Question

prove that 2-√2 ii's irrational.answer correctly​

Answer 1

Let us assume that 2-√2 is rational number...

Therefore, 2-√2 = a/b...

√2=a-2/b

Here a, b & 2 are integers & integers form a rational number....

But we know the fact that √2 is irrational...

This contradiction has arisen due to our incorrect assumption...

Therefore, 2-√2 is irrational...

Answer 2

To prove : $\tt{2 - \sqrt{2}}$ is irrational.

Proof:

Let's assume that $\tt{2 - \sqrt{2}}$ is rational.

So,

$\tt{2 - \sqrt{2} = \dfrac{a}{b}} \: \: - - (a \: and \: be \: are \: integers)$

$\tt{\longrightarrow} \: \tt{2 - \sqrt{2} = \dfrac{a}{b}}$

$\tt{\longrightarrow} \: \tt{ - \sqrt{2} = \dfrac{a}{b} - 2}$

$\tt{\longrightarrow} \: \tt{ \sqrt{2} = \dfrac{ - a + 2b}{b}}$

$\tt{\dfrac{a - 2b}{b}}$ is a rational number as a and b are integers. This means that $\tt{\sqrt{2}}$ is rational too. But this is a contradiction to the fact that $\tt{\sqrt{2}}$ is irrational.

The contradiction has arisen due to wrong assumption.

Hence proved, $\tt{2 - \sqrt{2}}$ is irrational.