Question

Prove that $\sqrt{2}+\sqrt{5}$ is an irrational number.

Answer 1

Let us assume that $\sqrt{2}$ be a rational number which can be expressed in the form of $\frac{p}{q}$ where $p$ and $q$ are integers, $q\ne 0$ and $p$ and $q$ are co prime that is $HCF(p,q)=1$.

We have,

$\sqrt{2}=\frac{p}{q}\Rightarrow \sqrt{2}q=p.....\left(1\right)\Rightarrow 2q^2=p^2\left(squaringbothsides\right)$

$\Rightarrow p^2$ is divisible by $2$

$\Rightarrow p$ is divisible by $2$$......\left(2\right)$

Therefore, for any integer $r$,

$p=2r\Rightarrow \sqrt{2}q=2r\left(from\right(1\left)\right)\Rightarrow 2q^2=4r^2\left(squaringbothsides\right)\Rightarrow q^2=\frac{4}{2}{r}^2\Rightarrow q^2=2r^2$

$\Rightarrow q^2$ is divisible by $2$

$\Rightarrow q$ is divisible by $2$$......\left(3\right)$

From equation 2 and 3, we get that $2$ is the common factor of $p$ and $q$ which is a contradicts that $p$ and $q$ are co prime. This means that our assumption was wrong.

Thus $\sqrt{2}$ is an irrational number.

Similarly, we can prove that $\sqrt{5}$ is an irrational number.

We know that the sum of two irrational numbers is an irrational number.

Hence $\sqrt{2}+\sqrt{5}$ is an irrational number.