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Prove that $(5\sqrt{3} + \frac{2}{3})$ is an irrational number given that $\sqrt{3}$ is an irrational number.
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Sol. Let $5\sqrt{3} + \frac{2}{3}$ be a rational number.
$\therefore 5\sqrt{3} + \frac{2}{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$.
$5\sqrt{3} = \frac{a}{b} - \frac{2}{3}$
$\sqrt{3} = \frac{3a - 2b}{15b}$
$3a - 2b$ and $15b$ are integers.
$\therefore$ RHS is rational.
But LHS $= \sqrt{3}$ is an irrational number which is contradiction to our supposition.
Hence $5\sqrt{3} + \frac{2}{3}$ is an irrational number.
$\therefore 5\sqrt{3} + \frac{2}{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$.
$5\sqrt{3} = \frac{a}{b} - \frac{2}{3}$
$\sqrt{3} = \frac{3a - 2b}{15b}$
$3a - 2b$ and $15b$ are integers.
$\therefore$ RHS is rational.
But LHS $= \sqrt{3}$ is an irrational number which is contradiction to our supposition.
Hence $5\sqrt{3} + \frac{2}{3}$ is an irrational number.