Question

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

Standard:- 10

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Answer 1

Let
$\sqrt{5} - \sqrt{3} = \frac{m}{n}$
where m, n are integers and n not equal to 0.

Here, we assumed √5-√3 is rational.

Then, squaring both sides.
$5 - 2 \times \sqrt{5} \times \sqrt{3} + 3 = \frac{ {m}^{2} }{ {n}^{2} } \\ = > 8 - 2 \sqrt{15} = \frac{ {m}^{2} }{ {n}^{2} } \\ = > 8 - \frac{ {m}^{2} }{ {n}^{2} } = 2 \sqrt{15} \\ = > 4 - \frac{ {m}^{2} }{2 {n}^{2} } = \sqrt{15}$

Here, RHS is ir- rational as given in question.
But,
LHS is rational cause subtraction of two rational numbers is always a rational. ( m/n is rational as we assumed)

This causes a contradiction.
Hence, Our Assumption is wrong.
=> (√5 -√3 ) is irrational.