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 z1 = x1 + iy1 and z2 = x2 + iy2 be elements of .
Suppose that z1 z2 so that x1x2 = 5(y1y2). Then

so z2 z1. Hence the relation is symmetric.
E3  Let z3 = x3 + iy3, and suppose that z1 z2 and z2 z3 so that

and

Then

so z1 z3. Hence the relation is transitive.

Therefore this is an equivalence relation.

Two complex numbers x1 + iy1 and x2 + iy2 are related by this relation if

that is, if

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

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