Describing a partition for an equivalence relation?

Question

Describe the partiton for the equivalence relation.

For each $x,y\in \mathbb{R}$ xRy $\iff$ $x-y\in \mathbb{Z}$

Now I am not sure how to find a partition for this I guess one could have negative integers or positive integers

I know a the set in a partition family must not overlap, and every element in the original set must be in the partition, and all the element in the partition must be in the original set.

Have you written down all the members of the partition containing, say, $\frac{1}{2}$ to get some idea what's happening? — Eric Towers, Jul 04 '14 at 20:11
The fact that $2 - \frac{1}{2} = \frac{3}{2}$ is not an integer tells you that $2$ is not related to $\frac{1}{2}$. No problem. Now, can you describe all of the numbers which are related to $2$? And all of the numbers which are related to $\frac{1}{2}$? — Lee Mosher, Jul 04 '14 at 20:22
All number releated to 2 are natural number and the one releated 1/2 are all real between 0 and 1 — Fernando Martinez, Jul 04 '14 at 20:28
No, the only real between $0$ and $1$ that is related to $\frac{1}{2}$ is $\frac{1}{2}$ itself. — Lee Mosher, Jul 05 '14 at 12:40

score 3 · Accepted Answer · answered Jul 04 '14 at 20:37

Given a number $x$, we have $yRx \iff y=x+n$ for some integer $n$. So the equivalence class of $x$ is $x+\mathbb{Z}$, that is, {$...,x-2,x-1,x,x+1,x+2,...$}.

The set of all such equivalence classes is the partition formed on $\mathbb{R}$. Now, $xRy$ iff $x$ and $y$ have the same decimal part. For example, $1.23R5.23$. So the distinct equivalence classes correspond precisely to the elements of $[0,1)$. That is, each distinct equivalence class contains exactly one element of this interval.

In short, this relation partitions $\mathbb{R}$ into sets of numbers having the same decimal part.

Describing a partition for an equivalence relation?

1 Answers1