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!

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- Oct 16th 2011, 07:42 AMmathproblemsEquivalence 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 AMSironRe: 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 AMmathproblemsRe: 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 AMSironRe: Equivalence relations X x X
- Oct 16th 2011, 08:54 AMmathproblemsRe: 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 AMPlatoRe: Equivalence relations X x X
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 AMmathproblemsRe: 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 AMPlatoRe: Equivalence relations X x X
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 AMmathproblemsRe: Equivalence relations X x X
thank you. i understood.