Question

prove that 7 is an
irrational number
​

Answer 1

Step-by-step explanation:

prove that 7 is an

irrational number

Answer 2

CORRECT QUESTION :

Prove that $\sf \sqrt {7}$ is an irrational number.

SOLUTION :

To prove $\sf \sqrt {7}$ as irrational number.

Let consider the $\sf \sqrt {7}$ as rational number.

We know that,

Rational number are written in the form of $\sf \dfrac {a}{b}$

$\implies \sf \sqrt {7} \ = \ \dfrac {a}{b}$

$\implies \sf a \ = \ b \sqrt {7}$

Now,

Squaring on the both sides.

$\sf a^{2} \ = \ 7b^{2}$

Consider, a = 7p, for some integer p.

Substitute the a,

we get,

$\implies \sf 49 \ p^{2} \ = \ 7 \ b^{2}$

$\implies \sf b^{2} \ = \ 7 \ p^{2}$

Now, Here it seems that the both a and b have the least common factor 7.

$\therefore$ Our contradiction or assumption that $\sf \sqrt {7}$ is a rational is false.

$\therefore$ $\sf \sqrt {7}$ is an irrational number

•°• Hence proved ✔︎