Question
The relation R is defined on ordered pairs by
\[(a,{\text{ }}b)R(c,{\text{ }}d){\text{ if and only if }}ad = bc{\text{ where }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in {\mathbb{R}^ + }.\]
(a) Show that R is an equivalence relation.
(b) Describe, geometrically, the equivalence classes.
▶️Answer/Explanation
Markscheme
(a) Reflexive: \((a,{\text{ }}b)R(a,{\text{ }}b)\) because \(ab = ba\) R1
Symmetric: \((a,{\text{ }}b)R(c,{\text{ }}d) \Rightarrow ad = bc \Rightarrow cb = da \Rightarrow (c,{\text{ }}d)R(a,{\text{ }}b)\) M1A1
Transitive: \((a,{\text{ }}b)R(c,{\text{ }}d) \Rightarrow ad = bc\) M1
\((c,{\text{ }}d)R(e,{\text{ }}f) \Rightarrow cf = de\)
Therefore
\(\frac{{ad}}{{de}} = \frac{{bc}}{{cf}}{\text{ so }}af = be\) A1
It follows that \((a,{\text{ }}b)R(e,{\text{ }}f)\) R1
[6 marks]
(b) \((a,{\text{ }}b)R(c,{\text{ }}d) \Rightarrow \frac{a}{b} = \frac{c}{d}\) (M1)
Equivalence classes are therefore points lying, in the first quadrant, on straight lines through the origin. A2
Notes: Accept a correct sketch.
Award A1 if “in the first quadrant” is omitted.
Do not penalise candidates who fail to exclude the origin.
[3 marks]
Total [9 marks]
Examiners report
Part (a) was well answered by many candidates although some misunderstandings of the terminology were seen. Some candidates appeared to believe, incorrectly, that reflexivity was something to do with \((a,{\text{ }}a)R(a,{\text{ }}a)\) and some candidates confuse the terms ‘reflexive’ and ‘symmetric’. Many candidates were unable to describe the equivalence classes geometrically.
Question
The relation R is defined on \(\mathbb{Z} \times \mathbb{Z}\) such that \((a,{\text{ }}b)R(c,{\text{ }}d)\) if and only if a − c is divisible by 3 and b − d is divisible by 2.
(a) Prove that R is an equivalence relation.
(b) Find the equivalence class for (2, 1) .
(c) Write down the five remaining equivalence classes.
▶️Answer/Explanation
Markscheme
(a) consider \((x,{\text{ }}y)R(x,{\text{ }}y)\)
since x – x = 0 and y – y = 0 , R is reflexive A1
assume \((x,{\text{ }}y)R(a,{\text{ }}b)\)
\( \Rightarrow x – a = 3M\) and \(y – b = 2N\) M1
\( \Rightarrow a – x = – 3M\) and \(b – y = – 2N\) A1
\( \Rightarrow (a,{\text{ }}b)R(x,{\text{ }}y)\)
hence R is symmetric
assume \((x,{\text{ }}y)R(a,{\text{ }}b)\)
\( \Rightarrow x – a = 3M\) and \(y – b = 2N\)
assume \((a,{\text{ }}b)R(c,{\text{ }}d)\)
\( \Rightarrow a – c = 3P\) and \(b – d = 2Q\) M1
\( \Rightarrow x – c = 3(M + P)\) and \(y – d = 2(N + Q)\) A1
hence \((x,{\text{ }}y)R(c,{\text{ }}d)\) A1
hence R is transitive
therefore R is an equivalence relation AG
[7 marks]
(b) \(\left\{ {(x,{\text{ }}y):x = 3m + 2,{\text{ }}y = 2n + 1,{\text{ }}m,{\text{ }}n \in \mathbb{Z}} \right\}\) A1A1
[2 marks]
(c) \(\{ 3m,{\text{ }}2n\} {\text{ \{ }}3m + 1,{\text{ }}2n\} {\text{ }}\{ 3m + 2,{\text{ }}2n\} \)
\(\{ 3m,{\text{ }}2n + 1\} {\text{ \{ }}3m + 1,{\text{ }}2n + 1\} {\text{ }}m,{\text{ }}n \in \mathbb{Z}\) A1A1A1A1A1
[5 marks]
Total [14 marks]
Examiners report
Stronger candidates had little problem with part (a) of this question, but proving an equivalence relation is still difficult for many. Equivalence classes still cause major problems and few fully correct answers were seen to this question.
