October 2021 Paper 1 Q1
1 Beth states that for all real numbers \(p\) and \(q\), if \(p^2 > q^2\) then \(p > q\).
Prove that Beth is not correct. [2]
| Scheme | Marks | AO |
|---|---|---|
| For example \((-3)^2 = 9 > 2^2 = 4\) and \((-3) < 2\) | M1 | 2.1 |
| So Beth is not correct | E1 | 2.2a |
| [2] |
Notes
M1: Stating any pair of numbers where \(p^2 > q^2\) and \(p < q\)
E1: Fully convincing argument – do not allow for only disproving the converse
Also accept general statement about a negative number [for \(p\)]