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.
 A: It is realtively easy to prove that if $q\in\Bbb Q$ is such that $q^2\in \Bbb Z$, then $q\in\Bbb Z$ as well. Once you have established that, you only need to prove that there is no integer with square $13$, and there must be three, four or perhaps ninety-seven trite ways to prove it.
A: Two other proof techniques you can use are :

*

*The rational root theorem : Let $f(x) = a_0 + a_1x + \cdots + a_nx^n$ be a polynomial, and let $\frac pq$ be a rational root of $f$. Then $p$ divides $a_0$ and $q$ divides $a_1$. For using this to prove that $\sqrt {13}$ is irrational, we create the contrapositive : if $f(x)$ is a polynomial as above, and for every $p$ dividing $a_0$ and $q$ dividing $a_n$ we have $f(\frac{p}{q}) \neq 0$, then every root of $f$ is irrational. Now, $x_0 = \sqrt{13}$ satisfies $f(x)= x^2-13$. You can see that $f$ has no rational roots using the rational root theorem, so $x_0$ must be irrational.


*Let $S = \{n \in \mathbb N : n\sqrt 13 \in \mathbb N\}$. If $\sqrt {13}$ is rational then $S$ is non-empty because $\sqrt{13} > 0$. Let $n \in S$ and let $q = n(\sqrt{13}-3)$. Then $q<n$, $q \in \mathbb N$ , and $q\sqrt{13} = 13n - 3n\sqrt{13} \in \mathbb N$ by $n \in S$. Therefore, $S$ has no minimal element despite being a subset of the natural numbers, a contradiction.
A: \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.
A: Suppose $\sqrt{13} = \dfrac k \ell$ where $k,\ell$ are positive integers and the fraction is in lowest terms.
Then $$\frac{65\ell -18 k}{5k - 18\ell} = \sqrt{13}$$ but here the numerator and denominator are positive integers smaller than $k,\ell$ respectively.
A: We can use a primitive Pythagorean triple (PPT).
By the Pythagorean theorem, there is a right triangle with hypotenuse
$\,\sqrt{13}\,$ and legs $\,2,3.\,$
Suppose that $\,\sqrt{13}=a/b\,$ for some $\,a,b\in\mathbb{Z}^+\,$
where $\,a,b\,$ are relatively prime. The given right triangle is similar
to one with hypotenuse $\,a\,$ and legs $\,2b,3b\,$ which is primitive
since the sides are relatively prime.
Use the known parametrization of PPTs to get relatively prime $\,m,n\,$ such that
$$a = m^2+n^2,\,\, 2b = 2mn,\,\, 3b = m^2-n^2.$$ The
last two equalities give $\,3mn=m^2-n^2,\,\,\,n(n+3m)=m^2,\,$ and
$\,n\,$ divides $\,m^2\,$ but
$\,m,n\,$ are relatively prime which implies a contradiction. Hence
$\,\sqrt{13}\,$ is irrational.
