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Prove that $4 - 2\sqrt{5}$ is an irrational number given that $\sqrt{5}$ is irrational.
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Sol. Let $4 - 2\sqrt{5}$ be a rational number. (I Mark)
$\therefore 4 - 2\sqrt{5} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (II Mark)
$\sqrt{5} = \frac{4b - a}{2b}$
RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $4 - 2\sqrt{5}$ is an irrational number. (III Mark)
$\therefore 4 - 2\sqrt{5} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (II Mark)
$\sqrt{5} = \frac{4b - a}{2b}$
RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $4 - 2\sqrt{5}$ is an irrational number. (III Mark)