   # Prove each statement in 39-44. For every positive integer n , if A and B 1 , B 2 , B 3 , are any sets, then A ∩ ( ∪ i = 1 n B i ) = ∪ i = 1 n ( A ∩ B i ) . ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
Publisher: Cengage Learning,
ISBN: 9781337694193

#### Solutions

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

5th Edition
EPP + 1 other
Publisher: Cengage Learning,
ISBN: 9781337694193
Chapter 6.2, Problem 40ES
Textbook Problem
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## Prove each statement in 39-44.For every positive integer n, if A and B 1 , B 2 , B 3 , are any sets, then A ∩ ( ∪ i = 1 n B i ) = ∪ i = 1 n ( A ∩ B i ) .

To determine

To Prove:

For all integers n1, if A and B1,B2,B3,...... are any sets then A(i=1nBi)=i=1n(ABi).

### Explanation of Solution

Given information:

Let A and B1,B2,B3,........,Bn are any sets.

Concept used:

:Union of sets:Intersection of sets

: subset of set

: Null set

Calculation:

Let A and B1,B2,B3,........,Bn are any sets.

Prove that A(i=1nBi)=i=1n(ABi).

First prove that A(i=1nBi)i=1n(ABi).

Suppose there is an element x in A(i=1nBi), by definition of intersection,

xA and xi=1nBi.

Since x be the element in i=1nBi, by general union implies that x be the element in any one of the sets B1,B2,B3,........,Bn. That is xBi, i=1,2,.....,n.

Since, xA and xBi, by definition of intersection xABi.

Thus, by definition of general union  xi=1n(ABi).

Thus xA(i=1nBi)xi=1n(ABi).

Therefore A(i=1nBi)i=1n(ABi)

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