Let $R$ be the relation on $X=\mathbb Z\times \mathbb N$ such that $(a,b)R(c,d)$ if and only if $ad=bc$.
Define an operation $\bullet$ on $X/R$ as follows: for $x=(a,b)$ and $y=(c,d)$ let: $$[x]\bullet[y]=[(ad+bc,cd)]$$
Is $\bullet$ well defined?
I'm struggling to understand what this question is asking and how to go about answering it. Any input appreciated. Thanks!
If $X$ is a set and $R\subseteq X\times X$ is an equivalence relation on it then we can define operations on $X/R$ (i.e. the collection of equivalence classes) by making use of representatives of the classes. If $[x]$ denotes the class represented by $x\in X$ and $f:X\times X\rightarrow X$ then we can 'define' (I would use the term 'predefine') an operation $\bullet$ by stating: $$[x]\bullet[y]:=[f(x,y)]$$ But wait a minute..., if $x'Rx$ and $y'Ry$ then we have $[x]=[x']$ and $[y]=[y']$ so we need: $$[f(x,y)]=[x]\bullet[y]=[x']\bullet[y']=[f(x',y')]$$ This shows that we can only speak of a well defined operation if it is guaranteed that under these conditions $[f(x,y)]=f(x',y')]$.
Shortly the definition is okay if we have: $$x'Rx\wedge y'Ry\implies f(x,y)Rf(x',y')$$
Proving that this is indeed the case is the same as answering the question "is $\bullet$ well defined?" with a justified "yes".
Proving that this is not the case is the same as answering the question "is $\bullet$ well defined?" with a justified "no".
So that's how you are supposed to deal with that question.