# Equivalence relation - Congruence modulo

• Nov 12th 2012, 10:26 PM
aprilrocks92
Equivalence relation - Congruence modulo
• Nov 13th 2012, 01:38 AM
Siron
Re: Equivalence relation - Congruence modulo
What have you tried so far? You should prove by definition that the given relation is reflexive, symmetric and transitive. Can you do that?
• Nov 13th 2012, 01:48 AM
aprilrocks92
Re: Equivalence relation - Congruence modulo
Thank you. I am familiar with the properties reflexive, symmetric and transitive, but not when it comes to modulo. I have never seen it before, and simply do not know where to start.
• Nov 14th 2012, 05:38 AM
Siron
Re: Equivalence relation - Congruence modulo
The relation $\displaystyle \equiv_5$ is defined as $\displaystyle \forall x,y \in \mathbb{S}_7: x\equiv_5 y \Leftrightarrow x \mod 5 = y \mod 5$
To check if the relation is reflexive you have to check $\displaystyle \forall x \in \mathbb{S}_7: x \equiv_5 x$ which is true because $\displaystyle x \mod 5 = x \mod 5$.

Can you check the symmetric and transitive property now?
• Nov 14th 2012, 06:01 AM
Plato
Re: Equivalence relation - Congruence modulo
Quote:

Originally Posted by aprilrocks92

I have a different but equidistant way of describing that relation.
Say $\displaystyle x\mathcal{R}y$ if and only if $\displaystyle x~\&~y$ have the same remainder when divided by $\displaystyle 5$

Thus it should be clear that $\displaystyle 2\mathcal{R}7$.

The three needed properties are easily checked.

There are five equivalence classes.