# Rigour in mathematics

Mathematics is very rigorous and everything must be proven properly even things that may seem true and obvious.

Can you give me examples of conjectures/theories that seemed true but through rigorous mathematical proving it was shown otherwise?

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You might also want to consider: math.stackexchange.com/questions/279347/…. Even when things are 'proven', you can run into issues. – Amzoti Aug 11 '13 at 17:07
A continuous function is differentiable, except perhaps at isolated points. This was believed by Cauchy and "nearly all mathematicians of his era" according to Morris Kline. Of course, later examples, the first due to Bolzano and the most famous of which due to Weierstrass, shows this is not the case. – David Mitra Aug 11 '13 at 17:07
Should this be CW? – Git Gud Aug 11 '13 at 17:11
@AndresCaicedo I undeleted it :) – please delete me Aug 12 '13 at 2:08
There's quite a good MO post on this: mathoverflow.net/questions/37610/… – Donkey_2009 Aug 12 '13 at 9:07

Finding the roots of a linear polynomial is trival. Already the Babylonians could find roots of quadratic polynomials. Methods to solve cubic polynomials and forth degree polynomials were discovered in the sixteenth century, all using radicals (i.e. $n$th roots for some $n$). Isn't it obvious that finding the roots of higher degree polynomials is also possible using radicals and that we have not found the formulas yet is only because they become more and more complicated with higher polynomial degrees?

Galois theory shattered this belief.

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Indian mathematicians find the roots of quadratic polynomials very first. This was a work of Bhaskaracharya (Bhaskara-2) en.wikipedia.org/wiki/Bh%C4%81skara_II – Dutta Aug 12 '13 at 5:31
@Samprity Ah, the Babylonians were just my guess out of the blue – Hagen von Eitzen Aug 12 '13 at 6:47
@HagenvonEitzen I'm not sure if Indian mathematicians found the roots of quadratic polynomials first. The cited reference was from the 12th century and as you pointed out the Babylonians (and also ancient Egyptians) were doing this well before the 12th century. – toypajme Aug 12 '13 at 17:23
@Samprity was that who you intended to link to, or were you mistaken about what he was first at? He lived several thousand years after the Babylonian civilization ended and they did have a solution for quadratic equations. The article you linked to indicates that he could have priority for developing, but not realizing the value of, some introductory calculus concepts. – Dan Neely Aug 12 '13 at 17:38
Do you have any historical evidence that it was widely believed that all polynomial equations could be solved by radicals prior to the work of Galois (and Abel)? – Rob Arthan Aug 12 '13 at 22:45

Berry's Phase was discovered after a lack of rigor in the proof of the Adiabatic theorem was discovered around 1980. It now appears in standard Quantum Mechanics texts and has produced at least 3000 papers since 1980 (actually, that number is about 10 years old, I'm not sure how many by now).

While this appears in physics, the mistake was mathematical. In particular, topological. There is an integration over parameter space in the proof which is assumed to be trivial. That would be fine if the parameter space was one-dimensional, but for higher-dimensional parameter spaces there might be a singularity in the domain which obstructs the vanishing of the integral. Once this mistake was uncovered, physicists then gained insight from the corrected theorem to modify phase with ease. It would be exciting to see similar developments in other physical arenas where adhoc mathematics is utilized. However, it seems this is the abberation from the norm. Much like the case with math. I think it's fair to say that most often the heursitic proof has turned out to be correct once the details are fleshed out. This is why this thread is interesting.

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First thing that comes to mind: not every smooth function is equal to its Taylor series over the series' region of convergence. As a counterexample, consider the function $$f(x) = \begin{cases} 0 & x=0\\ \exp\left(-\frac1{x^2}\right) & x\neq0 \end{cases}$$ whose Taylor series centered at $x=0$ is simply $0$, with an infinite radius of convergence.

