Question

prove that ✓5 +✓3 is an irrational number .​

Answer 1

To prove : $\sf \sqrt{5 } + \sqrt{3}$is irrational.

Let us assume it to be a rational number.

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

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

$\sf \sqrt{5} = \frac{p}{q} - \sqrt{3}$

squaring on both sides,

$\sf( \sqrt{5} )^{2} = \frac{p ^{2} }{q ^{2} } - ( \sqrt{3}) ^{2} \: \: \: \: \: \: \: \: \: using \:( a - b) ^{2}$

$\sf 5 = \frac{p ^{2} }{q ^{2} } - \frac{2 \sqrt{3} p}{q} + 3$

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

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

$\sf⇒ \sqrt{3} = \frac{ - 2q ^{2} + p ^{2} }{q ^{2} } \times \frac{ \ q}{2p}$

$\sf⇒ \sqrt{3} = \frac{ - 2q ^{2} + p ^{2} }{2pq}$

As p and q are integers RHS is also rational.

As RHS is rational LHS is also rational i.e $\sf\sqrt{3}$is rational.

But this contradicts the fact that $\sf \sqrt{3}$ is irrational.

This contradiction arose because of our false assumption.

so, $\sf \sqrt{5} + \sqrt{3}$ irrational.