Question

Prove that $\sqrt{3}+\sqrt{8}$ is an irrational number.

Answer 1

Let $\sqrt{8}$ be a rational number

Then $\sqrt{8}=\frac{m}{n}$ where m, n are integers and m, n are coprimes and $n\ne 0$

$\Rightarrow m=\sqrt{8}n$

Squaring both sides we get

$m^2=8n^2$

$\Rightarrow \frac{m^2}{8}=n^2$ ……………(iv)

$\Rightarrow 8$ divides $m^2$ i.e., $8$ divides m

Then m can be written as

$m=8$k for some integer k.

Substituting value of m in (iv) we get

$\Rightarrow \frac{(8k{)}^2}{3}=n^2$

$\Rightarrow 8k^2=n^2$

$\Rightarrow k^2=\frac{n^2}{8}$

$\Rightarrow 8$ divides $n^2$ i.e., $8$ divides n

Thus we get that $8$ is a common factor of m and n but m and n are co-primes which is a contradiction to our assumption.

Hence $\sqrt{8}$ is an irrational number.

Now consider $\sqrt{3}+\sqrt{8}$ to be an rational number

Then $\sqrt{3}+\sqrt{8}=\frac{a}{b}$ where a, b are integers, co-primes and $b\ne 0$

$\Rightarrow \sqrt{3}=\frac{a}{b}-\sqrt{8}$

Squaring both sides we get

$\Rightarrow 3=\frac{a^2}{b^2}+(\sqrt{8}{)}^2-2\frac{a}{b}\sqrt{8}$

$\Rightarrow 3=\frac{a^2}{b^2}+8-\frac{2a}{b}\sqrt{8}$

$\Rightarrow 3-\frac{a^2}{b^2}-8=-\frac{2a}{b}\sqrt{8}$

$\Rightarrow \frac{3b^2-a^2-8b^2}{b^2}=\frac{-2a}{b}\sqrt{8}$

$\Rightarrow \sqrt{8}=\frac{(3b^2-a^2-8b^2)}{-2ab}$

$\Rightarrow \sqrt{8}=\frac{a^2+8b^2-3b^2}{2ab}$

$\Rightarrow \sqrt{8}=\frac{a^2+5b^2}{2ab}$

Where $a^2+5b^2$ is an integer and $2ab$ is also an integer, but $\sqrt{8}$ is an irrational number, which is a contradiction.

Hence $\sqrt{3}+\sqrt{8}$ is an irrational number.