# Equivalence relations X x X

• Oct 16th 2011, 07:42 AM
mathproblems
Equivalence relations X x X
Let X ={1,2, ...,10} Define a relation R on X x X by (a,b) R (c,d) if a+d=b+c
How can I show that R is an equivalence relation on X x X?
Thank you!
• Oct 16th 2011, 07:44 AM
Siron
Re: Equivalence relations X x X
What's the definition of an equivalence relation R on X? (or even better: for which conditions is a relation an equivalence relation?)
• Oct 16th 2011, 07:55 AM
mathproblems
Re: Equivalence relations X x X
A relation that is reflexive, symmetric, and transitive on a set is called an equivalence relation.
• Oct 16th 2011, 08:38 AM
Siron
Re: Equivalence relations X x X
Quote:

Originally Posted by mathproblems
A relation that is reflexive, symmetric, and transitive on a set is called an equivalence relation.

Indeed! So use them to prove that R is an equivalence relation on X.
• Oct 16th 2011, 08:54 AM
mathproblems
Re: Equivalence relations X x X
I am not sure how to start, and how to use a+d=b+c
Do I plug numbers between 1 to 10 in there?

Let (a,b) R (c,d) be symmetric
therefore R { (a,b) , (c,d) ,(b,a), (d,c)}

Let (a,b) R (c,d) be reflexive
therefore R {(a,b), (c,d)}

Let (a,b) R (c,d) be transitive
therefore R {(a,b) (b,c) (c,d)}

So the Equivalence relation of (a,b) R (c,d) will be {(a,b) , (c,d) ,(b,a), (d,c) (b,c)}

where shall I use a+d= b+c?
Thank you.
• Oct 16th 2011, 09:20 AM
Plato
Re: Equivalence relations X x X
Quote:

Originally Posted by mathproblems
I am not sure how to start, and how to use a+d=b+c
Do I plug numbers between 1 to 10 in there?
Let (a,b) R (c,d) be symmetric
therefore R { (a,b) , (c,d) ,(b,a), (d,c)}
Let (a,b) R (c,d) be reflexive
therefore R {(a,b), (c,d)}
Let (a,b) R (c,d) be transitive
therefore R {(a,b) (b,c) (c,d)}
So the Equivalence relation of (a,b) R (c,d) will be {(a,b) , (c,d) ,(b,a), (d,c) (b,c)}
where shall I use a+d= b+c?

Once again, that sort of example has absolutely nothing to do with proof.
Do you understand what constitutes a proof?

I will get you started.
We know that $\displaystyle a+b=b+a$ therefore $\displaystyle (a,b)\mathbf{R}(a,b)$.
Therefore, $\displaystyle \mathbf{R}$ is reflexive.
• Oct 16th 2011, 09:49 AM
mathproblems
Re: Equivalence relations X x X
could you give me a hint on the numbers, how do I use them a+d=b+c? Thank you.
• Oct 16th 2011, 10:11 AM
Plato
Re: Equivalence relations X x X
Quote:

Originally Posted by mathproblems
could you give me a hint on the numbers, how do I use them a+d=b+c? Thank you.

Numbers have nothing to do with proof.
Did you understand the proof the $\displaystyle \mathbf{R}$ is reflexive?

Now suppose $\displaystyle (a,b)\mathbf{R}(c,d)$.
To prove symmetry we must show that $\displaystyle (c,d)\mathbf{R}(a,b)$.

We know that $\displaystyle (a,b)\mathbf{R}(c,d)$ means that $\displaystyle a+d=b+c$.
But we can rearrange that to get $\displaystyle c+b=d+a$.
That means $\displaystyle (c,d)\mathbf{R}(a,b)$.
Thus it is symmetric.
• Oct 16th 2011, 10:53 AM
mathproblems
Re: Equivalence relations X x X
thank you. i understood.