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.
 A: 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. 
A: $y>x$ means $y=x+M$ for some $M>0$.
$1/y>0$ is obvious: it's a ratio between positive numbers.
Then
$$
\frac1{y}=\frac1{x+M}=\frac1{x(1+\frac{M}{x})}=\frac1x\cdot\frac{1}{1+\frac{M}{x}}
$$
Next observe that $M,x>0$ hence $M/x>0$, thus $1+\frac{M}{x}>1$.
Now, by axioms of real numbers, the sign of an inequality is preserved is you multiply both sides by a positive quantity $c>0$. Let's multiply both sides of $1+\frac{M}{x}>1$ by the positive quantity $c=\frac{1}{1+\frac{M}{x}}$. We obtain then
$$
\frac{1}{1+\frac{M}{x}}<1.
$$
Hence you know that multiplying a positive quantity by a factor in $]0,1[$ you'll obtain something smaller (exercise). Thus
$$
\frac1x\cdot\frac{1}{1+\frac{M}{x}}<\frac1x
$$
as wanted.
A: We will be working entirely with positive numbers in this answer.
$\quad x \lt y \text{ iff } (\exists \, u) \; x + u = y$
Multiplying both sides of this equation by the multiplicative inverse of $x$ and using the law of distributivity,
$\tag 1 x^{-1}x + x^{-1}u =  x^{-1}y$
Multiplying both sides of the equation $\text{(1)}$ by the multiplicative inverse of $y$ and using the law of distributivity,
$\tag 2 y^{-1} (x^{-1}x) + y^{-1}(x^{-1}u) =  y^{-1}(x^{-1}y)$
Using the laws of associativity and commutativity for multiplication,
$\tag 3 y^{-1} (x^{-1}x) + y^{-1}(x^{-1}u) =  x^{-1}(y^{-1}y)$
So,
$\tag 4 y^{-1} (1) + y^{-1}(x^{-1}u) =  x^{-1}(1)$
So,
$\tag 5 y^{-1}  + y^{-1}(x^{-1}u) =  x^{-1}$
With $v = y^{-1}(x^{-1}u)$ we have $y^{-1}  + v =  x^{-1}$, so, indeed, $y^{-1}  \lt x^{-1}$.
