finding equivalence classes

**issacnewton** · September 14th 2011, 03:09 AM

Hi

Here's a problem I am trying to solve
Find the equivalence classes of this relation

$S=\{(x,y)\in \mathbb{R}\times \mathbb{R} \vert \;x-y \in \mathbb{Q}\}$

now this is an equivalence relation. If we choose any rational number x , then we can see that

$[x]_{S}=\mathbb{Q}$

so Q itself is an equivalence class of S. Now consider irrational number like pi

$\pi \in \mathbb{R}$

$\pi-\pi = 0 \in \mathbb{Q}$

so $(\pi,\pi)\in S$

$\therefore \pi \in [\pi]_{S}$

if we subtract any other irrational or rational number from pi , it will not be in
$\mathbb{Q}$ (is that correct ?) So

$[\pi]_{S} =\{\pi\}$

similar conclusions can be drawn for any other irrational number. so other equivalence classes would correspond to irrational numbers where the set will
contain only that irrational number. for example , since $\mathit{e}$ , the base of natural logarithm is an irrational number ,

$[\mathit{e}]_{S}=\{\mathit{e}\}$

so the equivalence classes are $\mathbb{Q}$ and set corresponding to
each irrational number.

$\blacksqaure$

is my reasoning correct ?

**emakarov** · September 14th 2011, 04:44 AM

Originally Posted by issacnewton

if we subtract any other irrational or rational number from pi , it will not be in
$\mathbb{Q}$ (is that correct ?)

No, $\pi-1$ and $\pi+2/3$ are irrational, but $\pi-(\pi-1)=1$ and $\pi-(\pi+2/3)=-2/3$ are rational. So, $[\pi]_S=\{\pi+r\mid r\in\mathbb{Q}\}$ , and similarly for other numbers.

There is no much easier way to describe the equivalence classes, AFAIK. If you pick exactly one point in [0, 1] from every equivalence class (so that any two points belong to different classes and thus their difference is irrational), you'll get a Vitali set, which is interesting because it is non-measurable.

**issacnewton** · September 14th 2011, 11:42 AM

So since $\pi+r \in [\pi]_S$ for all $r\in \mathbb{Q}$ , it means
infinite number of irrational numbers belong to this particular equivalence class.
That means we can't come up with different equivalence class for each irrational
number. So question arises , how many equivalence classes are there
apart from $\mathbb{Q}$ ?

**emakarov** · September 14th 2011, 11:52 AM

Originally Posted by issacnewton

So since $\pi+r \in [\pi]_S$ for all $r\in \mathbb{Q}$ , it means infinite number of irrational numbers belong to this particular equivalence class.

Yes.

Originally Posted by issacnewton

That means we can't come up with different equivalence class for each irrational number.

This does not follow.

Originally Posted by issacnewton

So question arises , how many equivalence classes are there
apart from $\mathbb{Q}$ ?

I think the set of equivalence classes has the same power as the set of reals, i.e., continuum $\mathfrak{c}$ . Each class is countable, i.e., its cardinality is $\aleph_0$ , and $\mathfrak{c}\cdot\aleph_0=\mathfrak{c}$ .

**issacnewton** · September 14th 2011, 11:59 AM

since equivalence relation partitions the set, and since $\pi-1 \in [\pi]_S \Rightarrow [\pi-1]_S=[\pi]_S$
how can $[\pi-1]_S$ be different from $[\pi]_S$ ?

**emakarov** · September 14th 2011, 12:08 PM

Why do $[\pi-1]_S$ and $[\pi]_S$ have to be different? They are the same.

**issacnewton** · September 14th 2011, 12:14 PM

I was thinking that we will need to come up with different equivalence class for each irrational number. But as you showed, infinitely many irrational numbers can belong to the same equivalence class. So which means , each irrational number
doesn't have its own equivalence class. What I understood from your post #4 , is that even though this is the case, there are still infinitely many equivalence classes. Is that correct ? I have not yet studied about the alephs , so couldn't understand
the last sentence in your post #4

**emakarov** · September 14th 2011, 12:21 PM

Yes, each equivalence class is countable, i.e., has the same power as $\mathbb{N}$ or $\mathbb{Q}$ . One class is $\mathbb{Q}$ ; all others consist of irrational numbers. The number of classes, which is the same as the number of reals, is vastly greater; it cannot be put into one-to-one correspondence with $\mathbb{N}$ .

**issacnewton** · September 14th 2011, 12:23 PM

makes sense ........ thanks

$\smile$

Thread: finding equivalence classes

LinkBack

Thread Tools

Display

finding equivalence classes

Re: finding equivalence classes

Re: finding equivalence classes

Re: finding equivalence classes

Re: finding equivalence classes

Re: finding equivalence classes

Re: finding equivalence classes

Re: finding equivalence classes

Re: finding equivalence classes

Similar Math Help Forum Discussions

Question on equivalence relations and equivalence classes

equivalence relation and equivalence classes

Equivalence relation and Equivalence classes?

finding equivalence classes

equivalence classes