Question

show that √5-√3 is irrational number​

Answer 1

Answer:

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Answer 2

Answer:

Let $\sqrt{5} - \sqrt{3}$ be rational number.

$\sqrt{5} - \sqrt{3} = \frac{a}{b} \\ \sqrt{5} = \frac{a}{b} - \sqrt{3}$

Squaring on both the sides.

$( { \sqrt{5} )}^{2} = {( \frac{a}{b} - \sqrt{3} )}^{2} \\ 5 = { ( \frac{a}{b} )}^{2} - \frac{2a \sqrt{3} }{b} + 3 \\ 2 = { (\frac{a}{b}) }^{2} - 2 \sqrt{3} \frac{a}{b} \\ 2 \sqrt{3} \frac{a}{b} = {( \frac{a}{b} )}^{2} - 2 \\ 2 \sqrt{3}\frac{a}{b} = \frac{ {a}^{2} - {2b}^{2} }{ {b}^{2} } \\ \sqrt{ 3} = \frac{ {a}^{2} - { {2b}}^{2} }{2ab}$

$\sqrt{5} - \sqrt{3}$ is an irrational number.

Hence Proved