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Prove that $6 - \sqrt{7}$ is irrational number, given that $\sqrt{7}$ is an irrational number.
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Let us assume that $6 - \sqrt{7}$ is rational
$\therefore 6 - \sqrt{7} = \frac{p}{q}$ ; $q \neq 0$ and $p, q$ are integers ($\frac{1}{2}$)
$\Rightarrow \sqrt{7} = \frac{6q - p}{q}$ ($\frac{1}{2}$)
p, q are integers, $\therefore 6q - p$ is an integer
$\Rightarrow \frac{6q - p}{q}$ is a rational number ($\frac{1}{2}$)
$\Rightarrow \sqrt{7}$ is rational number which contradicts our assumption that $\sqrt{7}$ is an irrational number
$\Rightarrow 6-\sqrt{7}$ is an irrational number ($\frac{1}{2}$)
$\therefore 6 - \sqrt{7} = \frac{p}{q}$ ; $q \neq 0$ and $p, q$ are integers ($\frac{1}{2}$)
$\Rightarrow \sqrt{7} = \frac{6q - p}{q}$ ($\frac{1}{2}$)
p, q are integers, $\therefore 6q - p$ is an integer
$\Rightarrow \frac{6q - p}{q}$ is a rational number ($\frac{1}{2}$)
$\Rightarrow \sqrt{7}$ is rational number which contradicts our assumption that $\sqrt{7}$ is an irrational number
$\Rightarrow 6-\sqrt{7}$ is an irrational number ($\frac{1}{2}$)