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If $\sqrt{2}$ is given as an irrational number, then prove that $(5-2\sqrt{2})$ is an irrational number.
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Let us assume that $5 - 2\sqrt{2}$ be a rational number.
$\therefore 5 - 2\sqrt{2} = \frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
$\Rightarrow \sqrt{2} = \frac{5q-p}{2q}$
RHS is a rational number. So, LHS is also a rational number which contradict the given fact that $\sqrt{2}$ is an irrational number.
So, our assumption is wrong.
Hence, $5 - 2\sqrt{2}$ is an irrational number.
$\therefore 5 - 2\sqrt{2} = \frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
$\Rightarrow \sqrt{2} = \frac{5q-p}{2q}$
RHS is a rational number. So, LHS is also a rational number which contradict the given fact that $\sqrt{2}$ is an irrational number.
So, our assumption is wrong.
Hence, $5 - 2\sqrt{2}$ is an irrational number.