Question

prove that 3√3-7 is an irrational​

Answer 1

Step-by-step explanation:

Given:

3 – 3√7

To Prove:

3 – 3√7 is an irrational number.

Proof: Let us assume, to the contrary , that ( 3 – 3√7 ) is rational.

Then, there exists co-primes a and b ( b ≠ 0 ) such that

But this contradicts the fact that √7 is irrational. So, our assumption is incorrect.

Hence, ( 3 – 3√7 ) is irrational

Answer 2

$\: \: \sf{let \:3 \sqrt{3} - 7 \: be \: a \: rational \: number}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \sf{3 \sqrt{3} - 7 = \frac{a}{b} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{3 \sqrt{3} = \frac{a}{b} + \frac{7}{1} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ \sqrt{3} = \frac{a + 7b}{3b} }$

$\: \: \: \: \: \: \: \: \sf{but \: \sqrt{3} \: is \: irrational}$

$\: \: \: \: \: \: \sf{so \: \frac{a + 7b}{3b} \: is \: also \: irrational}$

$\sf{therefore \: our \: assumption \: is \: wrong}$

$\: \: \sf{so \: 3 \sqrt{3} - 7 \: is \: irrational \: number}$