A2 October 2021 Q2
2. A binary operation \(\bigstar\) on the set of non-negative integers, \(\mathbb{Z}_0^+\), is defined by
\[m \mathbin{\bigstar} n = |m - n| \qquad m, n \in \mathbb{Z}_0^+\]| Scheme | Marks | AO |
|---|---|---|
| For \(m, n \in \mathbb{Z}_0^+\) we have \(m - n \in \mathbb{Z}\) (difference of integers is an integer) and so \(|m - n| \in \mathbb{Z}_0^+\), hence closed under \(\star\). | B1 | 2.4 |
| (1) |
Notes
B1: Checks difference of two non-negative integers is an integer and hence its modulus is a non-negative integer and concludes closure. “Always positive” as a conclusion is B0 without consideration of the equal zero case.
| Scheme | Marks | AO |
|---|---|---|
| For \(m \in \mathbb{Z}_0^+,\ 0 \star m = |0 - m| = |-m| = m\) | M1 | 1.1b |
| and \(m \star 0 = |m - 0| = |m| = m\) Hence 0 is an identity*. | A1* | 2.1 |
| (2) |
Notes
M1: Checks that 0 is a left or a right identity. (Checks either side)
A1*: Checks 0 works both sides as an identity and makes conclusion it is an identity. (Checks both sides and makes conclusion.)
| Scheme | Marks | AO |
|---|---|---|
| For \(m \in \mathbb{Z}_0^+\), we need \(|m - n| = 0 \Rightarrow n = \ldots\) or shows \(|m - m| = |0| = 0\) | M1 | 2.2a |
| As \(|m - m| = 0\) for all \(m \in \mathbb{Z}_0^+\) each \(m\) is self-inverse. | A1 | 2.1 |
| (2) |
Notes
M1: Realises \(m\) must be its own inverse for each \(m\) – accept if just stated \(m\) is self-inverse with no proof, or if an attempt is made to show it is self-inverse, or for an attempt to solve \(|m - n| = 0\)
A1: Each element is self-inverse with a full proof given.
| Scheme | Marks | AO |
|---|---|---|
| Checks associativity – ie evaluates \(m \star (n \star p)\) and \((m \star n) \star p\) with letter or numbers. | M1 | 1.2 |
| E.g, \(1 \star (2 \star 3) = 1 \star |2 - 3| = 1 \star 1 = 0\) but \((1 \star 2) \star 3 = |1 - 2| \star 3 = 1 \star 3 = |1 - 3| = 2\) | M1 | 3.1a |
| \(1 \star (2 \star 3) \neq (1 \star 2) \star 3\) hence not associative so not a group. | A1 | 2.4 |
| (3) | ||
| (8 marks) |
Notes
M1: Realises associativity must be checked in some way – may be by producing a counter example, or by attempting to evaluate both sides of the associativity axiom for a general case. A statement of the correct identity is sufficient for the mark to be awarded.
M1: Produces a suitable counter example and evaluates both sides of associativity equation. Attempts at algebraic proofs are unlikely to succeed but allow the method for e.g consideration of. \(m \gt n \gt p\) giving \(\big||m - n| - p\big| = |m - n - p|\) and \(\big|m - |n - p|\big| = |m - n + p|\) but must have a correct reason to disambiguate the inner moduli. If in doubt use review.
A1: Must have provided a counter example. Deduces associativity does not hold and concludes \(\mathbb{Z}_0^+\) is not a group under \(\star\)