# Direct proof: $\sqrt{13}$ is irrational

Show that $\sqrt{13}$ is an irrational number.

How to direct proof that number is irrational number. So what is the first step.....

• Perhaps you should clarify what do you mean by "direct" proof - do you mean "not by contradiction"? – Luka Mikec Dec 26 '13 at 11:22
• possible duplicate of use contradiction to prove that the square root of $p$ is irrational – Doktoro Reichard Dec 26 '13 at 11:26
• @DoktoroReichard That question specifically says to provide a proof by contradiction while this specifically says not to. – Git Gud Dec 26 '13 at 11:27
• @DoktoroReichard The english is a bit dodgy, but the OP wrote 'How to direft proof...' – Git Gud Dec 26 '13 at 11:29
• @Git sorry, misread. Still, it might be relevant. – Doktoro Reichard Dec 26 '13 at 11:29

Consider the polynomial $x^2-13\color{grey}{\in \mathbb Z[x]}$. The rational root theorem guarantees its roots aren't rational and since $\sqrt {13}$ is a root of the polynomial, it is irrational.

• how good that would be to use Rational root theorem for an elementary number theory problem... Keeping that aside, I really love this way :D – user87543 Dec 26 '13 at 11:44
• You break a butterfly on a wheel, but it works. – ulead86 Dec 26 '13 at 11:46
• @PraphullaKoushik For another example see Bill's answer here. – Git Gud Dec 26 '13 at 11:50
• @ulead86 The rational root theorem isn't deep at all, it has a simple elementary proof. – Git Gud Dec 26 '13 at 11:51
• If by "direct" the question meant "not by contradiction", then actually this proof is not valid, if you'll forgive me the pun. The rational root theorem uses either Gauss's lemma or Euclid's lemma, both of which uses proof by contradiction. ;-) – Balarka Sen Jan 13 '14 at 16:41

The standard proof that $\sqrt{p}$ is irrational for any prime $p$ is as follows

Let $\sqrt{p} = \frac{m}{n}$ where $m,n\in\mathbb N.$ and $m$ and $n$ have no factors in common.

Now $\frac{m^2}{n^2} = p \Rightarrow m^2 = p \cdot n^2$

Since $p$ is prime and $m^2$ is a multiple of $p$ then $m$ is multiple of $p$

So substitute $m = p \cdot k$

Now $\frac{(p \cdot k)^2}{n^2} = \frac{p^2 \cdot k^2}{n^2} = p\Rightarrow n^2 = p \cdot k^2$

Since $p$ is prime and $n^2$ is a multiple of $p$ then $n$ is multiple of $p$

We now have a contradiction since $m$ and $n$ must have no common factors (except 1) but we have proved that if $\frac{m}{n}$ exits then $m$ and $n$ must have common factor $p$

So $\frac{m}{n}$ can not exist and the square root of any prime is irrational.

The equation $m^2=13n^2$ is a direct contradiction to the Uniqueness part of the Fundamental Theorem of Arithmetic, since the left side has evenly many $13$’s, while the right side has oddly many.

You can try it this way:

A number is irrational, if you can not find a finite continued fraction.

Now try to write $\sqrt{13}$ as continued fraction and you'll see, its periodic:

$$\sqrt{13}=[3;\overline{1,1,1,1,6}]$$

Hope that is what your you looking for.

As mentioned, one can quickly prove the irrationality of square roots using the Rational Root Test, or uniqueness of prime factorizations, or other closely related propeerties such as Euclid's Lemma or Bezout's gcd identity. Below is a simple proof using Bezout that I discovered as a teenager.

Theorem $\quad \rm r = \sqrt{n}\;\;$ is integral if rational, $\:$ for $\:\rm n\in\mathbb{N}$

Proof $\ \$ Note that $\rm\,\ \color{#0a0}{r = a/b},\ \ \gcd(a,b) = 1\ \Rightarrow\ \color{#C00}{ad\!-\!bc \,=\, \bf 1}\;$ for some $\:\rm c,d \in \mathbb{Z}\,\$ by Bezout.

$\rm\color{#C00}{That\,}$ and $\rm\: r^2\! = \color{orange}{\bf n}\:\Rightarrow\ \color{#0a0}{0\, =\, (a\!-\!br)}\, (c\!+\!dr) \ =\ ac\!-\!bd\color{orange}{\bf n} \:+\: \color{#c00}{\bf 1}\cdot r \ \Rightarrow\ r \in \mathbb{Z}\ \ \$ QED

The proof easily generalizes to roots of monic quadratic polynomials (and to higher degrees).