Question

Prove that $(5-\sqrt{3})$ is irrational.
Or, prove that $\frac{3\sqrt{3}}{5}$ is irrational.

Answer 1

Let us assume that $\left(5-\sqrt{3}\right)$ is rational. That is, we can find integers p and q(≠ 0) such that
$\left(5-\sqrt{3}\right)=\frac{p}{q}\mathrm{or}5-\frac{\mathrm{p}}{\mathrm{q}}=\sqrt{3}$
$\Rightarrow \sqrt{3}=5-\frac{\mathrm{p}}{\mathrm{q}}$
Since, p and q are integers, $5-\frac{p}{q}$ is rational; so, $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
So, we can conclude that $\left(5-\sqrt{3}\right)$ is irrational.

OR
Let us assume that $\frac{3\sqrt{3}}{5}$ is rational. That is, we can find co -prime integers p and q(≠ 0) such that $\frac{3\sqrt{3}}{5}=\frac{p}{q}\mathrm{or}\frac{5p}{3q}=\sqrt{3}$
$\Rightarrow \sqrt{3}=\frac{5p}{3q}$
Since, p and q are integers, $\frac{5p}{3q}$ is rational; so, $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
So, we can conclude that$\frac{3\sqrt{3}}{5}$ is irrational.