Question

7Under root 5 is an irrational number prove that

Answer 1

Step-by-step explanation:

Let us assume, to the contrary, that 7√5 is rational.

Then,

7√3 = $\frac{a}{b}$ [where a&b are co-primes &b≠0]

√5 =$\frac{a}{7b}$

Since, a & b are integers, so $\frac{a}{7b}$ is rational.

Thus, √5 is also rational.

But, this contradicts the fact that √5 is irrational. So, our assumption is wrong.

Hence, 7√5 is irrational.