Question

Let A=NxN and define a relation R on A by (a,b)R(c,d) iff ab=cd ...show that R is an equivalence relation on A

Answer 1

So, there's three properties you need to show for a relation to be an equivalence relation: reflexivity, symmetry, and transitivity.

Do you know the definitions of those three?

Answer 2

yes

Answer 3

Okay, do you know how to show any of those three?

Answer 4

no, thats what I am confused about.

Answer 5

Alright. I'll explain how to prove it's symmetric. The other two properties follow the same pattern.

So, first, the definition. A relation is symmetric if xRy => yRx.

So, let (a,b) be in A and let (c,d) be in A. Then, if (a,b)R(c,d), then that => ab = cd (by definition above) => cd = ab (just by definition of an equation) => (c,d)R(a,b).

And, from that, you have shown it is symmetric.

Try to do the other two here.

Answer 6

Reflexivity:
Definition: xRx
so let (a,b) and (c,d) be in A. Then, because (a,b)R(c,d) and ab=bc then it can be rewritten as (a,b)R(a,b)

Answer 7

Transitivity:
Definition: if xRy and yRz, then xRz

Let (a,b) (c,d) and (e,f) be in A. we know that (a,b)R(c,d) and we can assume that (c,d)R(e,f). Because we know that ab=cd we can replace (c,d) with (a,b) which will give (a,b)R(e,f) ..... idk this doesn't prove it... can you help?

Answer 8

For reflexitivity, you've got the right idea, just confused about what you're actually trying to prove. All you're trying to prove is that xRx is true, using the definition (or in this case, equation) you're given.

So, let (a,b) be in A. Then, since ab = ab, aRa. And, you're done.

For transitivity, you have the right idea. And, I'd really like you to get this one by yourself. So, you have (a,b)R(c,d) and (c,d)R(e,f). Using your equation, what does that mean? And, further, can you do some substituting with those equations to get (a,b)R(e,f)?

Answer 9

Should actually be xRx* up there, not aRa. But, oh well. It doesn't matter too much. It's an arbitrary variable.

Answer 10

right. okay... so we will assume (a,b)R(c,d) and (c,d)R(e,f), both in A. Then ab=cd and cd=ef (using definition and our assumption). Then ab=cd=ef and we can say ab=ef so (a,b)R(e,f)....?

Answer 11

Yep. That's it.

Answer 12

okay. Thank you!!! :)

Answer 13

You're welcome.

Answer 14

So one last question.. if we let B=RxR (the reals) and we extend R to an equivalence relation on B, what would E(9,2) look like?

Answer 15

we are using the same equivalence relation from above....So would that be (1,18) (18,1) (6,3) (3,6) (2,9) (9,2) plus all of their negations?

Answer 16

Since before it was just the Natural numbers but now it is being extended to the Reals?