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I'm having an argument with someone who thinks it's never justified to teach something that's not strictly correct. I disagree: often, the pedagogically most efficient way to make progress is to iteratively learn and unlearn along the way.

I'm looking for examples in mathematics (and possibly physics) where students are commonly taught something that's not strictly true, but works, at least in some restricted manner, and is a good way to understand a concept until one gets to a more advanced stage.

MGA
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    whenever I draw a continuous function for my calculus students, it is invariably a smooth line. We unintentionally reinforce the idea that functions $\mathbb{R}\rightarrow \mathbb{R}$ are extremely well behaved. – James Jul 23 '14 at 15:27
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    This might be a nice question for the Math Education stackexchange as well. – Semiclassical Jul 23 '14 at 15:42
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    Most anything with a "d" in it. – djechlin Jul 23 '14 at 16:17
  • @djechlin Are you referencing something? – Dustan Levenstein Jul 23 '14 at 16:59
  • @DustanLevenstein leibniz notation. – djechlin Jul 23 '14 at 17:00
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    ah. There's no "d" in Leibniz. :P – Dustan Levenstein Jul 23 '14 at 17:02
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    Newtonian mechanics is not strictly true since it breaks down near the speed of light. But it's good enough that for the most part we build cars, bridges, ships and skyscrapers using it instead of general relativity for calculations. In fact, general relativity is not strictly true because it breaks down when confronted by the statistical model of quantum mechanics. But it's the best that we have at the moment so unless you're OK with saying "we know nothing so we can't do or build anything" then it's all we've got right now. – slebetman Jul 23 '14 at 17:12
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    http://en.wikipedia.org/wiki/Bohr_model is not strictly true. But it's taught in every grade 10 science class AFAIK – Cruncher Jul 23 '14 at 17:24
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    I think what djechlin is referring to is the notion -- which I saw all the time in my first year physics classes -- that "dx", "dy", etc, can be treated the same as any other term in algebra -- that you can multiply by them, divide by them, and so on, when in fact they are operators -- you can no more "multiply" by "dx" than you can multiply by "+". – Eric Lippert Jul 23 '14 at 18:45
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    @EricLippert $d \over dx$ is usually thought of as an operator. I believe it's a much harder sell to say that $dx$ itself is an operator. – Eliah Kagan Jul 23 '14 at 19:13
  • @EliahKagan: You are of course correct; it's been a long time since first year physics. :-) – Eric Lippert Jul 23 '14 at 20:19
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    We have no theory in Physics which is strictly true (and that's fine). – jinawee Jul 23 '14 at 20:48
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    Probably better suited to http://matheducators.stackexchange.com. – lhf Jul 23 '14 at 21:03
  • I've always wondered if the existence of the monster model for a given theory (in mathematical logic) is conceptually correct. – Martin Brandenburg Jul 23 '14 at 23:30
  • At the moment I can't find the author of the quote that teachers must tell "the truth, nothing but the truth, but not the whole truth" (I have a vague recollection that it was Einstein) but I tend to agree. It's not clear to me whether your position agrees or disagrees with this dictum. – joeA Jul 24 '14 at 11:58
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    It is impossible to be a research mathematician, nor a research physicist nor chemist etc., unless you use conceptualizations that are not strictly true. How else would you discover true facts without first fumbling around and discovering almost-maybe-kinda-true facts? I suppose it is the teacher's choice whether to hide this reality from students. But it is not my choice. In fact that's how I believe an ideal classroom should function: students fumble around and discover the true facts for themselves. – Lee Mosher Jul 24 '14 at 15:10
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    This sounds like Terry Pratchett's "Lies to Children" concept that's used in the Science of Discworld books http://wiki.lspace.org/mediawiki/index.php/Lies-To-Children – TV's Frank Jul 25 '14 at 09:12
  • All of my applied mathematics classes at school assumed $F=ma$, but then again, it sufficed for the class of problems we were solving. – Neil Jul 26 '14 at 00:37
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    @EricLippert: Symbols like $dy$ and $dx$ can be defined as infinitesimal numbers. That's historically what they originally meant, and there's nothing wrong with it. See http://math.stackexchange.com/questions/21199/is-fracdydx-not-a-ratio?lq=1 –  Jul 26 '14 at 01:08
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    @BenCrowell: Indeed, I think this is a good example of what the original poster is looking for: a concept that is not strictly speaking well founded, but which leads to good insights. There are no numbers that are "infinitesimally small", but by pretending that there are, you can get correct insights about calculus. As the answer you linked to points out, those concepts were not addressed rigorously for hundreds of years, but they were useful for all that time. – Eric Lippert Jul 26 '14 at 13:55
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    Mathematicians are really sexy. – Daniel R Hicks Jul 26 '14 at 19:46
  • Seems like there are often cases where a mathematical result feels like it ought to be true, but isn't, because of some pathological counterexample. Mathematicians revel in these counterexamples. Whenever I see such things, my first thought is that the definitions are faulty. With a little more care in defining the concepts, perhaps we could ensure that intuitively true things are actually true. Then we could spend more time pondering the core of the concepts, and less on the corner-case counterexamples. – bubba Jul 28 '14 at 04:53
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    Seems more adequate for M.E.S.E.. – Git Gud Jul 28 '14 at 13:58
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    We now have quite a lot of answers to this question. There are surprisingly few that are actually restricted to mathematics as opposed its relationship to physics. – Keith Jul 29 '14 at 00:45
  • It seems like you're trying too hard. Your way doesn't have to be the "pedagogically most efficient" way to be justified. It could be the fourth, fifth, or thousandth most efficient way and it would still be a damn good way. – Marcel Besixdouze Jul 31 '14 at 05:06

31 Answers31

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How about this notorious one I remember from high school?

"$f(x)$ is just a fancy name for $y$."

Dustan Levenstein
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    Wh$f(x)$ would an$f(x)$bod$f(x)$ sa$f(x)$ something like that? What's $y$, anyway? – Daniel Fischer Jul 23 '14 at 18:10
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    @DanielFischer It's because $y$ is often used as a function of $x$ in basic grade-school algebra courses. It might work okay in those contexts, but then when the same students go on to calculus, it becomes evident pretty quickly that many of them never really learned what a function was in the first place. And, moreover, at least in my experience, many of those students even manage to slip through a calculus course without ever figuring out what a function is. – Dustan Levenstein Jul 23 '14 at 18:12
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    I guessed that ;) I have faint memories of seeing stuff like $y = x^2+2$ myself. I don't know an answer to "Why would anybody say something like that?", however, to this day. – Daniel Fischer Jul 23 '14 at 18:16
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    Another humorous variation on that is a question one can use to tell physics students apart from math students: "If $f(x,y)=x^2+y^2$, what is $f(r,\theta)?$ – Semiclassical Jul 23 '14 at 18:28
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    @Semiclassical The notion of $\theta^2$ gives me a bit of a headache. – Venge Jul 23 '14 at 23:36
  • @Venge $ r=\theta^2$ in polar coordinates is just a spiral. – Dustan Levenstein Jul 23 '14 at 23:45
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    Can we clarify? $f(r,\theta) = r^2$ is what a physicist would say, correct? – Brady Trainor Jul 24 '14 at 00:15
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    Exactly. The physicist would assume $(r,\theta)$ to be polar coordinates, not $(x,y)\mapsto(r,\theta)$. @BradyTrainor – Semiclassical Jul 24 '14 at 00:42
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    Similarly, it is common for the math people to write $f(x) = f(x, y(x)) = f(x, y)$. And then take the derivative, but I can't remember if or what can go wrong with the notation at that point. – Brady Trainor Jul 24 '14 at 00:53
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    @Brady what does $ f_x $ mean? There are two possible distinct answers. – Dustan Levenstein Jul 24 '14 at 02:05
  • @Semiclassical So what do mathematicians/physicists say? – BCLC Jul 29 '14 at 23:38
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    @BCLC a physicist would say $f(r,\theta)=r^2$ whereas a mathematician would say $f(r,\theta)=r^2+\theta^2$. – Chris Taylor Aug 01 '14 at 14:33
  • Sorry but can you/someone else clarify what's wrong with writing a function, say $f(x)=x^3$ as $y=x^3$? – HeWhoMustBeNamed Oct 27 '17 at 17:31
  • @MrReality My point is that many students never seem to learn intuitively what a function is, as an association from $x$-values to $y$-values. This gets them into trouble when they reach calculus, and can't figure out what needs to be plugged in where when solving a problem.

