• Support PF! Buy your school textbooks, materials and every day products Here!

R is symmetric iff R = R^-1

  • Thread starter Testify
  • Start date
  • #1
15
0

Homework Statement


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 [tex]R^{-1}\subseteq{R}[/tex]:

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

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

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

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,792
920
It would help if you would tell us what "R" is! A relation?

It seems to me that he is saying that since xRy[itex]\rightarrow[/itex] yRx and yRx, xRy, which makes no sense.
If R is a relation, then it is a set of ordered pairs. [itex]R^{-1}[/itex] is defined as the set of pairs [itex]\{(x, y)| (y, x)\in R\}[/itex].

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 [itex]R^{-1}\subset R[/itex].
 
  • #3
15
0
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!
 

Related Threads for: R is symmetric iff R = R^-1

Replies
1
Views
746
  • Last Post
Replies
0
Views
937
Replies
3
Views
1K
  • Last Post
Replies
11
Views
11K
  • Last Post
Replies
5
Views
8K
  • Last Post
Replies
3
Views
1K
Replies
12
Views
870
Top