Question

proof
$\sqrt{3} + \sqrt{5}$
is irrational​

Answer 1

$\underline{ Let\:us\:assume\: that\: \sqrt{3} \: + \: \sqrt{5} \: is \: rational }$

$\underline{number.}$

$\sqrt{3}$ + $\sqrt{5}$ = $\dfrac{a}{b}$

Here, a and b are co-prime numbers.

Now!

Squaring on both sides.

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

As we know that

(a + b)² = a² + 2ab + b²

3 + 2 $\sqrt{15}$ + 5 = ${( \dfrac{a}{b}) }^{2}$

8 + 2 $\sqrt{15}$ = ${( \dfrac{a}{b}) }^{2}$

2 $\sqrt{15}$ = $\dfrac{ {a}^{2} \: - \: 8 {b}^{2} }{ {b}^{2} }$

$\sqrt{15}$ = $(\dfrac{1}{2})$ $(\dfrac{ {a}^{2} \: - \: 8 {b}^{2} }{ {b}^{2} } )$

$(\dfrac{1}{2})$ $(\dfrac{ {a}^{2} \: - \: 8 {b}^{2} }{ {b}^{2} } )$ is a rational number.

$\underline{So, \:\sqrt{15} \: is\: also\: a \:rational \:number.}$

$\underline{But \:as\: we \:know\: that \:\sqrt{15} \: is \:irrational\: number.}$

$\underline{So,\: our\: whole\: assumption \:is \:wrong.}$

$\bold{ \sqrt{3} \: + \: \sqrt{5} \: is \: irrational \: number }$

$\boxed{Hence,\:proofed}$