Let $x$ and $y$ be two positive real numbers with $x < y$. Using only the axioms for real numbers, show that $0 < \frac{1}{y} < \frac{1}{x}$. 
Let $x$ and $y$ be two positive real numbers with $x < y$. Using only the axioms for real numbers, show that $0 < \frac{1}{y} < \frac{1}{x}$.

How can I prove this?
This is what I have so far:
$0 < x < y$ (definition of positive)
$0 < 1 < \frac{y}{x}$ (division by $x$)
$0 < \frac{1}{y} < \frac{1}{x}$ (division by $y$)
But I think it seems too simple.

Suppose $x,y >0$ so that $x<y$ and for the sake of getting a contradiction assume that $\frac{1}{x}$ $\leq$ $\frac{1}{y}$. Then you get $x \geq y$, which means either $x=y$ or $x > y$. In either case you have a contradiction. 
