Date | May 2013 | Marks available | 3 | Reference code | 13M.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 relation R is defined on {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} by aRb if and only if \(a(a + 1) \equiv b(b + 1)(\bmod 5)\).
Show that R is an equivalence relation.
Show that the equivalence defining R can be written in the form
\[(a - b)(a + b + 1) \equiv 0(\bmod 5).\]
Hence, or otherwise, determine the equivalence classes.
Markscheme
reflexive: \(a(a + 1) \equiv a(a + 1)(\bmod 5)\), therefore aRa R1
symmetric: \(aRb \Rightarrow a(a + 1) = b(b + 1) + 5N\) M1
\( \Rightarrow b(b + 1) = a(a + 1) - 5N \Rightarrow bRa\) A1
transitive:
EITHER
\(aRb{\text{ and }}bRc \Rightarrow a(a + 1) = b(b + 1) + 5M{\text{ and }}b(b + 1) = c(c + 1) + 5N\) M1
it follows that \(a(a + 1) = c(c + 1) + 5(M + N) \Rightarrow aRc\) M1A1
OR
\(aRb{\text{ and }}bRc \Rightarrow a(a + 1) \equiv b(b + 1)(\bmod 5){\text{ and}}\)
\(b(b + 1) \equiv c(c + 1)(\bmod 5)\) M1
\(a(a + 1) - b(b + 1) \equiv 0(\bmod 5);{\text{ }}b(b + 1) - c(c + 1) \equiv 0(\bmod 5)\) M1
\(a(a + 1) - c(c + 1) \equiv 0\bmod 5 \Rightarrow a(a + 1) \equiv c(c + 1)\bmod 5 \Rightarrow aRc\) A1
[6 marks]
the equivalence can be written as
\({a^2} + a - {b^2} - b \equiv 0(\bmod 5)\) M1
\((a - b)(a + b) + a - b \equiv 0(\bmod 5)\) M1A1
\((a - b)(a + b + 1) \equiv 0(\bmod 5)\) AG
[3 marks]
the equivalence classes are
{1, 3, 6, 8, 11}
{2, 7, 12}
{4, 5, 9, 10} A4
Note: Award A3 for 2 correct classes, A2 for 1 correct class.
[4 marks]
Examiners report
Candidates knew the properties of equivalence relations but did not show sufficient working out in the transitive case. Others did not do the modular arithmetic correctly, still others omitted the \(\bmod(5)\) in part or throughout.
Candidates knew the properties of equivalence relations but did not show sufficient working out in the transitive case. Others did not do the modular arithmetic correctly, still others omitted the \(\bmod (5)\) in part or throughout.
Candidates knew the properties of equivalence relations but did not show sufficient working out in the transitive case. Others did not do the modular arithmetic correctly, still others omitted the \(\bmod(5)\) in part or throughout.