Question

1) Prove that root2-root3 is
an irrational number.​

Answer 1

Step-by-step explanation:

$Let \: us \: assume \: \sqrt{2} - \sqrt{3} = \: a \: where \: a \: is \: an \: rational \: number$

$- \sqrt{3} = a + \sqrt{2}$

$Squaring \: on \: both \: sides$

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

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

$3 = {a}^{2} + 2 - 2a \sqrt{2}$

$2a \sqrt{2} = {a}^{2} + 2 - 3$

$\sqrt{2} = \frac{ {a}^{2} - 1}{2a}$

$It \: i \: contrary \: that \: \frac{ {a}^{2} - 1}{2a} \: is \: rational$

$and \: \sqrt{2} \: is \: irrational$

Hence, Our assumption is wrong.

$Therefore \: \sqrt{2} - \sqrt{3} \: is \:an \: irrational \: number.$