Question

show by 4 root 2 is an international number
$show \: that \: 4 \sqrt{2 \: } is \: an \: irrational \: number$
​

Answer 1

ANSWER

Let us assume that 4√2 is a rational number.

Rational numbers can be expressed in the form a/b, where a and b are co - prime and b ≠0

$\implies 4\sqrt{2}=\dfrac{a}{b}$

$\implies \sqrt{2} = \dfrac{a}{4b}$

The RHS is a rational number

=> LHS is also a rational number

=> √2 is also a rational number

But this contradicts to the fact that √2 is an irrational number.

Hence, our assumption is wrong.

$\boxed{\boxed{\bold{Therefore, \ 4\sqrt{2} \ is \ an \ irrational \ number}}}}}$

                                                         

Answer 2

Step-by-step explanation:

Given:

• 4√2

To Prove:

• 4√2 is an irrational number.

Proof: Let us assume , to the contrary ,that 4√2 is a rational number.

Then, there exists coprime positive integers p & q ( q ≠ 0 ) such that

➸ 4√2 = $\dfrac{p}{q}$

➸ √2 = $\dfrac{p}{4q}$ ( ∵ p & q are integers )

Since, 4 , p & q are integers , $\dfrac{p}{4q}$ is rational, and so √2 is rational.

But this contradicts the fact that √2 is irrational and our assumption is incorrect.

So we conclude that 4√2 is irrational.