So I have the following question. I basically have to decide whether or not it is an equivalence relation on the reals R. If yes, then prove it and describe [2/3] (the equivalence class). If no, prove that it is not:
"x~y means x-y is an integer."
My understanding on this topic is crude!
Thanks in advance!
You have to verify the three basic properties of equivalence relations. These are the following:
1) Reflexivity: For all reals $x$, $x \sim x$.
2) Symmetry: Whenever $x \sim y$, then $y \sim x$.
3)Transitivity: Whenever $x \sim y$ and $y \sim z$, then $x \sim z$.
Remark: It is important that the above is checked for all possible values of each variable. In your comment, you said you were checking it only for $\frac 23$, this is incorrect.
So we will verify each one:
1) $x \sim x$ is true since $x-x=0$, which is an integer for all values of $x$, not just $\frac 23$.
2) $x \sim y$ means that $x-y$ is an integer. But then $y-x = (-1)(x-y)$ is also an integer, so that $y \sim x$ is also true, for all values of $x$ and $y$.
3) If $x \sim y$ and $y \sim z$ are true, then this means that $x -y $ and $y-z$ are integers. But then, $(x-y) + (y-z) = (x-z)$ is also an integer. This means that $x \sim z$. Again, the above is true for all values of $x,y,z$.
Hence, our relation is an equivalence relation.
The definition of equivalence class of an element $x$ is all $y$ such that $x \sim y$. So an element $k$ is related to $\frac{2}{3}$ if and only if $k-\frac{2}{3}$ is an integer.
Thus, in set notation, we can write $\left[\frac 23 \right] = \left\{ k + \frac 23 : k \in \mathbb Z \right\}$. It can also be written as $\{ \ldots, \frac{-7}3,\frac{-4}{3}, \frac {-1}3,\frac 23,\frac 53,\frac 83,\ldots\}$.
In our case, it so happens that this equivalence relation has a special name : $\mathbb R$ is a group under addition, and $\mathbb Z$ is a subgroup. Knowing this much itself, it is easy to see, more generally for $H \leq G$ groups, that $x \sim y \iff x-y \in H$ is an equivalence relation on $G$, and the set of equivalence classes $\{[x]\}$ is denoted $\frac GH$, called the quotient of $G$ by $H$. It is not always a group, in fact precisely when $H$ is a normal subgroup of $G$. In our case, we are actually dealing with $\mathbb R \over \mathbb Z$, and it turns out that this is a group since $\mathbb R$ is abelian, so every subgroup is normal. The equivalence classes are denoted $[x]$, and we have $\{x\} = x+\mathbb Z$