Question

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

Answer 1

Let us assume that $\surd 2+\surd 3$ is a rational number.

So it can be written in the form $\frac{a}{b}$

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

Here $a$ and $b$ are coprime numbers and $b\ne 0$

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

$\surd 2=\frac{a}{b}-\surd 3$

On squaring both the sides we get,

$\Rightarrow$${(\surd 2)}^2={\left(\frac{a}{b}-\sqrt{3}\right)}^2$

We know that

${(a–b)}^2=a^2+b^2–2ab$

So the equation ${\left(\frac{a}{b}-\sqrt{3}\right)}^2$ can be written as

${\left(\frac{a}{b}-\sqrt{3}\right)}^2=\frac{a^2}{b^2}+3–2\left(\frac{a}{b}\right)\left(\sqrt{3}\right)$

Substitute in the equation we get,

$2=\frac{a^2}{b^2}+3–2\left(\frac{a}{b}\right)\left(\sqrt{3}\right)$

Rearranging the equation we get,

$\frac{a^2}{b^2}+3-2=2\left(\frac{a}{b}\right)\left(\sqrt{3}\right)$

$\frac{a^2}{b^2}+1=2\left(\sqrt{3}\right)\left(\frac{a}{b}\right)$

$\frac{a^2+b^2}{b^2}\times \frac{b}{2a}=\left(\sqrt{3}\right)$

$\frac{a^2+b^2}{2ab}=\left(\sqrt{3}\right)$

Since, a, b are integers, $\frac{a^2+b^2}{2ab}$ is a rational number.

$\surd 3$ is a rational number.

It contradicts to our assumption that is$\surd 3$ irrational.

Therefore, our assumption is wrong

Thus, $\surd 2+\surd 3$ is irrational.