Question

Prove that following are irrational
√2+√7​

Answer 1

ANSWER:

• √2 +√7 is an irrational number.

GIVEN:

• Irrational number = √2 +√7

TO PROVE:

• √2+√7 Is an irrational number.

SOLUTION:

Let √2 +√7 be a rational number which can be expressed in the form of p/q where p and q have no common factor other than 1.

$\implies \: \sqrt{2} + \sqrt{7} = \frac{p}{q} \\ \implies \: \sqrt{2} = \frac{p}{q} - \sqrt{7} \\ \: \: \: \: squaring \: both \: the \: sides \: \\ \implies \: ( \sqrt{2}) {}^{2} = ( \frac{p - \sqrt{7} q}{q} ) {}^{2} \\ \implies \: 2 = \frac{(p - \sqrt{7} q) {}^{2} }{q {}^{2} } \\ \implies \: 2 = \frac{p {}^{2} + 7q {}^{2} - 2 \sqrt{7} pq}{q {}^{2} } \: \\ \implies \: 2q {}^{2} = p {}^{2} + 7q {}^{2} - 2 \sqrt{7} pq \\ \implies \: (2q {}^{2} - 7q {}^{2} ) - p {}^{2} = - 2 \sqrt{7} pq \\ \implies \: - (p {}^{2} + 5q {}^{2} ) = - 2 \sqrt{7} pq \\ \implies \: \frac{p {}^{2} + 5q {}^{2} }{2pq} = \sqrt{7}$

This shows that √7 is an rational number but it is a irrational number.

Thus our contradiction is wrong so √2 + √7 is an irrational number.