Question

prove that √3+√5 is irrational​

Answer 1

if possible, let √3+√5 be rational

let √3+√5 =a,where a is rational

so, √3=a-√5

squaring both sides,we get

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

$⟹ {a}^{2} + 5 - 2a \sqrt{5}$

$⟹ {a}^{2} + 2 - 2a \sqrt{5} = 0$

$⟹ \sqrt{5} = \frac{ {a}^{2} + 2 }{2a} \: \: \: \: \: \: \: \: ..(5)$

$but \: \frac{ {a}^{2} + 2}{2a}$

is rational number

thus from (5), √5 is rational

THIS contradicts the fact that √5 is irrational

since,the contradiction arises by assuming √3+√5 is rational

hence, √3+√5 is irrational

Answer 2

To prove : √3+√5 is irrational.

Let us assume that it to be a rational number.

Rational numbers are the ones that can be expressed in $\frac{p}{q}$ form,where p,q are integers and q isn’t equal to 0.

$\sqrt{3} + \sqrt{5} = \frac{p}{q}$

$⇒ \sqrt{3} = \frac{p}{q} - \sqrt{5}$

$⇒ 3 = \frac{ {p}^{2} }{ {q}^{2} } - 2. \sqrt{5} ( \frac{p}{q} ) + 5\:\:\:\:\:\:\:\:(squaring\:on\:both\:side)$

$⇒ \frac{(2 \sqrt{5} p)}{q} = 5 - 3 + \frac{ {p}^{2} }{ {q}^{2} }$

$⇒ \frac{(2 \sqrt{5} p)}{q} = \frac{2 {q}^{2} - {p}^{2} }{ {q}^{2} }$

$⇒ \sqrt{5} = \frac{2 {q}^{2} - {p}^{2} }{ {q}^{2} } . \frac{q}{2p}$

$⇒ \sqrt{5} = \frac{2 {q}^{2} - {p}^{2} }{ 2pq }$

As p and q are integers RHS is also rational.

As RHS is rational LHS is also rational i.e √5 is rational.

But this contradicts the fact √5 is irrational.

This contradiction arose because of our false assumption.

So, √3+√5 is irrational.