I plotted for all real number $\displaystyle y$ as a diagonal line on x-y axis and then plotted $\displaystyle x= \frac{1}{y} $. By plotting $\displaystyle x= \frac{1}{y} $, I have made sure none of the real number y is excluded. I found that all $\displaystyle \forall y \in \mathbb{R}-\{0\}$ is possible on the diagonal line, but for x, I found it a very different story, namely, $\displaystyle x=\frac{1}{y}$ approaches neither of the axes--neither x nor y, which means that if $\displaystyle xy=1$, neither $\displaystyle x$ nor $\displaystyle y$ can be a zero.
Since the statement says $\displaystyle \sim \exists x \forall y:xy=1$, we choose any $\displaystyle y$ then find $\displaystyle x$; in other words, by choosing any x then find y would makes no sense.
If you take $\displaystyle y=2$, you need not stick to $\displaystyle x=1$ because it did not say for all $\displaystyle x$. Since $\displaystyle \exists x$ 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.
You are still confusing the order of the quantifiers :P
$\displaystyle \exists x \forall y:xy=1$ - 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 $\displaystyle xy \neq 1$.
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.