It seems to me a silly question, but I couldn't find the answer.

Consider the expression:


If $x = 1$, the result is $\frac{0}{0}$.

But we can change the expression to:

$\frac{(x+1)(x-1)}{(x-1)} = (x+1)$

than, if $x = 1$, result $= 2$

I'm wondering if $\frac{(x+1)(x-1)}{(x-1)}$ is in some sense a valid / invalid expression. It seem to be unnecessarily complex, and easily simplified. The unsimplified version creates a problem that shouldn't exist.

In linguistics, we could affirm it's wrong to say "the ball is not not black", meaning "the ball is black". But the double negation should be avoided. I'd argue there is "noise" in "the ball is not not black", and also in $\frac{(x^2-1)}{(x-1)}$.

Is there something as noise in mathematical expressions?

Edit: by noise I meant the unnecessary complexity that do not change the real meaning of the expression. But I’ve been instructed that the expressions are different (contrary to all the practical explanations I’ve heard from teachers, which for sure were more practical than firmly rooted in sound theory).

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    $\begingroup$ This is an interesting question but perhaps a bit too open ended for our format. What is your mathematical background? $\endgroup$
    – Pedro
    Commented Oct 3, 2023 at 12:17
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    $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ Commented Oct 3, 2023 at 12:17
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    $\begingroup$ Ignoring the concept of "noise" that you are trying to get across... this specific example is one of a removable discontinuity. Officially your original function is undefined at $x=1$ and so the allowable domain for the function can not include $x=1$ and so is a technically different function than the one you simplified it to (again, assuming we take a maximal domain for each) since it actually is defined at $x=1$. $\endgroup$
    – JMoravitz
    Commented Oct 3, 2023 at 12:29
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    $\begingroup$ Rest assured that this is not a silly question $\endgroup$
    – AakashM
    Commented Oct 3, 2023 at 13:14
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    $\begingroup$ There's no two ways about this: You wrote If $x=1$, the result is $\frac{0}{0}$, but that is not correct. What is correct is to write If $x=1$, the expression is undefined. $\endgroup$
    – Lee Mosher
    Commented Oct 4, 2023 at 15:42

7 Answers 7


The two expressions $(x^2 - 1)/(x-1)$ and $x+1$ are not equal. They do have the same value for all real $x \neq 1$, but at $x=1$ the former is undefined. In your proposed process of simplifying the former, you divide both the numerator and denominator by $x-1$, which, when $x=1$, amounts to trying to divide by zero. So your simplification is invalid. That’s why we would say that $$\frac{x^2-1}{x-1} = x+1 \bf\mbox{, when } x \neq 1.$$

As for the existence in mathematical notation of what you describe as noise, one might be tempted to consider the + sign that labels numbers as greater than zero to be unnecessary outside the teaching of negative numbers because, for instance, $x+1$ does equal $x+(+1)$ and $\cos\pi = \cos(+\pi)$ (which is to say, because +1 is 1 and $+\pi$ is $\pi$). But that depends on context. An author could legitimately choose to use a longer such version if their intent was to emphasize or disambiguate the constant’s sign. Likewise, although the 2 in something like $\root{2}\of{e}$ can generally be considered redundant, one might reasonably choose to include it to emphasize exactly which (fractional) exponent one means.

As another example, one answer to the question In how many ways can we pick 3 students from a class of 20 to receive ice cream cones? is 1140. But in an important sense, the answer $20 \choose 3$ is a better one. Although the two are equal and the former is simpler, the latter explains not only what the answer is—assuming one knows how to interpret binomial coefficients—but also why.

