Every good teacher knows that giving an informal explanation (that is, an explanation not based on strict definitions and maybe making use of metaphors, intuition, or everyday language) of a mathematical concept can be really helpful to students. And every good teacher has at least one of these informal explanations for each of the main mathematical concepts they teaches. However, I believe that some of these informal explanations happen to actually cause misunderstanding of the mathematical concept, to the point that it would be much better avoiding them (or replacing them with better informal explanations).

I think it would be interesting to list some examples of these bad informal explanations. Here my example:

Bad: "Continuous functions are those functions that can be drawn without lifting the pencil from the paper."

I heard this one many times from many teachers, and I find it causes misconceptions for multiple reasons: 1) It is actually about the arc connectedness of the graph of the function, not the continuity 2) It convinces the students that many continuous functions are not continuous because they can't be drawn without lifting the pencil, like 1/x on its domain of definition 3) It has no connection with the delta-epsilon definition of continuity that the student need to learn.

To make my point more clear, I also give an example of a good informal explanation:

Good: "The concept of group encodes the properties of transformations of an object. The identity element corresponds to 'doing nothing', the inverse element to 'reverting the transformation', associative laws means... etc."

What are your examples of informal explanations that cause misconceptions?

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    $\begingroup$ I start with the pencil idea that you call a bad explanation and then I show them y=xsin(1/x), to show why that informal idea isn't adequate. $\endgroup$ – Sue VanHattum Oct 6 '19 at 17:53
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    $\begingroup$ I'm not sure that I see why the pencil explanation is "bad" explanation. It basically asserts that a function is continuous if it satisfies the intermediate value property. This isn't quite right, but it (1) gives an intuition and (2) helps to emphasize the need for more rigorous definitions (and $\varepsilon$-$\delta$ is a heavy lift in an elementary class, so good examples which break the "draw with a pencil rule" are good to have). $\endgroup$ – Xander Henderson Oct 7 '19 at 1:59
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    $\begingroup$ @Jorssen Please read to the end of my comment. The parenthetical is important: the $\varepsilon$-$\delta$ definition is a heavy lift for students in elementary classes (and is particularly difficult to deal with if it is unmotivated). The motivation is that functions which can be drawn without lifting the pencil are "nice" functions, and we would like to have a good mathematical definition which captures this idea. The problem occurs if you present the "lift-the-pencil" spiel as the definition of a continuous function. This is a bad definition, but it is a good informal explanation. $\endgroup$ – Xander Henderson Oct 7 '19 at 11:35
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    $\begingroup$ Let me be clear: I do not think that an instructor should present definitions which they know to be wrong, and I think that "a function is continuous if its graph can be drawn without lifting your pencil" is a terrible definition. However, as an informal explanation, it essentially captures the intermediate value property, which is very closely related to continuity on $\mathbb{R}$. It is a good place to start in a discussion of continuity. $\endgroup$ – Xander Henderson Oct 7 '19 at 11:49
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    $\begingroup$ ... Since there are other informal explanations of continuity (input-output control, two players game...) that do not suffer this problem, I do not see why sticking to not-pencil-lifting. $\endgroup$ – Jorssen Oct 7 '19 at 21:09

A point (say, in $\mathbb{R}^n$) is a vector. Vectors and points are really no different. They are both $n$-tuples in $\mathbb{R}^n$.

The difference between two points (in $\mathbb{R}^n$) is a vector, but a vector has no fixed position. Points are positions in space. Vectors are displacements. It makes no sense to add two points, but it does make sense to add two vectors.

In spaces other than $\mathbb{R}^n$, points and vectors can be quite different.

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    $\begingroup$ I would call the say "vectors have a direction and a magnitude, but not a position'' is an informal explanation which causes a misconception. On a differentiable manifold, the basepoint of a vector is very important. You need a fancy gadget called a "connection" to even talk about moving a vector around. They only "are not'' important in Euclidean space because the connection is so easy to work with. $\endgroup$ – Steven Gubkin Oct 8 '19 at 14:24
  • $\begingroup$ @StevenGubkin: Good point! (haha) $\endgroup$ – Joseph O'Rourke Oct 8 '19 at 14:49
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    $\begingroup$ This is the distinction between an affine space and its vector space of translations stated in more prosaic language. It's a subtle point, but a good one. $\endgroup$ – Dan Fox Oct 16 '19 at 7:07

For a much lower-level topic, consider explaining to beginning algebra students why "like terms" can be combined. On a few occasions, I have resorted to reasoning with students that adding algebraic expressions is like adding quantities with units. [Our curriculum begins with units and geometry before algebra, so this is usually safe ground in my class.]

If they already believe that $5cm$ + $7cm$ makes physical sense, and $5cm^2$ + $7cm$ does not, then we can make an argument that $5x + 7x$ can reasonably be called $12x$, but $5x^2 + 7x$ cannot be combined nicely. However, I always regret this line of conversation, because I feel I have to backpedal for the sake of honesty. Since students recognize that $5cm^2 + 7cm$ isn't even computable (forgoing the possibility that $5$ and $7$ have convenient units), they will sometimes say that $5x^2 + 7x$ is also a "bad" expression that cannot represent anything real. This is obviously a problem. [Hence, I avoid this informal explanation if possible.]

