| Date | May 2013 | Marks available | 2 | Reference code | 13M.3srg.hl.TZ0.1 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Determine | Question number | 1 | Adapted from | N/A |
Question
The binary operation \( * \) is defined on \(\mathbb{N}\) by \(a * b = 1 + ab\).
Determine whether or not \( * \)
is closed;
is commutative;
is associative;
has an identity element.
Markscheme
\( * \) is closed A1
because \(1 + ab \in \mathbb{N}\) (when \(a,b \in \mathbb{N}\)) R1
[2 marks]
consider
\(a * b = 1 + ab = 1 + ba = b * a\) M1A1
therefore \( * \) is commutative
[2 marks]
EITHER
\(a * (b * c) = a * (1 + bc) = 1 + a(1 + bc){\text{ }}( = 1 + a + abc)\) A1
\((a * b) * c = (1 + ab) * c = 1 + c(1 + ab){\text{ }}( = 1 + c + abc)\) A1
(these two expressions are unequal when \(a \ne c\)) so \( * \) is not associative R1
OR
proof by counter example, for example
\(1 * (2 * 3) = 1 * 7 = 8\) A1
\((1 * 2) * 3 = 3 * 3 = 10\) A1
(these two numbers are unequal) so \( * \) is not associative R1
[3 marks]
let e denote the identity element; so that
\(a * e = 1 + ae = a\) gives \(e = \frac{{a - 1}}{a}\) (where \(a \ne 0\)) M1
then any valid statement such as: \(\frac{{a - 1}}{a} \notin \mathbb{N}\) or e is not unique R1
there is therefore no identity element A1
Note: Award the final A1 only if the previous R1 is awarded.
[3 marks]
Examiners report
For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.
For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.
For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.
For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.
