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Prove that $\frac{2 + 3\sqrt{5}}{7}$ is an irrational number, given that $\sqrt{5}$ is an irrational number.
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Let $\frac{2 + 3\sqrt{5}}{7}$ be a rational number. ($\frac{1}{2}$ Mark)
Then $\frac{2 + 3\sqrt{5}}{7} = \frac{p}{q}$, where $q \neq 0$ and $p$ and $q$ are integers. ($\frac{1}{2}$ Mark)
$\Rightarrow 2 + 3\sqrt{5} = \frac{7p}{q}$ (1 Mark)
$\Rightarrow \sqrt{5} = \frac{7p - 2q}{3q}$ ($\frac{1}{2}$ Mark)
Since '$p$' and '$q$' are integers.
$\therefore \frac{7p - 2q}{3q}$ is rational. ($\frac{1}{2}$ Mark)
But this contradicts the fact that $\sqrt{5}$ is irrational.
Hence, $\frac{2 + 3\sqrt{5}}{7}$ is an irrational number.
Then $\frac{2 + 3\sqrt{5}}{7} = \frac{p}{q}$, where $q \neq 0$ and $p$ and $q$ are integers. ($\frac{1}{2}$ Mark)
$\Rightarrow 2 + 3\sqrt{5} = \frac{7p}{q}$ (1 Mark)
$\Rightarrow \sqrt{5} = \frac{7p - 2q}{3q}$ ($\frac{1}{2}$ Mark)
Since '$p$' and '$q$' are integers.
$\therefore \frac{7p - 2q}{3q}$ is rational. ($\frac{1}{2}$ Mark)
But this contradicts the fact that $\sqrt{5}$ is irrational.
Hence, $\frac{2 + 3\sqrt{5}}{7}$ is an irrational number.