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    $\begingroup$ If they argue that $5\text{ cm}^2 + 7\text{ cm}$ is not computable, show them a $5 \text{ cm}^2$ sheet of paper, and a $7\text{ cm}$ piece of string. Put them next to eachother. There. They have been added. It just doesn't mean very much to add such quantities. ;) $\endgroup$ – Xander Henderson Oct 8 '19 at 16:08

I shall post my humble and incomplete list of bad explanations I've given or heard over the years:

A function is continuous if you can draw its graph without lifting your pencil

As you mentioned this is bad, but, depending on the level of the student, it can be a reason for big or small misunderstandings.

At a high school level, this simply ignores the fact that ony needs the domain of the function to be connected so as to imply that the graph of the function is also connected - which is the closest I can get to "draw by pencil" in terms of strict mathematics. However, as mentioned in the comments, $x\sin\dfrac{1}{x}$ blows this story up.

At an undergraduate level I consider such a description useless, since e.g. any sequence is a continuous function under the $\varepsilon-\delta$ definition - $\mathbb{N}$ has no real accumulation points - so, this explanation only causes some trouble. Also, the notion of "drawing with pencil" get's pointless if one considers a function which is everywhere continuous and nowhere differentiable.

A possibly better explanation is that involving the high-school "definition" of continuity at $x_0$:

$$\lim_{x\to x_0}f(x)=f(x_0).$$

Under that definition, I prefer to explain to my students that continuity may be seen as the case when the estimation of the value of $f$ at $x_0$ based on its values close to it (in other words $\lim\limits_{x\to x_0}f(x)$) agrees with the actual value of the function itself. Usually, an example of a smooth flow of some liquid is a nice add on such an interpretation of continuity - always as an introductory approach; pathologic cases are left out under such a scope. Also, note that such an approach refers mainly to high-school students.

Integration is the opposite of differentiation

Well, this almost deterministically leads to the misunderstanding that if $f'$ is the derivative of some function $f$, then $f'$ is integrable. Evidently, this is not true, either because $f'$ may not be bounded, as in $f(x)=\dfrac{1}{x^2}\sin\dfrac{1}{x^2}$, which is a nice high-school level example, or because $f$ may be too bad, as the Volterra or Pompeiu functions.

To dig a little deeper, this sentence seems to lead to the more crucial misunderstanding that a function having an anti-derivative is integrable, as well. Well, this makes the whole stuff pretty much messy, since, neither having an antiderivative implies integrability nor the opposite is correct - take any piecewise constant and discontinuous function for the opposite.

So, while when it comes to "technique", (Riemannian) integration seems to be the inverse action of differentiation, it actually isn't. The way I've decided to tackle this is using the following question:

Is it true that every derivative of a function is integrable on any closed interval of its domain?

After some discussion, counter-examples etc - always depending on the level of the students - the Fundamental Theorem of Calculus emerges as a necessity to put some order in the chaos we've created - if the derivative is continuous, then integration is exactly the inverse of differentiation.

Another approach would be to introduce the Henstock-Kurzweil integral into high shcool... (kidding)

An indefinite integral is the set of all antiderivatives

Well, this is possibly the most "true" sentence one could write about the symbol $\int f(t)dt$, but I've seen it often cause much trouble. Consider the following equation:

$$\int f(t)+g(t)dt=\int f(t)dt+\int g(t)dt.$$

Pretty much neat. But, are we adding sets on the RHS? This is a natural question that may emerge, especially if one fails to mention somehow that indefinite integrals also posess some algebraic structure - since $\int f(t)dt$ is not merely a set of functions but an equivalence class of functions - hence the algebraic structure.

A possible way of tackling it, if one wants to avoid talking about vector spaces etc in a high school context, would be to define the indefinite integral as an arbitrary antiderivative of a function, however this may cause more trouble than that it solves.

The definite integral $\int_a^bf(t)dt$, $f\geq0$ is the area under the curve

I was a fan of this until I found out where the flaw is. The definite integral of a non-negative integrable function is not the area, but it may be interpreted as such - in a similar fashion, it is not the work of some force moving a particle, but it may be interperted as such.

To make myself clear, take the following inequlity:


If we take the definite integral on the LHS to be area and the sum of the RHS to be a number, then the LHS is measured into square units, while the LHS in simple units, making the use of $<$ between them nonsense. Yet, the RHS can be intepreted as area - e.g. a partial Darboux upper sum - and make both quantities "comparable".

Thus, the definite integral is interpretable,under some circmustances, as area, as well as every number - it would seem naive to try to compare my height with my weight, even if both of them are numbers :P.

A differentiable function's graph has no edges and/or sharp points

Well, this is a classic. Take, for instance $$f(x)=x^2,\ x\in[0,1].$$ Isn't $A(1,1)$ an edge of the graph of $f$? Isn't $f$ differentiable there? The crucial point is that the point under discussion should have an $x-$coordinate that is internal point of the domain of $f$. Provided that, this is also a nice intuitive reasoning of why we prefer to discuss differentiability on open sets.

