0
$\begingroup$

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?

  • 0
    @BrianM.$S$cott: Alright, tha$n$ks. If you add that as an answer I'll mark it as correct.2011-10-12

2 Answers 2

2

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 $\exists x [\lnot p(x)]$.

  • 0
    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.2011-10-12
0

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.