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?
Best Answer
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 $$\frac13+\mathbb{Z}=\{\dots,\frac13-1,\frac13,\frac13+1,\dots\}$$