Discrete Mathematics with Graph Theory (3rd Edition) 58

Discrete Mathematics with Graph Theory (3rd Edition) 58 -...

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56 Solutions to Exercises 20. (a) [BB] The graph of 1 shown at the right makes it clear that 1 is one-to-one and onto. (b) [BB] Solution 1. Since I: R -+ R is one-to-one by part (a), it is also one-to-one as a function with domain Z. Here, however, 1 is not onto for we note that 1(0) = 0, 1(1) = 4 and 1 is increasing, so 1 is not in the range of I. X Solution 2. (This solution mimics that given in Problem 8 in our discussion of discrete functions in this section.) If I(xt} = I(X2), then 3xf+xI = 3X~+X2' so 3(xf-x~) = X2-XI and 3(XI-X2)(X~+XIX2+X~) = X2 - Xl. If Xl f:. X2, we must have x~ + XIX2 + x~ = -i, which is impossible for integers Xl, X2. Thus, Xl = X2 and 1 is one-to-one. On the other hand, 1 is not onto. In particular, 1 is not in the range of 1 since 1 = 1 ( a) for some a implies 3a 3 + a = 1; that is, a(3a 2 + 1) = 1. But the only pairs of integers whose product is 1 are the pairs 1,1 and -1, -1. So here, we'd have to have a = 3a 2 + 1 = 1 or a = 3a 2 + 1 = -1, neither of which is possible. 21. (a) The graph of
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