Question

a) Show that 3 + 5 V7 is an irrational number​

Answer 1

Let us assume that 3 + 5√7 is a rational number

Rational numbers are expressed in the form a/b, where a and b are co - prime and b ≠0

$\implies 3 + 5\sqrt{7} = \dfrac{a}{b}$

$\implies 5\sqrt{7} = \dfrac{a}{b} -3$

$\implies 5\sqrt{7} = \dfrac{a-3b}{b}$

$\implies \sqrt{7} = \dfrac{a-3b}{5b}$

The RHS is a rational number

=> √7 is also a rational number.

But this contradicts to the fact that √7 is an irrational number

Hence, our assumption is wrong

$\boxed{\boxed{\bold{Therefore, \ 3 + 5\sqrt{7} \ is \ an \ irrational \ number}}}}}$

                                                       

Answer 2

Given:

• 3 + 5√7

To Prove:

• 3 + 5√7 is an irrational number.

Proof: Let us assume, to the contrary ,that 3 + 5√7 is rational.

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

$\small\implies{\sf }$ 3 + 5√7 = a/b

$\small\implies{\sf }$ 5√7 = a/b – 3

$\small\implies{\sf }$ 5√7 = a – 3b/b

$\small\implies{\sf }$ √7 = a – 3b/5b

Since, a and b are integers , so (a – 3b/5b) is rational.

This √7 is also rational.

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

Hence, (3 + 5√7) is irrational.