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Discrete Mathematics with Graph Theory (3rd Edition) 71

Discrete Mathematics with Graph Theory (3rd Edition) 71 -...

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Section 3.3 69 x = !. If r =1= 0, the quadratic formula tells us that this equation has solutions r + 1 ± J(r + 1)2 - 2r r + 1 ± v'r2 + 1 x = 2r = ---2-r':""'--- . We claim that the solution r+ 1- Jr2 + 1 x = ----:---- 2r is always in (0,1). First note that if r > 0, then r + 1 > v'r2 + 1 since (r + 1)2 > r2 + 1. On the other hand, if r < 0, then r + 1 < 1 < v'r2 + 1. In either case, we have shown that r + 1- v'r2 + 1 2r > O. To show that this expression is also less than 1, we will prove the equivalent inequality l-r- Jr2 +1 2r < O. If r > 0, then 1 - r < 1 < v'r2 + 1 giving the result. If r < 0, then 1 - r > v'r2 + 1 because (1 - r)2 > r2+ 1, again giving the result. Since f is one-to-one and onto, 'tis a one-to-one correspondence. 18. Start at (0,0) and move as illustrated. 19. (a) 2, -2,4, -4, 8, -8, 16, -16, ... (b) 1=2°,2,!,4,i,8,~, ... (c) 1,4,7,10,13,16,19, ... · r-e-e · r r-e • . r r r r r --- r ------ • • -e-e-e -e-e 1 -e 1 1 • • (d) (1,1), (1, 2), (1,3), (2, 1), (2, 2), (3, 3), (3,
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