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.