IB DP Math MAA HL : IB Style Mock Exams – Set 5 Paper 1

To prove that $\sqrt{3}$ is irrational using proof by contradiction, we assume the opposite, namely that $\sqrt{3}$ is rational. That is, there exist integers $a$ and $b$ with no common factors other than 1 such that $\sqrt{3} = \frac{a}{b}$.

We can assume that $a$ and $b$ are positive integers (since if $\sqrt{3} = \frac{a}{b}$, we can always multiply both sides by $-1$ to obtain $-\sqrt{3} = -\frac{a}{b}$, which is equivalent to $\sqrt{3} = \frac{-a}{b}$).

Squaring both sides of $\sqrt{3} = \frac{a}{b}$, we get:

$$3 = \frac{a^2}{b^2}$$

Multiplying both sides by $b^2$, we get:

$$3b^2 = a^2$$

This means that $a^2$ is divisible by 3, and therefore $a$ must also be divisible by 3 (since the square of an integer is only divisible by 3 if the integer itself is divisible by 3).

Let $a = 3c$, where $c$ is also a positive integer. Substituting into the equation above, we get:

$$3b^2 = (3c)^2 = 9c^2$$

Dividing both sides by 3, we get:

$$b^2 = 3c^2$$

This means that $b^2$ is divisible by 3, and therefore $b$ must also be divisible by 3.

However, we have now shown that both $a$ and $b$ are divisible by 3, which contradicts our assumption that $a$ and $b$ have no common factors other than 1. Therefore, our initial assumption that $\sqrt{3}$ is rational must be false, and $\sqrt{3}$ is indeed irrational.