# Why $\sqrt{3}$ is an irrational number?

Given :

The given number is $\sqrt{3}$.

To do :

We have to prove, why $\sqrt{3}$ is an irrational number.

Solution :

Let p be a prime number. If p divides $a^2$, then p divides a, where a is 𝑎 positive integer.

Now,

Let us assume, to the contrary, that $\sqrt{3}$  is rational.

So, we can find integers a and b $(≠ 0)$ such that $\sqrt{3} = \frac{a}{b}$.

Where a and b are co-prime.

$⇒ (\sqrt{3})^2 = (\frac{a}{b})^2$

$⇒ 3= \frac{a^2}{b^2}$

$⇒ 3b^2 = a^2$

Therefore, 3 divides $a^2$.

Now, by the above statement, it follows that 3 divides a.

So, we can write $a = 3c$ for some integer c.

$⇒ a^2 = 9c^2$

$⇒ 3b^2 = 9c^2$         $(Using, 3b^2 = a^2)$

$⇒ b^2 = 3c^2$

Therefore, 3 divides $b^2$.

Now, by the above statement, it follows that 3 divides b.

Therefore, a and b have at least 3 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational.

So, we conclude that $\sqrt{3}$ is irrational.

Updated on: 10-Oct-2022

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