Question

prove that 2√3-3√2 is an irrational number.

Answer 1

Hey frnd....

Thanks for asking this question.
Here is your answer....


The solution is in the attachment.

Hope this would help you.

Answer 2

Answer:

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

Step-by-step explanation:

Assume, $2\sqrt{3} -3\sqrt{2}$ is rational.

Let,

     $2\sqrt{3} -3\sqrt{2}$ is rational and its simplest form be $a$

⇒  $2\sqrt{3} -3\sqrt{2} =a$

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

On squaring on both sides,

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

⇒                $12=a^{2} +18+6\sqrt{2} a$

⇒           $6\sqrt{2} a=-(a^{2} +6)$

⇒               $\sqrt{2} =-\frac{a^{2} +6}{6a}$

Here, $\frac{(a^{2} +6)}{6a}$ is a rational number but $\sqrt{2}$ is irrational

        Rational  $\neq$ Irrational

This is a contradiction.

Therefore, Our assumption is wrong.

Hence $2\sqrt{3} -3\sqrt{2}$ is irrational.

Hence proved.