# Is sqrt33 an irrational number?

Sep 29, 2016

$\sqrt{33}$ is irrational

#### Explanation:

A rational number is a real number which can be expressed as the ratio of two integers. That is, $x$ is rational if and only if $x = \frac{a}{b}$ where $a$ and $b$ are integers and $b \ne 0$.

Now, suppose $\sqrt{33}$ is rational. Then $\sqrt{33} = \frac{a}{b}$ for some integers $a$ and $b$. Then, squaring both sides, we get ${\left(\sqrt{33}\right)}^{2} = {\left(\frac{a}{b}\right)}^{2}$

$\implies 33 = {a}^{2} / {b}^{2}$

$\implies 33 {b}^{2} = {a}^{2}$

Now we consider how many times each side is divisible by $33$. Suppose $a$ is divisible by $33$ $m$ times, that is, $a = j \cdot {33}^{m}$ for some integers $j$ and $m$ where $j$ is not divisible by $33$, and similarly $b = k \cdot {33}^{n}$ for integers $k$ and $n$ where $k$ is not divisible by $33$.

Substituting those in, we get

$33 {\left(k \cdot {33}^{n}\right)}^{2} = {\left(j \cdot {33}^{m}\right)}^{2}$

$\implies 33 \cdot {k}^{2} \cdot {33}^{2 n} = j \cdot {33}^{2 m}$

$\implies k \cdot {33}^{2 n + 1} = j \cdot {33}^{2 m}$

As neither $k$ nor $j$ are divisible by $33$, we know that the left hand side is divisible by $33$ exactly $2 n + 1$ times, whereas the right hand side is divisible by $33$ $2 m$ times.

But $2 n + 1$ is odd and $2 m$ is even, this meaning they are divisible by $33$ a different number of times, and thus cannot be equal. Thus our initial assumption was wrong, and so $\sqrt{33}$ is not rational, meaning it is irrational.

As a side note, this method can show that $\sqrt{n}$ is irrational for any positive integer $n$ unless $n$ is a perfect square.