# Equivalence Relations

• snaidu228
In summary, to prove that RxS is an equivalence relation, you need to show that it satisfies the three properties of reflexivity, symmetry, and transitivity. This can be done by showing that the elements (a,b) (RxS) (a,b) satisfy these properties, and then extending it to all other elements in ExF.

## Homework Statement

I need a little help in understand this question:

Let E and F be two sets, R a binary relation on the set E and S a binary relation on the set F. We define a binary relation, denoted RxS, on the set ExF in the following way ("coordinate- wise"):
(a,b) (RxS) (c,d) <--> aRc and bSd.
If R and S are equivalence relations, prove that RxS is an equivalence relation.

unknown

## The Attempt at a Solution

I said aRc => (a,c) and bSd=> (b,d)

I assumed that aRc and bSd are from ExF.
I'm not sure that what I am doing is right

ExF is the set of ordered pairs (x,y) where x is in E and y is in F. You have (a,b) and (c,d) are elements in ExF, so this means a and c are elements of E while b and d are elements of F.

To show that RxS is an equivalence relation, you need to show it satisfies three properties: reflexivity, symmetry, and transitivity.

snaidu228 said:

## Homework Statement

I need a little help in understand this question:

Let E and F be two sets, R a binary relation on the set E and S a binary relation on the set F. We define a binary relation, denoted RxS, on the set ExF in the following way ("coordinate- wise"):
(a,b) (RxS) (c,d) <--> aRc and bSd.
If R and S are equivalence relations, prove that RxS is an equivalence relation.

unknown

## The Attempt at a Solution

I said aRc => (a,c) and bSd=> (b,d)

I assumed that aRc and bSd are from ExF.
I'm not sure that what I am doing is right