$xy^{-1}$ In order to prove the following:
$$x<y \iff x^{-1}>y^{-1}$$
*for $x>0$ and $y>0$
I tried this:
$$x<y\implies y-x>0$$
I have to prove that this, implies that $\frac{1}{x}>\frac{1}{y}$, that is, $\frac{1}{x}-\frac{1}{y}>0$.
Well, we know that $$\frac{1}{x}-\frac{1}{y} = \frac{y-x}{xy}$$
The numerator $y-x>0$ by assumption, the denominator $xy$ can be $>0$ or $<0$. Since $x$ and $y$ are positive, then $xy>0$ and its proved that the quotient is positive and thus $>0$.
Now, for the converse, suppose $x^{-1}>y^{-1}$, then:
$$\frac{1}{x}-\frac{1}{y} = \frac{y-x}{xy}>0 \implies $$
Well, if the quotient is $>0$, then either $y-x$ and $xy$ are both positive, or both negative. If they're positive, then $y>x\implies x<y$ which is what we wanted to prove. Now, if both are negative, then $y-x<0 \implies y<x$. What? This can't happen.
UPDATE: totally forgot that $x>0$ and $y>0$... So, $xy$ can't be negative.
Also, in order to prove $x>0 \iff x^{-1}>0$ I tried:
$x>0 \implies \frac{1}{x}>0$
but... what? this is too obvious...

Once you've proved it forwards, you've already proved it backwards
$$x\gt0\implies x^{-1}\gt 0$$
$$y\gt0\implies y^{-1}\gt 0$$
$$(x\lt y\implies x^{-1}\gt y^{-1})\implies(y^{-1}\lt x^{-1}\implies (y^{-1})^{-1}\gt x^{-1})^{-1})$$
