Question
The relation \(R\) is defined for \(x,y \in {\mathbb{Z}^ + }\) such that \(xRy\) if and only if \({3^x} \equiv {3^y}(\bmod 10)\) .
(i) Show that \(R\) is an equivalence relation.
(ii) Identify all the equivalence classes.
Let \(S\) denote the set \(\left\{ {x\left| {x = a + b\sqrt 3 ,a,b \in \mathbb{Q},{a^2} + {b^2} \ne 0} \right.} \right\}\) .
(i) Prove that \(S\) is a group under multiplication.
(ii) Give a reason why \(S\) would not be a group if the conditions on \({a,b}\) were changed to \({a,b \in \mathbb{R},{a^2} + {b^2} \ne 0}\) .
Answer/Explanation
Markscheme
(i) \({3^x} \equiv {3^x}(\bmod 10) \Rightarrow xRx\) so R is reflexive. R1
\(xRy \Rightarrow {3^x} \equiv {3^y}(\bmod 10) \Rightarrow {3^y} \equiv {3^x}(\bmod 10) \Rightarrow yRx\)
so \(R\) is symmetric. R2
\(xRy\) and \(yRz \Rightarrow {3^x} – {3^y} = 10M\) and \({3^y} – {3^z} = 10N\)
Adding \({3^x} – {3^z} = 10(M + N) \Rightarrow {3^x} \equiv {3^z}(\bmod 10)\) hence transitive R2
(ii) Consider \({3^1} = 3,{3^2} = 9,{3^3} = 27,{3^4} = 81,{3^5} = 243\) , etc. (M2)
It is evident from this sequence that there are 4 equivalence classes,
\(1\), \(5\), \(9\), … A1
\(2\), \(6\), \(10\), … A1
\(3\), \(7\), \(11\), … A1
\(4\), \(8\), \(12\), … A1
[11 marks]
(i) Consider \(a + b\sqrt 3 \) \(c + d\sqrt 3 \) \( = (ac + 3bd) + (bc + ad)\sqrt 3 \) M1A1
This establishes closure since products of rational numbers are rational. R1
Since if \(a\) and \(b\) are not both zero and \(c\) and \(d\) are not both zero, it follows that \(ac + 3bd\) and \(bc + ad\) are not both zero. R1
The identity is \(1( \in S)\) . R1
Consider \(a + b{\sqrt 3 ^{ – 1}} = \frac{1}{{a + b\sqrt 3 }}\) M1A1
\( = \frac{1}{{a + b\sqrt 3 }} \times \frac{{a – b\sqrt 3 }}{{a – b\sqrt 3 }}\) A1
\( = \frac{a}{{({a^2} – 3{b^2})}} \times \frac{b}{{({a^2} – 3{b^2})}}\sqrt 3 \) A1
This inverse \( \in S\) because \({({a^2} – 3{b^2})}\) cannot equal zero since \(a\) and \(b\) cannot both be zero R1
and \(({a^2} – 3{b^2}) = 0\) would require \(\frac{a}{b} = \pm \sqrt 3 \) which is impossible because a rational number cannot equal \(\sqrt 3 \) . R2
Finally, multiplication of numbers is associative. R1
(ii) If \(a\) and \(b\) are both real numbers, \(a + b\sqrt 3 \) would have no inverse if \({a^2} = 3{b^2}\) . R2
[15 marks]
Question
The relation \({R_1}\) is defined for \(a,b \in {\mathbb{Z}^ + }\) by \(a{R_1}b\) if and only if \(n\left| {({a^2} – {b^2})} \right.\) where \(n\) is a fixed positive integer.
(i) Show that \({R_1}\) is an equivalence relation.
(ii) Determine the equivalence classes when \(n = 8\) .
Consider the group \(\left\{ {G, * } \right\}\) and let \(H\) be a subset of \(G\) defined by
\(H = \left\{ {x \in G} \right.\) such that \(x * a = a * x\) for all \(a \in \left. G \right\}\) .
Show that \(\left\{ {H, * } \right\}\) is a subgroup of \(\left\{ {G, * } \right\}\) .
