# Thread: When R is antisymmetric, R inverse is antisymmetric

1. ## When R is antisymmetric, R inverse is antisymmetric

Hello,

I need to prove: when R is antisymmetric, then R inverse is antisymmetric. Can someone tell me whether I am correct or wrong?

Let R be antisymmetric, therefore $(a,b), (b,a) \in R$ where $a = b$. This gives us an inverse function which is $(b,a), (a,b) \in R^{-1}$ but because it is impossible to get $a \neq b$ when $a = b$ (in R), therefore $R^{-1}$ is antisymmetric.

Is this correct? I have the feeling that this is wrong... (If this is wrong can someone give me a counterexample?

Thanks for everything!

2. I would expect students to do the following.
If $(a,b)\in R^{-1}~\&~(b,a)\in R^{-1}$ then $a=b$.
So much of what you have done is correct, but not the starting place.

3. I'm sorry I do not understand why do you start with the inverse of R? Because normally I would assume R antisymmetric then try to prove R inverse antisymmetric.

4. Originally Posted by umbrella
prove: when R is antisymmetric, then R inverse is antisymmetric.
Here the given is: “R is antisymmetric”.
You are to prove that: “R inverse is antisymmetric”.

Originally Posted by umbrella
I'm sorry I do not understand why do you start with the inverse of R?
Because that is the object of the proof; you are proving that $R^{-1}$ is antisymmetric.

5. I disagree a bit, I see it as follow: R antisymmetric -> R inverse is antisymmetric. But starting from the conclusion, doesn't seem right to me, I think you should start from the assumption that R is antisymmetric and not from the assumption that R inverse is already antisymmetric.

6. Originally Posted by umbrella
I disagree a bit, I see it as follow: R antisymmetric -> R inverse is antisymmetric. But starting from the conclusion, doesn't seem right to me, I think you should start from the assumption that R is antisymmetric and not from the assumption that R inverse is already antisymmetric.
Well of course you do. I never said otherwise.
Here is the proof.
Suppose that $(a,b)\in R^{-1}~\&~(b,a)\in R^{-1}$.
Then by definition $(b,a)\in R~\&~(a,b)\in R$.
But we are given that $R$ is antisymmetric.
That means $a=b$. Thus $R^{-1}$ is also antisymmetric.