**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. Continue reading