# A Proof that Square Root of 2 Plus Square Root of 3 is Irrational

#### Introduction

In the previous post, we have learned that $\sqrt{6}$ is irrational.  In this post, we will use that theorem to show that $\sqrt{2} + \sqrt{3}$ 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 $\sqrt{2}$ is irrational and $\sqrt{3}$ 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, $\sqrt{2}$ is irrational, $-\sqrt{2}$ is irrational, but $\sqrt{2} + (-\sqrt{2}) = 0$ is rational.

#### Theorem

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

#### Proof

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

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

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

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

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

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

## 2 thoughts on “A Proof that Square Root of 2 Plus Square Root of 3 is Irrational”

1. 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.