Is a simplified function the same as the original?


Let $f(x) = \frac{ax}{x}$, and $g(x) = a$

where $a$ and $x$ are real numbers.

Does $g$ = $f$?

  • 4
    $\begingroup$ One has a "hole" (a removable discontinuity, that is), and the other doesn't have one. $\endgroup$ – J. M. is a poor mathematician May 16 '12 at 5:40

A common convention for real functions of real variables is that of the "natural domain of a function":

If a function $f$ is given by a formula, and a domain is not explicitly specified, then the domain is understood to be the "natural domain of $f$", which means, all real numbers for which the formula makes sense and yields a unique output.

Another common convention for this type of function is that two functions are considered to be equal if and only if they have the same domain, and they have the same value at each element of their domain.

If we take both conventions, then $$f(x)=\frac{ax}{x}\quad\text{and}\quad g(x) = a$$ are not equal, because the domain of $f(x)$ consists of all real numbers except $0$; whereas the domain of $g(x)$ is all real numbers, including $0$. Since $f$ and $g$ have different domains, they are not equal as functions.

On the other hand, $$f(x)=\frac{ax}{x}\qquad\text{and}\quad g(x)=a, x\neq 0$$ are equal, because "$x\neq 0$" after $g(x)$ is specifying the domain explicitly, which means that $g(x)$ now has the same domain as $f(x)$ (all $x\neq 0$), and they have the same value at every point in their common domain.

If you are operating under these conventions (which are the usual conventions in, for example, most calculus textbooks in the U.S. that I am familiar with), then when you simplify the formula for a function, you should check to see if you haven't implicitly changed the domain. Some simplifications don't matter; e.g., $f(x) = (x-1)^2-1$ and $g(x) = x^2-2x$ are equal as functions (same domain, all real numbers; and same value at every real number). Others do (such as the one you did).


It depends on what you take to be the domain of each. If they’re given the same domain (which of course cannot include $0$), then they’re the same function. For example, considered as functions with domain equal to the set of positive real numbers, $f=g$.

If you give them different domains, however, they are no longer the same function, though it is still true that $f(x)=g(x)$ for all $x\in\operatorname{dom}f\cap\operatorname{dom}g$, i.e., for all $x$ for which $f$ and $g$ are both defined.

In a first-year calculus course it’s often assumed that a function has as its implied domain the largest subset of $\Bbb R$ on which it is defined. With that convention $f$ and $g$ have different domains: the domain of $g$ is all of $\Bbb R$, but the domain of $f$ is just $(-\infty,0)\cup(0,\infty)$. Technically, then, they are different functions.


In contemporary mathematics, a function should always be defined by means of: (1) a set called "domain of definition"; (2) a set called co-domain; (3) a "rule" or a "law" that sends every element of the set (1) to one (and only one) element of the set (2). If you do not like to think in terms of subsets of cartesian products, a function is a triple. As you have been told, calculus is abundant of functions as they were used by pioneers: just rules.

This can be tolerated at very low levels of teaching, but should be replaced by a more precise definition rather soon. And we, teachers and professors, should stop asking our student "to find the domain of definition of the following function", since there is no function without a domain.


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