    An equation involving $x$ and $y$ defines a relation between $x$ and $y$, which can be graphed just as easily whether it's a function or not (e.g., $x^2+y^2=1$). But it's important to understand the special significance of a function.

     – Dustan Levenstein Oct 27 '17 at 18:18
  • @Dustan What sort of trouble? I mean I see the notation $y=$some expression involving x all the time. Like in a problem, if we're given that $y=x^3$ then we can easily compute $\frac{dy}{dx} = 3x^2$. In fact- at least at the level I'm at- in problems this is more common than $f(x)=x^3$. Could you clarify what you meant? – HeWhoMustBeNamed Oct 28 '17 at 12:05
  • @MrReality You clearly haven't had this trouble, which is good for you. It's hard for me to recall specific examples, but I have seen this repeatedly in my tutoring and teaching experience. For example, when computing the power series of $\arctan(x)$, I would try to guide the students through taking the derivative, expressing that power series, and then integrating. They get into trouble and stressed out because they don't label their power series by the corresponding functions... – Dustan Levenstein Oct 28 '17 at 13:12
  • ... And of course I have seen the notorious $\frac{\sin(x)}{x} = \sin$ that one should immediately see makes no sense if they understand what a function is. Those are the two examples I can think of off the top of my head. I also remember people having trouble with the difference quotient, but I can't remember precisely what. – Dustan Levenstein Oct 28 '17 at 13:13
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Thinking of Dirac delta function as a function works reasonably well up to a certain point. For example, every physicist knows that $$\int_{-\infty}^{\infty}e^{i\omega x}dx=2\pi\cdot\delta(\omega),$$ but only a small part of them really studied the theory of distributions.

Start wearing purple
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    Physics is a rich vein for examples like this, since we usually only worry about objects which are 'nice enough' for us not to worry about all the details. For instance, in many cases a physicist will simply assert that boundary terms vanish when doing integration by parts; they would almost never refer to the concept of weak differentiability or Sobolev spaces. (Myself included.) – Semiclassical Jul 23 '14 at 15:34
  • @Semiclassical Although vanishing or at least periodicity actually matters - say, for Noether theorem. – Start wearing purple Jul 23 '14 at 15:37
  • If I remember, another comes up when physicists, who want to forgo vector calculus in lieu of differential forms, try to give various visualizations in terms of surfaces and such. – Semiclassical Jul 23 '14 at 15:38
  • Yes, but in a physics course (e.g. classical mechanics or QFT) that vanishing would usually be affirmed on physical grounds. – Semiclassical Jul 23 '14 at 15:41
  • I think it works very well indefinitely, if only one suitably enlarges one's notion of "function", e.g., dropping a too strict demand for "pointwise values" and such. – paul garrett Jul 23 '14 at 16:58
  • @paulgarrett Surely if one gives up "pointwise values", one is working in the theory of distributions? – Max Jul 24 '14 at 08:50
  • @Max, not necessarily... even $L^p$ spaces of functions don't have pointwise values everywhere, etc. – paul garrett Jul 24 '14 at 12:31
  • That's not even the worst. Physics students (and Professors) handles derivates the worst, even phisics books throw around differentials like ordinary ratios, and they just seldom replace it with a ratio of two differences. – Evan Dark Jul 25 '14 at 19:32
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Here are two false concepts taught in lower grades (explicitly or implicitly):

1) Every plane figure has an area. Elementary kids are not told that some figures have no area (or might not have an area, if you leave out the Axiom of Choice).

2) A set is a collection of objects satisfying any particular relation. (Some middle-school books avoid this by always discussing sets within some particular universal set, but this approach just caused me to raise questions even before I studied formal set theory.)

Rory Daulton
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    I would have a hard time calling anything the existence of whose area depends on AC a 'figure'! At the elementary level (and well past it), it's true that every plane figure has an area because of how the concept itself is defined at those levels (informally, the interior of a closed rectifiable curve, or possibly the union/intersection/difference of a finite number of such curves). – Steven Stadnicki Jul 23 '14 at 18:12
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    @StevenStadnicki I just realized that there's something confusing me on this point. I know there are non-measurable sets whose construction depends on AC; are there sets whose existence can be proven within ZF but which can only be proven immeasurable with choice? – Dustan Levenstein Jul 24 '14 at 21:29
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    @Dustan: What this comes down to is whether there is any definable set the ZFC proves non-measurable at all -- i.e. whether there is a formula $\varphi(x)$ such that ZFC proves $$\exists_1 x.\varphi(x) ~\land~ \forall x.\varphi(x)\to x\text{ is not measurable}$$ If there is such a formula, then $\bigcup_{x\subseteq \mathbb R, \varphi(x)} x$ can (trivially) be proven to exist by ZF, whereas ZFC will additionally prove it is non-measurable. – hmakholm left over Monica Jul 26 '14 at 16:26
  • Ad (2): A relation on a set is a subset of the cartesian product of this set. Still a set is only a subset of another set. – Horst Grünbusch Jul 28 '14 at 13:43
  • Outer measure is a perfectly reasonable definition of "area", and agrees with the Lebesgue definition on the measurable sets. The existence of so-called "non-measurable sets" (terrible name: you can easily measure them using their outer measure, or their inner measure for that matter) just shows that the most obvious definitions of "area" fail to be sigma-additive. Which is not surprising, considering sigma-additivity is a ridiculously powerful thing to ask of a reasonable notion of area, certainly when you also want to have the Axiom of Choice. – Marcel Besixdouze Jul 31 '14 at 05:02
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    @MarcelT.: If you use outer measure, you not only lose sigma-addivity, you also lose simple additivity. That is a big deal for early students, partly because simple additivity is an axiom in Euclid, and partly because it is so intuitive. There is a reason the Banach-Tarski paradox is so famous! My point was that letting early students think that every plane figure has an area and that area is additive is a good practice: the student will learn better later if he continues far enough in mathematics. – Rory Daulton Jul 31 '14 at 12:25
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  • Multiplication is repeated addition. So $\sqrt{2}\cdot \sqrt{8}=4$. Similarly division is repeated subtraction.
  • Solve the differential equation $$\dfrac{dy}{dx}=y$$ Easy! Multiply both sides by $dx$, so $\dfrac{dy}{y}=dx$; now, integrating both sides, $\ln{y}=x+C$. I don't know how many people understand this multiplication by $dx$ (frankly, I don't). Even university students (at my university) take this step for granted. (Relevant: Is $\frac{dy}{dx}$ not a ratio?)
  • The cancellation trick: $\require{cancel}x\cdot y= y\cdot z\implies x\cdot\cancel{y}=\cancel{y}\cdot z\implies x=z$. People use this 'trick' to prove $1=2$ etc.

  • High school teacher: Square root of negative number is not defined. So $x^2+1=0$ has no roots.

    Secondary school teacher: define $\sqrt{-1}=i$ and $x^2+1=0$ has two roots $x=-i,+i$.

  • "$1/0$ is $\infty$" Refer to point 5 here, this is correct.
  • $\dfrac{\partial^2 f(x,y)}{\partial y \, \partial x}=\dfrac{\partial^2 f(x,y)}{\partial x \, \partial y}$. This can't be false in Electrodynamics.
  • $1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\cdots =2$. The equality is little confusing. It should rather be considered as a limit of the sum (or limit of partial sums).
  • $\sqrt{mn}=\sqrt{m}\cdot\sqrt{n}$. This is only true when at least one of $m$ and $n$ is positive. This 'trick' can be used to prove $1=\sqrt{1\cdot 1}=\sqrt{-1 \cdot -1}=\sqrt{-1}\cdot\sqrt{-1}=i^2=-1$.
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hrkrshnn
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    On the physics side, at least, the multiplication of both sides by $dx$ on both sides is usually taken for granted as some 'infinitesimal' physical process. This applies especially in the context of thermodynamics (which evidently doesn't actually complete mathematical sense unless you understand contact manifolds---which I don't!) – Semiclassical Jul 24 '14 at 14:43
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    @Semiclassical I don't even think it makes sense as an infinitesimal process at all since $\frac{dy}{dx}$ is a notation, not an actual fraction. (It's partly for this reason that I am not a fan of Leibniz notation.) The best way to do it is via chain rule, which is really what is going on. $$\frac{dy}{dx} = y \Longrightarrow \frac{1}{y}\frac{dy}{dx} = 1.$$ The left hand side is recognizable as $(\log y)'$ and from there you can integrate easily. This "trick" is stupid and abhorrent. – Cameron Williams Jul 24 '14 at 17:17
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    Well, I chiefly have things like rates of change and the like in mind. But if you don't care for that, then you really won't like this bit of thermodynamics. @CameronWilliams – Semiclassical Jul 24 '14 at 17:36
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    @CameronWilliams in many cases in physics, it is an "infinitesimal" process we're dealing with, and the use of the derivative is only an approximation to the true situation which involves a ratio of small quantities. So the interpretation of $dy/dx$ as a fraction is justified. – David Z Jul 25 '14 at 08:51
  • The cancellation trick does work, except in the case $y=0$. If your teacher forgot to mention that, then they're not doing their job properly.

