Proof by contradiction: $r - \frac{1}{r} =5\Longrightarrow r$ is irrational? Prove that any positive real number $r$ satisfying: 
$r - \frac{1}{r} = 5$ must be irrational. 
Using the contradiction that the equation must be rational, we set $r= a/b$, where a,b are positive integers and substitute:
$\begin{align*}
&\frac{a}{b} - \frac{1}{a/b}\\

&\frac{a}{b} - \frac{b}{a}\\

&\frac{a^2}{ab} - \frac{b^2}{ab}\\

&\frac{a^2-b^2}{ab}
\end{align*}$
I am unsure what to do next?

To complete your solution, note that you can, without loss of generality, set $a$ and $b$ to be coprime. So you have $a^2=b^2+5ab=b(b+5a)$. Hence $a$ divides $b(b+5a)$. Euclid's lemma now tells you that $a$ divides $b+5a$ (because $a$ and $b$ are coprime). But then $a$ must divide $b$, which is contradiction with the fact that $a$ and $b$ are coprime.
Here's an alternative: transform it into $r^2-5r-1=0$. What are the real (if any) roots of this equation? The quadratic formula gives you: $r_{12}=5/2\pm \sqrt{29}/2$. Since there are at most two different roots for a quadratic polynomial in $\mathbb{R}$, these are the roots. So your problem comes down to showing that $\sqrt{29}$ is irrational. (In fact, you can prove that the square root of any non-perfect-square number is irrational. This is the number-theoretic part).
