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 ${a,b \in \mathbb{R},{a^2} + {b^2} \ne 0}$ .

(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]