    Also it's a common misconception that you can take the square root of negative numbers. The equation $x^2+1=0$ does have the two solutions $\pm i$, but the expression $\sqrt{-1}$ is generally not allowed, since it would generate unnecessary ambiguity (of the kind $1=\sqrt{1}=\sqrt{-1\cdot-1}=\sqrt{-1}\sqrt{-1}=-1$, also used to prove that $1=2$).

     – Jonas Granholm Jul 25 '14 at 12:56
  • Your series appears to be missing the $\frac{1}{2}$ term. Also, 1/0 is confusing, even IEEE floating point has +0 equal -0 but 1/+0 is not equal to 1/-0. – Neil Jul 26 '14 at 00:15
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    @JonasGranholm, Why do you think we can't take square root of negative numbers? $\sqrt{mn}=\sqrt{m}\sqrt{n}$ can be applied when at least one of $m$ and $n$ is positive. – hrkrshnn Jul 26 '14 at 05:25
  • @Neil. Yes $1/0$ is definitely confusing. But I know people who still teach students(primary school) that $1/0$ is a 'huge' number and called $\infty$. – hrkrshnn Jul 26 '14 at 05:30
  • @boywholived, The root function cannot be extended to an analytic function on all complex numbers; the values a line from $0$ to $\infty$ must be kept undefined. This line is often chosen to be precisely the negative real line. – Jonas Granholm Jul 26 '14 at 10:35
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    @JonasGranholm it doesn't have to be undefined. You can choose values of any branch for this line. One way to define square root would be to choose values with the argument in e.g. $(-\frac\pi2,\frac\pi2]$. – Ruslan Jul 27 '14 at 08:53
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    @CameronWilliams: I believe the notation actually makes sense if you use one of the non-standard analysis frameworks as a foundation, which provide consistent rules for infinitesimals. So the trick doesn't work in traditional analysis, but I wouldn't call it "stupid and abhorrent" since we can make the trick logically consistent, if only we completely rewrite the entire foundation of calculus. – Dietrich Epp Jul 27 '14 at 17:37
  • @DietrichEpp Well I agree with that but if you're in the realm of standard calculus, it's a bad trick. However if you've developed things in nonstandard analysis, then it's okay. They're equivalent theories for many purposes but it's still wrong based on what students are taught. – Cameron Williams Jul 27 '14 at 19:08
  • Multiplication by $dx$ means considering each side as a differential 1-form, and then the integration is finding the function (0-form) whose exterior derivative is the given one. In short: it's a notational trick that helps you keep track of change of variables. – Ryan Reich Jul 31 '14 at 05:13
  • @JonasGranholm Note that this argument would also mandate that $\sqrt 0$ be undefined because it is a branch point, even though it (is the only point which) has a unique solution $\sqrt 0=0$. You can't extend $\sqrt x$ to a continuous/analytic function on all of $\Bbb C$, but you can certainly define it on $\Bbb C$. – Mario Carneiro Mar 11 '15 at 10:21
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We tell students in an introductory algebra class, say, that mathematicians invent axioms and then study the properties of the resulting object or theory. The story goes that some bright mathematician had the idea to write down the four group axioms before anyone knew anything about group theory, and then he and other mathematicians worked out all the known properties of groups from those four axioms.

In reality, it took mathematicians decades of working with concrete examples of groups -- permutation groups and groups arising from geometric symmetries -- to finally notice that all of those examples shared a few fundamental properties. These properties then became the axioms for the group, but only because the objects satisfying those axioms had already appeared naturally in other fields of mathematics.

I think it is very rare for a mathematician to define some random set of axioms and then have something interesting to say about them. Instead, axiomatizing a theory is usually one of the final steps in making that theory rigorous. Axioms are more a sign of rigor mortis than of the potential for new results. Moreover, one theory -- most notably set theory -- can have competing axiomatizations which are all studied.

And yet, I think it is pedagogically useful to introduce students to the idea of an axiom in this way. A student who has studied no formal mathematics will not yet appreciate the subtle interactions between hypotheses, conclusions, and observations that govern mathematical progress. It is easier to give him or her a concrete set of "rules" to work with, and the cleanness of the ensuing theory -- by which I mean that steady march of (possibly unmotivated) definitions, lemmas, and proofs across the pages of math textbooks -- shows him or her our ideal for formal mathematics.

Edit: I realized that what I have written so far can be interpreted to mean something with which I vehemently disagree. Even though I think expositions of mathematics at the level of a first- or second-year undergraduate can benefit from an axiomatic treatment, any expositions that are more advanced -- certainly graduate textbooks, for instance -- should complement the rigid axiomatic method with thorough motivation for all definitions. The motivation is as important, if not more important, than the definitions and theorems!

user134824
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    There are very valid pedagogical methods which entirely avoid this interpretation of the axiomatic methods. Regarding your group theory example, just as mathematicians spent many decades studying concrete examples before groups were axiomatized, one can spend a few lectures teaching students many examples before formally axiomatizing them. The concrete "rules" are then grounded in the reality of the examples. – Lee Mosher Jul 24 '14 at 15:18
  • This is an amazing point. I've become increasingly more frustrated with advanced math texts for reasons such as this. I was trying to figure out some stuff about group $C^$ algebras and so many of the definitions and ideas just seem .. counterintuitive. That is, until you step back and consider the finite group case. In this setting, everything falls out nicely and gives a very* clear guideline to go by. Proofs from the finite group case carry over to the locally compact group case perfectly naturally (with some alterations) and make complete sense in this light. – Cameron Williams Jul 24 '14 at 17:11
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    So many texts just drop the motivation or guiding principles for brevity but it makes the subject matter seem completely arbitrary and unmotivated. – Cameron Williams Jul 24 '14 at 17:12
  • "Rigor mortis"? I think that you're using it wrong. – Asaf Karagila Jul 25 '14 at 22:33
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    It was a bad joke: I was trying to say that mathematical theories which have been "satisfactorily axiomatized" are usually no longer active fields of study, or at the very least have had most of their low-hanging fruit picked. Rigor mortis refers to the stage of bodily decomposition when a corpse's limbs stiffen. – user134824 Jul 25 '14 at 23:03
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    I prefer to think of a "satsifactory axiomatization" as meaning that mathematicians are broadly agreed (a) that this is worth talking about, and (b) what "this" is. Calling it "rigor mortis", granted was just a joke, but does incorrectly imply the subject is dead, which is a little harsh on existing group theorists! It's likely true about the low-hanging fruit, since easy important results will have been found without the help of the axiomatization. But if the axiomatization makes the subject easier (as it should), then there might be a new wave of low fruit, you never know :-) – Steve Jessop Jul 26 '14 at 20:28
  • Point taken! I get carried away sometimes . . . – user134824 Jul 26 '14 at 20:40
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    I personally think the "rigor mortis" joke was pretty good, especially because it arises in the final stages of rigorisation. – michaelb958--GoFundMonica Jul 31 '14 at 21:36
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I'm surprised nobody mentioned square roots before: I am 17 now and have always been told the square root of a negative number does not exist and can not be taken. I am interested in mathematics, so I know this to be false (though I don't really know what imaginary numbers are).

To not teach this at first makes sense, because when one first learns square roots one is generally not ready for imaginary numbers. (Until you are in high school IMHO - though the maths classes I take didn't and won't cover them.1)

1: where I live you can choose three types of maths: A, B and C, where A is mostly statistics and meant for people not very good at or interested in maths, C is a simplified version of A for people who are really bad at maths and B is 'real' mathematics, heavy on algebra and proving things, I think with B you'd learn about imaginary numbers. (As pointed out in the comments, if B isn't enough maths for you you can choose Maths D as a kind of add-on. Basically, choosing D means you get extra tough maths, on top of the regular B programme.) I had to choose A because otherwise I would not have been able to choose certain other subjects (though I was advised by my math teacher to take B, I chose these other subjects over challenging maths - my mistake).

