Question

prove that root 2
subtract 3 is irrational number​

Answer 1

Let

$\sqrt{ 2} - 3 \: is \: an \: irrational \: no.$

$\sqrt{2} - 3 = \frac{p}{q \:} where \ \: p \: and \: q \: are$

co-prime meaning that they dont have any common factor

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

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

we know that root 2 is an irrational no. And irrational no. Can't be equal to rational no.

So our supposition is wrong

root 2-3 is an irrational no.

Hope this will help u

Mark it as the brainliest one

Answer 2

To prove: $\sqrt{2} - 3$ is irrational number.

Proof: Let $\sqrt{2} - 3$ be rational number i.e.

$\sqrt{2} - 3 = \dfrac{a}{b}$

where 'a' and 'b' are integers and 'b' is not equal to 0 and both are co- prime numbers.

$\sqrt{2} = \dfrac{a}{b} + 3 \\ \implies \sqrt{2} = \dfrac{a+3b}{b}$

But we know that $\sqrt{2}$ is an irrational number.

This contradicts the fact. This contradiction is arisen by assuming $\sqrt{2} - 3$ a rational number.

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