SPS SPS FM 2019 — Question 6 3 marks

Exam BoardSPS
ModuleSPS FM (SPS FM)
Year2019
Marks3
TopicProof
TypeContradiction proof about integers
DifficultyStandard +0.8 This is a number theory proof requiring students to set up a contradiction argument with modular arithmetic. While the logic is straightforward once you assume 4 divides both expressions and derive that 2a and 2b are divisible by 4 (contradicting oddness), it requires careful algebraic manipulation and understanding of divisibility properties beyond routine A-level exercises.
Spec1.01d Proof by contradiction

If \(a\) and \(b\) are odd integers such that 4 is a factor of \((a - b)\), prove by contradiction that 4 cannot be a factor of \((a + b)\). [3]

If $a$ and $b$ are odd integers such that 4 is a factor of $(a - b)$, prove by contradiction that 4 cannot be a factor of $(a + b)$.
[3]

\hfill \mbox{\textit{SPS SPS FM 2019 Q6 [3]}}