# What is the domain of a division of functions?

This question is about real functions of real variables.

I think that, in general, if the domain of some function $f(x)$ is A, and the domain of another function $g(x)$ is B, then the domain of $(f/g)(x)$ is A$\cap$B and where $g\neq0$.

Now, what happens if I have something like $f(x)=2$, $g(x)=1/x$? In this case, $(f/g)(x)=2x$, which seems to be defined for all real numbers. But my statement above (which I think is correct in general) implies that $x=0$ is not allowed. So I'm conflicted.

Can somebody tell me what the domain of $(f/g)(x)$ is in this case? Is it all real numbers, or all real numbers except $0$?

Thanks.

• Sorry, I don't understand your comment. If I have a function $f$ that is only defined for $x=6$ and another one ($g$) only defined for $x=3$, then $(f/g)$ is not defined anywhere. I don't see the problem with that. Commented Aug 15, 2018 at 23:39
• I agree with the first answer given, which reiterates your second sentence. This will clear up your 3rd block of text: $2/(1/x)$ and $2x$ are different functions. So if I give you the function $2x$, then that's that. And if I give you the function $2/(1/x)$, that's good too. They aren't the same. But if I give you $2x$ and tell you I want the domain to be all reals minus $0$, now the two functions are the same. Without stating the domain, it's implied $2x$ is over all the reals. I could have told you that $2x$ is over the domain $6 < x \leq 7$, or whatever else Commented Aug 15, 2018 at 23:47
• The domain of $f/g$ excludes $0$ based on the definition of $f/g$ you gave. But do not take this too seriously. Math is about communication of ideas more than definitions, so if it is inconvenient to exclude $0$, you can "fill it in" - as long as you do this in a way that is completely clear to yourself and whoever else is interested in your work. Commented Aug 16, 2018 at 0:58
• This is correct if you think about them as functions in the set-theoretic sense. But for sufficiently well-behaved families of functions (rational, analytic) you can actually ignore those singularities and it makes perfect sense, even if pointwise you have silly stuff like division by zero. Or, if you treat them as functions on a measurable space, you can ignore what happens on a set of measure $0$, and that's fine, too. In each case, you can think of dividing equivalence classes of functions, rather than actual functions. Commented Aug 16, 2018 at 8:41

Your mistake is in thinking that $$\frac2{1/x}\quad\hbox{and}\quad 2x$$ are always equal. They're not. To carefully prove that they are equal, we have $$\frac2{1/x}=\frac2{1/x}\,1=\frac2{1/x}\frac xx=\frac{2x}1=2x\ .$$ But this is not correct when $x=0$, because $\frac00$ is not equal to $1$. So we have to consider $x=0$ separately. In that case we have $2x=0$, but $$\frac2{1/x}=\frac2{1/0}=\frac2{\hbox{nonsense}}=\hbox{nonsense}\ .$$ So in your example, the domain of $f/g$ must exclude $0$.

• I loved your answer; super simple and clear. Great name too, btw. Commented Aug 16, 2018 at 1:25
• @DavidS: Unfortunately, this answer is wrong. mfl's answer is correct. The error is that "$\frac2{1/x}$" is not even defined when $x = 0$, simply because "$1/x$" is not defined when $x = 0$. One cannot manipulate expressions as if they are defined, when they are actually not. Commented Aug 16, 2018 at 10:50
• @user21820 Basically, that's exactly what I said in my answer. What did you think I meant by "$\frac2{1/0}=\cdots=\hbox{nonsense}$"? Just asking. Commented Aug 16, 2018 at 23:55
• @David: I object to your explanation, especially the way it implies that the problem is that "$0/0$ is not equal to $1$". But that is not the real problem. "$1/x$" and anything containing it is already meaningless (or syntactically malformed, or a type error) if $x = 0$. For the same reason, your second chain of equalities is not mathematical equalities, but rather reductions. In particular, the correct reasoning is that for "$\frac2{1/x}$" to be well-defined we must have "$1/x$" well-defined, which requires $x \ne 0$. Note the use of quotes to distinguish expressions from values. Commented Aug 17, 2018 at 5:55
• Your second chain of equalities is not mathematical equalities - well of course not, and that is exactly my point. The problem that many beginners have is that they think $0/0$ is equal to $1$. Any question on this site needs to be answered on a level appropriate to the question, and judging by the OP's comment, I have successfully done that. Commented Aug 17, 2018 at 6:23

If the domain of $f$ is $A$ and the domain of $g$ is $B$ then the domain of $f/g$ is $$A\cap B\setminus\{x:B|g(x)\ne 0\}.$$ (Of course, we must assume that $A\cap B\setminus\{x:B|g(x)\ne 0\}\ne \emptyset$. In other case $f/g$ doesn't make sense.)

In your example, $f(x)=2$ and $g(x)=1/x.$ We have that

$$\dfrac{f(x)}{g(x)}=2x, \forall x\in\mathbb{R}\setminus\{0\}.$$ Why? Note that $g(0)$ doesn't exist. So we can't consider

$$\dfrac{f(0)}{g(0)}.$$

It is all real numbers without zero i think. Check out this similar example: $$F(x)=\dfrac{(x-2)(x-1)}{(x-2)}$$ You can simplify this fraction to $F(x)=(x-1)$ ONLY when 2 is excluded from the domain So we say that $F$ is defined $$\forall x\in\mathbb{R}\setminus\{2\}.$$ Even though $F(x)=(x-1)$

Usually you define division of functions $$f:A\to\mathbb R$$ and $$g:B\to\mathbb R$$ as pointwise division, that is, $$(f/g)(x)=f(x)/g(x),$$ which obviously is only defined where $$g(x)\ne 0$$. Therefore the domain of the quotient is $$D=\{x\in A\cap B\mid x\ne 0\}$$.

However while this is the common choice, it is not the only possible choice. Another possible choice for continuous functions might be to define $$f/g$$ as $$(f/g)(x) = \lim_{y\to x} f(y)/g(y)$$ with the domain being all points in $$\mathbb R$$ where this limit exists.

With that definition and $$f(x)=2$$ (with domain $$\mathbb R$$) and $$g(x)=1/x$$ (with domain $$\mathbb R\setminus\{0\}$$), you would indeed find that $$(f/g)(x) = 2x$$ with domain $$\mathbb R$$.