Year: 2019
Paper: 1
Question Number: 7
Course: LFM Pure
Section: Proof
In order to get the fullest picture, this document should be read in conjunction with the question paper, the marking scheme and (for comments on the underlying purpose and motivation for finding the right solution-approaches to questions) the Hints and Solutions document; all of which are available from the STEP and Cambridge Examinations Board websites. The purpose of the STEPs is to learn what students are able to achieve mathematically when applying the knowledge, skills and techniques that they have learned within their standard A-level (or equivalent) courses … but seldom within the usual range of familiar settings. STEP questions require candidates to work at an extended piece of mathematics, often with the minimum of specific guidance, and to make the necessary connections. This requires a very different mind-set to that which is sufficient for success at A-level, and the requisite skills tend only to develop with prolonged and determined practice at such longer questions for several months beforehand. One of the most crucial features of the STEPs is that the routine technical and manipulative skills are almost taken for granted; it is necessary for candidates to produce them with both speed and accuracy so that the maximum amount of time can be spent in thinking their way through the problem and the various hurdles and obstacles that have been set before them. Most STEP questions begin by asking the solver to do something relatively routine or familiar before letting them loose on the real problem. Almost always, such an opening has not been put there to allow one to pick up a few easy marks, but rather to point the solver in the right direction for what follows. Very often, the opening result or technique will need to be used, adapted or extended in the later parts of the question, with the demands increasing the further on that one goes. So it is that a candidate should never think that they are simply required to 'go through the motions' but must expect, sooner or later, to be required to show either genuine skill or real insight in order to make a reasonably complete effort. The more successful candidates are the ones who manage to figure out how to move on from the given starting-point. Finally, reading through a finished solution is often misleading – even unhelpful – unless you have attempted the problem for yourself. This is because the thinking has been done for you. When you read through the report and look at the solutions (either in the mark-scheme or the Hints & Solutions booklet), try to figure out how you could have arrived at the solution, learn from your mistakes and pick up as many tips as you can whilst working through past paper questions. This year's paper produced the usual sorts of outcomes, with far too many candidates wasting valuable time by attempting more than six questions, and with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. Around one candidate in eight failed to hit the 30 mark overall, though this is an improvement on last year. Most candidates were able to produce good attempts at two or more questions. At the top end of the scale, around a hundred candidates scored 100 or more out of 120, with four hitting the maximum of 120 and many others not far behind. The paper is constructed so that question 1 is very approachable indeed, the intention being to get everyone started with some measure of success; unsurprisingly, Q1 was the most popular question of all, although under two-thirds of the entry attempted it this year, and it also turned out to be the most successful question on the paper with a mean score of about 12 out of 20. In order of popularity, Q1 was followed by Qs.3, 4 and 2. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the applied questions combined scoring fewer 'hits' than any one of the first four questions on its own. Though slightly more popular than the applied questions, the least successful question of all was Q5, on vectors. This question was attempted by almost 750 candidates, but 70% of these scored no more than 2 marks, leaving it with a mean score of just over 3 out of 20. Q9 (a statics question) was found only marginally more appetising, with a mean score of almost 3½ out of 20. In general, it was found that explanations were poorly supplied, with many candidates happy to overlook completely any requests for such details.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Consider the following steps in a proof that $\sqrt{2} + \sqrt{3}$ is irrational.
\begin{enumerate}
\item If an integer $a$ is not divisible by 3, then $a = 3k \pm 1$, for some integer $k$. In both cases, $a^2$ is one more than a multiple of 3.
\item Suppose that $\sqrt{2} + \sqrt{3}$ is rational, and equal to $\frac{a}{b}$, where $a$ and $b$ are positive integers with no common factor greater than one.
\item Then $a^4 + b^4 = 10a^2b^2$.
\item So if $a$ is divisible by 3, then $b$ is divisible by 3.
\item Hence $\sqrt{2} + \sqrt{3}$ is irrational.
\end{enumerate}
\begin{questionparts}
\item Show clearly that steps 1, 3 and 4 are all valid and that the conclusion 5 follows from the previous steps of the argument.
\item Prove, by means of a similar method but using divisibility by 5 instead of 3, that $\sqrt{6} + \sqrt{7}$ is irrational.
Why can divisibility by 3 not be used in this case?
