Question

Prove that $(3-\sqrt{5})$ is irrational.
Or, prove that $\frac{2\sqrt{2}}{3}$ is irrational.

Answer 1

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

OR
Let us assume that $\frac{2\sqrt{2}}{3}$ is rational. That is, we can find co -prime integers p and q(≠ 0), such that
$$\begin{gathered} \frac{2\sqrt{2}}{3}=\frac{p}{q} \\ \mathrm{or}\frac{3p}{2q}=\sqrt{2} \end{gathered}$$
$\Rightarrow \sqrt{2}=\frac{3p}{2q}$
Since, p and q are integers, $\frac{3p}{2q}$ is rational; so $\sqrt{2}$ is rational.
But this contradicts the fact that $\sqrt{2}$ is irrational.
So, we conclude that$\frac{2\sqrt{2}}{3}$ is irrational.