Question

Root 3 - root 2 prove is an irrational number

Answer 1

Hello,

To prove that
$\sqrt{3} - \sqrt{2} \: is \: an \: irrational \: number$
So,

Let root3-root2 be a rational number.
So, it means that it can be expressed in the form of p/q where p and q are integers and ate co prime.

So,
$\sqrt{3} - \sqrt{2} = \frac{p}{q} \\ squaring \: both \: sides \\ ( {( \sqrt{3 }- {\sqrt{2)} } }^{2} = \frac{p}{q} \\ 3 + 2 - 2 \sqrt{6} = \frac{p}{q} \\ - 2 \sqrt{6} = \frac{p}{q} - 5 = \frac{p - 5q}{q} \\ \sqrt{6 } = \frac{p - 5q}{ - 2q}$
Since we know that
$\frac{p - 5q}{ - 2q}$
Is a rational number while root 6 is an irrational number.
So,
A rational number can never be equal to an irrational number.

Hence, it's a contradiction
Our assumption was wrong that root3 - root2 is a rational number.

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

Hence, PROVED.

Hope this will be helping you ✌️