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Closely related: The belief that every function admits a trigonometric series expansion. Really, the underlying issue is the notion of "arbitrary function", that was just being clarified at the time. – Andrés Caicedo Aug 11 '13 at 17:00
This also runs into the idea that functions on $\mathbb{R}$ are influenced by their definition on $\mathbb{C}$. This can be seen on the effects of $\mathbb{C}$ on the radius of convergence of a taylor series and that functions on $\mathbb{C}$ are more well behaved. – toypajme Aug 11 '13 at 19:57
@toypajme Could you please explain what you mean? I do not understand what "effects of $\mathbb C$ on the radius of convergence of a taylor series" you are talking about. Also, I do not understand in what sense "functions on $\mathbb C$ are more well behaved" (Surely, not arbitrary functions). Finally, I do not know what the influence is that you are referring to, and what you are trying to imply here. – Andrés Caicedo Aug 11 '13 at 20:40
@AndresCaicedo as it ends up, differentiability is defined very differently over $\mathbb C$, and many functions that are differentiable in $\mathbb R$ are not differentiable in $\mathbb C$. For example, the counterexample in question is not considered differentiable at $x=0$ in the complex plane (nor is it even continuous). As it ends up, every function that is complex-differentiable over its domain can be written as a convergent power series in the neighborhood of each point in the domain (that is, the class of complex-analytic functions is precisely the class of holomorphic functions). – Omnomnomnom Aug 11 '13 at 23:52
Consider the function $f(x)=\frac{1}{1+x^2}$. If you consider it's taylor series, the radius of convergence is 1. Now there is no intuition on why this is true besides the fact that we know a formula that tells us the radius of convergence. However, if you look at the function over $\mathbb{C}$ it has singularities at $\pm i$. Therefore, it is clear that the function cannot be (complex) analytic in a ball greater than radius 1 around the origin. If we restrict its power series to the real line, we recover the taylor series of $f$. I hope this has demonstrated $\mathbb{C}'s$ influence. – toypajme Aug 12 '13 at 17:15

I think the simplest example is the answer to the question:

Are there more rational numbers or natural ones, or is there equally many of those?

Intuition says that "of course there are much, much more rationals". However, rigorous mathematical proof shows that there are exactly the same number of each.

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Of course, it depends on what you mean by "more." Cantor's definition is nice and all, but the rationals are dense in $\mathbb R$ while the naturals are nowhere dense. The real key to this "surprise" is in the new definition of more, and that there were actual differences in sizes of infinite sets under this definition. – Thomas Andrews Aug 11 '13 at 17:04
Of course, but I don't feel this question wants us to go that broadly. I use Cantor's definition, as it is something easy to explain even to non-mathematicians. – Vedran Šego Aug 11 '13 at 17:08
Yeah, I just find it less surprising when you realize the result is really "under a certain definition of 'more,' ...." :) – Thomas Andrews Aug 11 '13 at 17:10
If we keep to Cantor's definition (that two sets have the same number of elements if and only if there is a bijection between them), then there are more reals than rationals/integers/naturals. Actually, there are infinitely many infinities. A Wikipedia article on cardinal numbers might be a good place to start reading about this. – Vedran Šego Aug 11 '13 at 19:02
To Cantor, it seemed "true and obvious" at first that $\mathbb{R}$ had "less" elements (points) than $\mathbb{R}^2$, but after he tried to prove it he found out that it was false, $|\mathbb{R}|=|\mathbb{R}^2|$. – Jeppe Stig Nielsen Aug 12 '13 at 8:43

The one that currently bugs me is that exponentiation is Diophantine.

This means that there exists an integer polynomial (so, no variables in the exponents) $P(x,y,z,w_1,\dots,w_n)$ such that:

$$\forall x,y,z\in\mathbb N\,\left(z=x^y \iff \exists w_1,\dots,w_n\in\mathbb N\,\left(0=P(x,y,z,w_1,\dots,w_n)\right)\right)$$

I've read the proof. I believe the proof is correct. I still don't instinctively believe the result.

One of the surprising results of this is that first order number theory only needs to have multiplication and addition - you can still answer questions about exponentiation using the above polynomial.

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One small bit that may help is that the statement in the right hand side is not just that a polynomial is zero, but rather that some (polynomial) projection is zero. Projections tend to be much more complicated. One level up, we have Borel sets, and their projections, some of which are no longer Borel. (Of course, this in itself may be surprising as well.) – Andrés Caicedo Aug 11 '13 at 18:37
Oh, yes, I get that part about projections. I still just find it baffling. Obviously, there is no polynomial $P(x,y,z)$ so that $P(x,y,z)=0\iff z=x^y$. :) – Thomas Andrews Aug 11 '13 at 18:40
This reminds me of Kreisel's review of the paper by Davis, Putnam and Robinson, where they reduce the 10th problem to showing that exponentiation is Diophantine. "These results are superficially related to Hilbert's tenth problem [...] it is likely that the present result is not closely connected with Hilbert's tenth problem. Also it is not altogether plausible that all (ordinary) Diophantine problems are uniformly reducible to those in a fixed number of variables of fixed degree, which would be the case if all r.e. sets were Diophantine." – Andrés Caicedo Aug 11 '13 at 18:50
@vermiculus The statement says that there is a polynomial that captures exponentiation in the sense that for all (natural) numbers $x,y,z$, we have $x^y=z$ iff we can find solutions to the equation $P(x,y,z,\vec w)=0$. So: If $x^y=z$, then we can find some $\vec w$ that makes $P(x,y,z,\vec w)=0$. And if $x^y\ne z$, then for any $\vec w$ we have $P(x,y,z,\vec w)\ne0$. – Andrés Caicedo Aug 12 '13 at 19:57
@AndresCaicedo and I as well :-) thank you so much for taking the time to clarify. – Sean Allred Aug 12 '13 at 20:05

Maybe you should also consider statements that are obviously true and are indeed true, but their proof is far from straightforward. The Jordan curve theorem surely falls under this category.

