# Define a relation on the set of all real numbers $x,y \in \mathbb{R}$ as follows:

Define a relation on the set of all real numbers $x,y \in \Bbb{R}$ as follows:

$x \sim y$ if and only if $x - y \in \Bbb{Z}$

Prove this is an equivalence relation and find the equivalence class of number $\dfrac {1}{3}$.

ATTEMPT:

Reflexive: For any $x\in\Bbb Z$, ${x} - {x}=0$ and $0 \in \mathbb{Z}$.

Symmetric: For any $x,y \in \Bbb Z$, if $x - y \in \Bbb Z$ then $y - x \in \Bbb Z$.

Transitive: For any $x,y,z \in \Bbb Z$, if ${x} - {y} = 2k$ and ${y} - {z}= 2l$ for some $k,l \in \Bbb Z$, then ${x} - {z} = 2(k+l)$

Is the proof correct? And how do I find equivalence class?

• your proof is not correct as the relation is defined on $\mathbb{R}$ while you pick your $x, y$ and $z$ in $\mathbb{Z}$ Dec 14 '13 at 3:51
• Why the $2$s in your transitivity argument? They are unnecessary. Dec 14 '13 at 3:52
• You find the equivalence class of $1/3$ by finding all real $x$ such that $x-(1/3)$ is an integer. Dec 14 '13 at 3:53

Your ideas are good, but your proof contains some errors. First of all we are talking about $x,y\in\mathbb{R}$ not $x,y\in\mathbb{Z}$. Fixing your errors, the proof would look something like this:

Reflexive: $\forall x\in\mathbb{R}$, $x-x=0$ so $x\sim x$. Therefore $\sim$ is reflexive.

Symmetric: For $x,y\in\mathbb{R}$, if $x\sim y$ then $x-y\in\mathbb{Z}$ so $x-y=k$ for some $k\in\mathbb{Z}$. Since $-k\in\mathbb{Z}$ it follows that $y-x\in\mathbb{Z}$ and so $y\sim x$. Therefore $\sim$ is symmetric.

Transitive: For $x,y,z\in\mathbb{R}$, if $x\sim y$ and $y\sim z$ then $x-y\in\mathbb{Z}$ and $y-z\in\mathbb{Z}$. Since addition is closed in the integers $(x-y)+(y-z)=x-z\in\mathbb{Z}$ and so $x\sim z$. Therefore $\sim$ is transitive.

To find the equivalence class of $\frac13$ you need to find all $x\in\mathbb{R}$ such that $\frac13-x\in\mathbb{Z}$ i.e. $$\frac13-x=k,k\in\mathbb{Z}$$ Therefore $x=\frac13-k$ and so the equivalnce class of $\frac13$ is just $$\frac13+\mathbb{Z}=\{\dots,\frac13-1,\frac13,\frac13+1,\dots\}$$

• ...or as fractions, $\left\{ \frac{3k+1}{3} \mid k \in \Bbb{Z} \right\}$. Dec 14 '13 at 4:14

Here is another way to proceed - one that works quite generally. Notice that $\rm\,x\sim y\,$ iff $\rm\,x\,$ and $\rm\,y\,$ have equal fractional parts, i.e. $\rm\:x\sim y \iff f(x) = f(y)\$ for $\rm\:f(x) =$ fractional part of $\rm\,x.$

Now it is straightforward to prove that any relation of the above form is an equivalence relation.

More generally, suppose $\rm\ u\sim v\ \smash[t]{\overset{\ def}{\iff}}\, f(u) \approx f(v)\$ for a function $\rm\,f\,$ and equivalence relation $\,\approx.\, \$ Then the equivalence relation properties of $\,\approx\,$ transport (pullback) to $\,\sim\,$ along $\rm\,f\,$ as follows:

• reflexive $\rm\quad\ f(v) \approx f(v)\:\Rightarrow\:v\sim v$

• symmetric $\rm\,\ u\sim v\:\Rightarrow\ f(u) \approx f(v)\:\Rightarrow\:f(v)\approx f(u)\:\Rightarrow\:v\sim u$

• transitive $\rm\ \ \ u\sim v,\, v\sim w\:\Rightarrow\: f(u)\approx f(v),\,f(v)\approx f(w)\:\Rightarrow\:f(u)\approx f(w)\:\Rightarrow u\sim w$

Such relations are called (equivalence) kernels. One calls $\, \sim\,$ the $\,(\approx)\,$ kernel of $\rm\,f.$

Just to add that the collection of equivalence classes can be represented as a circle $S^1$ , since , as said by others here, $x$~$y$ iff $x,y$ have the same decimal expansion; if $x-y$ is in $\mathbb Z$ , then $x,y$ must have the same decimal representation. Then all the equivalence classes are represented in the segment $[0,1)$ . BUT: note that $0$~$1$ here, since $1-0=1$ is in $\mathbb Z$. After identifying $0,1$ to a point, we get a copy of $S^1$.