Show that \(x * y \in S\) for all \(x,{\text{ }}y \in S\).

Show that the operation \( * \) on the set \(S\) is commutative.

Show that the operation \( * \) on the set \(S\) is associative.

Show that 2 is the identity element.

Show that each element \(a \in S\) has an inverse.

\(x,{\text{ }}y > 1 \Rightarrow (x - 1)(y - 1) > 0\)     M1

\((x - 1)(y - 1) + 1 > 1\)     A1

so \(x * y \in S\) for all \(x,{\text{ }}y \in S\)     AG

[2 marks]

\(x * y = (x - 1)(y - 1) + 1 = (y - 1)(x - 1) + 1 = y * x\)     M1A1

so \( * \) is commutative     AG

[2 marks]

\(x * (y * z) = x * \left( {(y - 1)(z - 1) + 1} \right)\)     M1

\( = (x - 1)\left( {(y - 1)(z - 1) + 1 - 1} \right) + 1\)     (A1)

\( = (x - 1)(y - 1)(z - 1) + 1\)     A1

\((x * y) * z = \left( {(x - 1)(y - 1) + 1} \right) * z\)     M1

\( = \left( {(x - 1)(y - 1) + 1 - 1} \right)(z - 1) + 1\)

\( = (x - 1)(y - 1)(z - 1) + 1\)     A1

so \( * \) is associative     AG

[5 marks]

\(2 * x = (2 - 1)(x - 1) + 1 = x,{\text{ }}x * 2 = (x - 1)(2 - 1) + 1 = x\)     M1

\(2 * x = x * 2 = 2{\text{ }}(2 \in S)\)     R1

Note:     Accept reference to commutativity instead of explicit expressions.

so 2 is the identity element     AG

[2 marks]

\(a * {a^{ - 1}} = 2 \Rightarrow (a - 1)({a^{ - 1}} - 1) + 1 = 2\)     M1

so \({a^{ - 1}} = 1 + \frac{1}{{a - 1}}\)     A1

since \(a - 1 > 0 \Rightarrow {a^{ - 1}} > 1{\text{ }}({a^{ - 1}} * a = a * {a^{ - 1}})\)     R1

Note:     R1 dependent on M1.

so each element, \(a \in S\), has an inverse     AG

[3 marks]