# Equivalence classes

1. Jun 1, 2012

### look416

Given the set S, where aSb if and only if a - b $\in$ $Z$

It is asking for the equivalence class and the answer given is
S has only one equivalance class for each real number x such that 0 ≤ x < 1. the class [x] is given by {x + k : k $\in$ $Z$}

i dun get it, since S is a set of relation where a - b is an element of Z, then there should be a lot of equivalance class for S, however, it states that S has only one equivalence class and its {x + k}

2. Jun 1, 2012

### chiro

Hey look416.

The equivalence class seems to be defined in terms of the usual way (in terms of a modulus like definition) and since you are dealing with integers this seems appropriate (the final thing is going to be a set of integers).

3. Jun 1, 2012

### look416

but its stating it owns one equivalence class only, i thought there should be infinite sets of classes?

4. Jun 1, 2012

### Stephen Tashi

The answer said "one equivalence class for each real number....etc. ", not "one equivalence class".

5. Jun 1, 2012

### look416

ok,but that raise me to another problem
it says that for each real number x such that 0 ≤ x < 1. , which means that the whole relation only valids for that range?
requires some help on understanding the relation between the a-b and x

6. Jun 1, 2012

### micromass

Well, there is an equivalence class for 1.5 too (for example), but that is equal to the equivalence class of 0.5 (since 1.5-0.5 is in Z). So every equivalence class can be written as $\{x+k~\vert~k\in \mathbb{Z}\}$ with x in [0,1[.

7. Jun 1, 2012

### look416

oh, so the variable x is account for the decimals among the integers, i see