Start with .
That means that why?
Does that mean .
Suppose that .
If then why?
What does that mean?
Hi Math people. Can someone tell me what's wrong with my proof, below? My professor did not give me credit for it, but did not provide feedback, either. She posted the answers online and her proof took a completely different tack than mine.
I get that I did the proof differently. But I'm curious what is wrong with my proof, especially the first part which I was pretty confident of. But all feedback is appreciated.
My approach, for the 'left-to-right' proof, was to take the two possible cases (x=y and x not= y) and show that only the expected on was in R and R inverse. Seems pretty straightforward to me.
Thanks, everyone.
Prove: A relation on a set is antisymmetric if and only if .
Proof: `` '': Assume is antisymmetric. That means and implies that .
Suppose , and . That means and . Since , we have that , therefore .
Now, suppose . Since is antisymmetric, we can't have both and . WLOG, say , then and .
`` '': Assume .
Suppose .\\
.
So we have , and , therefore, is not antisymmetric. In other words, if and , and , then is not antisymmetric.
But since we have that only ``twin'' elements may be in , must be antisymmetric (that's my argument and I'm sticking to it!).
Since my homework had already been graded, this wasn't a high priority, so I'm returning to this, finally. I get the first half, but I'm not sure about the right to left proof:
Left to Right
Assume is antisymmetric, that means that if and , then .
Let and .
But and and antisymmetric implies . In other words, antisymmetric .
RtoL
Suppose .
Then, and .
But . So we have that and .
Is that right?
Your use of "=>" is a little off in a few places. Here's how I would write it:
[I'll use R* for the converse of R]
Assume appropriate relativization to set A.
Suppose R is anti-symmetric. Suppose <x y> in R/\R*. So <x y> in R and <y x> in R. So x=y. So <x y> in {<a a> | a in A}. So R/\R* subset of {<a a> | a in A}.
Suppose R/\R* subset of {<a a> | a in A}. Suppose <x y> in R and <y x> in R. So <x y> in R*. So <x y> in R/\R*. So <x y> in {<a a> | a in A}. So x=y.