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.

(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]

This question was generally well done, with the exception of part(b)(iii), showing that the operation is non-associative.