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    $\begingroup$ This doesn't really answer the question though? $\endgroup$
    – cliesens
    Commented Oct 4, 2023 at 11:07
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    $\begingroup$ @cliesens, I have added a paragraph to address the OP’s question head on. $\endgroup$ Commented Oct 4, 2023 at 11:35

The function \begin{align*} f : \mathbb{R} \setminus \{1\} & \longrightarrow \mathbb{R} \\ x & \longmapsto \frac{x^2-1}{x-1} \end{align*} can be extended to a function $g_y : \mathbb{R} \longrightarrow \mathbb{R}$ defined on all of $\mathbb{R}$ in many different ways, by choosing any value $y$ that you like and defining$$ g_y(x) = \begin{cases} f(x) & (x \neq 1), \\ y & (x = 1). \end{cases}$$

Your observation is that the particular choice $g_2$ is a much more natural (or "less noisy"?) choice to extend $f$.

Indeed, function $g_2$ is a much more natural extension of $f$, for two reasons:

  • geometrically because $$\lim_{x \to 1} f(x) = 2;$$
  • algebraically because if the simplification by $x-1$ is allowed then $$\frac{x^2-1}{x-1} = x+1.$$

I have two good news for you.

The first good news is that the concept of some functions being more "natural" than others has interested many mathematicians. There is a whole field of study of natural functions. In particular:

  • the discontinuity of $\frac{x^2-1}{x-1}$ at $x=1$ is called a "removable" discontinuity;
  • the discontinuity of $\frac{x}{ | x |}$ at $x = 0$ is called a "jump" discontinuity;
  • the discontinuity of $\frac{1}{x-1}$ at $x=1$ is called a "first-order pole";
  • the discontinuity of $\frac{1}{{(x-1)}^n}$ at $x=1$ is called an "$n$th-order pole";
  • the discontinuity of $\sin\left(\frac{1}{x}\right)$ at $x=0$ is called an "essential" discontinuity.

The second good news is that automatic manipulation of algebraic expressions has interested many computer scientists. There is a field of study of algorithms that manipulate algebraic expressions, usually called "symbolic expressions" to highlight the fact that they contain abstract symbols as opposed to just numbers; and many algorithms to automatically simplify expressions, which of course requires having an idea of what it means to be "more simple". In the software Sage I can just type ( (x**2 - 1)/(x - 1) ).simplify() and I get x+1, without having defined any value for x beforehand.

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    $\begingroup$ Small typo: $1/(x-1)$ does not have an "essential" discontinuity at $x=0$. $\endgroup$
    – Federico
    Commented Oct 4, 2023 at 17:29
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    $\begingroup$ Maybe it's different terminology in different areas of mathematics, but coming from complex analysis, I would not use the word "essential" for the singularity of $1/(x-1)$ at $x=1$. If a function $f(x)$ is undefined at some $x=x_0$, but there is some $n$th-order polynomial $p(x)$ such that $p(x)f(x)$ just has a removable singularity there, then $f(x)$ has a $n$-th-order pole at $x=x_0$. Only when there is no such polynomial, as in the case of $\sin(1/x)$, is the singularity called essential. $\endgroup$
    – Tengil
    Commented Oct 5, 2023 at 12:25
  • $\begingroup$ @Tengil Absolutely! $\endgroup$
    – Stef
    Commented Oct 5, 2023 at 12:32

You ask about "noise" in mathematics, but provide no definition. There are several that would make this an interesting question, even though it's not strictly about mathematics per se.

The particular example of a function with a removable discontinuity suggests that one can think of noise as an error term of some kind. If you think philosophically that you can only deal with rational numbers, not reals, then any rational approximation to, say, $\pi$ is "noisy". There's an error but it's one you can approximate. Then in your example, the expression $(x^2-1)/(x-1)$ makes no sense at $x=0$ but is a noisy approximation to $1$ when $x$ is near $0$. This interpretation would lead you to a way to rewrite the definition of a limit in terms of noise.

The "not not black" example suggests that you might think of noise as a kind of redundancy in writing mathematics: using more words than necessary. Some redundancy is really necessary when you think of mathematics as something that human beings reason about. We can't always convey what we mean in a minimal way and be understood. Nor should we. I think "not not black" has a different connotation than simply "black". In a mathematical proof it might provide a hint as to how we discovered the proof. There can be times when it's the right way to describe something even when we accept the law of the excluded middle.


