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

Question

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

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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).