SPS SPS SM Pure 2022 June — Question 15 6 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2022
SessionJune
Marks6
TopicProof
TypeContradiction proof of irrationality
DifficultyStandard +0.8 Part (a) requires understanding divisibility proofs and algebraic manipulation (factoring n³-1 and using n=3k+1), which is moderately challenging. Part (b) combines logarithm manipulation with proof by contradiction using fundamental theorem of arithmetic—a technique less routine than standard A-level proofs. Both parts demand mathematical maturity beyond typical exercises, but the steps are accessible with careful reasoning.
Spec1.01a Proof: structure of mathematical proof and logical steps1.01d Proof by contradiction1.06c Logarithm definition: log_a(x) as inverse of a^x1.06e Logarithm as inverse: ln(x) inverse of e^x

  1. Prove that $$n - 1 \text{ is divisible by } 3 \Rightarrow n^3 - 1 \text{ is divisible by } 9$$ [3 marks]
  2. Show that if \(\log_2 3 = \frac{p}{q}\), then $$2^p = 3^q.$$ Use proof by contradiction to prove that \(\log_2 3\) is irrational. [3 marks]

\begin{enumerate}[label=\alph*)]
\item Prove that
$$n - 1 \text{ is divisible by } 3 \Rightarrow n^3 - 1 \text{ is divisible by } 9$$ [3 marks]

\item Show that if $\log_2 3 = \frac{p}{q}$, then
$$2^p = 3^q.$$

Use proof by contradiction to prove that $\log_2 3$ is irrational. [3 marks]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM Pure 2022 Q15 [6]}}