Define the relation $\sim$ on $\mathbb{Z}\times\mathbb{Z}$ by $(a,b)\sim(c,d)$ if $a-c=b-d$. Show that $\sim$ is an equivalence relation. What is the equivalence class of $(1,2)$?
I'm not sure how to approach this since I've never had to deal with 4-tuples before.
On
After proving that $\sim$ is an equivalence relation, try to study $$(a,b)\sim (1,2) \Leftrightarrow a-1=b-2,$$ that is $$a-b=-2+1=-1.$$ So you are looking for pairs $(a,b)$ with $b=a+1$.
On
Maybe it would help to rewrite $a-c=b-d$ as $a-b=c-d$. Note that if $a=1$ and $b=2$, then $a-b=-1$. So the $(c,d)$ that are equivalent to $(1,2)$ are all $(c,d)$ such that $c-d=-1$.
Let's think of some: $(10,11)$; $(47,48)$: and so on. Also $(-17,-16)$, and so on.
Remark: It is not really $4$-tuples that you are dealing with here. It is ordered pairs (as usual), but this time it is ordered pairs of ordered pairs.
On
It’s no different from what you’ve done before; the notation is just a bit more complicated. For instance, to show that $\sim$ is symmetric, you have to show that if $\langle a,b\rangle,\langle c,d\rangle\in\Bbb Z\times\Bbb Z$, and $\langle a,b\rangle\sim\langle c,d\rangle$, then $\langle c,d\rangle\sim\langle a,b\rangle$. Go back to the definition of $\sim$ to see what this really means.
$\langle a,b\rangle\sim\langle c,d\rangle$ means that $a-c=b-d$; this is your hypothesis.
$\langle c,d\rangle\sim\langle a,b\rangle$ means that $c-a=d-b$; this is what you need to prove in order to show that $\sim$ is symmetric. And this is easy to do: $c-a=-(a-c)\overset{*}=-(b-d)=d-b$, where the starred step uses the hypothesis that $\langle a,b\rangle\sim\langle c,d\rangle$.
Now see if you can prove reflexivity and transitivity on your own.
The equivalence class of $\langle 1,2\rangle$ is the set of all $\langle a,b\rangle$ such that $\langle 1,2\rangle\sim\langle a,b\rangle$, i.e., such that $1-a=2-b$; solve this to find what $b$ must be in terms of $a$ in order for $\langle a,b\rangle$ to be in the equivalence class.
On
Hint $\rm\ \ (a,b)\sim (c,d)\smash[t]{\overset{\ def}{\iff}} a\!-\!b = c\!-\!d \iff f(a,b) = f(c,d)\ $ for $\rm\ f(x,y) = x\!-\!y$
More generally, suppose $\rm\: u\sim v\ \smash[t]{\overset{\ def}{\iff}}\, f(u) \approx f(v)\:$ for a function $\rm\,f\,$ and equivalence relation $\,\approx.\ $ Then the equivalence relation properties of $\,\approx\,$ transport to $\,\sim\,$ by using the definition:
reflexive $\rm\quad\ f(v) \approx f(v)\:\Rightarrow\:v\sim v$
symmetric $\rm\,\ u\sim v\:\Rightarrow\ f(u) \approx f(v)\:\Rightarrow\:f(v)\approx f(u)\:\Rightarrow\:v\sim u$
transitive $\rm\ \ \ u\sim v,\, v\sim w\:\Rightarrow\: f(u)\approx f(v),\,f(v)\approx f(w)\:\Rightarrow\:f(u)\approx f(w)\:\Rightarrow u\sim w$
Such relations are called (equivalence) kernels. One calls $\, \sim\,$ the $\,(\approx)\,$ kernel of $\rm\,f.$
For every $(a,b)\in\mathbb Z\times\mathbb Z$, $a-a=0=b-b$. Hence the relation is reflexive.
For every $(a,b),(c,d)\in\mathbb Z\times\mathbb Z$, if $a-c=b-d$, then $c-a=d-b$. Hence the relation is symmetric.
For every $(a,b),(c,d),(e,f)\in\mathbb Z\times\mathbb Z$, if $a-c=b-d$ and $c-e=d-f$, then $a-e=(a-c)+(c-e)=(b-d)+(d-f)=b-f$. Hence the relation is transitive.
The class of $(1,2)$ is simply the set of all ordered pairs of integers $(a,b)$ such that $a-1=b-2$. This means $\{(x,x+1):x\in\mathbb Z\}$.