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What are some examples of notations and words in maths which have been overused or abused to the point of them being almost completely ambiguous when presented in new contexts?

For instance, a function $f$:

$f^{-1}(x)$ can be an inverse and a preimage and sometimes even $\frac{1}{f(x)}$.

$f^2(x)$ can be $(f\circ f)(x)$ and $(f(x))^2$.

$f^{(2)}(x)$ on the other hand, is the second derivative, even though adding parentheses to a number usually does nothing.

And for some functions the parentheses for the argument are omitted: $f\:x = f(x)$.

So how should $f^{(2-3)}(x)$ be interpreted? $f^{(-1)}$, an integral of $f$? or a composition, $\left(f^{(2)}\circ f^{(-3)}\right)(x)$? Or just $f^{2-3}(x) = f^{-1}(x)$?

Another example is mathematicians notorious use of the word normal to describe... normal things?

Using similar symbols and expressions for different things is unavoidable, but it can create some ambiguity when first introduced to their other uses.

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Frank Vel
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    $f^{-1}$ can be $\frac{1}{f}$. When you look at rings of (continuous, maybe) real or complex-valued functions on a space $X$, then it's natural to denote the multiplicative inverse of $f$ (if it exists, i.e. $f(x)\neq 0$ for all $x\in X$) by $f^{-1}$. – Daniel Fischer Nov 16 '14 at 14:27
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    I hate it when symbols $\infty$, $\omega$ and $\aleph_0$ are misused. – user2345215 Nov 16 '14 at 14:31
  • @DanielFischer Didn't realise that; it's been corrected. – Frank Vel Nov 16 '14 at 14:37
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    "nice" or "good" are used (depending on the context) to ensure that the setting is sufficient to prove theorems. Also I don't like "properly" to much. In german it is even worse. – Daniel Valenzuela Nov 16 '14 at 14:44
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    There's "canonical form", which means exactly what it says, but what it says is "the form mathematicians have chosen to be canonical, whatever that is for this particular thing." – David K Nov 16 '14 at 14:52
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    Technically there is no unambiguous way to write plain scalar multiplication. * is function convolution, $\cdot$ is dot product, $\times$ is cross product, and just putting the terms next to each other could be function application. I suppose you could divide by the reciprocal, but that is so evil we'll forget I said it. – hacatu Nov 16 '14 at 18:17
  • In linear algebra: properties of linear transformations vs. the matrices themselves. For example, kernel vs. nullspace, etc. – wchargin Nov 16 '14 at 18:53
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    See also http://mathoverflow.net/questions/7389/what-are-the-most-overloaded-words-in-mathematics. – lhf Nov 17 '14 at 00:30
  • I have never seen a case where $f(x)$ without the use of parentheses isn't using $xf=f(x)$ –  Nov 17 '14 at 02:22
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    Most of the examples listed are perceived as problematic only because for some reason which I cannot even imagine people seem to think that a notation can have one meaning, or that there is anything that trumps convenience... The only possible sin with respect to notation is not being explicit about what one means with it. There is no sacrosant association between upper indices and exponentiation, say... – Mariano Suárez-Álvarez Nov 17 '14 at 03:00
  • Relevant: Strangest notation? – MJD Nov 17 '14 at 03:16
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    Moreover, the term «abuse of notation» is very often misunderstood: there is nothing wrong with it. The very act of announcing that one is going to make an abuse of notation dissolves whatever problem might be imagined. In fact, these «abuses» are often part of the driving force behind the creating of mathematical concepts: the evolution of the notion of congruence between integers to that of the actual quotient set is an amazing example of this. – Mariano Suárez-Álvarez Nov 17 '14 at 03:59
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    I wouldn't say different meanings of a superscript is an inconsistency or an abuse as much as it is an overuse. Using the superscript $-1$ to mean both function inverse and multiplicative inverse seems perfectly consistent in the context of abstract algebra. Both represent inverses of group elements in the group of invertible functions and the group of nonzero real numbers, for example. – Randy E Nov 17 '14 at 20:22
  • I think "ambiguity" is a better term than "inconsistency" for many of the examples given. – Randy E Nov 17 '14 at 20:40
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    "That is, ...." – Yiyuan Lee Nov 18 '14 at 03:08
  • @pbs, that is neither an abuso of notation nor a mistake: it is simply the fact that in most contexts the notation log is reserved for the principal branch of the logarithmic function. – Mariano Suárez-Álvarez Nov 19 '14 at 05:16
  • @Mariano Suárez-Alvarez The notation $\text{Log}$ is also sometimes reserved for the principal branch, often when $\log$ for the (general) multi-valued function is used. In whichever context, I've seen this cause confusion in obtaining results. – pshmath0 Nov 19 '14 at 09:06
  • But the confusion is not due to a problem with the notation itself but to either the failure to be explicit about if a specific branch is intended or which branch is intended. – Mariano Suárez-Álvarez Nov 19 '14 at 09:30
  • Since I used $\log$ instead of $\text{Log}$ and I did not specify a branch then to me this is clearly a mistake. – pshmath0 Nov 19 '14 at 09:47
  • What about saying that something is well defined? – Bman72 Nov 20 '14 at 19:17
  • @Mariano Suárez-Alvarez Is not your position (especially in the penultimate sentence) a little bit extreme? I mean, don't you think that "good" notation can help to understand, or even study and solve a mathematical problem? Or, if you prefer, don't you think that "bad" notation (although precisely established) can be a serious obstacle in doing math? (try, for example, to do arithmetic with roman numerals). – suitangi Nov 20 '14 at 21:49
  • @PBS, it is your mistake: sure. But it is not a problem with the notation itself, which has nothing to do with you misusing it... It is not evrn an abuse of notation in any sense: it is your error – Mariano Suárez-Álvarez Nov 20 '14 at 23:04
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    I hate it when $\tan^{-1}$ is used for $\arctan$. – hkBattousai Nov 23 '14 at 11:57
  • If you ask me, pretty much the entirety of Differential Geometry. Einstein's sum notation can take a while to get used to. – IAmNoOne Dec 05 '14 at 08:49

