Question

Proove that √3 Is A irrational number​

Answer 1

√3=1.732.........

its not in the form of p/q

Answer 2

Step-by-step explanation:

AnswEr:

Let us Consider that the $\sf\sqrt{3}$ is an rational Number. So, it can be written in the form of $\sf\dfrac{a}{b}$

Here, a & b are co primes number & b is not equal to zero.

Such that,

$\implies\sf \sqrt{3} = \dfrac{a}{b}$

$\implies\sf \sqrt{3}b = a$

Squaring Both Sides,

$\implies\sf \sqrt{3b}^2 = a^2$ -------- (1)

$\implies\sf 3b^2 = a^2$

$\implies\sf \dfrac{a^2}{3} = b^2$

We can see that, 3 is divisible by a. ------(2)

Now,

$\implies\sf \dfrac{a}{3} = c$

$\implies\sf a = 3c$

Putting the value of a in Equation (1).

$\implies\sf 3b^2 = 3c^2$

$\implies\sf 3b^2 = 9c^2$

$\implies\sf b^2 = \cancel\dfrac{9c^2}{3}$

$\implies\sf b^2 = 3c^2$

$\implies\sf \dfrac{b^2}{3} = c^2$

Here, 3 is divisible by b & 3 is also divisible by b². -----(3)

From Equations (2) & (3)

a & b both have 3 as a common factor but it shows that a & b are not co primes number.

Therefore, it arises Contradiction because of our wrong Assumption.

Hence, √3 is an irrational Number.