# Proof that Square Root of 6 is Irrational

#### Introduction

We have learned that $latex \sqrt{3}$ is irrational and thousands of sites on the internet will show you the proof that $latex \sqrt{2}$ is irrational. In this post, we show that $latex \sqrt{6}$ is irrational. This is a prerequisite for the next post which will show that $latex \sqrt{2} + \sqrt{3}$ is irrational.

You will probably argue that since $latex \sqrt{2}$ is irrational and $latex \sqrt{3}$ is irrational, then their product should be irrational. This is not always the case. The product of two irrational numbers is not always irrational.  For instance, $latex \sqrt{2}$ is irrational but $latex (\sqrt{2})(\sqrt{2}) = 2$ is rational. And now, we proceed the the theorem and its proof.

#### Theorem

$latex \sqrt{6}$ is irrational.

#### Proof

To prove the theorem above, we use proof by contradiction.

Assuming that $latex \sqrt{6}$ is rational. Then, $latex \sqrt{6} = \frac{a}{b}$ where  $latex a$ and $latex b$ are integers and $latex \frac{a}{b}$ is in lowest terms. This means that $latex a$ and $latex b$ cannot be both even (Why?) (*)

Squaring both sides, we have $latex \displaystyle 6 = \frac{a^2}{b^2}$.

Multiplying both sides by $latex b^2$, we have $latex 6b^2 = a^2$. It follows that $latex a^2$ is even and  $latex a$ is even. (**)

If $latex a$ is even, then it can be expressed as $latex 2c$ where $latex c$ is an integer. Substituting to the equation above, we have $latex 6b^2 = (2c)^2$ which simplifies to

$latex 6b^2 = 4c^2$.

Dividing both sides by $latex 2$ gives $latex 3b^2 = 2c^2$. This implies that $latex 3b^2$ is even which means that $latex b$ is even (Why?). (***)

From *, we assumed that $latex a$ and $latex b$ cannot be both even. But from ** and *** $latex a$ and $latex b$ are even. A contradiction!

Therefore, $latex \sqrt{6}$ is irrational.

## 2 thoughts on “Proof that Square Root of 6 is Irrational”

1. Joash Romain on said:

Probably the best explanation I’ve had yet