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 this by contradiction, let ( b ) be an irrational number and let ( a = \frac{c}{d} ) be a non-zero rational number, where ( c ) and ( d ) are integers and ( d \neq 0 ).

Assume ( ab ) is rational, so we can write:

[ ab = \frac{p}{q} \quad \text{for integers } p, q \text{ and } q \neq 0. ]

Substituting for ( a ), we get:

[ \frac{c}{d} \cdot b = \frac{p}{q} \implies b = \frac{p \cdot d}{c \cdot q}. ]

Since ( b ) is expressed as the ratio of two integers, this implies ( b ) is rational, which contradicts our initial assumption that ( b ) is irrational.

Thus, we conclude that ( ab ) must be irrational when ( a \neq 0 ).