Question

Prove that 2 - 3√3 is irrartional, where √3 is irrational.

Answer 1

Let , 2 - 3√3 is a rational number .

Therefore ,

$2 - 3 \sqrt{3} = \frac{p}{q}$, where p and q are integers , p and q are coprime and q is not equal to zero .


$= > 3 \sqrt{3} = \frac{p}{q} + 2$



$= >3 \sqrt{3} = \frac{p \: + \: 2q }{q}$



$= > \sqrt{3} = 3( \frac{p \: + \: 2q}{q} )$



$= > \sqrt{3} = \frac{3p \: + \: 6q}{q}$


Therefore ,

$\frac{3p \: + \: 6q}{q}$ is a rational number , but it isn't possible , because $\sqrt{3}$ is irrational . Therefore , our assumption is wrong .

So ,

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