Question

PROVE THAT 7-ROOT 5 IS IRRATIONAL.(please give all the procedure)

Answer 1

let 7√5 be rational
7√5=p÷q(p÷q is rational)
√5=(p÷q)÷7
here √5 is irrational but it is equal to a rational number this contradicts that 7√5 is a irrational number.
by gourav

Answer 2

$7-\sqrt{5}$ is Irrational.

To find:

Prove that $7-\sqrt{5}$ is irrational.

Solution:

Given: $7-\sqrt{5}$

Let us assume that $7-\sqrt{5}$ is rational number  

Therefore, $7-\sqrt{5}=\frac{a}{b}$    (a, b ∈ Z, where b ≠ 0)

$-\sqrt{5}=\frac{a}{b}-7$

$\sqrt{5}=7-\frac{a}{b}$

If “a, b are integers” then $7-\frac{a}{b}$ is ‘rational number’.  

So, $\sqrt{5}$ is also a ‘rational number’.

But the fact is that $\sqrt{2}$ is rational.

Hence, we conclude that “$\bold{7-\sqrt{5}}$ is irrational number”.