Question

3. Prove that
$\sqrt{5} - \sqrt{3} \: is \: irrational \: \\ using \: the \: fact \: that \: \ \\ \sqrt{15} \: is \: irrational.$
Standard: 10

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Answer 1

Hello dear ,

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Let us assume on the contrary that 5-root3 is irrational. Then , there exist co-prime positive integers a and b such that

$5 - \sqrt{3} = \frac{a}{b} \\ \\ = > 5 - \frac{a}{b} = \sqrt{3} \\ \\ = > \frac{5b - a}{b} = \sqrt{3} \\ \\ = > \sqrt{3} \: is \: rational \: \: \: \: \: \: (bcoz \: a \: and \: b \: are \: integers \: \: so \: \frac{5b - a}{b} \: is \: a \: rational \: number)$

This contradicts the fact that root3 is irrational. So, our assumption is incorrect. Hence, 5-root3 is an irrational number.

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Hope it helps u !!!

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# Nikky