Proving that the square root of 5 is irrational 
Prove that $\sqrt{5}$ is irrational.

I begin with the identity $(\sqrt{5} + 2 )(\sqrt{5} - 2 ) = 1$.
Then I am told to extract $\sqrt{5}$ from the first or second factor and consider it to be $\frac{m}{n}$ so I should replace it in both sides.
I have $$\frac{m}{n} = (\frac{1}{\frac{m}{n}} + 2) + 2.$$
I am also told to work on the right side until I have a denominator less than $n$ and I have to explain the reasoning.
Then I have to prove this is false by contradiction.
Right now my main problem is I can't get a denominator less than $n$.
 A: Consider $\sqrt{15}$ to be rational. Then we can express it into the form $\frac{p}{q}$, where p and q are integers with $gcd(p, q) = 1.$
Now:
$$\frac{p}{q}= \sqrt{15}$$
$$⇒\frac{p^2}{q^2} = 15$$
$$⇒p^2 = 15 q^2$$
$$⇒ 15|p^2$$
$$⇒ 15|p \tag{*}$$

Now let $p = 15m$, for some $m ∈ ℕ$ 
$$p= 15m$$
$$⇒p^2 = (15)^2 m^2$$
$$⇒15q^2 = (15^2) m^2 \text { since $p^2 = 15q^2$}$$
$$⇒ 15|q^2$$
$$⇒ 15|q \tag{**}$$

Hence, from $(*)$ and $(**)$, leads us to think that our original assumption that the $\gcd = 1$ is wrong. This is a contradiction. Thus, our original statement holds. Hope this helps (:
A: Hint: correcting and simplifying your RHS,
$$\frac{m}{n}=\frac{1}{(m/n)+2}+2=\frac{2m+5n}{m+2n}\ ,$$
but there is no way the RHS denominator is less than the LHS denominator.  Try doing something similar but starting with
$$\sqrt5=\frac{1}{\sqrt5-2}-2\ .$$
This will give you a proof that $\sqrt5$ is irrational.
A: Once again:
Here is a proof of mine
by contradiction that
if $n$ is a positive integer
that is not a perfect square
then
$\sqrt{n}$ is irrational:
$\sqrt{17}$ is irrational: the Well-ordering Principle