Question
The relations R and S are defined on quadratic polynomials P of the form
\[P(z) = {z^2} + az + b{\text{ , where }}a{\text{ , }}b \in \mathbb{R}{\text{ , }}z \in \mathbb{C}{\text{ .}}\]
(a) The relation R is defined by \({P_1}R{P_2}\) if and only if the sum of the two zeros of \({P_1}\) is equal to the sum of the two zeros of \({P_2}\) .
(i) Show that R is an equivalence relation.
(ii) Determine the equivalence class containing \({z^2} – 4z + 5\) .
(b) The relation S is defined by \({P_1}S{P_2}\) if and only if \({P_1}\) and \({P_2}\) have at least one zero in common. Determine whether or not S is transitive.
▶️Answer/Explanation
Markscheme
(a) (i) R is reflexive, i.e. PRP because the sum of the zeroes of P is equal to the sum of the zeros of P R1
R is symmetric, i.e. \({P_1}R{P_2} \Rightarrow {P_2}R{P_1}\) because the sums of the zeros of \({P_1}\) and \({P_2}\) are equal implies that the sums of the zeros of \({P_2}\) and \({P_1}\) are equal R1
suppose that \({P_1}R{P_2}\) and \({P_2}R{P_3}\) M1
it follows that \({P_1}R{P_3}\) so R is transitive, because the sum of the zeros of \({P_1}\) is equal to the sum of the zeros of \({P_2}\) which in turn is equal to the sum of the zeros of \({P_3}\) , which implies that the sum of the zeros of \({P_1}\) is equal to the sum of the zeros of \({P_3}\) R1
the three requirements for an equivalence relation are therefore satisfied AG
(ii) the zeros of \({z^2} – 4z + 5\) are \(2 \pm {\text{i}}\) , for which the sum is 4 M1A1
\({{\text{z}}^2} + az + b\) has zeros of \(\frac{{ – a \pm \sqrt {{a^2} – 4b} }}{2}\) , so the sum is –a (M1)
Note: Accept use of the result (although not in the syllabus) that the sum of roots is minus the coefficient of z.
hence – a = 4 and so a = – 4 A1
the equivalence class is \({z^2} – 4z + k{\text{ , }}(k \in \mathbb{R})\) A1
[9 marks]
(b) for example, \((z – 1)(z – 2)S(z – 1)(z – 3)\) and
\((z – 1)(z – 3)S(z – 3)(z – 4)\) but \((z – 1)(z – 2)S(z – 3)(z – 4)\) is not true M1A1
so S is not transitive A1
[3 marks]
Total [12 marks]
Examiners report
Most candidates were able to show, in (a), that R is an equivalence relation although few were able to identify the required equivalence class. In (b), the explanation that S is not transitive was often unconvincing.
Question
Let R be a relation on the set \(\mathbb{Z}\) such that \(aRb \Leftrightarrow ab \geqslant 0\), for a, b \( \in \mathbb{Z}\).
(a) Determine whether R is
(i) reflexive;
(ii) symmetric;
(iii) transitive.
(b) Write down with a reason whether or not R is an equivalence relation.
