The binary operation \( * \) is defined on the set S = {0, 1, 2, 3} by

\[a * b = a + 2b + ab(\bmod 4){\text{ .}}\]

(a)     (i)     Construct the Cayley table.

  (ii)     Write down, with a reason, whether or not your table is a Latin square.

(b)     (i)     Write down, with a reason, whether or not \( * \) is commutative.

  (ii)     Determine whether or not \( * \) is associative, justifying your answer.

(c)     Find all solutions to the equation \(x * 1 = 2 * x\) , for \(x \in S\) .

(a)     (i)

     A3

Note: Award A3 for no errors, A2 for one error, A1 for two errors and A0 for three or more errors.

(ii)     it is not a Latin square because some rows/columns contain the same digit more than once     A1

[4 marks]

(b)     (i)     EITHER

it is not commutative because the table is not symmetric about the leading diagonal     R2

OR

it is not commutative because \(a + 2b + ab \ne 2a + b + ab\) in general     R2

Note: Accept a counter example e.g. \(1 * 2 = 3\) whereas \(2 * 1 = 2\) .

(ii)     EITHER

for example \((0 * 1) * 1 = 2 * 1 = 2\)     M1

and \(0 * (1 * 1) = 0 * 0 = 0\)     A1

so \( * \) is not associative     A1

OR

associative if and only if \(a * (b * c) = (a * b) * c\)     M1

which gives

\(a + 2b + 4c + 2bc + ab + 2ac + abc = a + 2b + ab + 2c + ac + 2bc + abc\)     A1

so \( * \) is not associative as \(2ac \ne 2c + ac\) , in general     A1

[5 marks]

(c)     x = 0 is a solution     A2

x = 2 is a solution     A2

[4 marks]

Total [13 marks]

This question was generally well answered.