   Chapter 8.3, Problem 21ES ### 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
5 views

# In 19—31, (1) prove that the relation is an equivalence relation, and (2) describe the distinct equivalence classes of each relation. 21. R is the relation defined on Z as follows: For every m ,   n ∈ Z ,   m   R   n ⇔ 7 m − 5 n is even.

To determine

To prove:

The relation R is an equivalence relation and describe the distinct equivalence classes of R.

Explanation

Given information:

for every m,nZ, mRn7m5n is even.

Proof:

We know that mRn if and only if, m and n are both even or m and n are both odd, or in other words, if and only if both m and n have the same parity.

A relation R is an equivalence relation if the relation R is transitive, symmetric and reflexive.

The equivalence class of a is the set of all elements that are relation to a.

A=ZR={(m,n)R|7m5n is even}

Reflexive:

Case 1: If x and y are both even.

Let (x,y)R

By definition of R, then 7x5x is even.

Which implies that 5x7x is even because difference of two even number is even.

Case 2: If x and y are both odd.

Let (x,y)R

By definition of R, then 7x5x is odd.

Which implies that 5x7x is odd because difference of two odd number is odd.

Since, (x,x)R for all xA and thusR is reflexive.

Symmetric:

Case 1: If x and y are both even.

Let (x,y)R

By definition of R, then 7x5y is even.

Which implies that 5y7x is even because difference of two even number is even.

Case 2: If x and y are both odd.

Let (x,y)R

By definition of R, then 7x5y is odd

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 