(i)     Show that \( * \) is commutative.

(ii)     Find the identity element.

(iii)     Find the inverse of the element a .

The binary operation \( \cdot \) is defined on \(\mathbb{R}\) as follows. For any elements a , \(b \in \mathbb{R}\)

\(a \cdot b = 3ab\) . The set S is the set of all ordered pairs \((x,{\text{ }}y)\) of real numbers and the binary operation \( \odot \) is defined on the set S as

\(({x_1},{\text{ }}{y_1}) \odot ({x_2},{\text{ }}{y_2}) = ({x_1} * {x_2},{\text{ }}{y_1} \cdot {y_2}){\text{ }}.\)

Determine whether or not \( \odot \) is associative.

(i)     if \( * \) is commutative \(a * b = b * a\)

since \(a + b + 1 = b + a + 1\) , \( * \) is commutative     R1

(ii)     let e be the identity element

\(a * e = a + e + 1 = a\)     M1

\( \Rightarrow e = - 1\)     A1

(iii)     let a have an inverse, \({a^{ - 1}}\)

\(a * {a^{ - 1}} = a + {a^{ - 1}} + 1 = - 1\)     M1

\( \Rightarrow {a^{ - 1}} = - 2 - a\)     A1

[5 marks]

\(({x_1},{\text{ }}{y_1}) \odot \left( {({x_2},{\text{ }}{y_2}) \odot ({x_3},{\text{ }}{y_3})} \right) = ({x_1},{\text{ }}{y_1}) \odot ({x_2} + {x_3} + 1,{\text{ }}3{y_2}{y_3})\)     M1

\( = ({x_1} + {x_2} + {x_3} + 2,{\text{ }}9{y_1}{y_2}{y_3})\)     A1A1

\(\left( {({x_1},{\text{ }}{y_1}) \odot ({x_2},{\text{ }}{y_2})} \right) \odot ({x_3},{\text{ }}{y_3}) = ({x_1} + {x_2} + 1,{\text{ }}3{y_1}{y_2}) \odot ({x_3},{\text{ }}{y_3})\)     M1

\( = ({x_1} + {x_2} + {x_3} + 2,{\text{ }}9{y_1}{y_2}{y_3})\)     A1

hence \( \odot \) is associative     R1

[6 marks]

Part (a) of this question was the most accessible on the paper and was completed correctly by the majority of candidates.

Part (b) was completed by many candidates, but a significant number either did not understand what was meant by associative, confused associative with commutative, or were unable to complete the algebra.