11684
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    Would you mind telling me where you live? I'm interested in your education system. – Luka Horvat Jul 24 '14 at 09:17
  • The Netherlands. Here's an overview of the education system: first 8 years elementary school (age 4-12), then four, five or six years middle/high school (age 12-16/17/18) (this depends on the level of education, higher education takes longer) then university (or a similar school - university is only for those who have completed VWO, the six year middle school). Halfway through middle school (so after two or three years) you have to choose a 'profile', which consists of three parts: 1) the subjects everyone has (sports, Dutch, English) 2) some mandatory subjects for that profile and... – 11684 Jul 24 '14 at 09:26
  • ...3) about two semi-free choice subjects (semi, because most profiles have different but sometimes overlapping choice-subjects). AFAIK there are 6 different profiles, but parts 2 and 3 could be different between schools (but are checked and approved by the government anyway). Mathematics always is in category 2, so because I wanted a certain combination of subjects (Latin and Greek) I could only choose the least scientific profile, so no Maths B... @LukaHorvat – 11684 Jul 24 '14 at 09:29
  • "I think with B you'd learn about imaginary numbers." -- Not that I recall, but it's been a long while for me, so I cannot rule out either misremembering, or it having changed in the last >10 years. – hvd Jul 24 '14 at 14:56
  • I'm pretty sure I taught myself imaginary numbers (when playing around with the Mandelbrot set, as I recall), and never learned anything about them in school (US, not a mathematician) beyond "this is a thing that exists". – Brian S Jul 24 '14 at 15:55
  • I think this is a good point conceptually speaking (+1) but just to clarify, what your teachers taught you is technically right--because they are assuming the set of real numbers R. You have to operate over a different "rule set" in order to make things like imaginary numbers possible. If you were using the set of "Counting numbers" it would be considered impossible to subtract anything from 1, if you were using whole numbers you could not accurately divide an odd number by two. – Bill K Jul 24 '14 at 16:43
  • @BillK That makes sense. I knew of set theory but didn't connect the dots. Thanks! – 11684 Jul 24 '14 at 16:59
  • @hvd I don't know, I have A... :( – 11684 Jul 24 '14 at 16:59
  • I did VWO in 2000 and finished in 2006 and I'm pretty sure imaginary numbers was covered. We only had Wiskunde A and B (not C) but that was also divided in B1 and B2. I did B1 and 2. Maybe difference could also be in the education level whether it is covered or not – Ivo Jul 25 '14 at 10:42
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    Actually it's a common misconception that you can take the square root of negative numbers. The equation $x^2=-1$ does have the two solutions $\pm i$, but the expression $\sqrt{-1}$ is generally not allowed, since it would generate unnecessary ambiguity (of the kind $1=\sqrt{1}=\sqrt{-1\cdot-1}=\sqrt{-1}\sqrt{-1}=-1$). – Jonas Granholm Jul 25 '14 at 12:44
  • Actually, there is a fourth type of Maths you can take in the Netherlands: D. That is the one that teaches about imaginary numbers. Not all schools offer D and you can only take it if you also take B. – Jasmijn Jul 26 '14 at 13:27
  • @Robin Indeed, maths for maths-addicts. But because so little people take these classes (and because it isn't even a stand-alone subject) I thought I'd leave it out of my super-global overview. – 11684 Jul 26 '14 at 13:30
  • Sure. It was in this case the most relevant one, though. ;) – Jasmijn Jul 26 '14 at 13:39
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    @Robin Good point. I edited it in. – 11684 Jul 26 '14 at 14:01
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    @JonasGranholm That's a common misconception. You can take the square root of any complex number. But, since every nonzero complex number has two distinct square roots, you must (well, should) say which one you mean. In principle, that also applies to positive real numbers, but there we have the strong convention that unless explicitly specified otherwise, $\sqrt{\hphantom{z}}$ denotes the positive branch of the square root. – Daniel Fischer Jul 26 '14 at 14:17
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    @DanielFischer Ok, I was a bit overzealous. What I meant was that many believe that $\sqrt{-1}=i$ (or worse, $\sqrt{-1}=\pm i$) is a mathematical fact. You can make somewhat satisfactory extensions of the root function that allow square roots of negative numbers, but it's neither perfect nor extremely useful, so the square root of negative numbers is often left undefined. – Jonas Granholm Jul 26 '14 at 17:10
  • Euler's proof (source?) to find all positive integer solutions to $y^3 = x^2 + 2$ becomes incomprehensible once you say that $\sqrt{-1}$ has a value. – Rhymoid Jul 26 '14 at 17:41
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Often in physics you pretend like higher order differentials vanish. For example, you have $(x+dx)(x+dx)$, you multiply through and say that the $(dx)^2$ term is $0$. You could try to appeal to nonstandard analysis for manipulating the differentials, but of course the amount of work involved to develop such a theory is hidden.

CoffeeCat
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  • The problem gets even hairier when it gets to volume differentials and surface differentials. For instance, when computing something like an electric field you'd commonly see it explained as: "Find the contribution $d\mathbf{E}$ due to a differential charge $dq=\rho dV$ with charge density $\rho(\mathbf{x})$." A nicely physical-sounding route, but one which I imagine would raise objections on the math side. – Semiclassical Jul 23 '14 at 18:09
  • @Semiclassical Can you clarify the mathematical objections that would raise? I'm curious to know. – Shivam Sarodia Jul 23 '14 at 19:28
  • @Draksis: I mean in the following sense: A physicist might regard $dq$ simply as meaning a small amount of charge at one spot, and that integration just means adding up all these bits of charge. (It's a common way of speaking to intro physics students about electric field calculations.) But that doesn't quite match the mathematical concept of $dq$ as a differential 3-form. – Semiclassical Jul 23 '14 at 19:31
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    No physicist would say that the $(dx)^2$ term is zero in this argument. The term would be discarded if it is smaller than is relevant for the computation. – Jonathan Jul 24 '14 at 12:45
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    Dear @Jonathan: Do you have an example of a situation where $(dx)^2 \neq 0$? – Bruno Joyal Jul 24 '14 at 17:10
  • http://math.stackexchange.com/questions/822664/could-we-assign-a-numerical-value-to-an-infinitesimal –  Jul 25 '14 at 20:56
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    @BrunoJoyal Quantities like $(dx)^2$ become relevant when thinking about the second derivative, although it is very hard to find this made rigorous in the literature. One source is Michor's "Natural Operations in Differential Geometry". $(dx)^2$ would be seen as a map from the iterated tangent bundle $T(TM)$. For example, the formula $d^2f = f''(x)(dx)^2+f'(x)d^2x$ is invariant under coordinate changes (satisfies a "chain rule"). – Steven Gubkin Jul 26 '14 at 14:39
  • @BrunoJoyal When I studied introductory calculus I played around with the "quantity" (or, mish-mash of symbols if you will) $(d(u v)-u dv-v du)/(dx^2)$. http://mathb.in/14694 It's not so concrete an example but some might find it interesting. –  Jul 31 '14 at 04:17
13

Euclid's 5th postulate: two lines at right angles to the same line will never meet--and only if they are both at right angles. Well, in a perfect plane universe that might be true, but in the real world, non-Euclidean geometry is more "true": large triangles on the surface of the planet don't add up to 180 degrees, etc.

brainypirate
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    Why would large triangles on the surface of the planet be any more "in the real world" than large triangles, for example, between three stars? – JiK Jul 23 '14 at 21:12
  • A more fundamental example is space-time, which (sayeth special relativity) is a non-Euclidean geometry (with metric $ds^2=dt^2-d\mathbf{r}^2)$. General relativity goes further in endowing spacetime with curvature, and in both cases those have important physical ramifications. (time dilation, twin paradox, gravitational lensing...) @JiK – Semiclassical Jul 23 '14 at 21:40
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    @JiK, Three stars aren't in the world, they're in space. [/cheeky] :) – Brian S Jul 24 '14 at 15:52
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    I think the oversimplification here is not that the fifth postulate is true, but that Euclidean geometry can be applied to the curved surface of our Earth. (Although I don’t remember anyone making that particular assertion when I was in school.) – bdesham Jul 25 '14 at 22:29
  • @Jik A triangle between three distant stars will also have an angle sum different from 180 degrees. However, I think that this answer is misleading, because the geometric principle is mathematically true - only that kind of geometry does not work on astronomical scales. – Dave Jul 26 '14 at 08:38
  • In real life, straight lines and right angles don't exist. – jinawee Jul 27 '14 at 09:52
  • @JiK triangle between three stars is actually in the Minkowski spacetime, so in hyperbolic geometry actually. – the gods from engineering Feb 09 '15 at 11:27
11

For the physics case, and I say this as a physicist not a mathematician having a dig:

Almost all of physics is "not strictly correct".