25 Answers25

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  1. 'The function $f(x)$'. No, the function is $f$.
  2. Let $f$ and $g$ be real differentiable functions defined in $\mathbb R$. Some people denote $(f\circ g)'$ by $\dfrac{\mathrm df(g(x))}{\mathrm dx}$. Contrast with the above. I discuss this in greater detail here.
  3. The differential equation $y'=x^2y+y^3$. Just a minor variant of 1. Correct would be $y'=fy+y^3$ where $f\colon I\to \mathbb R, x\mapsto x^2$, for some interval $I$.
  4. This is one I find particularly disgusting. "If $t(s)$ is a function of $s$ and it is invertible, then $s(t)$ is the inverse", lol what? The concept of 'function of a variable' isn't even definable in a satisfiable way in $\sf ZFC$. Also $\left(\frac{\mathrm dy}{\mathrm dx}\right)^{-1}=\frac{\mathrm dx}{\mathrm dy}$. Contrast with 1.
  5. In algebra it's common to denote the algebraic structure by the underlying set.
  6. When $\langle \,\cdot\,\rangle$ is a function which takes sets as their inputs, it's common to abuse $\langle\{x\} \rangle$ as $\langle x\rangle$. More generally it's common to look at a finite set $\{x_1, \ldots ,x_n\}$ as the finite sequence $x_1, \ldots ,x_n$. This happens for instance in logic. Also in linear algebra and it's usual to go even further and talk about 'linearly independent vectors' instead of 'linearly independent set' — this is only an abuse when linear (in)dependence is defined for sets instead of 'lists'.
  7. 'Consider the set $A=\{x\in \mathbb R\colon P(x)\}$'. I'm probably the only person who reads this as the set being the whole equality $A=\{x\in \mathbb R\colon P(x)\}$ instead of $A$ or $\{x\in \mathbb R\colon P(x)\}$, in any case it is an abuse. Another example of this is 'multiply by $1=\frac 2 2$'.
  8. Denoting by $+$ both scalar addition and function addition.
  9. Instead of $((\varphi\land \psi)\to \rho)$ people first abandon the out parentheses and use $(\varphi\land \psi)\to \rho$ and then $\land$ is given precedence over $\to$, yielding the much more common (though formally incorrect) $\varphi\land \psi\to \rho$.
  10. Even ignoring the problem in 1., the symbol $\int x\,\mathrm dx=\frac {x^2}2$ is ambiguous as it can mean a number of things. Under one of the common interpretations the equal sign doesn't even denote an equality. I allude to that meaning here, (it is the same issue as with $f=O(g)$).
  11. There's also the very common '$\ldots$' mentioned by Lucian in the comments.
  12. Lucian also mentions $\mathbb C=\mathbb R^2$ which is an abuse sometimes, but not all the time, depending on how you define things.
  13. Given a linear map $L$ and $x$ on its domain, it's not unusual to write $Lx$ instead of $L(x)$. I'm not sure if this can even be considered an abuse of notation because $Lx$ is meaningless and we should be free to define $Lx:=L(x)$, there's no ambiguity. Unless, of course, you equate linear maps with matrices and this is an abuse. On the topic of matrices, it's common to look at $1\times 1$ matrices as scalars.
  14. Geometers like to say $\mathbb R\subseteq \mathbb R^2\subseteq \mathbb R^3$.
  15. Using $\mathcal M_{m\times n}(\mathbb F)$ and $\mathbb F^{m\times n}$ interchangeably. On the same note, $A^{m+ n}=A^m\times A^n$ and $\left(A^m\right)^n=A^{m\times n}$.
  16. I don't know how I forgot this one. The omission of quantifiers.
  17. Calling 'well formed formulas' by 'formulas'.
  18. Saying $\forall x(P(x)\to Q(x))$ is a conditional statement instead of a universal conditional statement.
  19. Stuff like $\exists yP(x,y)\forall x$ instead of (most likely, but not certainly) $\exists y\forall xP(x,y)$.
  20. The classic $u=x^2\implies \mathrm du=2x\mathrm dx$.
  21. This one disturbs me deeply. Sometimes people want to say "If $A$, then $B$" or "$A\implies B$" and they say "If $A\implies B$". "If $A\implies B$" isn't even a statement, it's part of an incomplete conditional statement whose antecedent is $A\implies B$. Again: mathematics is to be parsed with priority over natural language.
  22. Saying that $x=y\implies f(x)=f(y)$ proves that $f$ is a function.
  23. Using $f(A)$ to denote $\{f(x)\colon x\in A\}$. Why not stick to $f[A]$ which is so standard? Another possibility is $f^\to(A)$ (or should it be square brackets?) which I learned from egreg in this comment.
The Amplitwist
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Git Gud
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    C'mon haters, I welcome your down votes. – Git Gud Nov 16 '14 at 14:49
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    Let $f \colon \mathbb{R} \to C(\mathbb{R},\mathbb{R})$. The function $f(x)$ ... – Daniel Fischer Nov 16 '14 at 15:00
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    It took me a while to "get" the last item; then I realized "$A={x\in \mathbb R\colon P(x)}$" is technically an equation, not a set, which is your point, right? – David K Nov 16 '14 at 15:03
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    Loved every one of these, although I'm guilty of number 7 myself. I suppose that the "right" way to write it is "Consider the set $A$, ${x \in \mathbb R : P(x) }$, ... ", so that the set notation is an appositive for the "$A$"...but no one would understand that, either. :( – John Hughes Nov 16 '14 at 15:21
  • @DanielFischer I thought about that, but as with the rest of my points, I preferred not to enter into too much detail. – Git Gud Nov 16 '14 at 15:29
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    @JohnHughes Usually I write 'Consider the set $A$ where $A=\text{Whatever}$'. For whatever reason what you suggest had never occurred to me. I will use it from now on. – Git Gud Nov 16 '14 at 15:38
  • @DavidK That's my point, yes. I'm very syntactical in the way I think. To me it's more natural to read as explained on 7. I always have to translate it in my brain to what people usually want to say. – Git Gud Nov 16 '14 at 15:40
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    Please, Git Gud, don't treat my note as a suggestion, unless your goal is obfuscation; your "Consider the set $A$, where $A = ...$," is far more readable for the rest of us, and also works for your "peculiarly syntactic" reading. – John Hughes Nov 16 '14 at 15:42
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    $\ldots$ and let us not forget the classic $\mathbb{C=R^2}$. – Lucian Nov 16 '14 at 16:03
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    Proposing a (slight) remedy, rather than write "the function $x^2$", I like to write "the function $x \mapsto x^2$". It's true that the domain and codomain are still left to the reader's inference, but at least we know whether we're talking about a quadratic or about the constant function $t \mapsto x^2$. :) – John Hughes Nov 16 '14 at 16:12
  • @JohnHughes I use that notation a lot, but I avoid using it equations. I don't really wanna write $y'=(x\mapsto x^2)y$, but I do sometimes. – Git Gud Nov 16 '14 at 16:14
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    Agree completely. But it's great to use in English sentences, or in function descriptions like "$f : A \to B : x \mapsto x + 1$." The more functional programming I do, the more that things like what you wrote seem reasonable to me, though. – John Hughes Nov 16 '14 at 16:17
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    "The concept of 'function of a variable' isn't even definable in a satisfiable way in $\sf{ZFC}$" Do you have a reference that talks more about this?

