# Making $\exists x \forall y [xy = 1]$ provable

I am trying to prove or disprove the statement:

$\mathcal{U} = \mathbb{R} > 0$

$\exists x \forall y [xy = 1]$

However, I have not learned the rule on how to do so. Does it somehow follow the single quantifier rule where: $\forall x [p(x)]$ changes to $\exists x [ \neg p(x) ]$ ?

I am only looking for rule on how to solve this, not the solution. Can anyone enlighten me?

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What do you mean with $xy=1$? –  user12205 Oct 12 '11 at 0:45
I think you mean it changes to $\neg\exists x[\neg p(x)]$. –  anon Oct 12 '11 at 0:47
Out of which domain do $x$ and $y$ come from. Without knowing that we cannot really help you do that. –  sxd Oct 12 '11 at 0:47
First decide whether it’s true. Is there some specific number whose product with every number is $1$? –  Brian M. Scott Oct 12 '11 at 0:49
You don’t have to do anything formally until you decide whether to prove it or to disprove it (by finding a counterexample). –  Brian M. Scott Oct 12 '11 at 0:52

Before you do anything with formal rules of inference, you need to decide whether the statement is true or false. If it’s true, then you’ll have to worry about the rules of inference, but if it’s false, you need only provide a counterexample. So the first the thing that you should do is ask yourself whether there actually is some specific positive real number whose product with every positive real number is $1$.

As long as I’m answering, I should note that as anon pointed out in the comments, $\forall x [p(x)]$ is equivalent to $\lnot\exists x [\lnot p(x)]$, not to $\exists x [\lnot p(x)]$.

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@anon: That’s stronger than is required (or possible). Such an example would be a proof that $\exists y\forall x [xy\ne 1]$, which is false for the universe of positive reals. What’s needed is to show that for each $x>0$ there is a $y>0$ such that $xy\ne 1$. –  Brian M. Scott Oct 12 '11 at 1:04
Ah, I fudged my semantics. I mean the existence of a counterexample to the inside $\forall$ statement is guaranteed independent of $x$, though not a single positive $y$ is a counterexample for all $x$. I only commented because it seemed weird to simply speak of a counterexample to what I perceive is ultimately a $\exists$ statement. –  anon Oct 12 '11 at 1:11

Try contradiction (which I believe is trying to get at). If there is some x such that for all positive real numbers y, xy=1, and since $$1 \in \mathbb{R}$$, then, well, a contradiction quickly arises. Disproof.

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