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Does Fermat's last theorem fall under that category? – please delete me Aug 11 '13 at 17:13
@user2357 I'm not sure if FLT is obviously true. All we know is that Fermat cosidered it true, but already his supposed reasoning was not obvious enough to fit marginis exiguitas. – Hagen von Eitzen Aug 11 '13 at 17:24
I read that using computers it worked for millions of numbers but that didn't matter unless it was proven properly. I also read that many other conjectures/theorems work for many numbers when using a computer but they must be proven rigorously. – please delete me Aug 11 '13 at 17:28
Not at all. I do not find the Jordan curve theorem obvious. Sure it is, for "nice" curves. But most curves are not nice, and the obviousness disappears. Plus, there are things like the Lakes of Wada. – Andrés Caicedo Aug 11 '13 at 17:29
@DanNeely A good place to begin is the paper The Jordan curve theorem is non-trivial, by Fiona Ross and William T. Ross, Journal of Mathematics and the Arts, 5 (4), (2011), 213-219. (The paper should be downloadable from the link.) In a technical sense, the examples presented there are still too simple, but they may begin to convey some of the subtleties of the result. – Andrés Caicedo Aug 12 '13 at 14:51

The Axiom of Choice was believed to be true (in $\sf ZF$), but it turned out be be independent from $\sf ZF$.
It might be good to mention that not being true, doesn't necessarily imply being false.

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Maybe a similar point: The parallels axiom of Euclidean geometry was long considered a consequence of the other axioms by many (and be it that they wanted to get rid of it for its comparatively complex formulation) until its independency was shown almost in parallel (no pun intended) by Lobatschewsky and Bolyay. – Hagen von Eitzen Aug 11 '13 at 17:14
@HagenvonEitzen Nice catch. Why not add it to your answer? – Git Gud Aug 11 '13 at 17:16
Not being provable doesn't imply being false. For us Platonists, not being true (for a well-formed formula with no free variables) is still synonymous with being false. – Robert Israel Aug 11 '13 at 17:38
As a historical matter, the axiom of choice (AC) was, indeed, believed to be true (and was used without special comment) until Zermelo pointed out, in 1904, that it is a non-obvious assumption. That started a controversy about whether AC should be accepted. All of that predated the axiomatic system ZF, most of which Zermelo introduced in 1908; the final form dates from about 1922. I'm not sure that people ever thought AC could be proved in ZF. – Andreas Blass Aug 12 '13 at 1:19
@Perseids Not quite: If a statement holds in every model of a set of axioms, then it is provable from the axioms. This is the completeness theorem. The incompleteness theorem says something different. Anyway, Robert Israel is not talking about this, which is why he emphasized his platonist perspective. – Andrés Caicedo Aug 12 '13 at 17:21

Not really an answer, but here you go anyway: the number of all Turing machines is countable. Thus, no matter the encoding, almost all real numbers are not computable.

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Similarly. The set of all (finite-length) sentences in a language with a countable set of characters is countable. – fhyve Aug 23 '13 at 17:52

Intuitively, it's easy to speak about volumes or areas of sets (in $\mathbb R^n$), like e.g. "the smaller a set, the smaller it's volume" which would lead to the statement

$$A \subset B \Rightarrow vol(A) \leq vol(B)$$

Our error is in fact that we imply that all sets actually had some well-defined volume. On precisely defining what we understand as a measure $vol(\cdot)$, it turns out that we can't define one for all subsets as there are non-measurable sets.

Statements like the above are actually only valid given that the sets we talk about are measurable.

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The inequality is true in the sense of outer measure, which is defined for all sets in $\mathbb R^n$. – Andrés Caicedo Aug 11 '13 at 20:51

We can think there is universal set. But, mathematicians proved it isn't a set. I think existence of universal set might not be seemed obviously true for you. But for a non-mathematician obviously there is a universal set! We can think that, mean of "obvious" gets shape with people, and for the man who proved the theorem it isn't any longer obvious.

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Using constructivist mathematics, which allows us to define a function the way we intuitively think about them,

a function $f:A \to B$ is a procedure assigning for every value in $A$ a unique value in $B$

we can prove rigorously that all functions are continuous. A shockingly surprising statement. It turns out that all of our classic discontinuous functions aren't actually computable.

