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?
 A: Since you have already seen answers which complete your proof, here is another proof using continued fractions.
If $r$ was rational, it will have a finite continued fraction, say $[a_1, a_2, \dots, a_n]$, but that can be extended to $[5, a_1, a_2, \dots , a_n]$ and $[5,5,a_1, a_2, \dots a_n]$ etc, because
$$ r = 5 + \frac{1}{r} = 5 + \frac{1}{[a_1, a_2, \dots, a_n]} = [5, a_1, a_2, \dots, a_n]$$
In fact, you can easily give the infinite continued fraction of $r$
$$[5,5,5,5,\dots]$$
A: I think your method is sound:
if we assume by contradiction that $r=\frac{a}{b}$ with $a,b$ integers
we get that
$$\frac{a}{b}-\frac{b}{a}=5$$
since neither $b$ nor $a$ is 0, we can multiply by $ab$ and get
$$(a-b)(a+b)=5ab$$
now there are three cases to investigate, i'll just point them out so you can finish it off:


*

*If $a$ and $b$ are both odd- look at both sides of the equation (mod2).

*If either one of them is odd and the other even- try the same trick of (mod 2)

*lastly- if both are even- you can divide the entire equation by 2 and continue.


hope that helps
A: 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).
A: Below are six methods - whose variety may prove somewhat instructive.
$(0)\ $ By the Parity Root Test, $\rm\: x^2-5\:x-1\:$ has no rational roots since it has odd leading coefficient, odd constant term and odd coefficient sum.
$(1)\ $ By the Rational Root Test, the only possible rational roots of $\rm\ x^2 -5\ x - 1\ $ are $\rm\ x = \pm 1\:.$
$(2)\ $ Complete your proof: show that $\rm\ (a,b) = 1\ \Rightarrow\ (ab,\:a^2-b^2) = 1\:.\:$ For example, if the prime $\rm\ p\ |\ a,\ a^2-b^2\ $ then $\rm\ p\ |\ b^2\ \Rightarrow\ p\ |\ b\:.\ $ Alternatively, since $\rm\ a,\:b\ $ are coprime to $\rm\ a-b,\:a+b\ $ then their products $\rm\ a\:b,\ a^2-b^2\:,\: $ are also coprime, by Euclid's Lemma.
$(3)\ $ Suppose it has a rational root $\rm\: R = A/B\:.\ $ Put it into lowest terms, so that $\rm\: B\:$ is minimal.   $\rm\ R = 5 + 1/R\ \Rightarrow\ A/B = (5\:A+B)/A\:.\:$ Taking fractional parts yields $\rm\ b/B = a/A\ $ for $\rm 0\le b < B\:.\:$ But $\rm\ b\ne0\ \Rightarrow\ A/B = a/b\ $ contra minimality of $\rm\:B\:.\:$ So $\rm\:b = 0\:,\:$ i.e. $\rm\ A/B\ $ has fractional part $ = 0\:,\:$ so $\rm\ R = A/B\ $ is an integer. Then so too is $\rm\ 1/R = R-5\:.\:$ So $\rm\ R = \pm 1\:,\:$ contra $\rm\ R^2 - 1 = 5\:R\:.$
$(4)\ $ As in $\rm(3),\ \ R = A/B = C/A\:,\: $ with $\rm\:A/B\:$ in lowest terms, i.e.  $\rm\:B =\: $ least denominator of $\rm\:R\:.\:$ By unique fractionization, the least denominator divides every denominator, therefore $\rm\:B\ |\ A\:,\:$ which concludes the proof as in $(3)$.
For more on the relationship between $(3)$ and $(4)$, follow the above link, where you'll find my analysis of analogous general square-root irrationality proofs, and links to an interesting discussion of such between John Conway and I.
$(5)\ $ As Euclid showed a very long time ago, the Euclidean gcd algorithm works also for rationals, so they too have gcds, and such gcds enjoy the same laws as for integers, e.g. the distributive law. Thus $\rm\ (r,1)^2 = (r^2,r,1) = (5r+1,r,1) = (r,1)\ $ so $\rm\ (r,1) = 1\:,\:$ so $\rm\ 1\ |\ r \ $ in $\rm\:\mathbb Z\:,\:$ i.e. $\rm\ r\in\mathbb Z\:,\:$ and the proof concludes as above. This is - at the heart - the same proof hinted by Aryabhata using non-termination of the continued-fraction algorithm (a variant of the Euclidean gcd algorithm). Alternatively, scaling the prior equations by $\rm\:b^2\:,\:$ where $\rm\ r = a/b\:,\:$ converts it to one using only integer gcd's, namely $\rm\ 1 = (a,b)^2 = (a^2,ab,b^2) = (5ab+b^2,ab,b^2) = b\:(a,b)\:$ so $\rm\ b\ |\ 1\:.\:$
These are essentially special cases of gcd/ideal-theoretic ways to prove that $\rm\: \mathbb Z\:$ is integrally-closed, i.e. it satisfies the monic Rational Root Test. Namely, a rational root of a monic polynomial $\rm\in\mathbb Z[x]\:$ must be integral. Perhaps the slickest way to prove such results is the elegant one-line proof using Dedekind's notion of conductor ideal. Follow that link (and its links) for much further discussion (both elementary and advanced).
