What is the domain of definition of $f(x,y)=\frac{y-x}{1+\ln(x)}$? I have some confusion over the domain of definition of a function.
Clearly, in the case above, it must be true that $x>0$ in order for the function to be defined. Also there is a singularity when $x=e^{-1}$.
So am I correct in saying that the domain of definition is: $x>0$, $x\neq e^{-1}$ and $y\in \mathbb{R}$?
However, if we rearrange the function, we get: $$(1+\ln(x))f(x,y)=y-x$$
So $x=e^{-1}$ implies that $$0=y-e^{-1}$$
So this would imply that $f(x,y)$ takes infinitely many values at the point $(e^{-1},e^{-1})$
This is where my confusion lies - so does the point $(e^{-1},e^{-1})$ form part of the domain of definition?
 A: Strictly speaking, the expression for $f(x,y)$ is meaningless unless the precise domain of definition is provided. This is because a function, by definition, is a triple $(f,X,Y)$ where $X$ is the domain of definition, $Y$ is the range, and $f$ is a function.
However, mathematics is often not very strict, so that in "practice" we get an expression such as $f(x,y)$ for which we have to figure out the precise domain of definition. Of course, we must assume something about $(x,y)$. For example, here it is tacitly assumed that $(x,y)\in\mathbb{R}^2$. (but that's only because of convention; in truth, there is nothing to prevent an alien from interpreting $(x,y)$ as an element of the Cartesian product of any two sets). Therefore, we must find the largest possible subset of $\mathbb{R}^2$ for which the expression $f(x,y)$ is well defined. That will be our domain of definition. By inspection, we see that we must avoid nonpositive $x$ and such $x$ for which $\ln x=-1$. The first condition gives the set $A=\{(x,y)\in\mathbb{R}^2| x>0\}$ and the second condition gives $B=\{(x,y)\in\mathbb{R}^2| x\neq e^{-1}\}$. Thus, the domain of definition is $A\cap B$ - the set where both conditions are met.