A function is differentiable at $x_0$ if its graph has a tangent line there

And how about the case when $f$ has a vertical tangent line $(x=x_0)$ at some point $(x_0,f(x_0))$? In this case, $f$ is not differentiable at $x_0$ since:

$$\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\pm\infty,$$

yet it has a tangent line at $(x_0,f(x_0))$. This could be nicely coupled with the previous one as the two most misleading interpretations of derivative. I mean, the tangent line function is the historical problem that leads to the definition of the derivative, but tangent lines themselves are not equivalent to differentiability - differentiability implies the existence of a tangent line and not vice versa.

Two functions are equal when they share the same domain and have the same formula

Well, this is practical when it comes to high-school excercises, but it is unutterably misleading. A function may not have a "formula", in the sense that not all functions can be expressed in terms of elementary functions.

If the graph of $f$ lies below that of $g$ then $f<g$

Intuitively, this seems correct. However, a common misunderstanding is, as I see it, that the inequality:


does not imply that $f,g$ are comparable. For instance, consider:

$$f(x)=1-\sqrt{-1-x},\ g(x)=\sqrt{x}.$$

Since $D_f\cap D_g=\varnothing$, we cannot compare them, yet the former inequality holds. So, "below" seems to vague in some cases - note however that the inverse implication ($f<g$ implies that $f$ lies below $g$) is more clear, since we have established that $f,g$ are comparable.

The depictions of asymptotic behaviour of functions

Well, this is not some explicit explanation but an effect that seems to take place due to "overfitting". Most students depict the notion of a function asymptotically approaching a straight line in some way that implies monotonicity near $\pm\infty$. However, as, for instance, it happens with $f(x)=e^{-x}\sin x$ at $+\infty$, no such depiction of asymptotic behaviour is a priori correct - e.g. in the aforementioned example, $f$ constantly oscillates as $x\to+\infty$.

We add functions as we do with numbers: $(f+g)(x)=f(x)+g(x)$

Well, this equality never made sense to me as a high-schoold student. Are we saying something non-obvious? I mean, how else should addition between functions be defined? What I did not take into account was that what takes place in this expression is "function over-loading", as a programmer would call it. The LHS $"+"$ is a an addition that refers to the set of all functions while the RHS $"+"$ refers to the usual addition of the reals. That, what we are actually doing here is defining a new function, $f+g$, pointwisely.

I just wanted this included in the list, since I think it is never given the attention it needs, at least based on personal experience.

I may edit this list later on to add some more vague explanations.

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    $\begingroup$ "Defining the sum/product etc of functions" I don't understand why you included this among your list of bad informal explanations. It is a definition, not an informal explanation. $\endgroup$ – Jorssen Oct 8 '19 at 7:42
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    $\begingroup$ (1) I don't see anything wrong with the informal explanation that the antiderivative is a set of functions. The more precise explanation is that the antiderivative is an equivalence class of functions, where equivalence is up to addition by a constant. But an equivalence class is a set, and I see no reason to introduce the formalism of equivalence classes in an elementary calculus class. Students have some naive understanding of what a set is, so I see no harm caused by saying that $\int f(x)\,\mathrm{d}x$ is a set of functions. $\endgroup$ – Xander Henderson Oct 8 '19 at 15:05
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    $\begingroup$ (3) Have you actually seen the informal explanation "A function is differentiable at $x_0$ if its graph has a tangent line there" in the wild? I've seen the converse, i.e. if a function is differentiable at $x_0$, then the derivative gives the slope of the tangent line, but I'm not sure that I've ever seen the informal explanation you suggest. If so, I agree that it is a bad explanation, but it seems like a strawman. $\endgroup$ – Xander Henderson Oct 8 '19 at 15:16
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    $\begingroup$ (4) I don't see any problem with the explanation "If the graph of $f$ lies below that of $g$, then $f<g$." In the example you give, $f$ doesn't lie below $g$. It is lower, but also off to the side. I am not sure that students would automatically start thinking about the comparison of infima and suprema of functions. While I agree that "below" might be vague in some contexts, the whole point of an informal explanation is that it is imprecise, but should give some intuition. $\endgroup$ – Xander Henderson Oct 8 '19 at 15:20
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    $\begingroup$ (5) Finally, I see no problem with the informal explanation that functions are added like numbers. The precise, formal definition of the sum of two functions is something like "If $f$ and $g$ are two functions with a common domain $X$ and a common codomain $G$, where $(G,+_G)$ is a group, then the sum of $f$ and $g$, denoted $f+g : X\to G$, is a new function defined pointwise by the property that $(f+g)(x) = f(x) +_{G} g(x). $$" Informally, $(f+g)(x) = f(x) + g(x)$. It is definitely worth pointing out that the notation is being overloaded, but I don't see this as a bad informal script. $\endgroup$ – Xander Henderson Oct 8 '19 at 15:25

Modular arythmetic works just like a clock

I've seen several students expect "half-hours" making an appearance in $\mathbb{Z}_{12}$. This leads them to be confused about statements like $5^{-1}=5$ rather than $5^{-1}=0:12$

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