\end{questionparts}\begin{questionparts}
\item \textbf{Step 1:} There are only three possibilities for the remainder of $a$ when divided by $3$, ($0$, $1$, $2$). $a = 3m+r$. If $r = 0$ we are done, if $r = 1$ take $k = m$, and $r=2$ take $k=(m+1)$ and we have $a = 3k-1$ as required.
Then $a^2 = (3k\pm1)^2 =9k^2\pm6k+1 = 3(3k^32\pm2k)+1$ which is clearly $1$ more than a square.
\textbf{Step 3:} \begin{align*}
&& \frac{a}{b} &= \sqrt{2}+\sqrt{3} \\
\Rightarrow && \frac{a^2}{b^2} &= 5+2\sqrt{6} \\
\Rightarrow && \frac{a^2}{b^2}-5 &= 2\sqrt{6} \\
\Rightarrow && 24 &= \left ( \frac{a^2}{b^2}-5 \right)^2 \\
&&&= 25 + \frac{a^4}{b^4}-10\frac{a^2}{b^2} \\
\Rightarrow && -b^4 &= a^4-10a^2b^2 \\
\Rightarrow && a^4+b^4 &= 10a^2b^2
\end{align*}
\textbf{Step 4:} If $a$ is divisible by $3$ then $b^4 = 10a^2b^2-a^4$ is a multiple of $3$, but if $b$ was not a multiple of $3$ then $b^2$ would be $1$ more than a multiple of $3$ (by \textbf{Step 3}) and $b^4$ would also be $1$ more than a multiple of $3$, and we would have a contradiction.
\textbf{Step 5}: Follows since either $a,b$ are both divisible by $3$ (contradicting \textbf{Step 2}), or neither is, but then $a^2$ and $b^2$ are both one more than a multiple of $3$ and the RHS is one more than a multiple of $3$ but the LHS is $2$ more than a multiple of $3$ which is a contradiction.
\item \textbf{Step 1:} If $a$ is not divisible by $5$ then $a^2 \equiv \pm 1 \pmod{5}$
\textbf{Step 2}: Suppose $\frac{a}{b} = \sqrt{6}+\sqrt{7}$
\textbf{Step 3}: \begin{align*}
&& \frac{a}{b} &= \sqrt{6}+\sqrt{7} \\
\Rightarrow && \frac{a^2}{b^2} &= 13 + 2\sqrt{42} \\
\Rightarrow && 168 &= \left (\frac{a^2}{b^2} - 13 \right)^2 \\
&&&= 169 - 26 \frac{a^2}{b^2} + \frac{a^4}{b^4} \\
\Rightarrow && a^4+b^4 &= 26a^2b^2
\end{align*}
\textbf{Step 4}: If $a$ is a multiple of $5$ then so is $b^4$ and hence so is $b^2$ and $b$.
\textbf{Step 5}: But the left hand side is always $2 \pmod{5}$ and the right hand side is never $2 \pmod{5}$ contradiction.
Divisibility by $3$ doesn't work here since mod $3$ we can have $a = 1, b = 1$ and have a valid solution.
\end{questionparts}
Once again, the prospect of gaining an easy few marks at the question's opening drew in a lot of interest; but many of these attempts (a third of them, in fact) were of negligible success. For serious takers, there was a good spread of marks, with some sensible comments and explanations being offered in several places. In part (i), steps 1, 3 and 4 were generally done well, although a common mistake in step 4 was to obtain an expression for b that contained a "factor" of 3 but in which the other factor was not obviously an integer. Having shown step 4, however, a great many candidates thought that the contradiction obtained if a is a multiple of 3 was sufficient to conclude that √2 + √3 was irrational, and didn't consider the case where a is not a multiple of 3, thus losing most of the marks for step 5. Some of those who avoided this pitfall forgot to appeal to symmetry to rule out the case of b being a multiple of 3. It was sometimes difficult to interpret candidates' logic from their prose. Part (ii) was less well done in general, with many candidates giving very little detail. When considering squares of non-multiples of 5, a common error was to deal with the case 5k + 1 but not 5k + 2. A significant number guessed the correct relationship a⁴ + b⁴ = 26a²b², but no credit was given for simply writing this without any reasoning. Very few candidates got the final mark for explaining why divisibility by 3 was not sufficient for this case, although some mentioned the key fact that 26 is 2 more than a multiple of 3, without saying why that is relevant.