62 Solutions to Exercises 8. (a) First we show that f is one-to-one. Suppose then that f(xt} = f(X2). If XI and X2 are both negative, then 21xII = 21x21. Since Ixl = -x for x < 0, we have -2xI = -2X2 and hence, XI = X2. If XI and X2 are both ~ 0, we have 2xI + 1 = 2X2 + 1, so 2xI = 2X2 and again, XI = X2. Finally, we note that it is impossible for f(xt} = f(X2) with XI < 0 and X2 ~ 0 (or vice versa) since in this case, one of f(xt}, f(X2) is an even integer while the other is odd. Next we show that f is onto. Suppose that n E N. If n = 2k is even, then n = 21 -kl = f( -k) since -k < 0, while if n = 2k + 1 is odd, then n = f(k) since k ~ O. Since f is one-to-one and onto, it has an inverse. (b) Let f-I(2586) = a. We must solve the equation 2586 = f(a). Since 2586 is even, we see that 2586 = f( -2586/2) = f( -1293). 9. (a) [BB] maternal grandmother (d) mother-in-law (g) you (b) paternal grandmother (e) father . (h)
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