Question

prove that is not rational number
$\sqrt{5} - \sqrt{3}$
​

Answer 1

$\mathfrak{\large{\underline{\underline{Solution:-}}}}$

Let us assume that √5 - √3 is rational

i.e √5 - √3 = a/b (Where a and b are co - primes and b ≠ 0)

$\sf - \sqrt{3} = \dfrac{a}{b} - \sqrt{5}$

By squaring on both sides

$\sf {( - \sqrt{3})}^{2} = {( \dfrac{a}{b} - \sqrt{5} )}^{2}$

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

[Since (x - y)² = x² - 2xy + y²]

$\sf 3 = \dfrac{ {a}^{2} }{ {b}^{2} } - 2 \sqrt{5}. \dfrac{a}{b} + 5$

$\sf 3 + 2 \sqrt{5}. \dfrac{a}{b} = \dfrac{ {a}^{2} }{ {b}^{2} } + 5$

$\sf 2 \sqrt{5}. \dfrac{a}{b} = \dfrac{ {a}^{2} }{ {b}^{2} } + 5 - 3$

$\sf 2 \sqrt{5}. \dfrac{a}{b} = \dfrac{ {a}^{2} }{ {b}^{2} } + 2$

By taking LCM on Right Hand Side of the equation

$\sf 2 \sqrt{5}. \dfrac{a}{b} = \dfrac{ {a}^{2} + 2 {b}^{2} }{ {b}^{2} }$

$\sf 2 \sqrt{5} = \dfrac{ {a}^{2} + 2 {b}^{2} }{ {b}^{2} } \times \dfrac{b}{a}$

$\sf 2 \sqrt{5} = \dfrac{ {a}^{2} + 2 {b}^{2} }{b} \times \dfrac{1}{a}$

$\sf 2 \sqrt{5} = \dfrac{ {a}^{2} + 2 {b}^{2} }{ab}$

$\sf \sqrt{5} = \dfrac{ {a}^{2} + 2 {b}^{2} }{2ab}$

Since 'a' and 'b' are integers Right Hand Side i.e $\sf \dfrac{ {a}^{2} + 2 {b}^{2} }{2ab}$is a rational number.

So, Left Hand Side of the equation is a rational number.

But, this contradicts the fact that √5 is irrational.

This contradiction has arised because of our wrong assumption that √5 - √3 is rational.

So we can conclude that √5 - √3 is irrational or not a rational number.