   Chapter 9.7, Problem 42ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# Use mathematical induction to prove that for every integer n ≥ 1 , if S is a with n elements, then S has the same number of subsets with an even number of elements as with an odd number of elements. Use this fact to give a combinatiorial arguments. Use this fact to give a combinatorial argumaent to justify the identity of exercise 36.

To determine

To prove:

If S is a set with n elements, then S has the same number of subsets with an even number of elements as with an odd number of elements for all positive integers n and to give a combinatorial argument to justify the identity (n0)(n1)+(n2)...+(1)n(nn)=0 for every integer n1.

Explanation

Given information:

S is a set with n elements, n1.

Proof:

Let’s prove the result by mathematical induction.

Let P(n) be “If S is a set with n elements, then S has the same number of subsets with an even number of elements as with an odd number of elements.”

Basic step:

n=1

Let S be a set with 1 element x.

element.

We then can notice that there is 1 subset with an even number of elements and there is 1 subset with an odd number of elements.

Thus P(1) is true.

Induction step:

Let P(k) be true, thus “If S is a set with k elements, then S has the same number of subsets with an even number of elements as with an odd number of elements.”

We need to prove that P(k+1) is true.

Let S be a set with k+1 elements x1,x2,x3,...,xk,xk+1.

Since by induction hypothesis, P(k) is true, R={x1,x2,x3,...,xk} has the same number of subsets with an even number of elements as with an odd number of elements.

Let us assume that R has m subsets with an even number of elements (then R also has m subsets with an even number of elements).

The subsets of S that do not contain xk+1 are the subsets of R, thus there are m such subsets with an even number of elements and m such subsets with an odd number of elements.

The subset of S that do contain xk+1 can be obtained by adding the element xk+1 to each subset of R. Since there are m subset with an even number of elements in R, there will be m subset containing xk+1 with an odd number of elements in S. Since there are m subset with an odd number of elements in R, there will be m subset containing xk+1 with an even number of elements in S.

Hence, there are m+m=2m subsets with an odd number of elements in S and there are m+m=2m subsets with an even number of elements in S

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