Discrete Mathematics with Graph Theory (3rd Edition) 68

Discrete Mathematics with Graph Theory (3rd Edition) 68 -...

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66 Solutions to Exercises (c) To see that t is onto, let y E R. Then we have y = LyJ + b for 0 $; b < 1. If b = 0, then t(y) = y. Otherwise,O < 1 - b < 1. So, with x = LyJ + 1 + (1 - b), we have LxJ = LyJ + 1. Thus, t(x) = LxJ - (1 - b) = (LyJ + 1) - (1- b) = LyJ + b = y. In any case, we have an x such that t(x) = y. { X ifxEZ (d) rl:R-4Risgivenbyrl(x)= n+1+(1-b) ifx=n+b,nEZ,0<b<1 Exercises 3.3 1. [BB] Ask everyone to find a seat. 2. [BB] The two lists 1 2 ,2 2 ,3 2 ,4 2 , ... and 1,2,3,4, . .. obviously have the same length; a 2 f--+ a is a one-to-one correspondence between the set of perfect squares and N. 3. (a) x +-t 14, Y +-t -3, {a, b, c} +-t t. (b) The function f: 2Z -4 17Z defined by f(2k) = 17k for2k E 2Z. (c) [BB] The function f: N x N -4 C defined by f(m, n) = m + ni for all m, n E N. (d) The function f: N -4 Q defined by f(n) = n/2. 4. f-I(z) = {2Z if z > 0 1- 2z if z $; 0 5. (a) If g(ml' nl) = g(m2' n2), then (ml' f(nI}) = (m2' f(n2)), so ml = m2 and f(nl) = f(n2). Since
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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