Proof that Square Root of 6 is Irrational

Introduction

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

You will probably argue that since $\sqrt{2}$ is irrational and $\sqrt{3}$ 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, $\sqrt{2}$ is irrational but $(\sqrt{2})(\sqrt{2}) = 2$ is rational. And now, we proceed the the theorem and its proof.

Theorem $\sqrt{6}$ is irrational.

Proof

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

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

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

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

If $a$ is even, then it can be expressed as $2c$ where $c$ is an integer. Substituting to the equation above, we have $6b^2 = (2c)^2$ which simplifies to $6b^2 = 4c^2$.

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

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

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