I'm not qualified to answer questions about mathematics at all, but yours regarding [what you call] "noise" is really philosophical, and perhaps the more technical answers are missing this.

I agree with @EthanBolker that it sounds like you're talking about redundancy of representation. Often the tersest representation is not optimal for communicating, and dependent on what "extra" detail is included, different connotations can be implied as he says. I agree with that perspective of mathematics as a social endeavour, with its written representations ultimately being acts of human communication.

FWICT, there is no general way to ascertain what the "least redundant" representation of some bit of mathematics even is. Mathematical discovery seems to entail drawing equivalences between concepts or objects not previously known to be equivalent, and much of our collective knowledge about / the perceived power of mathematics amounts to such discoveries. That two apparently completely unrelated objects from different fields actually represent, in some sense, the "same thing".

I take from this that reductionism simply cannot apply to mathematics - there is no guaranteed direction of travel toward ever simpler representations until we "hit the bottom". One can only choose one's direction for exploration according to one's own present intent and needs. It's not just turtles all the way down, but in all directions.

This topic feels similar/adjacent to many others, to the extent I feel like my "no one true representation" claim has a name, and has already been confirmed. But if I start handwaving toward all that now without remembering said name, the crank status of this answer will be confirmed :)

  • $\begingroup$ It's a philosophical question, for sure, and although my question was about mathematics, and the technical answers were correct, this answer is was I was leaning more toward. We could assume there is noise in $2+2+2+x+x$, because it could be reduced to $2x+6$. The problem is if there is a better representation of the expression in question. But, at least in the example I gave, the two expressions seem to represent different things (at least when we talk about their respective domains). $\endgroup$ Commented Oct 5, 2023 at 9:03
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    $\begingroup$ @calebmiranda the question about simplification can be treated formally. It is similar to the word problem: given two words are they equivalent (under some definition of equivalence)? The world problem is undecidable. And so is the problem of "minimal expression" that you talk about here. So for example $x-x$ can be always reduced to $0$. But there cannot be a general method achieving this for all expressions. $\endgroup$
    – freakish
    Commented Oct 5, 2023 at 12:19
  • $\begingroup$ @calebmiranda There are contexts where $2x+6$ is better written as $2(x+3)$. In general, there isn't a unique "most-reduced" form of an expression. $\endgroup$
    – Nayuki
    Commented Oct 6, 2023 at 6:34

Perhaps by "noise" you are referring to the concept of "simplest form", as in, "This mathematical expression is noisy because it has not been reduced to its simplest form. It is full of unnecessary complexity."

An obvious illustration is

$f(x) = 5 + log(x) - log(x)$

Adding and subtracting the log is pointless. Using everyday language, a person could say, "That function has so much unnecessary noise in it that I can barely hear what it's trying to say!" A more mathematical way to say this is, "That function could be reduced to a simpler form."

BTW, eliminating those two log terms would not only make the expression of the function simpler, it would also have the happy side effect of giving the function a bigger range (i.e., all the real numbers could play instead of just the positive numbers).

  • $\begingroup$ True, but... what if the question is "find the domain of the function given by the expression..."? $\endgroup$ Commented Oct 5, 2023 at 8:21
  • $\begingroup$ @Martín-BlasPérezPinilla — I was answering the question as posed which, as I read it, is focused on the complexity of expressions. $\endgroup$
    – SlowMagic
    Commented Oct 5, 2023 at 15:24

I'm no mathematican, but $\frac{(x^2-1)}{(x-1)}$ is not simply a "more complicated version" of $x+1$, but a different function. The difference is that the original version excluded one point from the domain (I hope that's the correct phrase), while the "simplified" version does not.

Or maybe in a real-world example: You fill a full HD screen with while pixels, but set one pixel to black. At the first glance it might look like a completely while screen, but actually a completely white screen would be different (even when easier to create than the original one).

Completely independent of that is the question whether the original intent was to create a completely white screen or not.


Here is the visual difference between the two functions:

enter image description here

The functions are almost the same, except for a single point at $x=1$ where the function $(x^2-1)/(x-1)$ is undefined.


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