Question

Prove that (V3 -√ 2) is an Irrational number.​

Answer 1

Answer:

To proof :

$\sqrt{3}-\sqrt{2}$

As we know that,

As a rational number is in the form p /q, where p and q are integers and q ≠0.

Let consider,

$\sqrt{3} - \sqrt{2} = \frac{p}{q}$

On squaring both sides,

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

$(3 + 2 - 2 \sqrt{6} ) = \frac{ {p}^{2} }{ {q}^{2} }$

$(5 - 2 \sqrt{6} ) = \frac{ {p}^{2} }{ {q}^{2} }$

$5 - \frac{ {p}^{2} }{ {q}^{2} } = 2 \sqrt{6}$

$\frac{5 {q}^{2} - {p}^{2} }{ {2q}^{2} } = \sqrt{6}$

$as \: \sqrt{6} \: is \: irrational$

So

$\frac{5 {q}^{2} - {p}^{2} }{2 {q}^{2} } is \: irrational$

Therefore,

$\sqrt{5} - \sqrt{3} is \: irrational.$

Answer 2

Heyaa♡♡

Let √3 - √2 be rational.

So, √3 - √2 = x (where x is a rational no.)

Squaring both sides:-

$( \sqrt{3} - \sqrt{2} ) {}^{2} = x {}^{2} \\ = > 3 + 2 - \sqrt[2]{6} = x {}^{2} \\ = > 5 - \sqrt[2]{6} = x {}^{2} \\ = > \frac{5 \ -x {}^{2} }{2} = \sqrt{6}$

We know that √6 is a irrational no.

But this contradicts the fact that √3 - √2 is irrational.

Therefore, our hypothesis is wrong.

So, √3 - √2 is also a irrational no.

Hope it helped you.