**Introduction**

In the previous post, we have learned that $latex \sqrt{6}$ is irrational. In this post, we will use that theorem to show that $latex \sqrt{2} + \sqrt{3}$ is also irrational. This proof was explained by Ivan Niven in his book Numbers: Rational and Irrational.

A common argument that might arise is that since $latex \sqrt{2}$ is irrational and $latex \sqrt{3}$ is irrational, then their sum is irrational. Like multiplication, the set of irrational numbers is not closed under addition. The sum of two irrational numbers is not necessarily irrational. For instance, $latex \sqrt{2}$ is irrational, $latex -\sqrt{2}$ is irrational, but $latex \sqrt{2} + (-\sqrt{2}) = 0$ is rational.

**Theorem**

$latex \sqrt{2} + \sqrt{3}$ is irrational.

**Proof**

Just like the previous post, we use proof by contradiction to prove the theorem above.

Suppose $latex \sqrt{2} + \sqrt{3}$ is rational, say $latex r$ so that

$latex \sqrt{2} + \sqrt{3} = r$.

Squaring both sides, we have $latex 2 + 2 \sqrt{6} + 3 = r^2$ which means that $latex \sqrt{6} = r^2 – 5$.

Since the set of rational numbers is closed under multiplication and addition, $latex r^2 – 5$ is therefore rational. However, as we have proved in the previous post, $latex \sqrt{6}$ is irrational. A contradiction!

Therefore, $latex \sqrt{2} + \sqrt{3}$ is irrational.

Prove root 5 -root 3 is irrational

It should be 2sqrt(6) = r^2 – 5. I think you forgot the 2 would need to be divided out so sqrt(6) = (r^2 – 5) / 2.