Question

Prove that $\sqrt{3}$ is irrational. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Proof : 2 Marks

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

That is, we can find integers a and b ($\ne$ 0) such that $\sqrt{3}=\frac{a}{b}$

where $a$ and $b$ are co - primes

So, $b\sqrt{3}=a$

Squaring on both sides, we get

$3b^2=a^2$.

$\therefore$ $a^2$ is divisible by 3, and it follows that a is also divisible by 3.

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

Substituting for $a$, we get

$3b^2=9c^2\Rightarrow b^2=3c^2$.

This means that $b^2$ is divisible by 3, and so $b$ is also divisible by 3 (using p = 3).

$\therefore$ a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are co prime.

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

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