What I really mean is: "strictly correct" in mathematics means an entirely different thing to what it means in physics. In fact, in physics there is a much larger set of interpretations of the term than there is in mathematics. Mathematics as a discipline considers "correctness" as fundamental. Not so in physics.

People have different views as to how Physics works. Mostly it is about modelling the world such that experiments can be constructed that will show results consistent or inconsistent with the model.

What this means is that the mathematical correctness of the model is a completely different thing from its physical correctness.

Indeed, if you had a model known to be a bit flaky mathematically but able to be used to give experimentally valid results, then it is better than a total mathematically consistent one which does not.

Now you might say how can a "flaky" model not necessarily have a space of solutions that are non-physical and hence not experimentally valid. The answer is that those conditions may only appear in a situation we cannot actually observe or construct due to practical reasons.

So to come back to the claim almost all physics is "not strictly correct": there simply is not basis for ever saying that a theory is strictly correct. For example, the Lorentz velocity addition formula is "better" than the Newtonian one, but is a the end of the day a model applicable to specific situations only. We cannot for example, just go about adding velocities for bodies at different locations once we allow for General Relativity.

It is in this sense that the current state of the art of physics "is a good way to understand a concept until one gets to a more advanced stage". Only if and when we have a serious candidate for the Theory Of Everything can (maybe) we go beyond this.

So to cap my answer, I'd suggest excluding physics from the question such that this becomes a non-answer . . .

Keith
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  • Another physicist here. I definitely agree with your main thrust---that the standard of physics is not proof, but progress. But I disagree with one point of substance regarding relativity: While adding velocities may not make much sense in GR v. SR, I think that's a bit of cherry-picking. What matters in SR is space-time as something inherently non-Euclidean in its geometry, a notion which GR develops even further. – Semiclassical Jul 25 '14 at 05:12
  • Only picked on velocities as it was given in another answer and to me it shows nicely the meaning of "correctness". Both the simple and Lorentz formula are mathematically correct. The Lorentz formula is less physically "correct" than we are given the impression when we first learn SR. Importantly, even GR is also not "correct". The Dirac delta example is more interesting in being mathematically flaky as generally introduced, but physically useful. – Keith Jul 25 '14 at 05:24
  • Fair enough, and I agree regarding Lorentz velocity transforms (though if we were talking about Lorentz covariance, I'd have different words to say). – Semiclassical Jul 25 '14 at 05:28
  • To me the more pertinent example in SR to this discussion is the twin paradox. "Intution" might make that seem odd, but from the perspective of geometry it's just the triangle inequality as appropriate for Minkowski spacetime. That's where I think one shouldn't be too quick to dismiss the relevance of math to physics. – Semiclassical Jul 25 '14 at 05:30
9

Here's a technical one I ran across at the beginning of grad school which always stuck with me. During questions after a colloquium talk on condensed matter physics, the speaker mentioned the fact that defects in these systems could be classified by topology. In the process, he equated the notion of homotopy methods with the concept of winding numbers.

This raised my eyebrows, since in my last undergraduate semester I'd spent a good amount of time reading up on my own about homotopy v. homology in the context of Cauchy's theorem in complex analysis. The problem with conflating homotopy with winding number is that, at least as the term is used in complex analysis, winding numbers are additive (i.e. commutative) whereas the first homotopy group could perfectly well be nonabelian. (In other words, there's situations where contours are non-contractible yet still integrate to zero because the winding numbers sum to zero).

So if one interpreted winding numbers in the complex analysis way, the analogy of that with homotopy just doesn't work. That bothered me for quite a while, and at some level it still does: defects can be classified using homotopy groups, yes, but those groups are free to be nonabelian. (Now I justify that language to myself by noting that Grassman numbers aren't commutative either.)

Semiclassical
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I can think of two:

1) For teaching elementary calculus, treating $\dfrac{dy}{dx}$ as a fraction is a great way of working with things like the chain rule, separable differential equations etc.

2) "To multiply by $10$ add a zero to the end" works perfectly for integers, and fails spectacularly for decimals.

Mathmo123
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    Nevermind other number systems... binary, etc. – eatonphil Jul 25 '14 at 19:50
  • Appending a $0$ to multiply by $10$ also fails in algebra: $10x=x0$ is not generally true. A student who understands why "add a zero to the end" works, knows it works with a particular way of writing numbers. Decimals (i.e, numerals written with a decimal point) and general algebraic expressions aren't written this way; such a student wouldn't explicitly believe "add a zero to the end" applies. (They might still do it without thinking, but we all apply the wrong algorithm from time to time.) Teachers of the add-a-$0$ way should explain why it works; then there's no oversimplification. – Eliah Kagan Jul 25 '14 at 23:41
  • Of course I agree... on the other hand saying "increase everything by one place value" to an 8 year old may fall on false ears. – Mathmo123 Jul 26 '14 at 21:29
  • I was taught this as "$10 = 10.00000...$, to multiply by 10 move the decimal point to the right to get $100.00000...$" – undergroundmonorail Jul 27 '14 at 14:09
  • Much prefer the fraction example to the multiplication one. – Keith Jul 29 '14 at 00:13
8

Physics and mathematics are vastly different about this.

Physics: everything is not entirely true

For example,

Newtonian mechanics is "true" and good enough when objects are not too big, not too small and not moving fast enough,

Relativistic mechanics is "true" and good enough when objects are not too small,

Quantum mechanics is "true" and good enough when objects are not too fast and too big,

and fringe cases keep cropping up in which none of our theories about the universe work, so we make up new ones and each theory has its scope of application.

So no, you are not lying or oversimplifying when you're explaining Newtonian mechanics in the same way you're not lying or oversimplifying when you speak about quantum mechanics, you simply need to be clear about a theory's area of application.

Mathematics: always tell the Truth, nothing but the Truth, but never the whole Truth

For example,

For every two points there is a unique straight line connecting them - it's true, but it's not the whole truth,

the proper sentence starts with "In a euclidean space",

and then it's the whole truth, but try explaining that in elementary school and you'll just confuse students from the issues that you're trying to explain (i.e. how to draw a perpendicular line).

Ruslan
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bbozo
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7

The velocity addition:

The first semester of my study I've been taught that if a guy walks with veolicty $v_g$ in a train that has velocity $v_t$ then his velocity with comparison to someone out of the train is $$v_r=v_g+v_t.$$ During the second semester I have been taught (by the same teacher) that $$v_r= \frac{v_g+v_t}{1+\frac{v_gv_t}{c^2}},$$ using the Lorentz transform.

EDIT:

@vonPetrushev: Can you elaborate? I assume the teacher explained that they're talking about two different theories, and, if it was a good course, that the first theory is a niche in the second and accounting only for the cases where v_g v_t << c^2.

First semester we did classical mechanic (Newton's mechanic). The second semester we got an introduction to the special relativity (including Lorentz transform, etc...). For sure the teacher pointed out that, in facts, Newton was wrong but his approximation is already very powerful. Except if the guy walking in the train a superhero (and runs VERY fast), the difference is negligible. This comment marked me because it was also showing that the models we are actually using (even the most complicated and complete ones) may still be wrong. The whole "art" is to see which theory will provide a result sufficiently precise for the expectations. It makes less sens to calculate the velocity of an apple falling on a head with special relativity tools, however these tools can be relevant when computing some spaceship trajectories, etc.