     – Christian Chapman Nov 16 '14 at 16:47
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    @enthdegree No, it's just me saying. The reason why I say it is basically because if you want to introduce a variable, then the function will be dependent of that variable. Thus, given two different variables $x$ and $t$, we'll have $x\mapsto \sin(x)$ different from $t\mapsto \sin(t)$. It's not all that problematic, one could argue that we can focus on the range of the functions (those are equal), but why would we even do that? This is why I claim it's not doable in a satisfying way. A lot of hassle for no reason. – Git Gud Nov 16 '14 at 16:55
  • Oh ok, makes sense! I was just wondering if there was some deep technical reason. – Christian Chapman Nov 16 '14 at 16:57
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    I've always interpreted constructions like #7 to mean "Consider the set $A$, which equals ${ x \in \mathbb{R} : P(x) }$," where the equals sign itself introduces an appositive. In my mind, $``="$ is a word that takes on two grammatical roles, acting as a verb ($2+2$ equals $4$) or a subordinating conjunction (as above), depending on the context. – David Zhang Nov 16 '14 at 17:20
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    This interpretation also makes chained relational operators read quite nicely, so that "observe that $x_0 \in I \subset \mathbb{R}$" reads as "observe that $x_0$ is a member of $I$, which is a subset of $\mathbb{R}$." – David Zhang Nov 16 '14 at 17:28
  • @DavidZhang Yes, that's how people usually read it and I can understand why one would such expressions. However mathematics must be parsed 'all at once' and this is why I claim it is an abuse. – Git Gud Nov 16 '14 at 17:36
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    1 is done SO, SO, SO often that it is SO annoying. Actuarial textbooks especially. Ugh. – Clarinetist Nov 16 '14 at 19:00
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    It irks me when a roman 'd' is used for the differential in a mathematical setting. – Jon Nov 16 '14 at 19:03
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    @Jon: Do you prefer the Greek $\delta$ or the German Blackletter $\frak d$? :-) – Asaf Karagila Nov 16 '14 at 20:52
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    Using $f(x)=O(x^2)$ is certainly an abuse of notation similar to #10 on your list, since $O(x^2)$ is a set of functions. (This is, of course, ignoring that $f(x)$ and $x^2$ refer to $f$ and $x\mapsto x^2$, not to the evaluation thereof, in a similar abuse to #2) – Milo Brandt Nov 16 '14 at 21:14
  • @Meelo Yes, I added your example to that point since it is something I think people are more familiar with. – Git Gud Nov 16 '14 at 21:23
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    Wow, Git Gud, it must be really painful for you to read any mathematics at all! How do you stand it? – TonyK Nov 16 '14 at 22:26
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    @TonyK I did suffer in my first year. I took a liking to logic because of this. Also algebra had less abuse so I liked it more. As time went by and my mathematical maturity grew, I learned how to deal with it. Now it's easy for me to translate stuff into something I understand. – Git Gud Nov 17 '14 at 00:11
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    What would make this answer much more interesting is if you could show which of these abuses of notation are outright wrong rather than merely un- or ill-defined. For example, #8 might make you freak out because you haven't seen functions being added with +, but if you just "play along", it can be used consistently, without leading to contradictions. You might call it an abuse or find it otherwise undesirable, but it's not wrong (in the sense of leading to incorrect conclusions), and (IMO) not as interesting for the purposes of this question. – user541686 Nov 17 '14 at 00:12
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    @Mehrdad You have a point there. – Git Gud Nov 17 '14 at 00:14
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    Btw, what's wrong with saying "linearly independent vectors"? Two vectors are linearly independent iff neither is a linear combination of the other, three vectors are linearly independent iff neither is a linear combination of the others, etc... – user541686 Nov 17 '14 at 02:15
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    @Mehrdad You can interpret it that way if you want, but in any expository text the concept will be defined for sets, not for sequences of vectors. – Git Gud Nov 17 '14 at 02:22
  • @GitGud: I don't have a textbook handy, but just looking at Wikipedia, I see the following: the statement "In general, n linearly independent vectors are required to describe any location in n-dimensional space." and the caption "Linearly independent vectors in R³". – user541686 Nov 17 '14 at 02:24
  • @Mehrdad I just browsed through the three first lecture notes on linear algebra I found on Google. One of them actually defines it for lists of vectors. This is surprising to me, regardless of that it seems that indeed this approach is used. I don't like it because you need to deal with the empty list separately, but that doesn't really matter. I will consider removing that part of my answer. – Git Gud Nov 17 '14 at 02:34
  • @Mehrdad I didn't remove it, but I corrected it according to what you brought up. – Git Gud Nov 17 '14 at 02:45
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    @GitGud,it is important for the definition of linear dependence to include cases where, for example, the vectors are repeated. Otherwise, you cannot evensay that a non-zero vector is linearly dependent with itself. The statement «the set ${(1,0,0),(1,0,0)}$ is a linearly dependent subset of $\mathbb R^3$» is true. – Mariano Suárez-Álvarez Nov 17 '14 at 18:09
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    @MarianoSuárez-Alvarez I find that a compelling reason to define the concept for lists instead of sets. Thanks. – Git Gud Nov 17 '14 at 18:30
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    If someone writes $A^{m \times n} = A^m \times A^n$, they are definitely abusing something, but it isn't notation. :) –  Nov 18 '14 at 12:11
  • @AsalBeagDubh Thanks, hadn't noticed the typo. – Git Gud Nov 18 '14 at 12:12
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    what is wrong with 20? – user140161 Nov 18 '14 at 15:33
  • @user140161 Read the first four paragraphs of this answer. Edit: I edited the link to go straight to the answer I meant to link you to in the first place. – Git Gud Nov 18 '14 at 15:35
  • Don't forget that if $x\mapsto x^2$ squares reals or complexes, then why does $\Bbb R^2=\Bbb R\times\Bbb R$ or the set of functions ${0,1}\to\Bbb R$ (according to your faith)? – Mario Carneiro Nov 18 '14 at 19:27
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    @MarioCarneiro I can't understand your question. I can't understand what comes after'then' and I'm having a hard time relating the map $x\mapsto x^2$ to what comes after 'then'. – Git Gud Nov 18 '14 at 21:13
  • Number 5 still does my heading in algebra class. Great to know someone else is annoyed by it! – Pabce Nov 18 '14 at 21:16
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    Based on #7 then is it also an abuse to say "$0<r<1$"? Meaning that we generally intend to say $r$ is a number between $0$ and $1$, rather than saying that $0$ is less than the truth value of "$r<1$"? – Matt Nov 19 '14 at 00:32
  • @GitGud I am pointing out the notational inconsistency in the fact that on the one hand the $\square^2$ function is a function on the reals while at the same time we use the notation unchanged for the cartesian product of a set with itself (which is itself ambiguous, because what then is $\Bbb R^3$? This touches on #15). In set theory, everything is a set; so then is $2^2$ a cartesian product or an algebraic operation? – Mario Carneiro Nov 19 '14 at 01:26
  • @MarioCarneiro Ah, I see what you mean. – Git Gud Nov 19 '14 at 02:15
  • No, the difference between your example an mine is that you're parsing the mathematics ($0<r<1$) is two different ways, where as in my example I'm parsing as it should be done, i.e., with priority over natural language. However $0<r<1$ is just short for $0<r\land r<1$, see this answer. – Git Gud Nov 19 '14 at 11:58
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    This is getting slightly ridiculous. There are conventions and we're a mathematical society. We can interpret things. If everyone was pedantic like this, we would have already killed each other. – Pedro Nov 20 '14 at 01:49
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    @PedroTamaroff What's getting ridiculous? – Git Gud Nov 20 '14 at 01:50
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    Numbers 1,3-8,13,14, particularly: "The concept of 'function of a variable' isn't even definable in a satisfiable way in ZFC." Surprise: most people's heads don't parse in ZFC. – Pedro Nov 20 '14 at 01:53
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    @PedroTamaroff The question asks for abuses. Are they not abuses? – Git Gud Nov 20 '14 at 01:54
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    No, I hardly think they are abuses. – Pedro Nov 20 '14 at 01:56
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    For 7, I usually add a colon, so "The set $A := {x \in S \mid p(x)}$" reads as "The set A, which is defined as..." – fhyve Nov 20 '14 at 07:53
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    "22. Saying that $x=y\implies f(x)=f(y)$ proves that f is a function." ... "16. I don't know how I forgot this one. The omission of quantifiers." ... simply because you practice it yourself ;) – Nikolaj-K Nov 20 '14 at 09:02
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    @NikolajK I never claimed to not commit some of this abuses. I just meant to compile a list. But the omission of quantifiers certainly isn't something I do. – Git Gud Nov 20 '14 at 10:42
  • The one that's acutally bothering me is calling $\langle G,*\rangle$ as well as $G$ the group. People go on and say "A monoidal category $\langle C,\otimes, I\rangle$ is a category together with... "Whaa, no, a monoidal category isn't a category, you've written down a tuple with 3 entries!" :) The "everybody know what's meant" argument just runs against implementation on a computer. – Nikolaj-K Nov 20 '14 at 10:45
  • How about $f(\color{red}{t})=\int\limits_{0}^{\color{red}{t}}{\color{red}{t},\mathrm{d} \color{red}{t}}$ – hrkrshnn Nov 20 '14 at 12:05
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    @HatMan That's not an abuse people commit. I mean, students make that mistake but that's because they don't know better. I've never seen anyone write that seriously. – Git Gud Nov 20 '14 at 12:18
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    @Hat Man: I wouldn't say that's wrong, just hard to read. The binder is $\mathrm d$ at the end of the expression, and the scope is between the $\int$ and the $\mathrm d$. So the later two $t$'s can be substituted for any variable and the thing is formally equivalent to $\int_0^tx,\mathrm d x$. – Nikolaj-K Nov 20 '14 at 15:29
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    Re #6: Please never think that linear independence is about sets of vectors (I strictly prefer families (or lists)). Else one ends up e.g. with square matrices having nontrivial kernel but linearly independent column vectors. There is less problem with "linearly independent vectors" than one micght strctly think, even though this is not about a property of each single vector; cf. "two distinct objects" where distinctness is not a property of the single objects but a relatin among them. – Hagen von Eitzen Nov 20 '14 at 17:34
  • $f(x)$ and $x^2$ are not a problem if $x = id$ or if we're talking about $\mathbb R[x]$ and similar constructs. Why did you decide that $x$ is a real constant? Likewise $\frac {d} {dx}$ makes no sense whatsoever, unless you assume $x = id$, and if you do then $\frac {df(g(x))} {dx}$ is reasonable. – Karolis Juodelė Nov 22 '14 at 14:47
  • I didn't decide it, people decide it because they want to be free to replace $x$ with constants, thus treating $x$ as both a formal variable and as a number. It is an abuse. I don't understand what you mean $\frac{\mathrm d}{\mathrm dx}$ not making sense, but $\dfrac{\mathrm d}{\mathrm d,id}$ making. – Git Gud Nov 22 '14 at 14:55
  • "Can be replaced by a constant" and "is a constant" are different things. As for $\frac d {dx}$, consider the chain rule $\frac {dz} {dx} = \frac {dz} {dy} \cdot \frac {dy} {dx}$. Surely $y$ is a function. If $x$ is a constant, then $\frac d {dx}$ and $\frac d {dy}$ have to be different kind of operators. – Karolis Juodelė Nov 22 '14 at 15:07
  • "Can be replaced by a constant" and "is a constant" are only different in formal systems. While writing mathematics informally $x$ in, say $2x+3$, is a constant and it is different from $x\mapsto 2x+3$. That form of the chain rule you mention is itself abusive. We want $fg$ to denote the function $t\mapsto f(t)g(t)$, that form of the chain rule abuses that. – Git Gud Nov 22 '14 at 15:15
  • I agree that my example was a bit ugly, but are you always opposed to notation $\frac {dg} {df}$ or, similarly, $\int g df$ where both $f, g$ are functions? If no, you have to make $x$ a function. – Karolis Juodelė Nov 22 '14 at 15:22
  • I am always opposed to $\frac {\mathrm dg} {\mathrm df}$ (until I see it defined properly). I've never seen the concept of 'derivative with respect to a function', but that's me. I'm not saying it can't be done. As for $\int g,\mathrm df$, this is meaningful to me as per Riemann-Stieltjes, but in this context I claim that $\int x,\mathrm dx$ is an abuse and correct would be $\int \mathrm d,id$. – Git Gud Nov 22 '14 at 15:31
  • Now I'm wondering. Do you consider $\lim_{x \to 0} f(x)$ abuse? If $x$ is a constant then this might as well say $\lim_{a\to 0} b$ with $a, b$ completely unrelated constants. And if you consider this $x$ to be typed as variable or function, then why not do the same in $x^2$? – Karolis Juodelė Nov 22 '14 at 15:40
  • @KarolisJuodelė I don't think it is in an abuse, $x$ is binded, it's a dummy variable and $\lim \limits_{x\to 0}\left(f(x)\right)=c$ is short for a well known formula, it's just an abbreviation, but I would prefer the notation $\lim \limits_0(f)$. Edit: And you need to treat both $x$'s the same way, so they are definitely related. – Git Gud Nov 22 '14 at 15:46
  • I don't know why you include $u=x^2\implies \mathrm du=2x\mathrm dx$, it's merely another notation for $\tfrac{{\rm d}u}{{\rm d}x}=2x$. – Hakim Nov 22 '14 at 16:09
  • @Hakim If you look at it that way, it's OK. But then you lose the right to integrate each side with respect to the appropriate variables and this is often the point of that notation. – Git Gud Nov 22 '14 at 16:11
  • I still can't tell what you will or won't allow to abbreviate. Can't $x^2$ be a shorthand for $x \mapsto x^2$? And can't $f = n^{O(g)}$ be shorthand for $\exists h\in O(g), f = n^h$. And how is the type of $x^2$ different from the type of the left hand side of $x \mapsto x^2$? – Karolis Juodelė Nov 22 '14 at 16:13
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    @GitGud Nope, using the $\tfrac{{\rm d}u}{{\rm d}x}$ we can employ the reverse chain rule, which is equivalent to the classic $u$-substitution method that uses the notation ${\rm d}u=\ldots{\rm d}x$. – Hakim Nov 22 '14 at 16:14
  • @KarolisJuodelė I consider $x^2$ as an abbreviation of $x\mapsto x^2$ an abuse because $x^2$ is already taken, it already means something. There's ambiguity. The same goes for the rest. – Git Gud Nov 22 '14 at 16:15
  • @Hakim Differentials have no business here. In the usual context (first year calculus) it is an abuse. I'm very well aware that notation can be made meaningful, but it isn't in this context. – Git Gud Nov 22 '14 at 16:17
  • @GitGud I'm talking about the usual context. – Hakim Nov 22 '14 at 16:47
  • @Hakim OK, then I misunderstood you. I'm not following your last comment. If you wish, explain everything carefully, otherwise I can't reply. I do understand if you don't want to do it, though. – Git Gud Nov 22 '14 at 16:49
  • About 8. $a + b$ can be scalar addition for $a, b$ scalars and function addition for $a, b$ functions. To require that every algebraic structure would have a different symbol for $+$ seems unreasonable and unnecessary. Also, for $a$ function, $b$ scalar, $a+b$ can be defined as $a + (x \mapsto b)$. Are you bothered that such definition is usually not given explicitly or is there something else? Likewise $x^2$ can be defined differently based on the type of $x$. If $x$ is a new variable, not assigned a function or a constant, it is reasonable to consider $x^2$ a function. I see no ambiguity. – Karolis Juodelė Nov 22 '14 at 16:56
  • @KarolisJuodelė I think we're not on the same page. I never said anything about these things being reasonable or unreasonable. I agree that most are unreasonable. I consider 8. to be an abuse because $+$ has two different meanings. You don't think using the same symbol for two different things is an abuse? Well, I consider it an abuse. "A rose by any other name would smell as sweet". If you don't want to call it an abuse, don't. But I think we agree on the fact that $+$ is used with two different meanings in the same context. That's what I'm claiming. – Git Gud Nov 22 '14 at 17:02
  • "Are you bothered that such definition is usually not given explicitly or is there something else?" I wouldn't say I'm bothered, but that's my point, yes. But it's more than that. Usually people are thinking in two different sums, one for scalars, one for functions. What you propose may have solved this by generalizing the sum, but that's not what people are usually doing and hence I consider it an abuse. – Git Gud Nov 22 '14 at 17:10
  • @GitGud I'll try to clarify myself, one can use $\tfrac{\mathrm du}{\mathrm dx}$ instead of $\mathrm du=\ldots\mathrm dx$ using the reverse chain rule. – Hakim Nov 22 '14 at 17:11
  • I agree that abuse is having two meanings in one context. I think we don't agree on what that context is. I wish we had a proper formalism to reason about. – Karolis Juodelė Nov 22 '14 at 17:30
  • How about using square brackets or using parentheses for writing matrices? Gets very interesting for 1x1 and 2x1 matrices since they start looking like classes of equality or binomial coefficients. How about using ( or < for writing open ended intervals? Especially fun when talking about functions in $\mathbb R^2$ and saying "for every (a, b)"... – Luka Horvat Nov 23 '14 at 01:24
  • @GitGud Here's an example clarifying my point, for instance let's say you want to integrate $e^x\sin(e^x)$ with respect to $x$, the classical way you substitute $u=e^x\implies{\rm d}u=e^x{\rm d}x$, then it suffices to integrate $\sin u$ and will get $-\cos u+{\rm C}=-\cos(e^x)+{\rm C}$. But there's an equivalent way using the reverse chain rule which tell us that $$\int f'(g(x))\cdot g'(x),{\rm d}x=f(g(x))+{\rm C}.$$ We can define $g(x)=e^x$ (just like in the substitution $u=e^x$) and $f'(x)=\sin x$, thus the integral evaluates to $-\cos(e^x)+{\rm C}$. – Hakim Nov 23 '14 at 14:55
  • @Hakim I didn't know that was called the 'reverse chain rule', to me that's simply integration by substitution and it is usually what I use. If you make a parallel between the reverse chain rule and $u$-substitution, OK, I can accept it as not being an abuse. However I don't think that parallel is what people are doing consciously and implicitly when they use $u$-substitution. – Git Gud Nov 23 '14 at 15:04
  • @GitGud You're right on the last point. – Hakim Nov 23 '14 at 15:07
  • Using + for adding scalars and functions isn't problem for me. As long as it's defined what objects it's working on. scalar+scalar = scalar, and function + function = function. No parsing problems here. – Calmarius Jan 16 '15 at 16:17
  • @GitGud :) $$\int e^x \sin\left(e^x\right) \operatorname{d}x=$$ $$=\int \sin\left(e^x\right) \operatorname{d}\left(e^x\right)=$$ $$=-\cos\left(e^x\right)+\text{C}.$$ – dbanet Jun 18 '15 at 17:37
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    ‘The function f(x)’ is simply an abbreviation for ‘The function f, whose bound variable I am going to designate as ‘x’’. If you don’t acknowledge, and accept, the role of abbreviations in mathematics, and ordinary language, there is no end to the ‘anomalies’ that you can come up with. As so often happens, a bit of humor can establish the reality that we may be ignoring: a customer, holding a can of bug spray, asks the sales clerk “Is this good for wasps?” and the sales clerk answers, “No, it kills them.” – clearly “Is this good for wasps?” is an abb for “Is this good for killing wasps?” –  Nov 19 '20 at 17:24
64

