I had to read this a couple of times myself as I thought it was an existence statement, and was starting to question my reading comprehension at the "for all".
The devil is in the details.
Writing with logic symbols may help sometimes, though in this case it comes really close: the statement $\displaystyle \neg\exists x\,\forall y P(x,y)$ is logically equivalent to the
statement $\displaystyle \forall x\,\exists y\,\neg P(x,y)$ , so the difference, imo, between what it is and what was being understood by some is only the negation of P(x,y) in the last formula...
Tonio
This really doesn't help: either I (and perhaps some others) am wrong or you are. I'm not trying to impose my opinion by authority or, much less, by force. I gave some reasons, using logic symbols, and I think I made my point clear. If you think it is otherwise you must then deffend your stand with arguments, not by dismissing others' opinions saying that they're confusing the OP which, again imo, was precisely what you did.
It wouldn't be the first time, and most probably neither the last one, I could be wrong, so I've no problem at all in being shown where my mistake is, in case it exists.
Tonio
I came up with a new proof:
Prove: there does not exist a real number x such that for all real numbers y, xy=1
Contradiction:
Suppose yes, that is supppose that there does exist a real number x such that for all real numbers y, xy=1.
Let x=z for any real number z and let y=0.
Then,
xy=z*0 by substitution
=0 by algebra
This is a contradiction. We were trying to prove that for all real numbers y there is a real number x where xy=1. This is NOT the case when we use the real number 0 for y. Therefore, the original statement is true. There does NOT exist a real number x such that for all real numbers y, xy=1. **END OF PROOF**
Why?
His proof is correct, although a bit redundant - there is no need to write "Let x=z for any real number z". Instead, you (matthayzon) should write: "Let $\displaystyle x \in \mathbb{R}$", or "Let x be a real number". Then let $\displaystyle y=0$ and procceed as before.
I think tonio summed it up pretty well. There was some confusion as to what the poster was actually asked to prove. In my opinion it is rather apparent that what we need to show is that there is no constant $\displaystyle x \in \mathbb{R}$ such that for ANY $\displaystyle y \in \mathbb{R}$, $\displaystyle xy=1$, which is exactly what tonio said.
Your proof is correct, the only thing you need to fix is what I said in this post. It should look like this:
Theorem: There does not exist a real number $\displaystyle x \in \mathbb{R}$ such that for any $\displaystyle y \in \mathbb{R}$, xy=1.
Proof: Assume, by contradiction, that there does exist such x, ie. that for any real number $\displaystyle y$, $\displaystyle xy=1$. Now, choose $\displaystyle y=0$. Since y (0) is a real number, by our assumption we get that $\displaystyle xy=1$. However, we know that $\displaystyle x * 0 = 0$ for any real number x. Therefore, $\displaystyle 0=1$, in contradiction.
y=0 contradicts it, but the claim that such x exists is false even if 0 is excluded.
what he was asked to prove\disprove is that you can use the SAME x for ALL y. In your example, x depends on y.
For example, could have taken y=1 at first, to get x*1=1 -> x=1, and then take y=2 and get 1*2 = 1, which is obviously a contradiction.
This, I tihnk is what people have been trying to communicate - sometimes unsuccessfully (although some did put it in symbols and there was still debate). The positioning of the existence and for all statements has been muddled a bit, however it does appear from the OP's original statement, the the original order is there exists, followed by for-all.