The relation \({R_2}\) is defined for \(a,b \in {\mathbb{Z}^ + }\) by \(a{R_2}b\) if and only if \((4 + \left| {a – b} \right|)\) is the square of a positive integer. Show that \({R_2}\) is not transitive.
Answer/Explanation
Markscheme
(i) Since \({a^2} – {a^2} = 0\) is divisible by n, it follows that \(a{R_1}a\) so \({R_1}\) is reflexive. A1
\(a{R_1}b \Rightarrow {a^2} – {b^2}\) divisible by \(n \Rightarrow {b^2} – {a^2}\) divisible by \(n \Rightarrow b{R_1}a\) so
symmetric. A1
\(a{R_1}b\) and \(b{R_1}c \Rightarrow {a^2} – {b^2} = pn\) and \({b^2} – {c^2} = qn\) A1
\(({a^2} – {b^2}) + ({b^2} – {c^2}) = pn + qn\) M1
so \({a^2} – {c^2} = (p + q)n \Rightarrow a{R_1}c\) A1
Therefore \({R_1}\) is transitive.
It follows that \({R_1}\) is an equivalence relation. AG
(ii) When \(n = 8\) , the equivalence classes are
\(\left\{ {1,3,5,7,9, \ldots } \right\}\) , i.e. the odd integers A2
\(\left\{ {2,6,10,14, \ldots } \right\}\) A2
and \(\left\{ {4,8,12,16, \ldots } \right\}\) A2
Note: If finite sets are shown award A1A1A1.
[11 marks]
Associativity follows since G is associative. A1
Closure: Let \(x,y \in H\) so \(ax = xa\) , \(ay = ya\) for \(a \in G\) M1
Consider \(axy = xay = xya \Rightarrow xy \in H\) M1A1
The identity \(e \in H\) since \(ae = ea\) for \(a \in G\) A2
Inverse: Let \(x \in H\) so \(ax = xa\) for \(a \in G\)
Then
\({x^{ – 1}}a = {x^{ – 1}}ax{x^{ – 1}}\) M1A1
\( = {x^{ – 1}}xa{x^{ – 1}}\) M1
\( = a{x^{ – 1}}\) A1
so \( \Rightarrow {x^{ – 1}} \in H\) A1
The four group axioms are satisfied so \(H\) is a subgroup. R1
[12 marks]
Attempt to find a counter example. (M1)
We note that \(1{R_2}6\) and \(6{R_2}11\) but 1 not \({R_2}11\) . A2
Note: Accept any valid counter example.
The relation is not transitive. AG
[3 marks]
Question
The set \(S\) consists of real numbers r of the form \(r = a + b\sqrt 2 \) , where \(a,b \in \mathbb{Z}\) .
The relation \(R\) is defined on \(S\) by \({r_1}R{r_2}\) if and only if \({a_1} \equiv {a_2}\) (mod2) and \({b_1} \equiv {b_2}\) (mod3), where \({r_1} = {a_1} + {b_1}\sqrt 2 \) and \({r_2} = {a_2} + {b_2}\sqrt 2 \) .
Show that \(R\) is an equivalence relation.
Show, by giving a counter-example, that the statement \({r_1}R{r_2} \Rightarrow r_1^2Rr_2^2\) is false.
Determine
(i) the equivalence class \(E\) containing \(1 + \sqrt 2 \) ;
(ii) the equivalence class \(F\) containing \(1 – \sqrt 2 \) .
Show that
(i) \({(1 + \sqrt 2 )^3} \in F\) ;
(ii) \({(1 + \sqrt 2 )^6} \in E\) .
Determine whether the set \(E\) forms a group under
(i) the operation of addition;
(ii) the operation of multiplication.
