Discrete Mathematics with Graph Theory (3rd Edition) 161

Discrete Mathematics with Graph Theory (3rd Edition) 161 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 5 159 7. Let S = {a, b} be a set of two elements. The only 2-element subset of S is S itself. Also, for n = 2, ~n( n -1) = 1, so the result is true. Now assume that k 2:: 2 and the result is true for sets of k elements. Let S be a set of k + 1 elements and let a be an element of S. By the induction hypothesis, there are ~k(k - 1) 2-element subsets of S" {a}. The only other 2-element subsets of S contain a, and there are k of these. Altogether, the number of 2-element subsets of S is ~k(k -1) + k = ~k(k + 1), which is the desired formula for n = k + 1. By the Principle of Mathematical Induction, we conclude that the result is true for all n 2:: 2. 8. (a) Any logarithm (0, 00) ~ R is an example. (b) For n = 1, f(XI) = f(XI), which is the desired result. Now assume that k 2:: 1 and that f(XIX2 ... Xk) = f(XI) + f(X2) + . .. + f(Xk) for any real numbers Xl, X2, ... , Xk E A. Given Xl, X2,···, Xk+l E A, we have f(XIX2 ... Xk+l) = f((XIX2'" Xk)Xk+I) since we are given that f(ab) = f(a)
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online