Year: 2015
Paper: 3
Question Number: 5
Course: LFM Pure
Section: Proof
A very similar number of candidates to 2014 once again ensured that all questions received a decent number of attempts, with seven questions being very popular rather than five being so in 2014, but the most popular questions were attempted by percentages in the 80s rather than 90s. All but one question was answered perfectly at least once, the one exception receiving a number of very close to perfect solutions. About 70% attempted at least six questions, and in those cases where more than six were attempted, the extra attempts were usually fairly superficial.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item In the following argument to show that $\sqrt2$ is irrational, give proofs appropriate for steps 3, 5 and 6.
\begin{enumerate}
\item Assume that $\sqrt2$ is rational.
\item Define the set $S$ to be the set of positive integers with the following property:
\begin{center} $n$ is in $S$ if and only if $n \sqrt2$ is an integer.
\end{center}
\item Show that the set $S$ contains at least one positive integer.
\item Define the integer $k$ to be the smallest positive integer in $S$.
\item Show that $(\sqrt2-1)k$ is in $S$.
\item Show that steps 4 and 5 are contradictory and hence that $\sqrt2$ is irrational.
\end{enumerate}
\item Prove that $2^{\frac13} $ is rational if and only if $2^{\frac23}$ is rational.
Use an argument similar to that of part (i) to prove that $2^{\frac13}$ and $2^{\frac23}$ are irrational.
\end{questionparts}\begin{questionparts}
\item For step 3, since we have assumed $\sqrt{2}$ is rational we can write it in the form $p/q$ with $p, q$ coprime with $q \geq 1$. Then $q \in S$ since $q\sqrt{2} = p$ which is an integer.
For step 5, notice that $(\sqrt{2}-1)k$ is an integer (since $\sqrt{2}k$ is an integer and so is $-k$. It is also positive since $\sqrt{2} > 1$. We must check that $(\sqrt{2}-1)k \cdot \sqrt{2} = 2k - \sqrt{2}k$ is also an integer, but clearly it is as both $2k$ and $-\sqrt{2}k$ are integers. Therefore $(\sqrt{2}-1)k \in S$.
For step 6, notice that $(\sqrt{2}-1) < 1$ and therefore $(\sqrt{2}-1)k < k$, contradicting that $k$ is the smallest element in our set. (And all non-empty sets of positive integers have a smallest element)
\item Claim: $2^{\frac13}$ is irrational $\Leftrightarrow 2^{\frac23}$ is irrational.
Proof: Since $2^{\frac13} \cdot 2^{\frac23} = 2$ if one of them is rational, then the other one must also be rational. Which is the same as them both being irrational at the same time.
\begin{enumerate}
\item Assume that $\sqrt[3]{2}$ is rational, ie $\sqrt[3]{2} = p/q$ for some integers.
\item $S := \{ n \in \mathbb{Z}_{>0} : n \sqrt[3]{2} \text{ and } n \sqrt[3]{4}\in \mathbb{Z}\}$
\item Suppose $k$ is the smallest element in $S$ (which must exist, consider $q^2$
\item Consider $(\sqrt[3]{2}-1)k$ then clearly this is an integer, and $(\sqrt[3]{2}-1)\sqrt[3]{2}k = \sqrt[3]{4}k - \sqrt[3]{2}k \in \mathbb{Z}$ and $(\sqrt[3]{2}-1)\sqrt[3]{4}k = 2 k -\sqrt[3]{4}k \in \mathbb{Z}$.
\item But this is a smaller element of $S$, contradicting that $k$ is the smallest element. Therefore, we have a contradiction.
\end{enumerate}
\end{questionparts}
Marginally less successful than question 2, a lot of candidates earned about half of the marks. Unfortunately, many candidates approached this on the basis of their knowledge of the standard irrationality proof for root two employing rational numbers expressed in lowest terms rather than observing the specified argument. In part (i), proving step 5 was frequently beset with omissions, and simple steps like 0 < √2 − 1 < 1 were not acknowledged let alone justified. The first result of part (ii) caused few problems except to those that did not appreciate 'if and only if', but defining a suitable set in order to construct a similar argument to prove the irrationality of the cube roots of 2 and 2 squared was beyond most leading to mostly spurious logic.