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Prove that $2 + 3\sqrt{5}$ is an irrational number given that $\sqrt{5}$ is irrational number.
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Let $2 + 3\sqrt{5}$ be a rational number.
$\therefore 2 + 3\sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and $p$ and $q$ are integers. (I) ($\frac{1}{2}$ Mark)
$\Rightarrow \sqrt{5} = \frac{p-2q}{3q}$ (II) ($\frac{1}{2}$ Mark)
As $\frac{p-2q}{3q}$ is a rational number, so $\sqrt{5}$ is rational.
But we know that $\sqrt{5}$ is irrational.
$\therefore$ Our assumption is wrong. Hence, $2 + 3\sqrt{5}$ is an irrational number. (III) (1 Mark)
$\therefore 2 + 3\sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and $p$ and $q$ are integers. (I) ($\frac{1}{2}$ Mark)
$\Rightarrow \sqrt{5} = \frac{p-2q}{3q}$ (II) ($\frac{1}{2}$ Mark)
As $\frac{p-2q}{3q}$ is a rational number, so $\sqrt{5}$ is rational.
But we know that $\sqrt{5}$ is irrational.
$\therefore$ Our assumption is wrong. Hence, $2 + 3\sqrt{5}$ is an irrational number. (III) (1 Mark)