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.

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]