$xy^{-1}$ 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...
 A: This is much too complex. You only have to know two rules for inequalities,


*

*The reciprocal of a number has the same sign as this number (this results from the rule of signs),

*You can multiply both sides of an inequality by the same positive number.


Now, if $x, y$ are positive, so are $x^{-1}, y^{-1}$ by point 1 , hence also their product. By point 2, we deduce
$$(x^{-1}y^{-1})x<(x^{-1}y^{-1})y, \quad\text{i.e.}\quad y^{-1}<x^{-1}$$
by the properties of multiplication
A: You know that $x$ multiplied by $\frac{1}{x}$ gives $1$, a positive number. So if $x$ is positive, can $\frac{1}{x}$ be negative?
A: Lemma: if $0<a<b$ and $0<c<d$, then $ac<bd$.
Proof: Let $b=a+m$ and $d=c+n$, where $m>0$ and $n>0$. Then, $bd=ac+cm+an+mn$, meaning that $bd-ac=cm+an+mn>0$, meaning that $ac<bd$.
Theorem: if $0<x<y$, then $x^{-1}>y^{-1}$.
Proof: Assume the contrary, that $x^{-1}<=y^{-1}$.
The equality case is trivially rejected, leaving us with $x^{-1}<y^{-1}$.
Then, using the lemma, we obtain $x\cdot x^{-1}<y\cdot y^{-1}$, meaning that $1<1$, which is obviously false.
Therefore, $x^{-1}>y^{-1}$.
A: Some of the answers here threw me off. I think it's a lot easier than some of the answers here. Simply use the properties of multiplication. Another answer here showed that the statement is true with one step, but you can step through the intermediary "calculations" to see what is going in.
If $x < y$ then $1 < x^{-1}y$. Here, I have simply left multiplied by $x^{-1}$ on both sides. Similarly, if $1 < x^{-1}y$ then $y^{-1} < x^{-1}$, which was to be shown. You have to be careful not to include elements for which no inverse exists, but you have given that $x, y >0$. The proof in the other direction follows the exact same pattern.
A: Once you've proved it forwards, you've already proved it backwards
$$x\gt0\implies x^{-1}\gt 0$$
$$y\gt0\implies y^{-1}\gt 0$$
$$(x\lt y\implies x^{-1}\gt y^{-1})\implies(y^{-1}\lt x^{-1}\implies (y^{-1})^{-1}\gt x^{-1})^{-1})$$