▶️Answer/Explanation
Markscheme
(a) (i) \({a^2} \geqslant 0\) for all \(a \in \mathbb{Z}\), hence R is reflexive R1
(ii) \(aRb \Rightarrow ab \geqslant 0\) M1
\( \Rightarrow ba \geqslant 0\) R1
\( \Rightarrow bRa\), hence R is symmetric A1
(iii) \(aRb{\text{ and }}bRc \Rightarrow ab \geqslant 0{\text{ and }}bc \geqslant 0,{\text{ is }}aRc?\) M1
no, for example, \( – 3R0\) and \(0R5\), but \( – 3R5\) is not true A1
aRc is not generally true, hence R is not transitive A1
[7 marks]
(b) R does not satisfy all three properties, hence R is not an equivalence relation R1
[1 mark]
Total [8 marks]
Examiners report
Although the properties of an equivalence relation were well known, few candidates provided a counter-example to show that the relation is not transitive. Some candidates interchanged the definitions of the reflexive and symmetric properties.
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.
a. Show that R is an equivalence relation.[6]
b. Identify the three equivalence classes.[4]
▶️Answer/Explanation
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.
Question
The group G has a subgroup H. The relation R is defined on G by xRy if and only if \(x{y^{ – 1}} \in H\), for \(x,{\text{ }}y \in G\).
Show that R is an equivalence relation.
The Cayley table for G is shown below.
The subgroup H is given as \(H = \{ e,{\text{ }}{a^2}b\} \).
(i) Find the equivalence class with respect to R which contains ab.
(ii) Another equivalence relation \(\rho \) is defined on G by \(x\rho y\) if and only if \({x^{ – 1}}y \in H\), for \(x,{\text{ }}y \in G\). Find the equivalence class with respect to \(\rho \) which contains ab.
▶️Answer/Explanation
Markscheme
\(x{x^{ – 1}} = e \in H\) M1
\( \Rightarrow xRx\)
hence R is reflexive A1
if xRy then \(x{y^{ – 1}} \in H\)
\( \Rightarrow {(x{y^{ – 1}})^{ – 1}} \in H\) M1
now \((x{y^{ – 1}}){(x{y^{ – 1}})^{ – 1}} = e\) and \(x{y^{ – 1}}y{x^{ – 1}} = e\)
\( \Rightarrow {(x{y^{ – 1}})^{ – 1}} = y{x^{ – 1}}\) A1
hence \(y{x^{ – 1}} \in H \Rightarrow yRx\)
hence R is symmetric A1
if xRy, yRz then \(x{y^{ – 1}} \in H,{\text{ }}y{z^{ – 1}} \in H\) M1
\( \Rightarrow (x{y^{ – 1}})(y{z^{ – 1}}) \in H\) M1
\( \Rightarrow x({y^{ – 1}}y){z^{ – 1}} \in H\)
\( \Rightarrow {x^{ – 1}}z \in H\)
hence R is transitive A1
hence R is an equivalence relation AG
[8 marks]
(i) for the equivalence class, solving:
EITHER
\(x{(ab)^{ – 1}} = e{\text{ or }}x{(ab)^{ – 1}} = {a^2}b\) (M1)
\(\{ ab,{\text{ }}a\} \) A2
OR
\(ab{(x)^{ – 1}} = e{\text{ or }}ab{(x)^{ – 1}} = {a^2}b\) (M1)
\(\{ ab,{\text{ }}a\} \) A2
(ii) for the equivalence class, solving:
EITHER
\({x^{ – 1}}(ab) = e{\text{ or }}{x^{ – 1}}(ab) = {a^2}b\) (M1)
\(\{ ab,{\text{ }}{a^2}\} \) A2
OR
\({(ab)^{ – 1}}x = e{\text{ or }}{(ab)^{ – 1}}x = {a^2}b\) (M1)
\(\{ ab,{\text{ }}{a^2}\} \) A2
[6 marks]
Examiners report
Stronger candidates made a reasonable start to (a), and many were able to demonstrate that the relation was reflexive and transitive. However, the majority of candidates struggled to make a meaningful attempt to show the relation was symmetric, with many making unfounded assumptions. Equivalence classes still cause major problems and few fully correct answers were seen to (b).
Stronger candidates made a reasonable start to (a), and many were able to demonstrate that the relation was reflexive and transitive. However, the majority of candidates struggled to make a meaningful attempt to show the relation was symmetric, with many making unfounded assumptions. Equivalence classes still cause major problems and few fully correct answers were seen to (b).
