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.

(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]