Surb
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  • Can you elaborate? I assume the teacher explained that they're talking about two different theories, and, if it was a good course, that the first theory is a niche in the second and accounting only for the cases where v_g v_t << c^2. – vonPetrushev Jul 24 '14 at 10:13
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    @vonPetrushev Is it the kind of elaboration you expected? – Surb Jul 24 '14 at 12:09
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    One way to state the point is that Galilean relativity is perfectly fine for a certain version of spacetime. It's the model of spacetime that's wrong---and even there, they're a perfectly coherent approximation in the classical domain to the special relativity required for Minkowski spacetime. – Semiclassical Jul 24 '14 at 14:25
  • This is not a conceptualization that works in mathematics but not strictly true. It's just equations of two different physical theories, both wrong if we speak strictly (Newtonian mechanics approximates SR in $c\to\infty$, SR approximates GR when curvature $\to0$). – Ruslan Jul 27 '14 at 09:00
  • @Ruslan I don't get your problem... Your description of my answer fits exactly the second paragraph of the OP. – Surb Jul 27 '14 at 11:01
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This is more of a collection than a single answer, but otherwise each of this would be too short for an answer:

  • "convex function can hold water": I frequently heard this while tutoring maths as something people get taught in highschool. Easy counterexample would be $e^{x}$.

  • "if a bounded function do not converge, it must oscillate": well there is precise mathematical meaning of oscillation, but it is never explained in calculus class. Most people when heard "oscillate" will think of something that go up, then down, then up, then down, indefinitely.

  • "Turing machine is just like usual computer": I only heard it once, presumably due to some poor explanation in high school, from a freshman who don't understand why we can take array access to be constant time.

  • "since the denominator is 0, the amplitude must increase to infinity": almost always what the teacher say while solving wave with driving force assuming the solution to be a stationary sinusoidal solution to deal with the resonance case. I have the misfortune to have taken 12 different courses that end up solving this at some point, so I should know.

  • "the error surface is like a terrain and the algorithm is like a drop of water, so it will get to the local minima", "to find fixed point, keep applying $f$ until adjacent term are close together": I have heard this in computational classes from science department, but luckily not in numerical analysis class in maths department.

  • "the gradient is the direction of fastest rate of increase": not strictly wrong by itself, but combining with the usual issue of high schooler being only taught about gradient by its formula (rather than its actual definition) and trouble occur when function without gradient come up.

  • "if 2 space is homeomorphic we can turn one into another with just twisting, bending and stretching, but no tearing, cutting or gluing": disregard the issue of what "twisting" and other action could means in an arbitrary topological space, this is still wrong even on nice simple subspace of Euclidean space. For example, a trefoil knot and an unknot are still homeomorphic.

  • "semiproduct is like direct product, except that the group being extended end up wrecking its helper in the process": I just find the image too funny to not include. Well strictly speaking whether this is strictly wrong or not depends on how it is going to get misinterpreted.

  • "nilpotent group still incur cost whenever you want to swap, but the cost get smaller the more you try": while this might be a nice way to show why the concept of nilpotent is close to abelian, this probably is going to be a bad way to think about nilpotent group in general.

  • "the pdf of this distribution is similar to that distribution when the parameter are sufficiently large, so we can just replace this with that for all practical purpose": commonly used argument, but this is a classic confusion on method of convergence (I'm not even sure if there is even a notion of "convergence in pdf").

Gina
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    I think the second dot point is correct, though. – goblin GONE Jul 24 '14 at 23:38
  • Is "a convex function can hold water if you put it the right way up" true, though? Rotate $e^x$ through any small angle clockwise and you're good to go. The problem is that you've laid your bucket on its side... – Steve Jessop Jul 26 '14 at 20:35
  • @SteveJessop: well then any concave function can also hold water too once you rotate it enough. – Gina Jul 26 '14 at 20:37
  • @Gina: true, I'd say that the intuition about water is that a function that is neither of those things in some sense "can't hold water" any way up. It might be able to hold water in an interval, but not on the whole domain. Maybe I'm over-thinking this, and it's supposed to just be a mnemonic to distinguish convex from concave? In which case I agree with you, it's inaccurate but "works". Anyway I guess there comes a point where to answer this question one has to nitpick between falsehoods commonly taught because they "work", and falsehoods commonly taught because of common teacher errors. – Steve Jessop Jul 26 '14 at 20:41
  • $x+e^x$ might be a more “interesting” counterexample to the characterization of convex functions, in that it isn’t bounded below. Still, all that means is that you can rectify it by rotating it $(45+\epsilon)$°. Conversely, $x^4-2x^2+1$ is very bucket-like, but is not convex. – Scott - Слава Україні Jul 28 '14 at 20:35
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"If $f'(a)>0$ then $f$ is increasing near $a$."

Counterexample: $f(x)=x+2x^2\sin\bigl({1\over x}\bigr)$ for $x\neq 0$, $f(0)=0$, $a=0$. We have $f'(0)=1$, but there are points arbitrarily close to $0$ with $f'(x)<0$.

Per Erik Manne
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  • More generally, it is very often implicitly assumed that all of the derivatives of a function are bounded in the neighborhood of interest. – user7530 Jul 27 '14 at 07:29
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Not really about mathematics, but a case where a highly respected mathematician, computer scientist and author believes (as you do) that it's useful when we are "taught something that is not strictly true, but works, at least in some restricted manner, and is a good way to understand a concept until one gets to a more advanced stage".

Donald Knuth writes, in the preface to the TeXbook:

"Another noteworthy characteristic of this manual is that it doesn't always tell the truth. When certain concepts of TeX are introduced informally, general rules will be stated; afterwards you will find that the rules aren't strictly true. In general, the later chapters contain more reliable information than the earlier ones do. The author feels that this technique of deliberate lying will actually make it easier for you to learn the ideas. Once you understand a simple but false rule, it will not be hard to supplement that rule with its exceptions."

I heartily agree with this approach. In my view, many mathematics and computer science texts spend far too much time fretting about odd-ball corner cases that do little to elucidate the general concepts. I find it annoying and unhelpful, personally.

bubba
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Usually you are taught that topological spaces provide a nice framework for doing topology. But when you really sit down and try to prove some of the "obvious homeomorphisms" (for example $|X \times Y| \cong |X| \times |Y|$ for simplicial sets $X,Y$ with geometric realizations $|X|,|Y|$, or the exponential law $C(X \times Y,Z) \cong C(X,C(Y,Z))$), you will see that you will need more sophisticated models, for example compactly generated weak hausdorff spaces. See here for more details. In my impression many lectures gloss over this deficiency of $\mathsf{Top}$.

anomaly
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Martin Brandenburg
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  • I'm not entirely sure I agree that topological spaces are introduced as a "nice" framework for doing topology. A first course on point-set, even if it doesn't focus on pathologies, usually instills a sense that the bare minimal axioms of topology are pretty insufficient. – jxnh Jul 24 '14 at 16:25
  • Sure, but even adding separation axioms etc. is not enough. We really need to take different products, different function spaces etc. – Martin Brandenburg Jul 24 '14 at 22:49
6

I learned that the parabolic motion of projectiles is "wrong" in the short physics course I took at university. The path is not parabolic, it's elliptic, as the object that you throw is put in orbit the Earth until it hits the ground a few meters away.

Of course the parabole works perfectly well for throwing stones, since the stone never gets far enough to experience a pull along the x axis, but maybe you need to work with ellipses for throwing passive projectiles longer (though I don't know who uses trebuchets these days).

TV's Frank
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Geometry, and differential geometry, are full of these kinds of shortcuts, for the simple reason that geometry is a very visual, accessible, and venerable topic, resting on a deceptively deep body of theory required to make it rigorous. You are taught concepts in high school that will require many years of topology, linear algebra, measure theory, real and functional analysis, etc. before you have the full story.

Some examples:

  • You are taught to calculate the volume and surface areas of solids by two kinds of slicing. Rarely is it explained why surface area requires approximating a surface with slices that are generalized cones, while generalized cylinders are good enough for volume, and I've never seen discussed why and for which kinds of surfaces these techniques don't work at all. More generally, it is often assumed that if you want to measure something about a smooth surface, you can do so by looking at a limiting sequence of discrete surfaces that converge to the smooth surface under "reasonable" refinement. Defining "reasonable" requires great care (beyond the scope of high school geometry) to avoid counterexamples like the Schwarz lantern.

  • Concepts like arc length, surface area, curvature, etc. are not defined in a complete or rigorous way. On the one hand you have some formulas that clearly require a certainly level of differentiability, and on the other you have no problem calculating e.g. the surface area of a polyhedron, and probably don't bat an eye at the fact that polyhedra satisfy a meaningful version of Gauss-Bonnet, despite ostensible missing two degrees of differentiability! In addition to regularity, subtleties about orientability, compactness, etc. are likely not discussed.

