| Date | May 2011 | Marks available | 6 | Reference code | 11M.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 for a , \(b \in {\mathbb{Z}^ + }\) such that aRb if and only if \({a^2} - {b^2}\) is divisible by 5.
Show that R is an equivalence relation.
Identify the three equivalence classes.
Markscheme
reflexive: aRa because \({a^2} - {a^2} = 0\) (which is divisible by 5) A1
symmetric: let aRb so that \({a^2} - {b^2} = 5M\) M1
it follows that \({a^2} - {b^2} = - 5M\) which is divisible by 5 so bRa A1
transitive: let aRb and bRc so that \({a^2} - {b^2} = 5M\) and \({b^2} - {c^2} = 5N\) M1
\({a^2} - {b^2} + {b^2} - {c^2} = 5M + 5N\) A1
\({a^2} - {c^2} = 5M + 5N\) which is divisible by 5 so aRc A1
\( \Rightarrow \) R is an equivalence relation AG
[6 marks]
the equivalence classes are
{1, 4, 6, 9, …} A2
{2, 3, 7, 8, …} A1
{5, 10, …} A1
Note: Do not award any marks for classes containing fewer elements than shown above.
[4 marks]
Examiners report
Many candidates solved (a) correctly but solutions to (b) were generally poor. Most candidates seemed to have a weak understanding of the concept of equivalence classes and were unaware of any systematic method for finding the equivalence classes. If all else fails, a trial and error approach can be used. Here, starting with 1, it is easily seen that 4, 6,… belong to the same class and the pattern can be established.
Many candidates solved (a) correctly but solutions to (b) were generally poor. Most candidates seemed to have a weak understanding of the concept of equivalence classes and were unaware of any systematic method for finding the equivalence classes. If all else fails, a trial and error approach can be used. Here, starting with 1, it is easily seen that 4, 6, … belong to the same class and the pattern can be established.
