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Quiz

# MAT102 quiz

3 Pages
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Department
Mathematics
Course
MAT102H5
Professor
Shay Fuchs
Semester
Fall

Description
1. Let A;B two nonempty sets and f : A ! B a function. a) [2 marks] Prove that for any two subsets C ▯ A and D ▯ A, we have f(C \ D) ▯ f(C) \ f(D): Solution: Suppose that y 2 f(C \ D). Then y = f(x) for some x 2 C \ D. Since x 2 C, we have that y 2 f(C) and since x 2 D, we have that y 2 f(D). It follows that y 2 f(C) \ f(D). b) [1 mark] Let A = f1;2;3g, B = fx;yg and f : A ! B the function de▯ned by its values f(1) = y;f(2) = x;f(3) = y: Find two subsets C;D of A for which f(C) \ f(D) = fx;yg and f(C \ D) = fxg. Justify your answer brie y. Solution: Let C = f1;2g and D = f2;3g. Then f(C) = fx;yg and f(D) = fx;yg, so that f(C) \ f(D) = fx;yg. Since C \ D = f2g, we have f(C \ D) = ff(2)g = fxg. Note: This example shows that that the inclusion in part a) of the question need not be an equality. 1 2. For real numbers x;y let P(x;y) be the assertion \x = y " and Q(x;y) the assertion \x = y". Consider the following proposition: \For any two real numbers x;y satisfying x = y , we have x = y." (a) [2 marks] Write the proposition using P(x;y);Q(x;y), and logical symbols. No explanation is required. Solution: (8x;y 2 R)(P(x;y) ) Q(x;y)) (b) [2 mark] Show that the proposition is FALSE. Solution: It su▯ces to give a counterexample.
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