Question

Prove that root of 2 plus root of 3 is irrational

Answer 1

Solution. We prove that the number is irrational by contradiction. Let $\sqrt{2}+\sqrt{3}$ be rational.

Therefore, there exist coprime integers $a$ and $b$:

$$\sqrt {2}+\sqrt{3} = \frac{a}{b}$$

Raised to a power 2 both sides of the equation one gets

$$2+2\sqrt6+3=\frac{a^2}{b^2}$$

$$\sqrt6=\frac{a^2}{2b^2}-\frac{5}{2}$$

$$\sqrt6=\frac{a^2-5b^2}{2b^2}$$

$a^2-5b^2$ and $2b^2$ are integers because $a$ and $b$ are integers. Hence, the number $\sqrt{6}$ is rational. Thus, we can write

$$\sqrt6=\frac{p}{q}$$

where $p,q$ are coprime integers. From the last equation it follows that

$$6q^2=p^2$$

The left side of the equation is divisible by 2 therefore the right side is divisible by 2, hence $p$ is an even number. As a result,

$$p=2k$$

where $k$ is integer. Therefore,

$$6q^2=4k^2 \implies 3q^2=2k^2$$

The right side of the last equation is divisible by 2, hence the left side is divisible by 2. As a result,

$$q=2n,$$

where $n$ is integer. Therefore, one gets $p, q$ are not coprime integers. A contradiction occurred. Hence, $\sqrt{6}$ is irrational and

the assumption that 'the number $\sqrt{2}+\sqrt{3}$ is rational' is false. We proved that the number $\sqrt{2}+\sqrt{3}$ is irrational by contradiction.