Question

Prove that ${\displaystyle 5\sqrt{2}-3}$ is an irrational number, given that ${\displaystyle \sqrt{2}}$ is an irrational number.

Answer 1

Let us assume that ${\displaystyle 5\sqrt{2}-3}$ is an irrational number.

Then we can find integers a and b(b ≠ 0) such that

${\displaystyle 5\sqrt{2}-3=\frac{a}{b}}$

${\displaystyle 5\sqrt{2}=\frac{a}{b}+3}$

${\displaystyle 5\sqrt{2}=\frac{a+3b}{b}}$

${\displaystyle \sqrt{2}=\frac{a+3b}{5b}}$

This is a contradiction because the right-hand side is a rational number while given that ${\displaystyle \sqrt{2}}$ is an irrational. So, our assumption that ${\displaystyle 5\sqrt{2}-3}$ is rational is wrong. Hence, ${\displaystyle 5\sqrt{2}-3}$ is an irrational number.