# Thread: Prove or Disprove

1. Originally Posted by Defunkt
Well, both proofs are correct, so I don't see any reason to prefer one over the other.
Ah, I didn't see his proof actually changed. Yes it is correct (qlthough a bit risky to take zero as an example). These proofs are all alike, I must be tired

2. I plotted for all real number $y$ as a diagonal line on x-y axis and then plotted $x= \frac{1}{y}$. By plotting $x= \frac{1}{y}$, I have made sure none of the real number y is excluded. I found that all $\forall y \in \mathbb{R}-\{0\}$ is possible on the diagonal line, but for x, I found it a very different story, namely, $x=\frac{1}{y}$ approaches neither of the axes--neither x nor y, which means that if $xy=1$, neither $x$ nor $y$ can be a zero.

Originally Posted by Defunkt

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.
Since the statement says $\sim \exists x \forall y:xy=1$, we choose any $y$ then find $x$; in other words, by choosing any x then find y would makes no sense.

If you take $y=2$, you need not stick to $x=1$ because it did not say for all $x$. Since $\exists x$ means at least one, I do have one, namely x=1/2.

At any rate, it's very nice of you to help me. I enjoy reading many of you posts. You are quite good at math.

3. Originally Posted by novice
If you take $y=2$, you need not stick to $x=1$ because it did not say for all $x$. Since $\exists x$ means at least one, I do have one, namely x=1/2.
You are still confusing the order of the quantifiers :P
$\exists x \forall y:xy=1$ - this means that there is a single x, such that for every y that you pick, xy=1. So, logically, how do you disprove such a statement? you simply find one such y, for which $xy \neq 1$.

This is the essence of the proof - you assume by contradiction that there exists such an x. That means that for every y that you pick, xy=1 should be true.

Specifically, it should be true for y=1. That means 1*x = 1 -> x = 1 (remember, one x for all y).
So you have that x*y = 1*y = 1 for all real y. To prove that is not true, simply take any other y - for example y2 = 2. Your assumption implies that x*y2 = 1*y2 = 1, but 1*y2 = 1*2 = 2. So you get 2=1, which is obviously a contradiction.

4. Emakarove, where are you? We need to exterminate this flea about the order of quantifiers.

5. Originally Posted by novice
Emakarove, where are you? We need to exterminate this flea about the order of quantifiers.

There is no flea, imo: I already explained it, Defunkt already explained it (read his last post: cyrstal clear)...

Tonio

6. Originally Posted by tonio
Writing with logic symbols may help sometimes, though in this case it comes really close: the statement $\neg\exists x\,\forall y P(x,y)$ is logically equivalent to the

statement $\forall x\,\exists y\,\neg P(x,y)$ , 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
Quite clear. Grazie tanto!

7. So why is setting y=0 so bad when trying to prove the statement is TRUE using contradiction?

8. Originally Posted by matthayzon89
So why is setting y=0 so bad when trying to prove the statement is TRUE using contradiction?
It is not bad; your proof is valid.

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