# Is one form of a function more 'true' than another?

Here's a function: $f(x) = \frac{x^2}{x}$

Now, if we were to look for the $0$ value, we would end up with a division by zero situation.

By simplifying it to an equivalent function: $f(x) = x$, this is no longer a problem.

Does this mean that the latter function is, in a sense, a more 'true' form of the function? Do both of these forms have the same domain?

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I'm not a pro, but it seems to me that if two functions are not equivalent, then neither one is "truer", they're just different. You have the premise that the two functions are equal, but you then go on to show that they're not–shouldn't that be exactly the answer you're looking for? – zneak Oct 20 '13 at 16:55

Good question.

Firstly, let me say that the concept of an "implicit domain" is one of my pet peeves! The day we universally reject such antiquated conventions will be a glorious day for mathematics.

Thus, my answer is that we should never define a function with just with a formula; always define it by specifying a particular domain and codomain.

For example:

Let $f$ denote the unique function $\mathbb{R}\setminus\{0\} \rightarrow \mathbb{R}$ subject to the following condition. $$f(x) = \frac{x^2}{x}.$$

This is a (not-at-all ambiguous, and therefore perfectly acceptable) shorthand for the following, more long-winded formalization.

Let $f$ denote the unique function $\mathbb{R}\setminus\{0\} \rightarrow \mathbb{R}$ such that for all $x \in \mathbb{R}\setminus\{0\}$, we have the following. $$f(x) = \frac{x^2}{x}.$$

Now lets focus our attention on the expression $x$. There's (at least!) two ways of defining a function using this formula. One way defines a function $g : \mathbb{R} \setminus \{0\} \rightarrow \mathbb{R}$, the other a function $h : \mathbb{R} \rightarrow \mathbb{R}$. These are different functions. The first coincides with $f$; that is, $g=f$. The second does not.

That's basically all there is to it.

Actually, there's one more thing that should really be addressed. Both $f$ and $g$ (they're the same function, after all) can be extended in a unique manner to a continuous function $\mathbb{R} \rightarrow \mathbb{R},$ and this function turns out to be $h$.

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This was an enlightening answer. Thanks for looking up this relatively old question. :) – Jimmy C Jan 12 '14 at 10:00
@JimmyC, you're welcome, although I actually didn't look it up. Questions automatically bounce to the top of the list every so often, so perhaps that is what happened. – goblin Jan 12 '14 at 10:22

The function $f(x) = \dfrac{x^2}{x}$ has largest domain $\mathbb{R}\setminus\{0\}$ whereas the function $g(x) = x$ has largest domain $\mathbb{R}$. For all $x \neq 0$, $\dfrac{x^2}{x} = x$; as $g|_{\mathbb{R}\setminus\{0\}} = f$ we can say that $g$ is an extension of $f$. But for any $a \in \mathbb{R}$,

$$h_a(x) = \begin{cases} x &\ \text{if}\ x\neq 0\\ a &\ \text{if}\ x = 0 \end{cases}$$

is also an extension of $f$. However, there is only one way to extend $f$ to a continuous function, and that extension is $h_0 = g$.

In the context of your question, $f(x) = \dfrac{x^2}{x}$ and $g(x) = x$ are not 'equivalent functions' unless we take as understood that $g$ has domain $\mathbb{R}\setminus\{0\}$. The function $g$ is not a 'truer' form of $f$, but rather its unique continuous extension.

You may think that this is incredibly pedantic; why don't we automatically extend functions continuously, just like we automatically take the largest domain of a function if it is not specified? One reason is that a continuous extension doesn't always exist. For example, $f(x) = \dfrac{1}{x}$ has largest domain $\mathbb{R}\setminus\{0\}$ but has no continuous extension defined on $\mathbb{R}$.

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No. The domain of the $f(x)$ you gave does NOT include $0$.

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$\frac{x^2}{x}$ and $x$ are equal as polynomials, but not as functions.

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but $\frac{x^2}{x}$ is not a polynomial. – Leox Oct 20 '13 at 17:01
@ Leox: Can I divide the polynomial $x^2$ by $x$? Will I get a polynomial as result of division? – Boris Novikov Oct 20 '13 at 17:47
yes, of cource. AFTER divison you will get a polynomial but before division you have a rational expression. – Leox Oct 20 '13 at 18:09
@ Leox: OK. But a rational expression differs from a function. The formers constitute a field, and the ring of polynomials is embedded in this field. So $\frac{x^2}{x}$ and $x$ are identified. – Boris Novikov Oct 20 '13 at 19:32
@Michael Albanese: I think it's a matter of opinion and agreement. – Boris Novikov Oct 21 '13 at 6:39