The inconsistent treatment of raising trig functions to powers: $$ \sin^n x \,.$$

Seriously, starting ab inito $$\sin^2 x$$ could mean either $$\sin( \sin(x) )$$ if you are a quantum mechanic and like to see everything as an operator or as $$(\sin x)^2$$ which is the conventional meaning.

So why is $$\sin^{-1} x$$ used for $$\arcsin x$$ (which is vaguely consistent with the former) instead of $$(\sin x)^{-1} $$ in keeping with the latter.

dmckee --- ex-moderator kitten
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    This is an inconsistent treatment of that notation if you mix two or more of the meanings. There is nothing abusive in the notation itself. – Mariano Suárez-Álvarez Nov 17 '14 at 04:40
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    @Mariano It's pretty common to see $\sin^2 x = (\sin x)^2$ and $\sin^{-1} x = \arcsin x$ both used in precalculus or calculus textbooks. So, one or both of them (definitely including the second one) is an abuse. – Ryan Reich Nov 17 '14 at 15:04
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    @RyanReich, What is wrong in that is the superposition of the two different notations, not the notations themselves --neither of them. – Mariano Suárez-Álvarez Nov 17 '14 at 15:45
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    @Mariano Even if it's just the superposition that's wrong, that superposition is itself a common convention. Also: $\sin^{-1} x$ is definitely an abuse of notation. It is not the inverse function, but refers to the inverse of a specific restriction of $\sin x$ to a domain on which it's injective. Note that $\cos^{-1}$ and $\tan^{-1}$ are also used, and their restrictions are different. This notation can only be defined in an extremely kludgy way. You seem to believe that notation can't be abused if it's well-defined: well, this notation can't be well-defined. – Ryan Reich Nov 17 '14 at 18:47
  • This is the one abuse of notation that I refuse to take part in. Many of the ones mentioned by Git Gud can genuinely simplify your writing, but is it really so hard to just write $(\sin x)^2$? It's two parentheses! – Jack M Nov 18 '14 at 00:05
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    Maybe it's time to realize trig functions are functions. If we define a function f(x) = 3x, you would never write f40 = f(40), yet people will gladly write sin40 = sin(40). The parenthesis are (should be, at least) mandatory, which means anything outside them should be evaluated after the function. So sin(40)^2 = sin(40)sin(40). I could be wrong, but isn't f(40)^2 commonly interpreted as f(40)f(40)? – Dan Watkins Nov 18 '14 at 04:28
  • Well, we could adopt putting the operation into the exponent and write $\sin^{ 2}x=(\sin x)^2$, and $\sin^{\circ2}x=\sin(\sin(x))$ and $\sin^{\circ-1}x$ (multiplication symbol is invisible, so you can't see it in the exponent). – Joker_vD Nov 18 '14 at 09:16
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    @DanWatkins, I don't understand why you think $f(x)$ is not very often written simply $fx$... That happens all the time! Why would anyone dream of making those parenthesis that you want mandatory? What possible sensible interpretations of $\sin40$ can you think of apart from the intended one? Even if you can think of another one, just as with anything related to human communication what is mandatory is that you become accustumed with the usual traditions... – Mariano Suárez-Álvarez Nov 19 '14 at 09:33
  • Yeah, I wasn't sure about the fx notations, I've just always seen people write f(x) only. The only goal with mandatory parenthesis is consistent interpretation rules, but if it's common to write fx then my argument is over. – Dan Watkins Nov 19 '14 at 16:35
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    @JackM The $\sin^2x$ notation is historical; in trigonometric formulas it's common to have those powers and people devised that way for getting less clumsy formulas. The concept of composition of function as an operation came much later and the notation remained, which is surely more practical; how many times did you find in your life $\sin(\sin x)$ compared to $(\sin x)^2$? What I find abusive is $\sin^{-1}$ for the reasons well expressed by Ryan Reich. – egreg Nov 20 '14 at 11:38
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Double factorial $n!!=n(n-2)(n-4)\cdots$, where the product run through positive integers.
At the first time this notation confused me a lot because it looks the same as $(n!)!$ .
Similar argument about multifactorial.

Dan Oak
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41

In the first year at college, I was really confused with the notion of a sequence

$$\{ x_n : n \in \mathbb N \}$$

because this could also be a set! Then I discovered

$$(x_n)_{n \in \mathbb N}$$

And now I am fine with sequences.

