Show that \(R\) is an equivalence relation.

Determine the equivalence classes of \(R\).

Determine the number of equivalence classes of \(S\).

METHOD 1

reflexive: \({4^a} - {4^a} = 0\) which is divisible by 7 (for all \(a \in \mathbb{¢}\))     R1

so \(aRa\) therefore reflexive

symmetric: Let \(aRb\) so that \({4^a} - {4^b}\) is divisible by 7     M1

it follows that \({4^b} - {4^a} = - ({4^a} - {4^b})\) is also divisible by 7     A1

it follows that \(bRa\) therefore symmetric

transitive: let \(aRb\) and \(bRc\) so that \({4^a} - {4^b}\) and \({4^b} - {4^c}\) are divisible by 7     M1

it follows that \({4^a} - {4^b} = 7M\) and \({4^b} - {4^c} = 7N\) so that \(({4^a} - {4^b}) + ({4^b} - {4^c}) = {4^a} - {4^c} = 7(M + N)\)     A1

therefore \(aRb\) and \(bRc \Rightarrow aRc\)     R1

so that \(R\) is transitive

 

Note:     For transitivity, award A0 if the same variable is used to express the multiples of 7; R1 is dependent on the M mark.

 

since \(R\) R is reflexive, symmetric and transitive, it is an equivalence relation     AG

METHOD 2

reflexive: \({4^a} - {4^a} \equiv 0\bmod 7\) (for all \(a \in \mathbb{Z}\))     R1

so \(aRa\) therefore reflexive

symmetric: let \(aRb\). Then \({4^a} - {4^b} \equiv 0\bmod 7\)     M1

it follows that \({4^b} - {4^a} \equiv - ({4^a} - {4^b}) \equiv 0\bmod 7\)     A1

it follows that \(bRa\) therefore symmetric

transitive: let \(aRb\) and \(bRc\), ie, \({4^a} - {4^b} \equiv 0\bmod 7\) and \({4^b} - {4^c} \equiv 0\bmod 7\)     M1

so that \({4^a} - {4^c} \equiv ({4^a} - {4^b}) + ({4^b} - {4^c}) \equiv 0\bmod 7\)     A1

therefore \(aRb\) and \(bRc \Rightarrow aRc\)     R1

so \(R\) is transitive

 

Note:     For transitivity, award A0 if mod 7 is omitted; R1 is dependent on the M mark.

 

since \(R\) is reflexive, symmetric and transitive, it is an equivalence relation     AG

 

[6 marks]

attempt to solve \({4^a} \equiv 4\bmod 7\) or \({4^a} \equiv {4^2} \equiv 2\bmod 7\) or \({4^a} \equiv {4^3} \equiv 1\bmod 7\)     (M1)

the equivalence classes are

\(\{ 1,{\text{ }}4,{\text{ }}7,{\text{ }} \ldots \} ,{\text{ \{ }}2,{\text{ }}5,{\text{ }}8,{\text{ }} \ldots \} \) and \(\{ 3,{\text{ }}6,{\text{ }}9,{\text{ }} \ldots \} \)     A2

 

Note:     Award (M1)A1 for one or two correct equivalence classes.

 

[3 marks]

starting with 1, we find that 2, 3, 4, … all belong to the same equivalence class or \({4^c} - 4 \equiv 4({4^{c - 1}} - 1) \equiv 4({2^{c - 1}} - 1)({2^{c - 1}} - 1) \equiv 0\bmod 6\) or \({4^c} \equiv 4\bmod 6\)     (M1)

therefore there is one equivalence class     A1

[2 marks]

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