Question

Prove that 3-root 3 is irrational number

Answer 1

Heya!

Here is yr answer.....

Let us assume 3-√3 is rational

let 3-√3 = a/b (a,b are any integers)

=> 3 - a/b = √3

=> √3 = 3 - a/b

=> √3 = 3b-a/b

For any two integers, RHS (3b-a/b) is rational

But, LHS(√3) is irrational

A rational and irrational are never equal

So, our assumption is false

Therefore, 3-√3 is irrational

Hope it hlpz...

Answer 2

Hey there!

$let \: 3 - \sqrt{3} \: be \: rational \\ \\ so \: \: 3 - \sqrt{3} = \frac{p}{q} \: \: \: where \: p \: nd \: q \: are \: coprimes \\ \\ - \sqrt{3} = \frac{p}{q} - 3 \\ \sqrt{3} = - \frac{p + 3q}{q} \\ as \: \sqrt{3 } \: is \: irrational \: and - \frac{p + 3q}{q} is \: rational \\ \\ therefore \: our \: consumption \: is \: wrong \\ \\ so \: lhs = rhs \\ \\ hence \: 3 - \sqrt{3} \: is \: irrational$

Hope it help