Year: 2008
Paper: 1
Question Number: 1
Course: LFM Pure
Section: Proof
There were significantly more candidates attempting this paper this year (an increase of nearly 25%), but many found it to be very difficult and only achieved low scores. The mean score was significantly lower than last year, although a similar number of candidates achieved very high marks. This may be, in part, due to the phenomenon of students in the Lower Sixth (Year 12) being entered for the examination before attempting papers II and III in the Upper Sixth. This is a questionable practice, as while students have enough technical knowledge to answer the STEP I questions at this stage, they often still lack the mathematical maturity to be able to apply their knowledge to these challenging problems. Again, a key difficulty experienced by most candidates was a lack of the algebraic skill required by the questions. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many students were simply unable to progress on some questions because they did not know how to handle the algebra. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was also pleasing that one of the applied questions, question 13, attracted a very large number of attempts. However, the examiners were again left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers and other available resources to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
What does it mean to say that a number $x$ is \textit{irrational}?
Prove by contradiction statements A and B below, where $p$ and $q$ are real numbers.
\begin{itemize}
\item \textbf{A}: If $pq$ is irrational, then at least one of $p$ and $q$ is irrational.
\item \textbf{B}: If $p+q$ is irrational, then at least one of $p$ and $q$ is irrational.
\end{itemize}
Disprove by means of a counterexample statement C below, where $p$ and $q$ are
real numbers.
\begin{itemize}
\item \textbf{C}: If $p$ and $q$ are irrational, then $p+q$ is irrational.
\end{itemize}
If the numbers $\e$, $\pi$, $\pi^2$, $\e^2$ and $\e\pi$ are irrational, prove that at most one of the numbers $\pi+\e$, $\pi -\e$, $\pi^2-\e^2$, $\pi^2+\e^2$ is rational.\begin{itemize}
\item \textbf{A}: Suppose for sake of contradiction that neither $p$ nor $q$ is irrational, then $pq$ is the product of two rational numbers, ie is also rational. Therefore $pq$ is rational. Contradiction.
\item \textbf{B}: Suppose for the sake of contradiction both $p$ and $q$ are rational, but then $p+q$ is also rational, contradicting $p+q$ is irrational.
\item \textbf{C}: Note that $\sqrt{2}$ and $-\sqrt{2}$ are both irrational, but $\sqrt{2}+(-\sqrt{2}) = 0$ which is rational.
\end{itemize}
Since $(\pi + \e) + (\pi - \e) = 2\pi$ is irrational, at most one of $\pi+\e$ and $\pi - \e$ can be rational.
Since $(\pi+\e)(\pi-\e) = \pi^2 - \e^2$ is is the product of a (non-zero) rational and an irrational, $\pi^2 - \e^2$ cannot be rational.
Therefore for two of these numbers to be irrational, we need $\pi^2 + \e^2$ to be rational. But then squaring whichever of $\pi \pm \e$ is rational and subtracting $\pi^2+\e^2$ we obtain $\pm 2\pi \e$ which is irrational. But the product and sum of rationals is irrational. Therefore it cannot be the case that more than one of these numbers is rational.
This question was primarily about logical thinking and structuring an argument. While it was a very popular question, the marks were disappointing: only 30% of candidates gained more than 6 marks. Most candidates could describe vaguely what is meant by the term irrational, though only a handful gave a precise, accurate definition. The popular offering of 'a number with an infinite decimal expansion' was not acceptable. It was pleasing to see, though, that the majority of candidates were capable of using proof by contradiction to prove statements A and B, and they then went on to provide a counterexample to statement C. A small number of very strong candidates justified their counterexamples by proving that the numbers they presented were in fact irrational, though any well-known irrational examples were given full marks without the need for justification. It is important to stress the difference between proving a statement and disproving one; while a single (numerical) counterexample is adequate to disprove a statement, a proof of the truth of a statement requires a general argument. Too many candidates wrote things such as: 'If pq = 3, then p = 3 and q = 1, so . . . .' Also, it is unknown what an irrational number 'looks like', so the frequently occurring arguments such as 'We know that e + π must be irrational because the numbers are not of the same form' (when comparing this example to something like (1 − √2) + √2) are spurious. Sadly, very few candidates made any significant progress on the main part of the question. Several attempted (unsuccessfully) to prove that all four of the given numbers are irrational. Others asserted that since π and e are both irrational, π + e must also be, despite having just disproved statement C. A number of candidates successfully showed that π + e and π - e cannot both be irrational by appealing to B, but then could not see how to continue. The best attempts proceeded by using A and B repeatedly to show that no pair of π ± e and π² ± e² could simultaneously be rational (that is, they considered all six cases separately).