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Prove that $7 - 3\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is an irrational number.
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Assuming $7-3\sqrt{5}$ to be a rational number.
Let $7 - 3\sqrt{5} = \frac{a}{b}$ where $a$ and $b$ are integers & $b \neq 0$
$\Rightarrow \sqrt{5} = \frac{7b-a}{3b}$
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence, $7 - 3\sqrt{5}$ is an irrational number.
Let $7 - 3\sqrt{5} = \frac{a}{b}$ where $a$ and $b$ are integers & $b \neq 0$
$\Rightarrow \sqrt{5} = \frac{7b-a}{3b}$
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence, $7 - 3\sqrt{5}$ is an irrational number.