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Discrete Mathematics with Graph Theory (3rd Edition) 76

# Discrete Mathematics with Graph Theory (3rd Edition) 76 -...

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74 Solutions to Exercises 17. Let A = {a1,a2, ... ,an} and B = {b1,b2, ... ,b m }. Since A n B = 0, ai =f bj for any i,j. Let C = {I, 2, ... , n + m}. If we can establish a one-to-one correspondence between A U B and C, then we will have IA U BI = ICI = n + m. Define f: AU B --+ C by f(aI) = 1, f(a2) = 2, ... , f(a n) = n, f(b1) = n + 1, f(b2) = n + 2, ... , f(b m ) = n + m. Since f(x) = f(y) --+ x = y, f is one-to-one. Clearly f is onto, so it is a one-to-one correspondence. 18. The function f: (-2,2) --+ (1,9) defined by f(x) = 2x + 5 is a bijection. 1 1 19. Define f: (1,3) --+ R by f(x) = -- - -2 and suppose f(X1) = f(X2). x-I 1 1 1 1 1 1 Then -- - - = -- - -, so -- = --, Xl - 1 = X2 - 1 and Xl = X2. Therefore, f Xl - 1 2 X2 - 1 2 Xl - 1 X2 - 1 . (0) I 2y + 3 1S one-to-one. lVen y E ,00, et X = 2y + We leave it to the student to verify that f(x) = y. Also 1 < ~::~ < 3. Finally, rng f = (0,00), so f is a one-to-one correspondence between (1,3) an.d (0,00). 20. (a) Assume to the contrary that S is countable. In that case, we can list the elements of S as a1, a2, a3, .... Writing each of these using their decimal expansions, we have a1 = O.an a12a13a14 .. .
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