**Introduction**

In the previous post, we have learned that is irrational. In this post, we will use that theorem to show that 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 is irrational and 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, is irrational, is irrational, but is rational.

**Theorem**

is irrational.

**Proof**

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

Suppose is rational, say so that

.

Squaring both sides, we have which means that .

Since the set of rational numbers is closed under multiplication and addition, is therefore rational. However, as we have proved in the previous post, is irrational. A contradiction!

Therefore, 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.