ThePortakal
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    I don't get it, ${ x_n : n \in \mathbb N }$ is unmistakably the image of the sequence. What's the abuse I'm missing here? – Git Gud Nov 17 '14 at 00:50
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    @GitGud ${ x_1, x_2, x_3, \dots } = { x_2, x_1, x_3, \dots }$ as sets, but not as sequences. (And I think I abused $\dots$ here) – ThePortakal Nov 17 '14 at 00:56
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    This is a real mistake. It is more problematic in uses like «the set ${x_1,x_2}$ is linarly independent,» which is usually not intended to include the claim that when $x_1=x_2\neq0$, the one-element set mentioned in linearly indepenent. – Mariano Suárez-Álvarez Nov 17 '14 at 04:41
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    A sequence in the set $X$ should be defined as a function $x:\mathbb{N} \rightarrow X$ where we denote the image $x(n)$ of $n \in \mathbb{N}$ by $x_{n}$. – Vincent Pfenninger Nov 17 '14 at 17:29
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    I see ${x_n:n\in \mathbb{N}}$ for more often than $(x_n)_{n\in \mathbb{N}}$, but I've never cared for it. Set notation to me implies unordered list without repeated elements, but a sequence is ordered and may have repeats. Of course, there is an implicit order inherited from the order on $\mathbb{N},$ but using set notation still seems like an abuse. – Randy E Nov 17 '14 at 20:02
  • Why not use $(x_n:n\in\Bbb N)$? – Akiva Weinberger Jul 03 '17 at 02:34
  • Chevrons to denote sequences – ryang Jul 01 '22 at 19:10
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$\mathbb N\;\;\;\;\;\;\;\;\;\;\;\;$

TonyK
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  • What's the reasoning behind excluding/including 0 in this set? – Frank Vel Nov 19 '14 at 18:46
  • @fvel I think the reasoning is that it's ambiguous. Some odd people believe $0\notin \mathbb{N}$. – Daniel Fischer Nov 19 '14 at 19:41
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    @DanielFischer I propose we let $\mathbb{N}$ denote all natural numbers including $0$, and let $\mathbb{C}$ denote all counting numbers. – Frank Vel Nov 19 '14 at 19:48
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    What the [censored] are "counting numbers"? – Daniel Fischer Nov 19 '14 at 19:57
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    @fvel: yeah, and $\mathbb{K}$ all complex numbers, and $\mathbb{F}$ some generic field, and... – leftaroundabout Nov 19 '14 at 19:58
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    @fvel Apart from the obvious, "You can have zero of a thing," it gives $\mathbb{N}$ an identity under addition, which is nice. On the other hand, people (well, the ones who aren't programmers, anyway) seem to like to count from one, and you don't have to worry about dividing by zero in $\mathbb{Z}^+$. – Stuart Olsen Nov 20 '14 at 02:27
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    Then $+$ would be an example too. – egreg Nov 20 '14 at 11:32
  • @fvel: The most widely accepted definition I have seen is that the nonnegative integers are called whole numbers, and the positive integers are called natural numbers. – user541686 Nov 23 '14 at 09:57
40

We often write $f(n) = O(g(n))$, when in fact $O(g(n))$ is a set, and should be written as $f(n) \in O(g(n))$. Similarly for other asymptotic notation, such as $\Theta$ and $\Omega$.

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    Do you mean $f \in O(g)$? From the typical usage of these symbols, I'd expect $f(n)$ and $g(n)$ to be numbers, and things like $4 \in O(8)$ do not make sense. – Prateek Nov 17 '14 at 10:39
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    Yeah, and that major inconsistency that became second nature to us now. –  Nov 17 '14 at 11:01
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    @Prateek Identifying a number with the constant function attaining that value (which is a useful abuse of notation often employed), the statement $4\in O(8)$ makes perfect sense (and is correct). – Daniel Fischer Nov 17 '14 at 11:37
  • @DanielFischer True that. Then $f(n) \in g(n)$ is going to be a rather uninteresting claim. :-) – Prateek Nov 17 '14 at 11:41
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    @Prateek: If you find that annoying, I bet you'll really hate things like $\frac{\rm d}{{\rm d}x}x^2=2x$ (since, if you insisted on assigning a value to $x$ and evaluating from the inside out, you'd end up with something totally nonsensical like $\frac{\rm d}{{\rm d}5}25=10$). It makes more sense if you allow "unevaluated expressions in one or more variables" as first-class objects in your mathematical framework (and not just when buried inside function definitions). If you don't, you'll have to go for something like $f\in O(n\mapsto n^2)$ to make asymptotical notation rigorous. – Ilmari Karonen Nov 17 '14 at 11:42
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    There is really no problem in writing a function as $x^2$. Define $x$ to be the identity function. Define the ring operations on functions as usual. Then $x^2$ is the function which takes $t$ to $t^2$, etc. – Martin Brandenburg Nov 18 '14 at 09:02
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In financial mathematics one encounters index numbers, such as consumer price indexes, where a base year (such as 1992) is commonly specified by truly ghastly expressions such as

1992=100.

And no, they're not working modulo a divisor of 1892: they're referring to the fact that the index number for the base year is 100. It feels wrong to even bring this up in polite company.

Shane O Rourke
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    If you don’t make a distinction between background and foreground, things like this will seem anomalous. What is foregrounded (i.e., in the real-world stuff) is necessarily abbreviated. The practice of saying ‘a = b’, where a != b is UNIVERSAL, and so, necessarily correct as a linguistic practice, analogous to the fact that 100% of the speakers of a language cannot be wrong about a point of usage of the language. Otherwise, avert your eyes as you walk down to paper-towel aisle at the supermarket, so that you won’t be offended at the Bounty claim ‘6 = 12’. –  Nov 19 '20 at 17:10
  • This is just a widely understood abbreviation that can be squeezed into (in often small) diagrams and figures. This is like complaining that anyone ever uses any abbreviations whatsoever. –  Aug 16 '23 at 05:59
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Einstein summation convention is a self-explanatory example.

Fourier transforms

I feel the majority of people (myself included) abuse notation when describing Fourier transforms.

For example, it's common to see:

$$\mathcal{F}\{{f(x)}\} = F(\omega)$$

to which my natural response is: uhm, no, I believe you mean

$$\mathcal{F_x}\{{f}\} = F$$

or perhaps

$$\mathcal{F}\{f(\,\cdot\,)\} = F(\,\cdot\,)$$

The original is clearly incorrect because $f(x)$ is the value of $f$ at some point $x$, and its Fourier transform $F$ is defined as $$F(\omega) = f(x) \delta(\omega)$$ which is clearly not what is intended.

