Determining Equivalence Relation on $\Bbb{Z}$ Alright, I have a homework problem which I have researched, read up on and I (think) solved. I just need someone to either confirm my answer (and re-affirm my knowledge) or explain why I am wrong.
Here is the question:


*

*Is $\{(x,y) | |x|=|y|\}$ an equivalence relation on $\Bbb{Z}$?


I believe this is true. I created a sample relation (is this the correct term?)
consisting of $\{(3,3),(2,2),(1,1),(1,1),(1,1)\}$ just to help my visualize the question.
From what I understand, and what I can see in my example, the proposed realation is
Reflexive, Symmetric, and Transative. Am I correct? 
 A: Here is a nice way to proceed - one that works quite generally.  Notice that $\rm\,x\sim y\,$ iff $\rm\,x\,$ and $\rm\,y\,$ have equal $ $  magnitude: $\rm\ \ x\sim y {\overset{\ def}{\color{#c00}\iff}} f(x) = f(y)\ $ for $\rm\:f(x) = \rm\,|x|.$
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}{\color{#c00}\iff}}\, f(u) \approx f(v)\ $ for a function $\rm\,f\,$ and equivalence relation $\,\approx.\, \ $ Then the equivalence relation $\rm\color{#0a0}{properties\ (E)}\,$  of $\,\approx\,$ transport (pullback) to $\,\sim\,$ along $\rm\,f$ as follows


*

*reflexive  $\rm\quad\ \color{#0a0}{(E)}\Rightarrow f(v) \approx f(v)\:\color{#c00}\Rightarrow\:v\sim v$

*symmetric  $\rm\,\ u\sim v\:\color{#c00}\Rightarrow\ f(u) \approx f(v)\:\color{#0a0}{\overset{(E)}\Rightarrow}\:f(v)\approx f(u)\:\color{#c00}\Rightarrow\:v\sim u$

*transitive $\rm\ \ \ u\sim v,\, v\sim w\:\color{#c00}\Rightarrow\: f(u)\approx f(v),\,f(v)\approx f(w)\:\color{#0a0}{\overset{(E)}\Rightarrow}\:f(u)\approx f(w)\:\color{#c00}\Rightarrow u\sim w$
Such relations are called (equivalence) kernels. One calls $\, \sim\,$ the $\,(\approx)\,$ kernel of $\rm\,f.$ 
Yours is the special case when $\,\approx\,$ is the equivalence relation of equality.
