of Calgary, B is the set of professors at the University of Calgary, and C is the set of courses currently being offered at the University of Calgary. Under what conditions is each of the following function? {(a, c) | a is taking a course from b} Solution: Not a function. Each student is taking courses handled by b1 and b2, then the sets have (a, b1) and (a, b2) where a represents each student. {(a, c) | a’s first class each week is in c} Solution: This is a function. We assume that every student takes at least one course {(a, c) | a has a class in c on Friday afternoon} Solution: Not a function 12. Let S = {1, 2, 3, 4, 5} and define f : S ⟶ Z by f x=x2+1 If x is even 2x-5 if x is odd Express f as a subset of S x Z. Is f one-to-one? Solution If f(x) for each value is computed, and if f(x) ≠ f(y) for all x ≠ y, then it is 1-1. The subset is given by {(1,f(1)), (2,f(2)),...,(5, f(5))}. Since 1 is odd, f(1)

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