Question

8. Prove that 3 - 3√7 is an irrational number.​

Answer 1

Let  3 - 3√7 be a rational number.​

Rational numbers are of the form p/q, where p and q are co-prime and q≠0

$3-3\sqrt{7}= \dfrac{a}{b} \\\\-3\sqrt{7}= \dfrac{a}{b} -3\\\\-3\sqrt{7} = \dfrac{a-3b}{b} \\\\-\sqrt{7} = \dfrac{a-3b}{3b } \\\\\sqrt{7} = \dfrac{3b-a}{3b }$

The RHS is a rational number

=> √7 is a rational number.

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

Hence, our assumption is wrong.

$\boxed{\boxed{\bold{Therefore, \ 3-3\sqrt{7} \ is \ an \ irrational \ number}}}}$

                                                               

Answer 2

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

$= > 3 - 3 \sqrt{7} = \frac{a}{b} \\ \\ = > - 3 \sqrt{7} = \frac{a}{b} - 3 \\ \\ = > - 3 \sqrt{7} = \frac{a - 3b}{b} \\ \\ = > \sqrt{7} = \frac{a - 3b}{ - 3b} \\ \\ since \: a \: and \: b \: are \: integers \: so \: \frac{a - 3b}{ - 3b} \: is \: rational \\ \\ thus \: \sqrt{7} \: is \: also \: rational.$

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

Hence, ( 3 – 3√7 ) is irrational.