Question

prove that 4 - 3 under root 2 is irrational number​

Answer 1

This is your answer dude

Answer 2

$\underline{\huge\mathfrak{Question:}}$

Prove that $4 - 3\sqrt{2}$is an irrational number .

$\underline{\huge\mathfrak{Answer:}}$

Let us assume that ,$4 - 3 \sqrt{2}$is a rational number .

Therefore ,

$4 - 3 \sqrt{2} = \frac{p}{q}$, where$\:p$and$\:q$are integers , $p$and$\:q$are integers and $\:q$is not equal to zero .

$= > - 3 \sqrt{2} = \frac{p}{q} - 4$

$= > - 3 \sqrt{2} = \frac{p - 4q}{q}$

$\sqrt{2} = \frac{p - 4q}{ - 3q}$

Since ,

$p$and $q$are integers , so $\frac{p - 4q}{ - 3q}$is a rational number .

But , it is not possible because$\sqrt{2}$is irrational .

Therefore ,

$4 - 3 \sqrt{2}$is an irrational number .

Note :- Here , ( in proving all the stuffs related to this type ) the value of the square term is always taken as irrational .