Answer/Explanation
Markscheme
reflexive: if \({r_{}} = {a_{}} + {b_{}}\sqrt 2 \in S\) then \(a \equiv a(\bmod 2)\) and \(b \equiv b(\bmod 3)\)
\(( \Rightarrow rRr)\) A1
symmetric: if \({r_1}R{r_2}\) then \({a_1} \equiv {a_2}(\bmod 2)\) and \({b_1} \equiv {b_2}(\bmod 3)\) , and M1
\({a_2} \equiv {a_1}(\bmod 2)\) and \({b_2} \equiv {b_1}(\bmod 3)\) , (so that \({r_2}R{r_1}\) ) A1
transitive: if \({r_1}R{r_2}\) and \({r_2}R{r_3}\) then
\(2|{a_1} – {a_2}\) and \(2|{a_2} – {a_3}\) M1
\( \Rightarrow 2|{a_1} – {a_2} + {a_2} – {a_3} \Rightarrow 2|{a_1} – {a_3}\) M1A1
\(3|{b_1} – {b_2}\) and \(3|{b_2} – {b_3}\)
\( \Rightarrow 3|{b_1} – {b_2} + {b_2} – {b_3} \Rightarrow 3|{b_1} – {b_3}( \Rightarrow {r_1}R{r_3})\) A1AG
[7 marks]
consider, for example, \({r_1} = 1 + \sqrt 2 \) , \({r_2} = 3 + \sqrt 2 \) \(({r_1}R{r_2})\) M1
Note: Only award M1 if the two numbers are related and neither \(a\) nor \(b = 0\) .
\(r_1^2 = 3 + 2\sqrt 2 \) , \(r_2^2 = 11 + 6\sqrt 2 \) A1
the squares are not equivalent because \(2 \ne 6(\bmod 3)\) A1
[3 marks]
(i) \(E = \left\{ {2k + 1 + (3m + 1)\sqrt 2 } :k,m \in \mathbb{Z} \right\}\) A1A1
(ii) \(F = \left\{ {2k + 1 + (3m – 1)\sqrt 2 } :k,m \in \mathbb{Z} \right\}\) A1
[3 marks]
(i) \({(1 + \sqrt 2 )^3} = 7 + 5\sqrt 2 \) A1
\( = 2 \times 3 + 1 + (3 \times 2 – 1)\sqrt 2 \in F\) R1AG
(ii) \({(1 + \sqrt 2 )^6} = 99 + 70\sqrt 2 \) A1
\( = 2 \times 49 + 1 + (3 \times 23 + 1)\sqrt 2 \in E\) R1AG
[4 marks]
(i) \(E\) is not a group under addition A1
any valid reason eg \(0 \notin E\) R1
(ii) \(E\) is not a group under multiplication A1
any valid reason eg \(1 \notin E\) R1
[4 marks]
Question
\(S\) is defined as the set of all \(2 \times 2\) non-singular matrices. \(A\) and \(B\) are two elements of the set \(S\).
(i) Show that \({({A^T})^{ – 1}} = {({A^{ – 1}})^T}\).
(ii) Show that \({(AB)^T} = {B^T}{A^T}\).
A relation \(R\) is defined on \(S\) such that \(A\) is related to \(B\) if and only if there exists an element \(X\) of \(S\) such that \(XA{X^T} = B\). Show that \(R\) is an equivalence relation.
