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Prove that $14 - 2\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is irrational.
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Sol. Let $14 - 2\sqrt{3}$ be a rational number. (1/2 Mark)
$14 - 2\sqrt{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (1/2 Mark)
$\sqrt{3} = \frac{14b - a}{2b}$ (1 Mark)
RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $14 - 2\sqrt{3}$ is an irrational number.
$14 - 2\sqrt{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (1/2 Mark)
$\sqrt{3} = \frac{14b - a}{2b}$ (1 Mark)
RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $14 - 2\sqrt{3}$ is an irrational number.