| Date | November 2017 | Marks available | 2 | Reference code | 17N.3srg.hl.TZ0.4 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Show that | Question number | 4 | Adapted from | N/A |
Question
The set \(S\) is defined as the set of real numbers greater than 1.
The binary operation \( * \) is defined on \(S\) by \(x * y = (x - 1)(y - 1) + 1\) for all \(x,{\text{ }}y \in S\).
Let \(a \in S\).
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.
Markscheme
\(x,{\text{ }}y > 1 \Rightarrow (x - 1)(y - 1) > 0\) 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]
