1) Give the grade of A(correct), C(part correct), F(incorrect because claim is wrong or proof is wrong)

Claim: The relation T on RxR given by (x,y)T(r,s) iff x+y=r+s is symmetric.
Proof: Suppose (x,y) belongs to RxR. Then (x,y)T(y,x) because x+y=y+x. Therefore, T is symmetric.

What is the grade and why?

I say it is A since the proof generally seems correct and it makes logical sense.

2) Give the proof grade A (correct), C(part correct), F(wrong because claim or proof wrong)

Claim: the relation W on RxR given by (x,y)R(r,s) iff x-r=y-s is symmetric.

Proof: Suppose (x,y) and (r,s) are in RxR and (x,y)W(r,s). Then x-r=y-s. Therefore, r-x=s-y, so (r,s)W(x,y). Thus W is symmetric.

What is the grade and why?

I say it is F since x-r is not the same as r-x, so the claim itself is wrong...i.e proof of something false.

2. ## Symmetric Relations

Hello zhupolongjoe
Originally Posted by zhupolongjoe
1) Give the grade of A(correct), C(part correct), F(incorrect because claim is wrong or proof is wrong)

Claim: The relation T on RxR given by (x,y)T(r,s) iff x+y=r+s is symmetric.
Proof: Suppose (x,y) belongs to RxR. Then (x,y)T(y,x) because x+y=y+x. Therefore, T is symmetric.

What is the grade and why?

I say it is A since the proof generally seems correct and it makes logical sense.

2) Give the proof grade A (correct), C(part correct), F(wrong because claim or proof wrong)

Claim: the relation W on RxR given by (x,y)R(r,s) iff x-r=y-s is symmetric.

Proof: Suppose (x,y) and (r,s) are in RxR and (x,y)W(r,s). Then x-r=y-s. Therefore, r-x=s-y, so (r,s)W(x,y). Thus W is symmetric.

What is the grade and why?

I say it is F since x-r is not the same as r-x, so the claim itself is wrong...i.e proof of something false.
Sorry, but I don't think you're understanding what a symmetric relation is. A relation $\displaystyle R$ on a set $\displaystyle S$ is symmetric if $\displaystyle \forall x, y \in S, xRy \Rightarrow yRx$.

You're getting confused because the set in this question is a set of ordered pairs of real numbers (rather than single numbers) and you're confusing the individual numbers in an ordered pair with two ordered pairs that may or may not be related. (Sorry if that sentence sounds complicated!)

So let me try and simplify it for you. Take the first claim, and represent the ordered pair $\displaystyle (x,y)$ by $\displaystyle p$, and the second ordered pair $\displaystyle (r,s)$ by $\displaystyle q$. Then if you're going to decide whether or not the relation $\displaystyle T$ is symmetric, you must decide whether $\displaystyle p\,T\,q \Rightarrow q\,T\,p$. Can you now see that this has nothing to do with $\displaystyle (x, y)\,T\,(y,x)$? So the answer to the first question is F - the proof is wrong.

Now apply the same thinking to the second question. Does $\displaystyle p\,W\,q \Rightarrow q\,W\,p$? Yes, and the proof is correct, because $\displaystyle (x-r = y-s) \Rightarrow (r-x = s-y)$. So the grade is A (correct).