June 2025 Paper 3 Q12
12 Given that the real numbers \(a\), \(b\), and \(c\) are such that
\[a \times b = c\]Use proof by contradiction to show that if \(c\) is irrational then at least one of \(a\) or \(b\) is irrational. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| States \(a\) and \(b\) are rational | B1 | 2.1 |
| Forms the product of two rational numbers \(\dfrac{m}{n} \times \dfrac{p}{q}\) | M1 | 3.1a |
| Concludes the argument fully correctly by stating that \(c\) is rational and contradicts \(c\) is irrational and if \(c\) is irrational at least one of \(a\) or \(b\) is irrational. To be awarded R1, marks B1M1 must be scored with their \(\dfrac{mp}{nq}\) seen | R1 | 2.1 |
| (3 marks) |
Typical solution
Let \(a\) and \(b\) be rational
\[a = \frac{m}{n} \text{ and } b = \frac{p}{q}\]\[c = ab = \frac{m}{n} \times \frac{p}{q} = \frac{mp}{nq}\]Hence \(c\) is rational which contradicts that \(c\) is irrational.
Therefore, if \(c\) is irrational at least one of \(a\) or \(b\) is irrational.