"Take a 6-sided die and place it on your desk with one side facing you. Each possible orientation defines a function from $\{front, back, left, right, top, bottom\}$ to ${1, 2, 3, 4, 5, 6}$. $f(x)$ is the number that is showing on face $x$. Let $S$ be the set of all such functions.
Define a relation ∼ on S by $f$ ∼ $g$ ⇔ $f(top) = g(top)$. In other words, ~ is a relation on the set of all possible die orientations; orientations are related if they share the same number on the top face.
Prove that this is an equivalence relation."
Work So Far
The actual proof for this is relatively simple; I can define$front_0 = a, back_0=b, left_0=c, right_0=d, top_0=e, bottom_0 = f.$ Then just establish that $top_0=top_{90}=top_{180}=top_{270}$, $left_0 = right_{180}$, etc.
The difficulty I'm running into is figuring out exactly how this function is supposed to be represented. At first glance, it seems to me like it should be a set of 6 ordered pairs, a sample for one orientation would be:
$\{(front,5),(back,2),(left,3),(right,4),(top,1),(bottom,6)\}$
The problem is, it's impossible for a relation made from 4 of these functions to be equivalent; you're never going to get the point $(5, front)$, for example, which would be required to show symmetry with $(front, 5)$. So clearly, I'm representing it wrong, but am not sure about the correct alternative. What am I getting wrong here?
There are 24 orientations in this scheme, four for each number that can appear at the top of the die, but I don’t think that’s really directly relevant. It seems to me that $\sim$ being an equivalence relation is a simple consequence of equality being an equivalence relation. Take symmetry, for instance. If $f\sim g$, this means that $f(\text{top})=g(\text{top})$. By symmetry of equality, $g(\text{top})=f(\text{top})$ and so $g\sim f$. Can you prove reflexivity and transitivity now?