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.[6]
Show that the equivalence defining R can be written in the form
\[(a – b)(a + b + 1) \equiv 0(\bmod 5).\][3]
Hence, or otherwise, determine the equivalence classes.[4]
▶️Answer/Explanation
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.
Question
Let \((H,{\text{ }} * {\text{)}}\) be a subgroup of the group \((G,{\text{ }} * {\text{)}}\).
Consider the relation \(R\) defined in \(G\) by \(xRy\) if and only if \({y^{ – 1}} * x \in H\).
(a) Show that \(R\) is an equivalence relation on \(G\).
(b) Determine the equivalence class containing the identity element.
▶️Answer/Explanation
Markscheme
(a) \(R\) is reflexive as \({x^{ – 1}} * x = e \in H \Rightarrow xRx\) for any \(x \in G\) A1
if \(xRy\) then \({y^{ – 1}} * x = h \in H\)
but \(h \in H \Rightarrow {h^{ – 1}} \in H\), ie, \(\underbrace {{{({y^{ – 1}} * x)}^{ – 1}}}_{{x^{ – 1}}{\text{*}}y} \in H\) M1
therefore \(yRx\)
\(R\) is symmetric A1
if \(xRy\) then \({y^{ – 1}} * x = h \in H\) and if \(yRz\) then \({z^{ – 1}} * y = k \in H\) M1
\(k * h \in H\), ie, \(\underbrace {({z^{ – 1}} * y) * ({y^{ – 1}} * x)}_{{z^{ – 1}} * x} \in H\) A1
therefore \(xRz\)
\(R\) is transitive A1
so \(R\) is an equivalence relation on \(G\) AG
[6 marks]
(b) \(xRe \Leftrightarrow {e^{ – 1}} * x \in H\) M1
\(x \in H\) A1
\([e] = H\) A1 N0
[3 marks]
Examiners report
Part (a) was fairly well answered by many candidates. They knew how to apply the equivalence relations axioms in this particular example. Part (b) however proved to be very challenging and hardly any correct answers were seen.
Question
Consider the set S defined by \(S = \{ s \in \mathbb{Q}:2s \in \mathbb{Z}\} \).
You may assume that \( + \) (addition) and \( \times \) (multiplication) are associative binary operations
on \(\mathbb{Q}\).
(i) Write down the six smallest non-negative elements of \(S\).
(ii) Show that \(\{ S,{\text{ }} + \} \) is a group.
(iii) Give a reason why \(\{ S,{\text{ }} \times \} \) is not a group. Justify your answer.
The relation \(R\) is defined on \(S\) by \({s_1}R{s_2}\) if \(3{s_1} + 5{s_2} \in \mathbb{Z}\).
(i) Show that \(R\) is an equivalence relation.
(ii) Determine the equivalence classes.
▶️Answer/Explanation
Markscheme
(i) \({\text{0, }}\frac{{\text{1}}}{{\text{2}}}{\text{, 1, }}\frac{{\text{3}}}{{\text{2}}}{\text{, 2, }}\frac{{\text{5}}}{{\text{2}}}\) A2
Notes: A2 for all correct, A1 for three to five correct.
(ii) EITHER
closure: if \({s_1},{\text{ }}{s_2} \in S\), then \({s_1} = \frac{m}{2}\) and \({s_2} = \frac{n}{2}\) for some \(m,{\text{ }}n \in {\text{¢}}\). M1
Note: Accept two distinct examples (eg, \(\frac{1}{2} + \frac{1}{2} = 1;{\text{ }}\frac{1}{2} + 1 = \frac{3}{2}\)) for the M1.