Community
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user541686
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    In what possible way is Einstein summation inconsisten and or an abused notation?! – Mariano Suárez-Álvarez Nov 17 '14 at 03:54
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    @MarianoSuárez-Alvarez: I can't tell if you're being sarcastic or serious... – user541686 Nov 17 '14 at 04:00
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    Serious, of course. I have never ever heard or read anyone call a perfectly well-defined, precise and immensely useful notation such as Einstein's convention an abuse of notation... – Mariano Suárez-Álvarez Nov 17 '14 at 04:01
  • @MarianoSuárez-Alvarez: Did you see the article I linked to though? One of the reasons is explained literally right there: "The upper indices are not exponents* but are indices of coordinates, coefficients or basis vectors."* That's... hella confusing, put it mildly. – user541686 Nov 17 '14 at 04:05
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    Huh? Why would upper indices have one and only one meaning? If in some context you are never going to use exponents and you will use thousands of times indices of coordinates, why would you not use upper indices for the latter role? The notation, moreover, makes it impossible to write things which simply do not make sense it in the context it is used (for example: powers of coordinates, which make no sense in pretty much any situation...) – Mariano Suárez-Álvarez Nov 17 '14 at 04:07
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    Of course, if you ar not familiar with the notation, you will have to get used to it —just as if you are not familiar with English you are going to get some familiarity before trying to read Shakespeare. – Mariano Suárez-Álvarez Nov 17 '14 at 04:09
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    @MarianoSuárez-Alvarez: I don't follow your reasoning. Every widely-used abuse of notation happens for a reason; just because there's a reason or context behind it doesn't mean it's not an abuse of notation. When people write $\mathcal{F}{f(x)} = F(\omega)$ it's 100% clear from the context that they don't intend to take the FT of a single number, but that doesn't make it any less abusive of the notation. – user541686 Nov 17 '14 at 04:09
  • That's absurd. If a notation makes it clear to the reader (with the level of familiarity with the notation that the write assumes when writing) what is meant, there is absolutely abusive about nothing – Mariano Suárez-Álvarez Nov 17 '14 at 04:11
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    @MarianoSuárez-Alvarez: Could you give me a list of what you consider to be abuses of notation? I feel like I can't find any common ground with you to base my reasons off. – user541686 Nov 17 '14 at 04:12
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    The only possible abuse of notation is unexplained notation. The act of announcing, à la Bourbaki, that one is going to make an abuse of notation of of nomenclature is just a colorful way of introducing new notation or new terms. And the new notation need not be different from existing notation: context may be needed to tell one from the other —that is a perfectly correct practice. – Mariano Suárez-Álvarez Nov 17 '14 at 04:14
  • That notation or the meaning of terms sometimes depends on context and/or requires a certain background and ingenuity from the reader is not an abuse: it is a reflextion of the fact that mathematical writing is a means of communication between humans, which are amazing machines. – Mariano Suárez-Álvarez Nov 17 '14 at 04:15
  • You should spend some time reading the two wonderful volumes that Florian Cajori wrote on the history of mathematical notations. It will provide perspective —half of the first volume is about notations for exponents, in fact. – Mariano Suárez-Álvarez Nov 17 '14 at 04:19
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    @MarianoSuárez-Alvarez: So if I told you I want to denote the sum of A and B as $A^B$ you wouldn't call it an abuse of notation? If you would really call that "perfectly correct practice" just because its meaning was "explained" to you I have to say I think you're a little crazy! – user541686 Nov 17 '14 at 04:20
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    Of course, I would call it perfectly correct. It often happens that one writes $x+y$ to mean the product in a group, if for some reason the group was earlier views as multiplicative (say, the multiplicative group of a field) and then used in a context where groups are additive. likewise, the trivial group is written $0$ or $1$, depending on what you are doing, and often the two notations are used side by side. – Mariano Suárez-Álvarez Nov 17 '14 at 04:24
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    In contexts where group actions are discussed, one often writes $x^f$ (or ${}^fx$, depending on who and in what context) instead of $f(x)$. This is standard notation, and has been for a century... You may say I am crazy, but along with me you'll have to call a whole ton of other people crazy, too :-) – Mariano Suárez-Álvarez Nov 17 '14 at 04:24
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    The standard notation for Legendre's symbol is $\left(\frac ab\right)$, and this is used in contexts where fractions are also discussed and where of course parentheses appear all the time. After the initial 3 minutes, this does not confuse anyone. – Mariano Suárez-Álvarez Nov 17 '14 at 04:27
  • @MarianoSuárez-Alvarez: I think you're dodging my question. I'm not talking about group theory. I was talking about something simple, the sum of two variables (think middle and high school algebra, not PhD level abstract algebra). If I wrote a book denoting it (consistently throughout, mind you!!) as $A^B$ and instead chose $A + B$ to mean (say) exponentiation, you would be perfectly fine with it just because I explained it to you and stayed consistent throughout the whole book?! – user541686 Nov 17 '14 at 04:28
  • I would be fine with it. I would say, though, that it is a terrible idea, because an elementrary textbook should familiarize the reader with standard notations, and then that choice of notation would only show poor judgement on the part of the author. The problem is not at all in any perceived «abuse of notation». – Mariano Suárez-Álvarez Nov 17 '14 at 04:29
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    In any case, my problem with your post is the mention of Einstein's summation convention, which is not by any standards an abuse of anything (and it surely is not used in highschool textbooks...) – Mariano Suárez-Álvarez Nov 17 '14 at 04:32
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    In Russian literature it's common to write Fourier transform as $\mathcal Ff$. At least we do so in our courses. – Тимофей Ломоносов Nov 17 '14 at 05:27
  • @ТимофейЛомоносов: Yeah I've seen that one too, I like that actually. – user541686 Nov 17 '14 at 05:35
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    @ТимофейЛомоносов That's very nice notation. I'll adopt it from now on! – Mark Fantini Nov 17 '14 at 05:46
  • The Fourier transform of the constant $c$ (interpreted as a tempered distribution) is not $\delta_c$ (or $\delta(\omega - c)$) but $c\cdot \delta$. – Daniel Fischer Nov 17 '14 at 11:34
  • @DanielFischer: Whoops, sorry, I wasn't thinking! I'll fix that, thanks. – user541686 Nov 17 '14 at 15:36
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    Einstein's notation is not an abuse of notation, it's a conflict of notation with the notation for raising to a power. Using exponents for indices happens a lot in mathematics, not just with Einstein's notation (cosimplicial sets, for example off the top of my head). – Najib Idrissi Nov 17 '14 at 15:45
  • @NajibIdrissi: I suggest you post your own answer. – user541686 Nov 17 '14 at 16:04
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    @Mehrdad, one of the purposes of comments is to signal errors of fact in existing answers. – Mariano Suárez-Álvarez Nov 17 '14 at 18:03
  • I don't see how my comment could be an answer. I'm pointing out that, in fact, Einstein's notation is not a good example. And I don't understand why you're so irate about all this. – Najib Idrissi Nov 17 '14 at 19:12
  • Maybe I'm in the minority, but I do find that Einstein summation has a issue with ambiguity (nothing to do with exponents, though). Namely, suppose I have a sum $a_ix^i$ and now I want to refer to a general summand: what do I call it? If I write $a_jx^j$, how do I "turn off" the summing? I seem to remember this ambiguity really does come up when you use the notation (although I haven't actually done that for ~10 years). Of course, needless to say, there are workarounds, and the notation is still extremely useful. –  Nov 18 '14 at 11:57
  • @MarianoSuárez-Alvarez I'm a little afraid to get caught in the crossfire here, but I would agree that Einstein notation is an abuse of notation, although as you point out in a comment on the OP this is not a derogatory term, and it is indeed a very useful abuse. Exponents are not "never used" - the EFE $G_{\mu \nu} + g_{\mu \nu} \Lambda = \frac{8 \pi G}{c^4} T_{\mu \nu}$ contains Einstein notation and an exponent on the $c$. A related abuse is physicists' tendency to use $e$ for Euler's number and electric charge, even in the same formula: $$\psi(r)\sim e^{-rm_ee^2\hbar^{-2}}$$ – Mario Carneiro Nov 19 '14 at 01:50
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Don't know whether these count as notational abuses as such, but a few common causes of confusion I have come across are

\begin{align} &\log(x)\text{ and }\ln(x)\\ &\sin^2(x)\text{ and }\sin(x)^2\\ &\sin^{-1}(x)\text{ and }\arcsin(x)\\ &\log_2(x)\text{ meaning }\log(\log(x)),\text{ & }\log_2(x)\text{ meaning base }2\\ &\log(\log(x))\text{ and }\log\log x \text{ etc. }\\ &\mathbb{S}^n\text{ in topology & }\mathbb{S}^n\text{ in geometry }\\ \end{align}

re last one, from MathWorld:

... A geometer would therefore regard the object described b $ x_1^2+x_2^2=R^2 $ as a $2$-sphere, while a topologist would consider it a $1$-sphere and denote it $\mathbb{S}^1$. Similarly, a geometer would regard the object described by $x_1^2+x_2^2+x_3^2=R^2 $ as a $3$-sphere, while a topologist would call it a $2$-sphere and denote it $\mathbb{S}^2$. Extreme caution is therefore advised when consulting the literature. Following the literature, both conventions are used in this work, depending on context, which is stated explicitly wherever it might be ambiguous.

This last one was included as a curiosity, but generally though, subscript and superscript is often abused / ambiguous unless explicitly stated.

... more interesting anecdotes and opinions.

Community
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martin
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20

Calculus I,II,III: The '$dx$'s in integrals and derivatives are just notation to help keep track of the important variables in a given problem, and they're otherwise meaningless in isolation.

Real Analysis: In fact, when the integration variable is unambiguous we may as well dispense with the differentials altogether and just denote the integral of $f$ over a region $R\subset \operatorname{dom}{f}$ as,

$$\int_{R}f.$$

Differential Geometry: Wait nevermind, $dx$ is a differential 1-form, $d$ is itself an operator, and the placement of '$dx$'s and so forth in notation for integrals couldn't be less optional, case and in point being the general Stokes' theorem:

$$\int_{R}d\omega=\oint_{\partial R}\omega.$$

Sorry we lied.

Edit: correctified for grammars.

David H
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    By "case and point" I believe you mean "case in point". – user541686 Nov 23 '14 at 10:00
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    @Mehrdad heh, oops. Finally corrected. Apparently automatic notifications aren't sent for comments on community wikis. – David H Mar 01 '15 at 22:45
  • I haven't gotten to Differential Geometry yet, but I assume that is the final and correct answer? – Ovi May 10 '17 at 23:19
  • In measure theory books they really use $ \ dx = d \mu (x) \ $ etc as just notation to help keep track of the important variables in a given problem and there is no manifold (or even a topology) involved. Examples found in Rudin's book of real and complex analysis in the section about Fubini's theorem. – Gustavo Nov 06 '17 at 14:13
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The word "trivial." The many uses of this word include:

1.) The colloquial usage as a synonym for "easy."

2.) The trivial group which consists only of the identity element.

3.) The trivial ring which consists only of the multiplicative and additive identities.

4.) A trivial solution to an equation, often when a variable equals 0 (or constant in the cases of differential equations).

5.) Similar to #2 and #3, any object which satisfies the bare minimum of some particular definition but has no further structure. Often some sort of identity or null element.

6.) A trivial application of a theorem can refer to a special case where the truth of the theorem is more or less self-evident, e.g. a theorem in set theory which is obviously true when applied to the empty set.

This list could go on forever, but I'll stop here.

Note that the converse, of course, is the word "non-trivial," which is just as ambiguous.

