SPS SPS SM 2020 June — Question 8 4 marks

Exam BoardSPS
ModuleSPS SM (SPS SM)
Year2020
SessionJune
Marks4
TopicProof
TypeContradiction proof about integers
DifficultyChallenging +1.2 This is a straightforward proof by contradiction requiring students to square both sides, rearrange to show a² must be even (contradicting a being odd), and conclude no such integers exist. The algebraic manipulation is direct and the contradiction is immediate once the equation is squared, making this easier than average proof questions but still requiring proper proof technique.
Spec1.01d Proof by contradiction

Prove by contradiction that there are no positive integers \(a\) and \(b\) with \(a\) odd such that $$a + 2b = \sqrt{8ab}$$ [4]

Prove by contradiction that there are no positive integers $a$ and $b$ with $a$ odd such that
$$a + 2b = \sqrt{8ab}$$ [4]

\hfill \mbox{\textit{SPS SPS SM 2020 Q8 [4]}}