Answer/Explanation
Markscheme
(i) \(A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\)
\({A^T} = \left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right)\) M1
\({({A^T})^{ – 1}} = \frac{1}{{ad – bc}}\left( {\begin{array}{*{20}{c}} d&{ – c} \\ { – b}&a \end{array}} \right)\;\;\;\)(which exists because \(ad – bc \ne 0\)) A1
\({A^{ – 1}} = \frac{1}{{ad – bc}}\left( {\begin{array}{*{20}{c}} d&{ – b} \\ { – c}&a \end{array}} \right)\) M1
\({({A^{ – 1}})^T} = \frac{1}{{ad – bc}}\left( {\begin{array}{*{20}{c}} d&{ – c} \\ { – b}&a \end{array}} \right)\) A1
hence \({({A^T})^{ – 1}} = {({A^{ – 1}})^T}\) as required AG
(ii) \(A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\;\;\;B = \left( {\begin{array}{*{20}{c}} e&f \\ g&h \end{array}} \right)\)
\(AB = \left( {\begin{array}{*{20}{c}} {ae + bg}&{af + bh} \\ {ce + dg}&{cf + dh} \end{array}} \right)\) M1
\({(AB)^T} = \left( {\begin{array}{*{20}{c}} {ae + bg}&{ce + dg} \\ {af + bh}&{cf + dh} \end{array}} \right)\) A1
\({B^T} = \left( {\begin{array}{*{20}{c}} e&g \\ f&h \end{array}} \right)\;\;\;{A^T} = \left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right)\) M1
\({B^T}{A^T} = \left( {\begin{array}{*{20}{c}} e&g \\ f&h \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {ae + bg}&{ce + dg} \\ {af + bh}&{cf + dh} \end{array}} \right)\) A1
hence \({(AB)^T} = {B^T}{A^T}\) AG
\(R\) is reflexive since \(I \in S\) and \(IA{I^T} = A\) A1
\(XA{X^T} = B \Rightarrow A = {X^{ – 1}}B{({X^T})^{ – 1}}\) M1A1
\( \Rightarrow A = {X^{ – 1}}B{({X^{ – 1}})^T}\) from a (i) A1
which is of the correct form, hence symmetric AG
\(ARB \Rightarrow XA{X^T} = B\) and \(BRC = YB{Y^T} = C\) M1
Note: Allow use of \(X\) rather than \(Y\) in this line.
\( \Rightarrow YXA{X^T}{Y^T} = YB{Y^T} = C\) M1A1
\( \Rightarrow (YX)A{(YX)^T} = C\) from a (ii) A1
this is of the correct form, hence transitive
hence \(R\) is an equivalence relation AG
Question
The set of all integer s from 0 to 99 inclusive is denoted by S. The binary operations \( * \) and \( \circ \) are defined on S by
\(a * b = \left[ {a + b + 20} \right]\)(mod 100)
\(a \circ b = \left[ {a + b – 20} \right]\)(mod 100).
The equivalence relation R is defined by \(aRb \Leftrightarrow \left( {{\text{sin}}\frac{{\pi a}}{5} = {\text{sin}}\frac{{\pi b}}{5}} \right)\).
Find the identity element of S with respect to \( * \).
Show that every element of S has an inverse with respect to \( * \).
State which elements of S are self-inverse with respect to \( * \).
Prove that the operation \( \circ \) is not distributive over \( * \).
Determine the equivalence classes into which R partitions S, giving the first four elements of each class.
Find two elements in the same equivalence class which are inverses of each other with respect to \( * \).
Answer/Explanation
Markscheme
\(a + e + 20 = a\)(mod 100) (M1)
\(e = – 20\)(mod 100) (A1)
\(e = 80\) A1
[3 marks]
\(a + {a^{ – 1}} + 20 = 80\)(mod 100) (M1)
inverse of \(a\) is \(60 – a\) (mod 100) A1
[2 marks]
30 and 80 A1A1
[2 marks]
\(a \circ \left( {b * c} \right) = a \circ \left( {b + c + 20} \right)\)(mod 100)
\( = a + \left( {b + c + 20} \right) – 20\)(mod 100) (M1)
\( = a + b + c\)(mod 100) A1
\(\left( {a \circ b} \right) * \left( {a \circ c} \right) = \left( {a + b – 20} \right) * \left( {a + c – 20} \right)\)(mod 100) M1
\( = a + b – 20 + a + c – 20 + 20\)(mod 100)
\( = 2a + b + c – 20\)(mod 100) A1
hence we have shown that \(a \circ \left( {b * c} \right) \ne \left( {a \circ b} \right) * \left( {a \circ c} \right)\) R1
hence the operation \( \circ \) is not distributive over \( * \) AG
Note: Accept a counterexample.
[5 marks]
{0,5,10,15…} A1
{1,4,11,14…} A1
{2,3,12,13…} A1
{6,9,16,19…} A1
{7,8,17,18…} A1
[5 marks]
for example 10 and 50, 20 and 40, 0 and 60… A2
[2 marks]