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]