Date | November 2008 | Marks available | 10 | Reference code | 08N.3srg.hl.TZ0.2 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | Construct and Determine | Question number | 2 | Adapted from | N/A |
Question
A binary operation is defined on {−1, 0, 1} by
\[A \odot B = \left\{ {\begin{array}{*{20}{c}}
{ - 1,}&{{\text{if }}\left| A \right| < \left| B \right|} \\
{0,}&{{\text{if }}\left| A \right| = \left| B \right|} \\
{1,}&{{\text{if }}\left| A \right| > \left| B \right|{\text{.}}}
\end{array}} \right.\]
(a) Construct the Cayley table for this operation.
(b) Giving reasons, determine whether the operation is
(i) closed;
(ii) commutative;
(iii) associative.
Markscheme
(a) the Cayley table is
\(\begin{gathered}
\begin{array}{*{20}{c}}
{}&{ - 1}&0&1
\end{array} \\
\begin{array}{*{20}{c}}
{ - 1} \\
0 \\
1
\end{array}\left( {\begin{array}{*{20}{c}}
0&1&0 \\
{ - 1}&0&{ - 1} \\
0&1&0
\end{array}} \right) \\
\end{gathered} \) M1A2
Notes: Award M1 for setting up a Cayley table with labels.
Deduct A1 for each error or omission.
[3 marks]
(b) (i) closed A1
because all entries in table belong to {–1, 0, 1} R1
(ii) not commutative A1
because the Cayley table is not symmetric, or counter-example given R1
(iii) not associative A1
for example because M1
\(0 \odot ( - 1 \odot 0) = 0 \odot 1 = - 1\)
but
\((0 \odot - 1) \odot 0 = - 1 \odot 0 = 1\) A1
or alternative counter-example
[7 marks]
Total [10 marks]
Examiners report
This question was generally well done, with the exception of part(b)(iii), showing that the operation is non-associative.