# R is symmetric iff R = R^-1

1. Mar 8, 2010

### Testify

1. The problem statement, all variables and given/known data
In Velleman's "How to Prove it", he gives a proof that "R is symmetric iff R = R-1, which I find to be confusing when he is proving that $$R^{-1}\subseteq{R}$$:

Now suppose $$(x,y)\in R^{-1}$$. Then $$(y,x)\in R$$, so since R is symmetric, $$(x,y)\in R$$. Thus, $$R^{-1}\subseteq R$$ so R=R-1

It seems to me that he is saying that since $$xRy\rightarrow yRx$$ and yRx, xRy, which makes no sense.

Basically my question is this: how this part of his proof could be correct?

Last edited: Mar 8, 2010
2. Mar 8, 2010

### HallsofIvy

Staff Emeritus
It would help if you would tell us what "R" is! A relation?

If R is a relation, then it is a set of ordered pairs. $R^{-1}$ is defined as the set of pairs $\{(x, y)| (y, x)\in R\}$.

What he is saying is that if (x,y) is in R-1, then (y, x) is in R. Since R is symmetric, (x, y) is in R and so $R^{-1}\subset R$.

3. Mar 8, 2010

### Testify

Ah, okay. I guess I was stuck thinking that (x,y) was in R and didn't consider that R being symmetric could mean that if yRx then xRy.

And yes, R was a relation haha.

Thanks!