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This, of course, depends on how we formalize "computable", and on why we consider this definition to capture some intuitive notion. – Andrés Caicedo Aug 11 '13 at 17:32
Certainly. In fact I find the definitions required to make this true wildly unintuitive. But it's always fun to say to people because it is so patently flies in the face of everything we know. – Devin Murray Aug 11 '13 at 17:37
I would love to see a reference for this, since we can have finite topological spaces with non-continuous functions between them, and I don't see how any sort of constructivism can avoid those. – Tobias Kildetoft Aug 11 '13 at 19:06
Ah yes it's for functions from subsets of R to R. For finite spaces I believe that you're ok. Here is a link looking at the question more in depth, with references. math.stackexchange.com/questions/176279/… – Devin Murray Aug 11 '13 at 19:42

It might seem obvious that every topological manifold has a triangulation; I mean manifolds are just spaces which are locally Euclidean, so we should be able to split up the manifold enough in to compact areas which are homeomorphic to disks, and then triangulate from there right?.

In fact, for dimensions 4, there are non-triangulable manifolds such as the $E_8$ manifold. In higher dimensions, the problem may still be open (although a recent preprint of Ciprian Manolescu may have proved that there are non-triangulable manifolds in all dimensions greater than $4$ as well - a great early review of the result and its implications can be seen here).

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Most useful example in this context may be considered Integration. Riemann Integration is used everywhere. Consider simplest function $f(x) = x$. Its antiderivative is $\frac{x^2}{2} + c$, definite integral can be calculated also very easily. But there is a theoretical part behind that, which is very much necessary for a logical understanding.

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I guess you're addressing fact that sometimes people think of integration merely as \int_a^b f =F(b)-F(a)$where$F'=f$? Yes, that is awful. – Pedro Tamaroff Aug 11 '13 at 18:45 Yes it is no doubt awful. But when we learnt integration in school we had this type of idea. The inner material is not so easy. – Dutta Aug 12 '13 at 5:24 This is an example of a statement that seems true and turns out that its true (like one of Hagen von Eeitzen's answers). Let$p(x)\in\mathbb{R}[x] $, then$p(x)=q_1(x)q_2(x)...q_n(x)$for some polynomials$q_1,q_2,...,q_n\in\mathbb{R}[x]$such that$\deg q_i\leq 2$. Before I knew its proof, I never thought that this fact was obvious. But I know a lot of people who think that it's obvious (they don't know the proof though, they just think its obvious!). - When I have lots of time (ha!) and discuss the fundamental theorem of algebra, I ask the students to try to convince me that a (real) polynomial of degree$6$has a complex root. After this, the appreciation for the theorem increases. – Andrés Caicedo Aug 12 '13 at 1:45 @AndresCaicedo Why is the degree 6 case harder to convince about than degree 4? – ronno Aug 12 '13 at 17:15 @ronno There is an explicit formula via radicals for the solutions of a degree$4$polynomial equation. And, at least if the coefficients are real, degree$5\$ polynomials always have at least one real root, by basic calculus. – Andrés Caicedo Aug 12 '13 at 17:17

Here is a case where you can easily be misled by working out some early examples. All groups up to order 60 are solvable (a nice definition of this available on Wikipedia) -- you can show that by checking them all out. With 59 consecutive examples to start with, you might think that all groups are solvable. But it's not true. There is at least one group of order 60 which is not solvable. This is directly connected to the insolubility in radicals of polynomial of degree ≥ 5; so it would be important to get it right.

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This will give you some idea of how deceiving and tantalizing even relatively simple things may be. The Wikipedia link tells the temporal history well enough but to appreciate the work that went into the proofs, you have to read at least one original paper...

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I feel the words "Shapiro inequality" ought to appear somewhere in your answer. – Rahul Aug 12 '13 at 21:02
They pop up as soon as you bother to click on any of the provided links and tell you nothing before you do so unless you know the story already. The point is not how it is called but how uncanny it is. The full inequality name should actually include from 5 to 9 personal names and I'm not even sure "Shapiro" should be among them because, as far as I can judge, his own contribution didn't go far beyond posting it in Monthly. However, if you tell me that he proved some non-trivial cases himself, I'll gladly take my last words back :) – fedja Aug 12 '13 at 21:52
Well, you can call the inequality whatever you like; I only suggested "Shapiro inequality" because that's what the articles you linked to said. But as a reader I dislike clicking on links which tell me nothing about what's on the other side. Besides, someone else who might be about to post about this inequality would not notice your answer upon scanning down the page. – Rahul Aug 12 '13 at 22:10
Now he certainly will :) Anyway, I agree that this is not an issue to pick a big fight about... – fedja Aug 12 '13 at 22:16

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