Date | May 2015 | Marks available | 2 | Reference code | 15M.3srg.hl.TZ0.3 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | Show that | Question number | 3 | Adapted from | N/A |
Question
The relation \(R\) is defined on \(\mathbb{Z}\) by \(xRy\) if and only if \({x^2}y \equiv y\bmod 6\).
Show that the product of three consecutive integers is divisible by \(6\).
Hence prove that \(R\) is reflexive.
Find the set of all \(y\) for which \(5Ry\).
Find the set of all \(y\) for which \(3Ry\).
Using your answers for (c) and (d) show that \(R\) is not symmetric.
Markscheme
in a product of three consecutive integers either one or two are even R1
and one is a multiple of \(3\) R1
so the product is divisible by \(6\) AG
[2 marks]
to test reflexivity, put \(y = x\) M1
then \({x^2}x - x = (x - 1)x(x + 1) \equiv 0\bmod 6\) M1A1
so \(xRx\) AG
[3 marks]
if \(5Ry\) then \(25y \equiv y\bmod 6\) (M1)
\(24y \equiv 0\bmod 6\) (M1)
the set of solutions is \(\mathbb{Z}\) A1
Note: Only one of the method marks may be implied.
[3 marks]
if \(3Ry\) then \(9y \equiv y\bmod 6\)
\(8y \equiv 0\bmod 6 \Rightarrow 4y \equiv 0\bmod 3\) (M1)
the set of solutions is \(3\mathbb{Z}\) (ie multiples of \(3\)) A1
[2 marks]
from part (c) \(5R3\) A1
from part (d) \(3R5\) is false A1
\(R\) is not symmetric AG
Note: Accept other counterexamples.
[2 marks]
Total [12 marks]
Examiners report
A surprising number of candidates thought that an example was sufficient evidence to answer this part.
Again, a lack of confidence with modular arithmetic undermined many candidates' attempts at this part.
(c) and (d) Most candidates started these parts, but some found solutions as fractions rather than integers or omitted zero and/or negative integers.
(c) and (d) Most candidates started these parts, but some found solutions as fractions rather than integers or omitted zero and/or negative integers.
Some candidates regarded \(R\) as an operation, rather than a relation, so returned answers of the form \(aRb \ne bRa\).