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 Mechanics - Question 16 - 2017 - Paper 1

We will use proof by contradiction:

1. Let b be an irrational number and let a be a non-zero rational number. We can express a as ( a = \frac{c}{d} ) where c and d are integers, and ( d \neq 0 ).
2. Assume that the product ( ab ) is rational. That is, we can write ( ab = \frac{p}{q} ) for some integers p and q, where ( q \neq 0 ).
3. Then, substituting for a, we have: $$$$

ab = \frac{c}{d}b = \frac{p}{q}$$ 4. Rearranging gives us:

b = \frac{pd}{qc}$$ 5. Since p, q, c, and d are all integers, this means b must be rational. 6. However, this contradicts our initial assumption that b is irrational. 7. Therefore, we conclude that our assumption must be false, and thus \( ab \) must be irrational if a is a non-zero rational number.