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Discrete Mathematics with Graph Theory (3rd Edition) 66

# Discrete Mathematics with Graph Theory (3rd Edition) 66 -...

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64 (c) (f 0 g)-1 = {(I, 4), (2,3), (3,2), (4, 1), (5, 5n g-1 0 f- 1 = {(I, 4), (2,3), (3,2), (4, 1), (5, 5n = (f 0 g)-I. f- 1 0 g-1 = {(I, 5), (2,3), (3,2), (4,4), (5, In =I (f 0 g)-I. Solutions to Exercises 20. (a)[BB]ForxEB,(fog)(x)=fC:~I)= 2~ = 2t )=22 x =x. x-I - 2 2x - 2 x-I 2(~) 2x 2x (b) [BB] For x E A, (g 0 f)(x) = g(x:'2) = ~--1 = x _ (x _ 2) = 2" = x and so, by x-2 Proposition 3.2.7, f and 9 are inverses. 21. (a) Suppose 0 E A and let a = f(O). Then 0 = f-l(a) = fta) which is not possible for any real number f(a). (b) Since f and f4 each have domain A, we have only to prove that f4(a) = a for all a E A. Let then a be some element of A. Let al = f(a), a2 = f(al) = f2(a), a3 = f(a2) = f3(a) and a4 = f(a3) = f4(a). We must show that a4= a. From al = f(a), we have a = f- 1 (al) = f(~J) and so a2 = f(al) = ~. From a3 = f(a2), we obtain a2 = f- 1 (a3) = f(~3) and so a4 = f(a3) = -.!.. = _1_ == a as desired. a2 l/a 22. (a) [BB] Suppose g(b 1) = g(b2) for b1 ,b2 E B. Since f is onto, b 1 = f(aJ) and b2 = f(a2) for some alo a2 E A. Thus, g(f(al)) = g(f(a2)); that is, 9 0 f(al) = go f(a2). Since 9 0 f is one-to-one, al = a2. Therefore, b 1 = f(al) = f(a2) = b2 proving that 9 is one-to-one.
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