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Prove that $5-2\sqrt{3}$ is an irrational number. It is given that $\sqrt{3}$ is an irrational number.
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Sol. Assuming $5 - 2\sqrt{3}$ to be a rational number.
Let $5 - 2\sqrt{3}= \frac{a}{b}$ where $a$ and $b$ are integers & $b\neq 0$
$\Rightarrow \sqrt{3} = \frac{5b-a}{2b}$
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence, $5 - 2\sqrt{3}$ is an irrational number.
Let $5 - 2\sqrt{3}= \frac{a}{b}$ where $a$ and $b$ are integers & $b\neq 0$
$\Rightarrow \sqrt{3} = \frac{5b-a}{2b}$
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence, $5 - 2\sqrt{3}$ is an irrational number.