  • Many theorems in plane geometry require objects to be in "general position" (usually left vaguely defined) and the corner cases are ignored. Similarly, pathological counterexamples to the formula for the Euler characteristics of polyhedra (e.g. polyhedra with odd Euler characteristic) are usually ignored.

  • Finally, while not the same as teaching students incorrect facts, many geometric arguments you see in early classes implicitly rely on unproven "obvious" results like the Jordan curve theorem, invariance of domain, Sard's lemma, existence of triangulations, etc. that are far from obvious.

user7530
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    Interesting answer. We have a bunch of other answers explaining why rough and ready maths is used in physics but that's OK because physics is fundamentally not maths anyway. This answer harks back to the origin of maths and Euclid's problem with the parallel postulate: at school we are taught geometry as being about the world, i.e. part of physics. This is the fundamental untruth - and more than just "not strictly true". Later on, we learn to split geometry into maths' presentation of axiomatic geometry and physics' presentation of possible geometries of the real world. – Keith Jul 29 '14 at 00:26
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This is more like a long comment.

I am having an argument with someone who thinks that it's never justified to teach something that is not strictly correct. I disagree: often, the pedagogically most efficient way to make progress is to iteratively learn and unlearn along the way.

In math, or in science?

If we're talking about science then everything we teach is (at some level) incorrect. For example, Newtons laws break down at high energies; they predict the non-existence of certain observed phenomena, etc. So obviously, it is okay when teaching science to say things that aren't strictly correct.

However, if we're talking about mathematics, I think a strong argument can be made that it is never okay to teach things that we know to be incorrect.

Definition. Let $\varphi$ denote a mathematical theorem. Let us say that $\varphi$ is known to be incorrect iff there is a consensus on how to formalize $\varphi$, and that formalization is known to be false.

For example, here are some examples of theorems that are known to be incorrect.

  1. No continuous function $f : \mathbb{R} \rightarrow \mathbb{R}^2$ fills the plane
  2. Every function $f:\mathbb{R} \rightarrow \mathbb{R}$ is analytic and/or has a Fourier transform
  3. The 2-sphere cannot be turned inside out without any creases (see e.g. Smale's paradox).

In my opinion, it would be inappropriate to teach any of the above "theorems."

goblin GONE
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    I'm fond of the following line: "Education is the art of telling smaller and smaller lies." It'd be nice if one could exclude math from that, but pedagogically I think that's just not possible. – Semiclassical Jul 23 '14 at 19:23
  • @Semiclassical: I disagree with you regarding excluding math. See my comment to the OP's original question. – Lee Mosher Jul 24 '14 at 15:21
  • @LeeMosher) If I'm following you: One never does research with absolutely stable concepts, so one should get accustomed sooner rather than later to having to work with what you've got? I agree with that, but I'd consider precisely part of the 'telling smaller and smaller lies' business: it's a good attitude for higher level courses, but at the introductory level I find myself skeptical. – Semiclassical Jul 24 '14 at 15:27
  • @Semiclassical: That's not quite my point. Mathematical concepts become stabilized because humans realize their importance, by a long process that includes trial and error. Probably aliens too. And maybe computers someday, but not yet, computers do not yet have judgement. – Lee Mosher Jul 24 '14 at 15:29
  • @LeeMosher: "Concept stabilization is a social process" seems like a good slogan for what you're saying. – Semiclassical Jul 24 '14 at 15:32
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  • A vector is just an arrow in space or an object which components transform in some particular way under rotations (see the question I asked in the Physics site).

    In a similar fashion, a tensor is just a matrix.

  • The fundamental theorem of calculus is the definition of a definite integral; it works, but it's wrong:

$$\int_a^b f(x)dx=F(b)-F(a)$$

  • Given a primitive $F$ of a function, to get the set of all primitives we just have to add a constant, when in fact, it can be a picewise constant function (if we integrate a discontinuous function).

  • When solving differential equations, the assumption that the derivate of a convergent series converges too.

  • Analytic is the same as differentiable/holomorphic (true in $\mathbb{C}$ but not in $\mathbb{R}$).

  • In Quantum Field Theory the usual definition of path integral has no mathematical rigour at all.

jinawee
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    This is relevant: http://38.media.tumblr.com/e6aed8fb200c0aa7af4d49273f1db88f/tumblr_mijphbfgbc1s5nl47o1_500.jpg –  Jul 24 '14 at 12:19
  • I believe you want a disconnected function, not discontinuous, to get a piecewise constant difference in primitives (a different constant for each connected component). The function $f(x)=x$ on $[0,1]\cup[2,3]$ is continuous, but the domain is disconnected. – Mario Carneiro Mar 11 '15 at 10:13
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In physics there is the point charge, and to a broader extent, any point particle, which is strictly an idealization.

The point charge is an idealized model. It is dimensionless and has infinite charge at the point.

Mars
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  • It's relevant to mention here that one of the ideas behind string theory is to generalize the concept of point particle to 1D strings and their higher dimensional counterparts ( branes). – Semiclassical Jul 23 '14 at 23:31
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    @Zaphod Are physics students misled to believe point charges are real things? I'm not sure I'd agree that we know for sure there's no such thing as a point charge; electrons are commonly modeled as point particles (quantum mechanically speaking, with other properties like spin), and that might even be accurate. I presume you mean in the context of classical electromagnetism, though. As far as I know, when point charges are taught, they're explicitly presented as an idealization, both in introductory and higher levels. Classical physics itself is usually explicitly presented as approximate. – Eliah Kagan Jul 24 '14 at 00:53
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    This can actually be mathematically correctly be described with the Dirac delta function. – vonPetrushev Jul 24 '14 at 10:14
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    Also, the attribute 'idealized' in front of model is superfluous. All models are idealized, even the QED. – vonPetrushev Jul 24 '14 at 10:15
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    Note that this notion of particles as being points may be a bit deceptive, since it leaves out the understanding that said particles are all states of an underlying field. (Which I think is utterly crucial to understanding the Pauli principle, for example.) – Semiclassical Jul 24 '14 at 14:27
  • @vonPetrushev. Let's not get pedantic. – Mars Sep 24 '14 at 21:21
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The shortest distance between two points is the unique line segment joining them.

This is not strictly true, since this line may even not exists, e.g. Taxicab geometry. enter image description here

(Wikipedia picture)

Surb
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I think the definition of a limit is a good example. In high school, I was taught that $\lim_{x \to a} f(x)$ is the value of $f(x)$ as $x$ gets closer and closer to $a$, or something equally nebulous like that. Similarly, $\lim_{x \to \infty} f(x)$ is the value of $f(x)$ as $x$ gets bigger and bigger. At the time, this was good enough; students in pre-calculus courses really don't need to understand limits much more rigorously than that.

Of course, later I learned that limits do in fact have a perfectly concrete definition:

$$ \begin{align} \lim_{x \to a} f(x) = L &\iff \forall \epsilon>0,\ \exists \delta > 0\ \text{such that}\ 0 < |x - a| < \delta \implies |f(x) - L| < \epsilon \\ \lim_{x \to \infty} f(x) = L &\iff \forall \epsilon > 0,\ \exists c\ \text{such that}\ x > c \implies |f(x)-L|<\epsilon \end{align}$$

I think it's clear that teaching most high school students these definitions would only confuse them.

michaelb958--GoFundMonica
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Joe K
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    I'm a high school student, and these definitions just confused the heck out of me. – michaelb958--GoFundMonica Jul 24 '14 at 23:19
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    In particular one could illuminate the ambiguity in the phrase "$\lim_{x \to a} f(x)$ is the value of $f(x)$ as $x$ gets closer and closer to $a$" by giving alternate formal interpretations, and giving examples to demonstrate that these alternate interpretations are wrong. – Dustan Levenstein Jul 25 '14 at 12:12
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Does the use of "wrong" definitions/formalizations count for your purposes?

For example, in high school it is probably reasonable (although not essential) to use the term "vector space" to mean "finite-dimensional vector space". This avoids the need to spend more time on the Axiom of Choice than is really called for by the curriculum, to justify the operation of taking a basis for your vector space.

The resulting mathematics might be correct in itself (in effect using an additional axiom in the definition of vector space), but the terminology contradicts standard.

More generally: foundations of mathematics are not taught early and I think your friend would have a hard time doing so. Pre-university education often appeals to naive set theory in a way that could easily be "abused" to demonstrate that the system in use is inconsistent.

