   # Complete the proof from Example 6.1.3: Prove that B ⊆ A where A = { m ∈ Z | m = 2 a for some integer a } and B = { n ∈ Z | n = 2 b − 2 for some integer b } ### 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 6.1, Problem 2ES
Textbook Problem
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## Complete the proof from Example 6.1.3: Prove that B ⊆ A where A = { m ∈ Z | m = 2 a   for   some integer  a } and B = { n ∈ Z | n = 2 b − 2   for   some   integer  b }

To determine

Complete the proof form example: prove that BA where,

A={mZ|m=2a for some integer a}

And

B={nZ|n=2b2 for some integer b}

### Explanation of Solution

Given information:

A={mZ|m=2a for some integer a}

And

B={nZ|n=2b2 for some integer b}

Concept used:

To prove this, both subset relations AB and BA must be proved.

Calculation:

Proof that AB.

Suppose x is a particular but arbitrarily chosen element A.

[We must show that xB. By definition of B, this means we must show that x=2. (Some inter) 2 ]

By definition of A, there is an integer such that x=2a.

[Given that x=2a, can x also be expressed as 2. (Some integer) 2 ? i.e., is there an integer, say b, such that 2a=2b2 ? Solve for b to obtain b=(2a+2)/2=a+1. Check to see if this work.]

Let b=a+1.

{First check that b is an integer.}

Then b is an integer because it is a sum of integers.

[Then check that x=2b2 ]

Also 2b2=2(a+1)2=2a+22a=2a=x.

Thus, by definition of B,x is an element of B.

[Which is what was to be shown]

Part 2, proof that BA: this part of the proof is left as exercise 2 at the end of this section

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