I'll also note that although you may consider this a trivial answer, but I have had students get tripped up by my usage of the word "trivial" in lecture, so the ambiguity must be somewhat non-trivial.

josh314
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    I feel like your answer needs a little more justification. Right now you've shown that uses of the term "trivial" are broad but not that they are ambiguous. To me they all fall under the umbrella of "nothing much to it": the "nothing much" in the colloquial usage being in terms of effort, while in the mathematical uses, it is "nothing much" in terms of structure or complexity. –  Nov 20 '14 at 18:04
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    I agree with you, that is answer doesn't really emphasise its ambiguity, but I feel that I am not the only one who has spend some considerable time to proof "trivial" statement, that turned out to be not so trivial at all... Some mathematicians have the tendency to call virtually every result trivial, which they have seen a proof for. – Stefan Mesken Nov 20 '14 at 20:13
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    There can be ambiguity when there's more than one degenerate object of a certain kind. E.g. if $V$ is a vector space, certainly ${0}\subset V$ is a trivial subspace. Is $V$ also a trivial subspace? – Julian Rosen Nov 22 '14 at 16:08
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    My pet peeve when a paper says things like "trivially easy" about a concept I don't have any grasp on... – Calmarius Jan 16 '15 at 15:27
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My favorite has always been with Fourier transforms.

Suppose a particle is in a potential $V$ given by $$V(x) = V_0\cos(x/a).$$ Then $$V(k) = V_0\sqrt{\dfrac{\pi}{2}} \left( \delta(k-a) + \delta(k+a) \right).$$

I know physicists do this a lot. I am not sure about mathematicians.

Brian Moths
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    The same goes for change of variables. E.g. you switch from $x$ to $\xi$ in $\psi(x)$, and then your equation suddenly is not for $\psi(x)$ but for $\psi(\xi)$ where $\psi$ is now a different function. – Ruslan Nov 21 '14 at 06:42
  • A well known mathematician, with a Field Medal, also did that. – Felix Marin Dec 07 '14 at 05:58
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Here is one which I think goes back to Euler: $$i^i = e^{-\frac{\pi}{2}}$$

AD - Stop Putin -
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  • What's wrong with this? Raising by an imaginary number?? – Frank Vel Nov 18 '14 at 22:50
  • @fvel Yes as far as I get it. This logic is basically: $i^i=(e^{i\frac \pi 2})^i=e^{-\frac \pi 2}$. But also $i^i=(e^{i\pi \frac 5 2})^i=e^{-\pi \frac 5 2}\ne e^{-\frac \pi 2}$. At least I think that's what it's referring to. –  Nov 18 '14 at 23:08
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    Ah, so the equality should be $i^i = \exp\left(-\frac{\tau}{4}+k\tau\right)$. Kinda like if you took $\int{1}\mathrm{d}x=x$ by assuming $C$ is $0$. – Frank Vel Nov 18 '14 at 23:14
  • @fvel I think you meant $i^i = \exp(-\pi/2+2\pi\tau)$. An identity of sets would be more accurate. – AD - Stop Putin - Nov 19 '14 at 06:49
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    @AD that was exactly what I said? $\tau := 2\pi$. – Frank Vel Nov 19 '14 at 08:13
  • @fvel Fine, I have never seen that notation. – AD - Stop Putin - Nov 19 '14 at 10:29
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    @AD. Probably not the most known constant, but it exists! I thought it'd be worth mentioning since this thread is about ambiguous notations. – Frank Vel Nov 19 '14 at 10:40
  • Note that even when you correct it, the most remarkable thing is still true: $i^i\in\mathbb{R}$. – Mark Hurd Feb 24 '15 at 02:13
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    @MarkHurd Now you did it again, the main reason I put the post in this list is that $i^i$ is not one number - it is a rather a set of real numbers (given in the comments above) - and yes! That is remarkable! – AD - Stop Putin - Feb 24 '15 at 07:05
10

This may be a regional thing, but when I started studying at a British university, so many of the lecturers wrote multiplication as a single . (full-stop). This got really confusing when, after having studied in the states I was used to $$ 0.5 + 0.5 = 1$$ whereas here it meant $$0.5 + 0.5 = 0 + 0 = 0$$

kinbiko
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    No. $$\color{red}\Huge \cdot$$ (\cdot) is used to represent multiplication. I've never seen a lower dot being used in this situation. – beep-boop Nov 22 '14 at 15:29
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    http://en.wikipedia.org/wiki/Interpunct#English - the interpunct used to have other uses in the UK, I assume this is why full stops were used. :) – kinbiko Nov 22 '14 at 15:42
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    @alexqwx It varies by country, but a lower dot is definitely used some places. Why are you presupposing that because you have never seen it, it isn't a real thing? – Caleb Stanford Nov 23 '14 at 00:38
  • @Goos It's definitely not widely used in British universities (I have studied at 3 different ones and have watched online lectures from several more, and not one of them has ever had a lecturer using a lower dot to represent multiplication). – beep-boop Nov 23 '14 at 10:57
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    @alexqwx It seems to be dying out, fortunately, but e.g. in Hardy/Wright, the full stop is being used for multiplication. But with some space on both sides, so it'd be $0, .,5 + 0 , .,5 = 0$ (or perhaps wider spacing), which in print is distinguishable well enough from $0.5 + 0.5$. In handwriting, it may be indistinguishable. – Daniel Fischer Nov 23 '14 at 12:41
  • @DanielFischer Fair enough. Maybe this notation died out before I went to school, then. – beep-boop Nov 23 '14 at 12:47
  • When I grew up in Britain, the decimal point was centralised, so the lower dot was used for multiplication. This is definitely different to US usage. – BudgieJane Feb 25 '15 at 12:18
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Outer measures aren't measures.

Martín-Blas Pérez Pinilla
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  • Just like how the imaginary part $y$ is neither imaginary nor a part of $x+yi$ for $x,y$ real (which might explain why some people seem to call $yi$ the imaginary part instead...). I recall there's a linguistic term (likely something along the lines of "generalised noun/term/usage/adjective") describing situations where "an X Y" does not mean "a Y that is X" but can't put my finger on it -- someone like Tim Gowers mentioned this in connection with pedagogy. – Vandermonde Nov 20 '14 at 06:37
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    When we place an adjective in front a noun (or after it if you are French!) it may signify something less than, or more than, or a variant of the object that noun symbolises. (Virtual keyboard is not a keyboard.) So, IMO, outer measure is a perfectly valid choice. – P Vanchinathan Nov 21 '14 at 01:27
  • Another example is that a multivalued function is not a function. See the entry, radial concept in A Handbook of Mathematical Discourse by Charles Wells. Note that Tim Gowers discusses multivalued functions in http://gowers.wordpress.com/2009/06/08/why-arent-all-functions-well-defined/. – J W Nov 22 '14 at 15:54
  • See also http://www.abstractmath.org/Word%20Press/?tag=multivalued-function in Wells' Gyre&Gimble blog. – J W Nov 22 '14 at 16:04
  • @J W: I'm fairly sure that that's the post I was looking for! I had also looked at abstractmath.org and related docs but forgotten the name this phenomenon went by there; "accumulation of attributes" looks to be spot-on. Thanks for scratching my two-year itch. – Vandermonde Dec 01 '14 at 06:12
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    For my own future reference as much as anyone else's, the archetypal example I couldn't remember of a noun's meaning being changed by an adjective as opposed to being restricted is "almost foo". – Vandermonde Dec 17 '14 at 21:06
  • Having finally tracked it down again and seeing how it's apropos here, I also must mention Polya's 'animal doctor' anecdote. It's pure (comedy) gold.

    (Happy anniversary of this post!)

     – Vandermonde Nov 18 '15 at 06:00
6

The inconsistency between the reading of "Negative" versus "Minus" has, in my opinion, been a thorn in the side of every teacher and student since their acceptance.

Community
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user 85795
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Lebesgue and (proper or improper) Riemann integrals written in the same way, ex.: $$\int_a^b f(x)dx.$$ I have even found "mixed integrals" where $\int_{A\cup B}f(x)+g(x)dx$ means (by writing the measure in the Lebesgue integral and marking the Riemann integral with $\mathscr{R}$) $\int_Af+gd\mu+\int_Afd\mu+\mathscr{R}\int_B g(x)dx$ (p. 422 here).

Lebesgue-Stieltjes and Riemann-Stieltjes integrals written both$$\int_a^bf(x)d\Phi(x).$$

Limit point can be found as meaning accumulation point, or adherent point, or point that is the limit of a subsequence of a sequence. I have found them used for all the three things even in the same textbook. I am not aware of other possible usages of the term.

Hilbert spaces sometimes intended to be separable, sometimes not necessarily separable.

Writing $A=B$ for isomorphisms $A\simeq B$ is something that can be confusing for less advanced students, especially if done to prove something without explaining why the isomorphism proves the desired result.

I find it particularly confusing when books do not specify the scalar field associated to a vector space which is beeing discussed, and I find it even more confusing when texts do not specify domains and codomains of maps. Such omissions are not too rare in engineering-oriented or older textbooks in general.

hardmath
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Self-teaching worker
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  1. $\mathbb N \subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C \subset \mathbb H$

  2. If $f \colon X \rightarrow Y$ is a map and $A \subseteq X$, then people often write $f(A)$ to denote $f'' A := \{ y \in Y \mid \exists x \in X : f(x) = y \}$, which can be quite confusing in cases where $A \in X$.

  3. "Canonical" ...

  4. $(a,b,c) = ((a,b),c) = (a,(b,c)) = f,$ where $f \colon \{0,1,2\} \rightarrow \{a,b,c\}$ is such that $f(0) = a$, $f(1) = b$ and $f(2) = c$.

  5. "regular", "dense", "dimension","rank", "computable", "recursive", "closed", "compatible", "compactification"... and other notions which, in a given context, may have several different meanings.

