Operations that are not well defined A basic example of an operation that is not well defined is the following: 
Let $r,s \in \mathbb{Q}$, then there exists $a,b,c,d \in \mathbb{Z}$ and $b,d \neq 0$ such that $r=\dfrac{a}{b}$ and $s=\dfrac{c}{d}$.
Define: 
\begin{align}
\oplus: \mathbb{Q} &\times \mathbb{Q} \rightarrow \mathbb{Q}\\
\dfrac{a}{b} &\oplus \dfrac{c}{d} = \dfrac{a+c}{b+d} 
\end{align} 
The reason why the operation is not well defined is clear, consider:
$r=\dfrac{2}{3}$ and $s=\dfrac{3}{4}$ then $r \oplus s=\dfrac{2+3}{3+4}=\dfrac{5}{7}$. 
By other hand, also $r=\dfrac{4}{6}$ and $s=\dfrac{9}{12}$ so $r \oplus s=\dfrac{4+9}{6+12}=\dfrac{13}{18}$
so $r \oplus s \neq r \oplus s$, contradiction.
A way to avoid the "not well defindness" of the orpeation is based on unique representation of a rational number in the form $\dfrac{p}{q}$ where $p \in \mathbb{Z}$, $q \in \mathbb{N}$ and $g.c.d.(p,q)=1$. Thus redefine $\oplus$ in terms of this unique representation and all works fine.
This is a situation where we can avoid the "not well definedness" using a certain property (unique representation of the rational numbers in this way)
I am interested in knowing what alternatives are known, so that an operation that initially was not well defined later can be redefined using a certain property and therefore be well defined.
In other words, I would like to know examples where, given an operation that initially is not well defined, some property has been used so that the operation is now well defined
thank you very much
 A: What is "really" $\mathbb{Q}$? $\mathbb{Q}$ are the pairs $(p,q) \in \mathbb{Z}\times\mathbb{Z}$ with the equivalence relation $(p,q) \equiv (r,s) \iff ps = rq$, so $\mathbb{Q} = \mathbb{Z}\times\mathbb{Z}/\equiv$. In algebra, there is the similar concept of localization.
But the concept of well definedness doesn't appear only in algebra: it appears when we talk about quotient sets and equivalence classes: we have two sets $A,B$ and an equivalence relation $\sim$ on $A$. We want to define a function $f:A/\sim\rightarrow B$ but defining it directly might be difficult. If we define a function $g:A \rightarrow B$ such that $a_1 \sim a_2 \Rightarrow g(a_1) = g(a_2)$ then we are done! If $\bar{a}$ is the equivalence class of $a$, define $f(\bar{a}) = g(a)$.
Everytime we can choose a representative of an equivalence class, we can define a function on $A/\sim$ from a function on $A$ (like in your example), but it might be difficult to prove that such a function is a homomorphism, continuous, ... 
A: Look at functions such as $\sqrt{}$ and $\log$ on the complex numbers.  To make them unique you need to define branches.
A: Consider the division of rational fractions. Let $k$ be a field and let $f(g),g(x)\in k(x)$, with $g(x)\neq0$. What is $\frac{f(x)}{g(x)}$? Well, it is just the map $x\mapsto\frac{f(x)}{g(x)}$, right?! Well, no, because then it would not be true that $\frac xx=1$ (the domain of $\frac xx$ would be $k\setminus\{0\}$). So, we change the meaning of “$=$”: we say that $\frac{f(x)}{g(x)}=\frac{f^\star(x)}{g^\star(x)}$ when $f(x)g^\star(x)=f^\star(x)g(x)$. Then, yes, $\frac xx=1\left(=\frac11\right)$, and everything works just fine.
