Define $\sim$ on $\Bbb R^2$ by $(x,y)\sim(z,w)$ iff $|x−y|=|z−w|.$ Show that $\sim$ is an equivalence relation. So I know we have to prove that the relation is transitive, reflexive, and symmetric. 
Reflexive: I have to prove that every element of the set is related to itself. By the definition of this relation, wouldn't $|x-y|$ have to equal $|x-y|$ and $|z-w| = |z-w|$? In this case, is it not a given that the relation is reflexive because this will be true for all $x,y$ and $z,w$?
Transitive: If $|x-y| = |z-w|$ and $|z-w| = |d-e|$, then $|x-y| = |d-e|.$ I'm unsure of where to go from here.
Symmetric: $|x-y| = |z-w| \implies |z-w| = |x-y|$. Again isn't this just a given because the order doesn't matter here?
Hopefully my logic isn't too far off. Any further guidance is much appreciated.
 A: For reflexivity, all you have to show is $(x,y)\sim(x,y)$ for all $x$ and $y$.  This is that $\lvert x-y\rvert=\lvert x-y\rvert$, which is evidently true.
For transitivity, all you need to follow up on is that $\lvert x-y\rvert = \lvert d-e\rvert$ implies $(x,y)\sim (d,e)$.
For symmetry, again, you just need to follow up and say that $\lvert z-w\rvert=\lvert x-y\rvert$ implies $(z,w)\sim(x,y)$.
This problem is pretty much an exercise to make sure you say everything you need to say to make everything as clear as possible.  As you've noticed, there are no "non-trivial" steps, by which I mean everything is direct application of definition expansion.

Broader context (and why I'm answering this question in the first place): consider the function $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ defined by $f(x,y)=\lvert x-y\rvert$.  Recall that equality is an equivalence relation, in particular equality for $\mathbb{R}$ is an equivalence relation.  There is a concept of "pulling back" an equivalence relation through a function.  Define $(x,y)\sim (z,w)$ if and only if $f(x,y)=f(z,w)$.  Since $=$ is an equivalence relation for $\mathbb{R}$, it follows that $\sim$ is an equivalence relation for $\mathbb{R}\times\mathbb{R}$.  Of course, that's an (easy) theorem you'd have to prove, but it shows that there's nothing special about $\lvert x-y\rvert$ itself.
A: Let $f: A\to B$ be a map. Define $a_1\sim a_2\iff f(a_1)=f(a_2)$. Show that $\sim$ is an equivalence relation. Apply this to your problem.
A: It is reflexive since, given any $(a,b)\in\Bbb R^2,$ we have $\lvert a-b\rvert$ equal to itself, and hence $(a,b)\sim(a,b)$.
For transitivity, note that $\sim$ is given by an "if and only if" statement. This allows us to conclude that $(x,y)\sim(d,e)$.
It is symmetric exactly when $(a,b)\sim(c,d)$ implies $(c,d)\sim(a,b)$. Thus, let $(a,b)\sim(c,d)$. Then $\lvert a-b\rvert=\lvert c-d\rvert,$ i.e. $\lvert c-d\rvert=\lvert a-b\rvert,$ from which we can conclude $(c,d)\sim(a,b)$.
A: Reflexive: |x-y| = |x-y|, which is true, so (x, y)~(x, y)
Transitive: |x-y| = |z-w| and |z-w| = |d-e|, then, |x-y|-|z-w| = 0 = |z-w|-|d-e|. So, |x-y|=|d-e|<=> (x, y)~(d, e)
Symmetric: |x-y| = |z-w| <=> |x-y|-|z-w| = 0 <=> -|z-w| = -|x-y| => |z-w| = |x-y| <=> (z, w)~(x, y)