  6. "$f \colon X \rightarrow Y$ smooth" or similar expressions, where "smooth" may refer to $f$, $X$ or $Y$.

  7. "pictures" and "diagrams" can sometimes be ambiguous to an extent where they don't mean anything to anyone - or maybe that's just me.

  8. the "constant" $c$.

  9. $a < b < c \in d$

  10. $1 = \left( \frac 2 7 \right) \neq \frac 2 7$

  11. $\prod_{i \in I} (X_i, \le_i) \subseteq \prod_{i \in J} (Y_j, \le_j)$

user132181
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Stefan Mesken
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    What's wrong with 1.? – Frank Vel Nov 20 '14 at 10:25
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    @fvel There are different ways to construct $\mathbb Z, \mathbb Q, \ldots$ from $\mathbb N$. Following the "standard" definitions in set theory, one has $\mathbb N = \omega$ such that each $n \in \mathbb N$ has cardinality $|n| = n < \aleph_0$, while each element $z \in \mathbb Z$ is a subset of $\mathbb N \times \mathbb N$ of size $|z| = \aleph_0$. These identifications basically forget about the underlying sets and deal only with the "induced" structures. – Stefan Mesken Nov 20 '14 at 10:46
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    @fvel Maybe, I should give an example how this identification can cause trouble when one ignores that fact, that they typically are not subsets: In the "standard" construction of $\mathbb R$, every element $x \in \mathbb R$ has size $|x| = 2^{\aleph_0}$. If we had $\mathbb N \subseteq \mathbb R$, then pick any $n \in \mathbb N$. As $n \in \mathbb N \subseteq \mathbb R$, $n$ itself has cardinality $2^{\aleph_0}$. Now, $\mathbb N = \omega$ is a transitive set, so $n \subseteq \mathbb N$. This yields $2^{\aleph_0} = |n| \le |\mathbb N| = \aleph_0$, which contradicts Cantor's Theorem. – Stefan Mesken Nov 20 '14 at 11:27
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    That is an absurd objection, really. Anything coming out of a specific construction of the real numbers is absolutely irrelevant... – Mariano Suárez-Álvarez Nov 20 '14 at 23:07
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    Anf, to be honest, I don't even see what the standard construction of the real numbers is for you! – Mariano Suárez-Álvarez Nov 20 '14 at 23:11
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"$\subset$" is called proper subset (http://mathworld.wolfram.com/ProperSubset.html) and for example if $A = \{1,2,3\}$ then by definition, $\{1,2,3\}$ is not a proper subset of $A$ so we cannot write $\{1,2,3\} \subset A$ or $A \subset A$ simply. Instead we use $\subseteq$ but for some of the people, use of $\subset$ is ambiguous and it can include the equality of the sets. In other words, if $A \subset B$ and $B \subset A$ then $A = B$ rather than a contradiction.

Here is an example of an ambiguity from MSE: Proof verification: prove $A\subseteq B$ if and only if $A\cap B=A$.

ArsenBerk
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"X is dense in itself" is not equivalent to "X is dense in X".

An ellipse is not an elliptical curve.

I absolutely detest the "infinity is not a number, it's just a concept".

I believe it's already been addressed - natural numbers, whole numbers, counting numbers.

Paul Foster
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It is true that 5 is a divisor of 0, and yet it is true that there are no divisors of 0.

1

$L/K$ for field extensions.

First, this is not a quotient, and second, this is not even some other kind of operation, just the statement that $K \subseteq L$ as fields.

Antoine
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  • It's worth pointing out that often when writing L/K you are thinking of L as a vector space over K. – Dastur Nov 08 '22 at 12:42
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When we list seasons it goes, summer, autumn, winter, spring, summer, autumn, and one says seasons occur cyclically.

In the group denoted (by those without broken pieces of chalk) as $\mathbf{Z}$ the elements are (half of them) go like this 1,2,3,4, etc without ever repeating and yet it is called the infinite cyclic group!

user729424
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P Vanchinathan
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    This has absolutely nothing to do with the question. – Mariano Suárez-Álvarez Nov 20 '14 at 23:09
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    We call crrtain graphs «trees» and they are not trees! :-/ – Mariano Suárez-Álvarez Nov 20 '14 at 23:10
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    @ Mariano Su'arez-Alvarez: The title of the post says "inconsistent phrase" and the phrase "cyclic group" for a group where there is no periodicity definitely qualifies for it. – P Vanchinathan Nov 21 '14 at 01:00
  • About trees: Mathematicians try to name things by analogy. Trees of botanical kind, after shedding their leaves, resemble those of combinatorial kind and are perfectly valid. I think OP wanted to know if the analogy behind the name actually contradicts the concept it tries to illustrate. – P Vanchinathan Nov 21 '14 at 02:33
  • Well, infinite cyclic groups are quite analogous to the finite ones, and I would say the analogy is infinitely closer than between trees and trees. – Mariano Suárez-Álvarez Nov 21 '14 at 02:49
  • You are right. $\mathbf{Z/nZ}$ and $ \mathbf{Z}$ are closer in the sense of being monogenic (generated by a single element) but not in the sense of exhibiting periodic (cyclic) behaviour. A crow and a human could not both be called mammals because crow is not a mammal. But they can both be called bipeds for having two feet. – P Vanchinathan Nov 21 '14 at 09:50
  • Clearly, $\textrm{the seasons}\cong \Bbb{Z}/4\Bbb{Z}$. Let $f(\mathrm{winter})=2$. We can see that $f(\mathrm{winter + winter})=2+2=0=f(\mathrm{summer})$ which is confirmed by the fact that my part of the country had six solid months of freezing weather and somehow skipped straight to a hot summer. QED – Patrick Shambayati Mar 31 '15 at 19:49
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  • $\sqrt{2}$ is usually read as "root two". The degree of root should be mentioned though (like "square root two").
  • $\tan^{-1}$ is used for $\arctan$ mostly in Physics and Electronics.
  • $j$ is used for $\sqrt{-1}$ in Electronics, while $i$ is used for it in Mathematics.
  • Dots are used for time derivative in Pyhsics and Control Engineering (e.g.; $\ddot{y}+4\dot{y}+3y = 3\dot{x} + 2x + 8$)
  • Vectors are sometimes denoted in bold ($\mathbf{x}$), with bar on ($\bar{x}$) or with arrow on ($\overset\rightarrow x$).
  • Maybe the most frustrating things is when someone uses the word "integral" in the meaning of "anti-derivative" (e.g.; "Integral is the inverse of derivation.").
  • The unit scalers are used in the place of units themselves (e.g.; "I bought two kilos of potatoes.")
  • The $\nabla$ (del) operator can be very confusing sometimes.
  • I have never understood the purpose of omitting the preceding zero in a decimal number (e.g.; writing $.5$ instead of $0.5$).
  • In C++ language, log() is used for ln() and log10() is used for log().
  • I still don't know what $2^{3^4}$ does equal to. $2^{(3^4)}=2^{81}$ or ${(2^3)}^4=8^4$?
  • The division operator $/$ is misused. Example: $N/A\cdot m$ is written for $N/(A\cdot m)$.
hkBattousai
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    Well, C++ isn't all bad. In mathematics, the prevailing convention is that $\log$ denotes the (a local) inverse of $\exp$, and other bases are explicitly indicated ($\log_{\text{base}}$). Only when speaking to/writing for physicists/engineers has $\ln$ somewhat widespread use. – Daniel Fischer Nov 23 '14 at 12:50
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    -1 for the C++ comment. $\log$ is the standard way mathematicians write the natural log, and mathematicians never write $\log$ for the base-10 log. – Charles Sep 10 '16 at 17:30
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$$\bigcup \{ A ,B, C\} = A \cup B \cup C $$ or $$\bigcup \mathcal A $$ where $\mathcal A $ is a set of sets. It should be

$$\bigcup_{X \in \mathcal A} X. $$ This can come up in topology sometimes (unioning over a collection of open sets), and can get rather confusing if you start doing things (edit:) like $\left(\bigcup \mathcal A\right) \cup U \cup V$ where $U$ and $V$ are just sets (eg. when adding open sets to a cover).

Edit: It's like defining $\sum S = \sum_{x\in S} x$ where $S \subset \mathbb R$ is finite (or ordered). Then usually you expect in something like $(\sum S) + x + y$ for $S$, $x$, and, $y$ to be the same sort of thing.

fhyve
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    This is just wrong, and rather a misunderstanding on your part. What do you mean by "the union of a single set is the set"? This is false in general. For sure, $\bigcup {x}=x$, but $\bigcup x$ does not need to be $x$. Consider for example $x={\emptyset}$. This is a singleton, so $x\ne\emptyset=\bigcup x$. You are probably more used to seeing $\cup$ as a binary relation, written in between two sets, while $\bigcup$ is unary, written preceding the set it acts on. Perhaps that's the source of the confusion. The notation is quite precise here, there is no ambiguity. – Andrés E. Caicedo Nov 20 '14 at 08:16
  • I edited to correct that out. I was mistakenly generalizing the case of the intersection of one $A$ set when considered as a subset of some ambient set $S$, in which case we can define it as a one object product as sets over $S$. – fhyve Nov 20 '14 at 08:43
  • I remember something that uses this sort of convention coming up in my topology class where it was definitely abusive, and where doing it rigorously would be a bit of annoying leg work, but I can't find it in my notes :( – fhyve Nov 20 '14 at 08:56