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 $a \in \mathbb{Z}$)     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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