How to prove statement with quantifiers? How would I prove something like this:
$ \forall x >0, \exists y\gt 0,(xy = 1) $
I read it as "For all $x$ greater than $0$, there exists a $y$ greater than $0$, such that $xy=1$."
Is this correct? 
I understand this is such an easy proof and I assume it is obviously true, but I don't really understand where to start it or how to prove it. I can see that I eventually need to show that $\ y=1/x $ but how do I show that? Should I do it by cases? Should I do it by contradiction?
 A: hint
When the statement begins by
$$\forall x\in E $$
the proof should begin by
Let $x $ be an element of $E $.
When the statement contains
$$\exists y\in B $$
there are two possibilities :
1.
the existence of $y $ is insured by a theorem (IVT,  MVT,  Rolle...)
2.
You construct your $y $ by 
Put $y=... $.
In your case, we will say :
Let $x $ be a real $>0$.
put $y=\frac {1}{x} $
then $xy=1$.
A: No need to do this case by case.  
All we need is to input any/every real positive number $x$, (for every $x\in \mathbb R$, such that $x>0$,) 
we can find a $y> 0, y\in \mathbb R$,  specifically $$(y> 0,\; \;y\in \mathbb R)\; \;y=\frac 1x$$ such that that 
 $$xy =x\cdot \frac 1x = 1$$
Why we consider $y= \dfrac 1x \gt 0$?: 
Given $x>0,x\in \mathbb R$, can you show that $y=\dfrac 1x$ ensures us that $y>0,$ and $y\in \mathbb R$?
A: Let's say we look at $x,y \in \mathbb{Q}$ for now.
Well, you assume that you have an arbitrary $x > 0$. So you are not allowed to use any "special" properties of this number except its positivity.
Now you need to find (at least) one $y > 0$ such that xy = 1. One way to do that is to note that $x$ can be written as $x = \frac{a}{b}$ with both $a,b \in \mathbb{N}$, i.e. $a,b > 0$. Then the $y$ you are looking for is given by $y = \frac{b}{a} = \frac{1}{x}$. This is again a positive number, and you can easily check that $xy = 1$.
So the idea is that you take a very general $x$ with no other properties, and then your task is to find an associated $y$ such that $xy = 1$. In finding $y$, you may, however, use what you know about $x$, as $x$ is fixed after you have "chosen" to consider one. Then if you look at a different $x$, if you have done the proof right, you should be able to find the corresponding $y$, which then completes the proof.
Hope that helps!
A: A common structure for a proof of a 'for all...there exists' statement is to start by saying: Let $x$ be an arbitrary... and then producing an object that satisfies the conditions.
So in your problem, I would say: Let $x$ be an arbitrary positive real number.  We can then take $y=\frac1x.$
Then $xy=x\cdot \frac1x=1$, as desired.
A: As the OP states, the question looks so simple that coming up with a proof looks daunting. But this is an opportunity to set up an interesting 'premise framework'   and to proceed with a demonstration that builds on these axioms.
(P1) Every nonzero real number $x$ has a multiplicative inverse $y$ that is also not equal to zero.
(P2) The 'take the sign' mapping $\pi$ on the set of nonzero real numbers, defined by
$\quad \pi(x) = 1 \text{ for } x \gt 0$
and
$\quad \pi(x) = -1 \text{ for } x \lt 0$
satisfies $\pi(xy) = \pi(x) \pi(y)$ for all nonzero real numbers $x$ and $y$
(the product of two nonzero numbers can't be equal to zero).
Proposition: $\forall x >0, \exists y\gt 0,(xy = 1)$
Proof
Let $x$ be a positive real number. By (P1) $x$ has a nonzero inverse $y$. By (P2) we can apply $\pi$,
$\quad \pi(xy) = \pi(1) = 1 = \pi(x) \pi(y) = 1 \pi(y) = \pi(y)$
so we conclude that $\pi(y) = 1$. But the inverse of $\{1\}$ under $\pi$ is precisely the set of positive numbers, so $y$ must be positive. $\quad \blacksquare$
A: Although the following in not how proofs are normally done in practice (in most math textbooks etc.), I just wanted to show how it could be proven with the rules of natural deduction and algebraic axioms:
\begin{array}{l}
& \{1\} & 1. & a \in \mathbb{R} \land a>0 & \text{ Assum. }\\
& - & 2. & a(\frac{1}{a})=a(\frac{1}{a}) & \text{ =I }\\
& \{1\} & 3. & a(\frac{1}{a})=1 & \text{ 1,2 Inverse (if $a \neq 0$) }\\
& - & 4. & 1>0 & \text{ Def. 1 }\\
& \{1\} & 5. & \frac{1}{a} > \frac{0}{a} & \text{ 1,4 Division (if $a > 0$) }\\
& \{1\} & 6. & \frac{1}{a} > 0 & \text{ 5 Mult. property of 0 }\\
& \{1\} & 7. & \frac{1}{a} \in \mathbb{R}  & \text{ 1,5 Closure on $\mathbb{R}$  }\\
& \{1\} & 8. & \frac{1}{a} \in \mathbb{R} \land \frac{1}{a}>0 \land a(\frac{1}{a})=1 & \text{ 3,6,7 $\land$I }\\
& \{1\} & 9. & \exists y[y \in \mathbb{R} \land y>0 \land ay=1] & \text{ 8 EI }\\
& - & 10. & (a \in \mathbb{R} \land a>0) \implies \exists y[y \in \mathbb{R} \land y>0 \land ay=1] & \text{ 1,9 CP }\\
& - & 11. & \forall x[(x \in \mathbb{R} \land x>0) \implies \exists y[y \in \mathbb{R} \land y>0 \land xy=1]] & \text{ 10 UI }\\
\end{array}