Anyway, the fact that something is commonly done in a particular way, even if we can prove that it's pedagogically more efficient to do in that way, won't shift your friend's position that it's "unjustified" ;-)

Steve Jessop
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When I was in primary school my teacher told me $\pi = 3$, then $\pi= 3.14$, then $\pi = \frac{22}{7}$, then $\pi$ was an irrational number and we could never write down $\pi$ completely. Then I was told $\pi$ was actually transcendental.

But if I'd been told when I was 8 that $\pi$ was a transcendental constant I don't think I would have understood much about what it really represented to me at the time: the ratio of circumference to diameter of a circle.

Eliah Kagan
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CameronJWhitehead
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    Presumably you could have been told that "$\pi$ is pretty close to..." and that would not have distracted you with the concept of transcendental numbers. None of these seem like examples of when it is "justified to teach something that is not strictly correct" (as the question puts it). – Eliah Kagan Jul 25 '14 at 20:21
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    Watch me write down $\pi$ completely: $\pi=4\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}$ – Mario Carneiro Mar 11 '15 at 10:33
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In real analysis courses, students are often taught the equation x^2 = -1 does not have solutions before they learn what complex numbers are : i is a solution for this eqaution.

Clément F
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    So you're saying sqrt(i) = -1? I don't think that's right. – Max Nanasy Jul 25 '14 at 20:17
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    As @MaxNanasy says, $(-1)^2 \neq i$; instead, $i^{1\over2}=\pm{{1+i}\over\sqrt2}$. $\pm i$ are instead solutions to $x^2=-1$. As a separate matter, at least in the United States, I don't think anyone takes anything called or characterized as a "real analysis" course before they know about imaginary and complex numbers, even if such knowledge is not actually required for the course's contents. Furthermore, as long as the question being asked is explicitly about real number solutions (which is typically the case in real analysis), it is strictly true that $x^2=-1$ has none. – Eliah Kagan Jul 25 '14 at 20:41
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How about e=mc^2? This is taught as fact in almost every physics class, and is commonly 'known' to the public, but is actually a simplified version of the real formula. You see, this says that the energy of an object is a product of it's mass and the speed of light squared. This completely disregards the energy that an object can have due to its velocity (kinetic energy).

You could go even further into how the speed of light is taught as if it's a constant, when in reality the speed of light is variable based on the medium that it is travelling through. If you ask a physics student what the speed of light is, they will rattle off a value without any indication that they are giving you the speed of light 'in a vaccuum'.

I hope this helps!

Jacob B.
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    That's a misconception about a physical concept, not of a mathematical one. Same goes for the index of refraction. – Semiclassical Jul 23 '14 at 17:52
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    No, it is a lack of understanding due to how it is taught, which is exactly what was asked for. e=mc^2 is not 'strictly correct'. Neither is stating the speed of light as a constant. Both of these things are taught in the classroom and in textbooks. – Jacob B. Jul 23 '14 at 17:53
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    I think you're stretching the intention of the question. (By contrast, I'd consider a misconception in the context of the hyperbolic geometry of spacetime to be in the spirit of the question.) Both are misconceptions, to be sure, but the former is a misconception about physics concepts whereas the other is about how mathematics is applid in physics. But @MGA can probably best clarify the matter as the asker. – Semiclassical Jul 23 '14 at 17:56
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    Actually, $E=mc^2$ is perfectly correct, as long as by $m$ we mean relativistic mass. i.e. $m=\gamma m_0$. The problem is not that the formula is wrong, but rather, inconsistent conventions regarding what $m$ means. – goblin GONE Jul 23 '14 at 18:55
  • Eh, as a physicist I can't really agree with that. Einstein himself found it objectionable: @goblin – Semiclassical Jul 23 '14 at 19:26
  • "It is not good to introduce the concept of the mass $M=m/\sqrt{1-v^2/c^2}$ of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ $m$. Instead of introducing $M$ it is better to mention the expression for the momentum and energy of a body in motion." – Semiclassical Jul 23 '14 at 19:27
  • @Semiclassical, okay, but perhaps he changed his mind later? I was under the impression that Einstein advocated using the word "mass" to mean "relativistic mass." – goblin GONE Jul 23 '14 at 19:29
  • That's from a 1948 letter (I grabbed it from wikipedia's page on mass in special relativity.) So I think the history there is the reverse of what you're suggesting. @goblin – Semiclassical Jul 23 '14 at 19:33
  • Some related misconceptions are that the refractive index will be always equal or bigger than one (which reflects the misunderstanding between signal speed and phase speed), that speed of light in vaccum is constant (it only applies for inertial frames) or that $E^2=p^2+m^2$ (which only applies for free particles, not 100% sure about that right now). – jinawee Jul 23 '14 at 20:53
  • It applies to interacting particles if you understand $p$ as the kinetic momentum plus the appropriate field momentum (i.e. the coupling to the vector potential $A$.) @jinawee – Semiclassical Jul 23 '14 at 21:10
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    @JacobB. The speed of light is a constant, it is a misconception that light slows down in a medium. In actuality, the distance the light has to travel in a medium (the path taken) increases do to various reflections from atom to atom. – Dane Bouchie Jul 23 '14 at 23:17
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    Also, one other argument for not introducing the the 'mass of a moving body' is that it does not apply for massless particles (e.g. photon), which have energy. The correct formula E^2 = p^2 c^2 + m^2 c^4 still applies, and becomes simply E = pc. – vonPetrushev Jul 24 '14 at 10:28
  • @goblin Most physicists nowadays have dropped the notion of relativistic mass because it is not that well-founded of an idea. The term is a relic of a bygone era. – Cameron Williams Jul 24 '14 at 17:04
  • @vonPetrushev, but photons do have a relativistic mass. That is surely the simplest explanation as to why they're able to impart momentum on solar sails. – goblin GONE Jul 24 '14 at 23:33
  • @goblin I would say the simplest explanation that photons impart momentum is because photons have momentum, not because they have mass. In relativity the notions are sufficiently decoupled that I don't see an issue with saying that a massless particle can have momentum. – Mario Carneiro Mar 11 '15 at 09:50
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What about all the basic rules of weight & motion--aren't they just simplifications of terribly complex rules that generally work as long as you don't deal with anything too small or going too fast?

It seems that EVERY problem in early physics/calculus is simplified to eliminate most of the variables because the problem would become impossibly complex if you added them? For instance, falling object calculations don't generally take into consideration wind resistance, and if they do they don't take wind and varying pressure into account. A ball rolling down a ramp always considers only perfect surfaces. A draining tub doesn't take into account the speed difference of the funnel created by the flow or how long it will take to form?

Most things are simplified so we can fit them into our heads in one way or another. I've always had lots of luck imagining electricity as water--I know it's inaccurate for many reasons, but it works extremely well even for things like induction (inertia), even though it's obviously "Wrong".

I also believe every description of what is going on in quantum mechanics is a best guess at this point, doesn't mean it isn't helpful though.

Bill K
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"There is such a thing as a set of goats."

WillO
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    Are you suggesting there are so many goats, they constitute a proper class? :) – Eliah Kagan Jul 25 '14 at 23:45
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    @EliahKagan: or just that Goedel, von Neumann, Grothendieck etc, in constructing their universes, all neglected to construct any goats. Leaving set theory unable to account for them and hence clearly unsuitable as a foundation for natural philosophy. – Steve Jessop Jul 26 '14 at 20:50
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The existence of Gabriel's Horn.

That is, there exists an object with a finite volume but with an infinite surface area.

Mathematically, we can show this to be true by taking a 360-degree revolution of the graph $y=\frac{1}{x}$ in the domain $[1,\infty)$ about the $x$-axis, and calculating its volume ($\pi$) and its surface area (infinite).

This is all well and good in theory, but in our physical world, no such object exists.

One of the reasons for this is that the thickness of the horn is limited by the size of the atom, so cannot be arbitrarily small.

beep-boop
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    How does that make the mathematical statement false, and how is this special? Presumably there's no such physical object as a point, line segment, surface, nor anything infinitely large, like a line or plane. $3$ is also not a physical object. Separately, we do know subatomic stuff exists, such as the particles that make up atoms and those that mediate physical interactions. The vast majority of "objects" in the universe seem to be smaller than any atom. Finally, going to the intent of this question: do people ever teach Gabriel's Horn without explaining that it's (presumably) non-physical? – Eliah Kagan Jul 24 '14 at 00:39
  • @EliahKagan It doesn't. The mathematical statement is true. The physical statement is false. – beep-boop Jul 24 '14 at 00:48