# 4 Equivalence relations

## 4.3 Further exercises

### Exercise 59

Let be the relation defined on by

Give counter-examples to show that is not reflexive, symmetric or transitive.

### Solution

is not reflexive because, for example, 1 1 since 2 × 1 − 1 = 1 is not divisible by 7.

is not symmetric because, for example, 5 3 since 2 × 5 − 3 = 7 which is divisible by 7, but 3 5 since 2 × 3 − 5 = 1 which is not divisible by 7.

is not transitive because, for example, 5 3 and 3 6 since 2 × 5 − 3 = 7 and 2 × 3 − 6 = 0 which are both divisible by 7, but 5 6 since 2 × 5 − 6 = 4 which is not divisible by 7.

### Exercise 60

Let *A* be the set of all functions with domain and codomain , and let be the relation defined on *A* by

Show that this is an equivalence relation and describe the equivalence classes.

### Solution

We show that properties E1, E2 and E3 hold.

- E1 For any function
*f*: → ,*f*(0) =*f*(0), so*f**f*and hence the relation is reflexive. - E2 If
*f**g*so that*f*(0) =*g*(0), then*g*(0) =*f*(0), so*g**f*and hence the relation is symmetric. - E3 If
*f**g*and*g**h*so that*f*(0) =*g*(0) and*g*(0) =*h*(0), then*f*(0) =*h*(0), so*f**h*and hence the relation is transitive.

Therefore this is an equivalence relation.

Each equivalence class consists of all functions in *A* that take a particular value at 0; that is, each equivalence class is of the form

for some *r* .

### Exercise 61

Let be the relation defined on by

where

Show that this is an equivalence relation and describe the equivalence classes.

### Solution

We show that properties E1, E2 and E3 hold.

- E1 Let
*z*=*x*+*iy*. Then - so
*z**z*. Hence the relation is reflexive. - E2 Let
*z*_{1}=*x*_{1}+*iy*_{1}and*z*_{2}=*x*_{2}+*iy*_{2}be elements of . - Suppose that
*z*_{1}*z*_{2}so that*x*_{1}−*x*_{2}= 5(*y*_{1}−*y*_{2}). Then - so
*z*_{2}*z*_{1}. Hence the relation is symmetric. - E3 Let
*z*_{3}=*x*_{3}+*iy*_{3}, and suppose that*z*_{1}*z*_{2}and*z*_{2}*z*_{3}so that - and
- Then
- so
*z*_{1}*z*_{3}. Hence the relation is transitive.

Therefore this is an equivalence relation.

Two complex numbers *x*_{1} + *iy*_{1} and *x*_{2} + *iy*_{2} are related by this relation if

that is, if

Hence the equivalence classes are the lines *x* − 5*y* = *r*, for each real number *r*; that is, the lines with gradient
.

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