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Discrete Mathematics with Graph Theory (3rd Edition) 91

Discrete Mathematics with Graph Theory (3rd Edition) 91 -...

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Section 4.3 89 (c) We must prove that f is one-to-one and onto. One-to-one: Supposef(ml/n1) = f(m2/n2). Assumefirstthatm1 = m2 = 1,n1 =p~l ... p~t and n2 = q{l ... q£", where P1, ... ,Pi are distinct primes, q1, ... ,qk are distinct primes, €i > 0 and fi > 0 for all i. Then p~el-1 ... p~et-1 = q~h-1 ... q~j,.-l. By the Fundamental Theorem of Arithmetic, we conclude that n1 = n2, so ml/n1 = m2/n2. The same approach can be applied to the other eight possible cases for the pair (m1 / n1 , m2 / n2), showing that f is one-to-one. Onto: Given YEN, if y = 1, then y = f(1). Otherwise, y is a natural number> 1 and so, by the Fundamental Theorem of Arithmetic, is the unique product of powers of distinct primes. Separating the powers into even and odd powers, we can write y = p~el ... p%e" if there are no odd powers, y = q~h -1 ... q~ft-1 if there are no even powers, and y = p~el ... p%e"q~h -1 ... q~!t-1 if there are both odd and even powers. In the first case, Y = f(m) where m = p~l ... P~", in the second case, y = f( ~) where n = q{l ... qft and, in the third case, y = f( ~) where m = p~l ...
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