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Prove that $2 - 5\sqrt{3}$ is an irrational number given that $\sqrt{3}$ is irrational.
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Sol. Let $2 - 5\sqrt{3}$ be a rational number.
$2 - 5\sqrt{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (I) (1/2)
$ \sqrt{3} = \frac{2b-a}{5b}$ (II) (1/2)
RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $2 - 5\sqrt{3}$ is an irrational number. (III) (1)
$2 - 5\sqrt{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (I) (1/2)
$ \sqrt{3} = \frac{2b-a}{5b}$ (II) (1/2)
RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $2 - 5\sqrt{3}$ is an irrational number. (III) (1)