Prove that $\sqrt{13}$ is irrational in 3 ways. I am asked how to prove that the $\sqrt{13}$ is irrational in 3 ways. I only know of one, where we assume $\sqrt{13} = \frac{a}{b}$ where $a, b$ are coprime, and we prove with contradiction that $13 \mid a$ and $13 \mid b$.
However, I am not sure where to begin with the other two proofs. I thought that this was the only way to prove irrationality of a number. I would really appreciate some help on where to start/look for to start the two other proofs.

\begin{align}
\sqrt{13} & = 3 + \text{a fractional part, since } 3^2<13<4^2 \\[8pt]
& = 3 + (\sqrt{13}-3) \\[8pt]
& = 3 + \frac{4}{\sqrt{13}+3} \text{ by rationalizing the numerator} \\[8pt]
& = 3 + \frac 1 {(\sqrt{13}+3)/4}
\end{align}
Since $3<\sqrt{13}<4,$ we have $1 < (\sqrt{13}+3)/4 < 2,$ so $(\sqrt{13}+3)/4$ is between $1$ and $2.$
So we have
\begin{align}
\sqrt{13} & = 3 + \frac 1 {1 + \left( \frac {\sqrt{13}-1} 4 \right)} \\[12pt]
& = 3 + \frac 1 {1 + \cfrac 3 {\sqrt{13} + 1}} \text{ by rationalizing the numerator as before}
\end{align}
Continuing in this way, we get
$$
\sqrt{13} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1{\sqrt{13}+3}}}}}.
$$
In the steps before this, the expression in the denominator was $\dfrac{\sqrt{13} + (\text{an integer})}{\text{an integer}}$ and the denominator in this last fraction was more than $1.$ But this time, the denominator is $1,$, i.e. we have $\dfrac{\sqrt{13}+3} 1.$ This means that now in place of $\sqrt{13}$ at the end of this long expression one can put this entire continued fraction, and we get
$$
\sqrt{13} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1{6 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1{\sqrt{13}+3}}}}}}}}}}.
$$
And then we can do that again. And again. We get this repeating sequence:
$$
3, \, \overbrace{1,1,1,1,\,6,}\, 1,1,1,1,\,6,\, 1,1,1,1,\,6,\, \ldots
$$
The part under the $\overbrace{\text{overbrace}}$ must keep repeating. This could not happen if this number were rational, as we will argue below. To see, why, suppose we start with a ration approximation to $\sqrt{13}$:
\begin{align}
\sqrt{13}\approx\frac{119}{33} & = 3 + \frac{20}{33} = 3 + \cfrac 1 {1 + \cfrac{13}{20}} \\[12pt]
& = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 7 {13}}} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 6 7}}} \\[12pt]
& = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 7 {13}}} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 6}}}}.
\end{align}
But we cannot keep getting smaller positive integers forever. So if we start with a rational number, this process cannot go on forever; it must end. But as shown above, if we start with $\sqrt{13},$ it does not end; it just keeps repeating.
