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.
Best Answer
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.