\({s_1} + {s_2} = \frac{{m + n}}{2} \in S\) A1
OR
the sum of two half-integers A1
is a half-integer R1
THEN
identity: 0 is the (additive) identity A1
inverse: \(s + ( – s) = 0\), where \( – s \in S\) A1
it is associative (since \(S \subset \S\)) A1
the group axioms are satisfied AG
(iii) EITHER
the set is not closed under multiplication, A1
for example, \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\), but \(\frac{1}{4} \notin S\) R1
OR
not every element has an inverse, A1
for example, 3 does not have an inverse R1
[9 marks]
(i) reflexive: consider \(3s + 5s\) M1
\( = 8s \in {\text{¢}} \Rightarrow \) reflexive A1
symmetric: if \({s_1}R{s_2}\), consider \(3{s_2} + 5{s_1}\) M1
for example, \( = 3{s_1} + 5{s_2} + (2{s_1} – 2{s_2}) \in {\text{¢}} \Rightarrow \)symmetric A1
transitive: if \({s_1}R{s_2}\) and \({s_2}R{s_3}\), consider (M1)
\(3{s_1} + 5{s_3} = (3{s_1} + 5{s_2}) + (3{s_2} + 5{s_3}) – 8{s_2}\) M1
\( \in {\text{¢}} \Rightarrow \)transitive A1
so R is an equivalence relation AG
(ii) \({C_1} = {\text{¢}}\) A1
\({C_2} = \left\{ { \pm \frac{1}{2},{\text{ }} \pm \frac{3}{2},{\text{ }} \pm \frac{5}{2},{\text{ }} \ldots } \right\}\) A1A1
Note: A1 for half odd integers and A1 for ±.
[10 marks]
Examiners report
[N/A]
[N/A]
Question
The relation \(R=\) is defined on \({\mathbb{Z}^ + }\) such that \(aRb\) if and only if \({b^n} – {a^n} \equiv 0(\bmod p)\) where \(n,{\text{ }}p\) are fixed positive integers greater than 1.
a. Show that \(R\) is an equivalence relation.[7]
b.Given that \(n = 2\) and \(p = 7\), determine the first four members of each of the four equivalence classes of \(R\).[5]
▶️Answer/Explanation
Markscheme
since \({a^n} – {a^n} = 0\), A1
it follows that \((aRa)\) and \(R\) is reflexive R1
if \(aRb\) so that \({b^n} – {a^n} \equiv 0(\bmod p)\) M1
then, \({a^n} – {b^n} \equiv 0(\bmod p)\) so that \(bRa\) and \(R\) is symmetric A1
if \(aRb\) and \(bRc\) so that \({b^n} – {a^n} \equiv 0(\bmod p)\) and \({c^n} – {b^n} \equiv 0(\bmod p)\) M1
adding, (it follows that \({c^n} – {a^n} \equiv 0(\bmod p)\)) M1
so that \(aRc\) and \(R\) is transitive A1
Note: Only accept the correct use of the terms “reflexive, symmetric, transitive”.
[7 marks]
we are now given that \(aRb\) if \({b^2} – {a^2} \equiv 0(\bmod 7)\)
attempt to find at least one equivalence class (M1)
the equivalence classes are
\(\{ 1,{\text{ }}6,{\text{ }}8,{\text{ }}13,{\text{ }} \ldots \} \) A1
\(\{ 2,{\text{ }}5,{\text{ }}9,{\text{ }}12,{\text{ }} \ldots \} \) A1
\(\{ 3,{\text{ }}4,{\text{ }}10,{\text{ }}11,{\text{ }} \ldots \} \) A1
\(\{ 7,{\text{ }}14,{\text{ }}21,{\text{ }}28,{\text{ }} \ldots \} \) A1
[5 marks]
Examiners report
Most candidates were familiar with the terminology of the required conditions to be satisfied for a relation to be an equivalence relation. The execution of the proofs was variable. It was grating to see such statements as \(R\) is symmetric because \(aRb = bRa\) or \(aRa = {a^n} – {a^n} = 0\), often without mention of \(\bmod p\), and such responses were not fully rewarded.
This was not well answered. Few candidates displayed a strategy to find the equivalence classes.
