A student argues that when a rational number is multiplied by an irrational number the result will always be an irrational number - AQA - A-Level Maths Pure - Question 16 - 2017 - Paper 1

To prove that multiplying any non-zero rational number by an irrational number yields an irrational number, we use proof by contradiction.

Let:

• $b$ be an irrational number,
• $a$ be a non-zero rational number, expressed as $a = \frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \neq 0$.

Assume, for the sake of contradiction, that the product $ab$ is rational, such that: $ab = c,$ where $c$ is a rational number. Then we have: $a b = \frac{p}{q} b = c.$
This implies that: $b = \frac{cq}{p}.$
Since both $c$ and $q$ are rational, the fraction $\frac{cq}{p}$ is also rational (given that $p \neq 0$).

This leads to a contradiction since $b$ was assumed to be irrational. Thus, our assumption that $ab$ is rational must be false. This means that for all non-zero rational numbers $a$, the product $ab$ must be irrational.