Question: Determine all pairs of integers A,B such that (m,n)∼(u,v)⟺m−An=Bu−v
is an equivalence relation on the set of all pairs of integers.
My attempt: I had worked out that there are 2 pairs of A, B that satisfies the above equivalence relation, namely (A, B) = (v/n, m/u) and (m/n, v/u). But it seems that it isn't correct.
Why it is wrong and how to solve the question? Many thanks!
Well it's reflexive so we must have $(m,n)\sim(m,n)$ or $m-An=Bm-n$ for all $m,n$. So $(1-B)m=(A-1)n$ for all $m,n$. If $1-B \ne 0$ then $\frac {A-1}{1-B} = \frac mn$ for all $m,n; n\ne 0$. But that's impossible as $\frac {A-1}{1-B}$ would be a constant. So $1-B=0$ and $B=1$ and $(A-1)n=0$ for all $n$. So $A-1=0$ and $A=1$.
Those are the only case that makes $\sim$ reflexive. So $(m,n)\sim(u,v) \iff m-n=u-v$.
So that is the only possible answer. But we have to show that that is an answer and that $(m,n)\sim(u,v) \iff m-n = u-v$ is an equivalence.
We have to show that is an equivalence relation. It's reflexive and $m-n= m-n$ for all $m,n$.
Is it symmetric? If $m-n = u-v \implies u-v = m-n$? Yes. So it is symmetric.
Is it transitive? If $m-n = u-v$ and $u-v = w-z$ does that imply $m-n=w-z$. Yes it does. so it is transitive.