Question
The relation R is defined on \(\mathbb{R} \times \mathbb{R}\) such that \(({x_1},{\text{ }}{y_1})R({x_2},{\text{ }}{y_2})\) if and only if \({x_1}{y_1} = {x_2}{y_2}\).
a. Show that R is an equivalence relation.[5]
b. Determine the equivalence class of R containing the element \((1,{\text{ }}2)\) and illustrate this graphically.[4]
▶️Answer/Explanation
Markscheme
R is an equivalence relation if
R is reflexive, symmetric and transitive A1
\({x_1}{y_1} = {x_1}{y_1} \Rightarrow ({x_1},{\text{ }}{y_1})R({x_1},{\text{ }}{y_1})\) A1
so R is reflexive
\(({x_1},{\text{ }}{y_1})R({x_2},{\text{ }}{y_2}) \Rightarrow {x_1}{y_1} = {x_2}{y_2} \Rightarrow {x_2}{y_2} = {x_1}{y_1} \Rightarrow ({x_2},{\text{ }}{y_2})R({x_1},{\text{ }}{y_1})\) A1
so R is symmetric
\(({x_1},{\text{ }}{y_1})R({x_2},{\text{ }}{y_2})\) and \(({x_2},{\text{ }}{y_2})R({x_3},{\text{ }}{y_3}) \Rightarrow {x_1}{y_1} = {x_2}{y_2}\) and \({x_2}{y_2} = {x_3}{y_3}\) M1
\( \Rightarrow {x_1}{y_1} = {x_3}{y_3} \Rightarrow ({x_1},{\text{ }}{y_1})R({x_3},{\text{ }}{y_3})\) A1
so R is transitive
R is an equivalence relation AG
[5 marks]
\((x,{\text{ }}y)R(1,{\text{ }}2)\) (M1)
the equivalence class is \(\{ (x,{\text{ }}y)|xy = 2\} \) A1
correct graph A1
\((1,{\text{ }}2)\) indicated on the graph A1
Note: Award last A1 only if plotted on a curve representing the class.
[4 marks]
Examiners report
[N/A]
[N/A]
Question
The relation R is defined on \({\mathbb{Z}^ + }\) by aRb if and only if ab is even. Show that only one of the conditions for R to be an equivalence relation is satisfied.
The relation S is defined on \({\mathbb{Z}^ + }\) by aSb if and only if \({a^2} \equiv {b^2}(\bmod 6)\) .
(i) Show that S is an equivalence relation.
(ii) For each equivalence class, give the four smallest members.
▶️Answer/Explanation
Markscheme
reflexive: if a is odd, \(a \times a\) is odd so R is not reflexive R1
symmetric: if ab is even then ba is even so R is symmetric R1
transitive: let aRb and bRc; it is necessary to determine whether or not aRc (M1)
for example 5R2 and 2R3 A1
since \(5 \times 3\) is not even, 5 is not related to 3 and R is not transitive R1
[5 marks]
(i) reflexive: \({a^2} \equiv {a^2}(\bmod 6)\) so S is reflexive R1
symmetric: \({a^2} \equiv {b^2}(\bmod 6) \Rightarrow 6|({a^2} – {b^2}) \Rightarrow 6|({b^2} – {a^2}) \Rightarrow {b^2} \equiv {a^2}(\bmod 6)\) R1
so S is symmetric
transitive: let aSb and bSc so that \({a^2} = {b^2} + 6M\) and \({b^2} = {c^2} + 6N\) M1
it follows that \({a^2} = {c^2} + 6(M + N)\) so aSc and S is transitive R1
S is an equivalence relation because it satisfies the three conditions AG
(ii) by considering the squares of integers (mod 6), the equivalence (M1)
classes are
{1, 5, 7, 11, \( \ldots \)} A1
{2, 4, 8, 10, \( \ldots \)} A1
{3, 9, 15, 21, \( \ldots \)} A1
{6, 12, 18, 24, \( \ldots \)} A1
[9 marks]
Examiners report
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[N/A]