# Equivalence classes

• Feb 20th 2009, 09:27 AM
ninano1205
Equivalence classes
This is a question from my professor that I couldnt understand.
Let X=R, x~y if x-y is a subset of Z(integers).
Describe the set of equivalence classes.
He said it's just a circle.
I have no idea what he meant.
Anyone can understand what he meant?
THanks.
• Feb 20th 2009, 09:38 AM
Plato
Quote:

Originally Posted by ninano1205
This is a question from my professor that I couldnt understand.
Let X=R, x~y if x-y is a subset of Z(integers).
Describe the set of equivalence classes.

Just what are you saying, "x-y is a subset of Z".
Are you saying that the difference is a integer?
• Feb 20th 2009, 03:09 PM
ninano1205
subset of integers
is integers?? right?
• Feb 20th 2009, 04:09 PM
Plato
Quote:

Originally Posted by ninano1205
is integers?? right?

I thought as much. In that case, the mention of circle has no relevance.
As an aside, this fits into a different problem that I have worked recently.
Now one equivalence class is simply the set of integers: $\left[ 0 \right] = \mathbb{Z}$.

Do you know about the floor function?
$\left\lfloor z \right\rfloor$ is the greatest integer not exceeding $z$.
We can show that this function is well defined on $\Re$ having the property that:
$\left\lfloor x \right\rfloor \leqslant x < \left\lfloor x \right\rfloor + 1\, \Rightarrow \,0 \leqslant x - \left\lfloor x \right\rfloor < 1$.

Now back to your question. Define a function on $\Re$ as $\mathbb{D}(x) = x - \left\lfloor x \right\rfloor$ (decimal part) .
Take note that $\left( {\forall x \in \Re } \right)\left[ {0 \leqslant \mathbb{D}(x) < 1} \right]$.
For the general equivalence classes: $\left( {\forall x \in \Re } \right)\left[ {\left[ x \right] = \bigcup\limits_{k \in \mathbb{Z}} {\left\{ {\mathbb{D}(x) + k} \right\}} } \right]$.