Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) sin(z) is a bounded function;
(iii) sin(z) is defined and analytic everywhere on C;
(iv) sin(z) is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of sin(z) to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset U of R must be the whole of R. The "proof" of this statement is that every point x is arbitrarily close to a point u in U, so when you put a small neighbourhood about u it must contain x.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

share|improve this question
I have to say this is proving to be one of the more useful CW big-list questions on the site... –  Qiaochu Yuan May 6 '10 at 0:55
The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. –  Unknown May 22 '10 at 9:04
wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. –  Suvrit Sep 20 '10 at 12:39
It's a thought -- I might consider it. –  gowers Oct 4 '10 at 20:13
Meta created tea.mathoverflow.net/discussion/1165/… –  quid Oct 8 '11 at 14:27

188 Answers 188

The false belif could be:

A sequence is a net. Then a subnet of a sequence is a subsequence

Counterexample: The sequence $sin(nx)$is a sequence in the compact topological space $[-1, 1]^{\mathbb{R}}$. So this net has a convergence subnet. But it is well known that the above sequence has no a subsequence which is point wise convegent(See the last page of the book of Walter Rudin Principle of mathematical Analysis). So in this example the convergent subnet can not be counted as a subsequence.

share|improve this answer
This seems essentially a dupe as it is contained explictly in mathoverflow.net/a/40091 –  quid Nov 12 '14 at 13:31
@quid Thank you. I did not read it already. however this one introduce an explicte counter example. –  Ali Taghavi Nov 12 '14 at 13:59

For $p$ prime and the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \cdots$ given by multiplication by $p$, then $\bigcup_n \mathbb{Z}/p^n\mathbb{Z}$ is not the group of $p$-adic integers $\mathbb{Z}_p$, but its Pontryagin dual, the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$.

share|improve this answer
Is that actually a common false belief? After all, $\mathbb{Z}_p$ is uncountable, as everyone realizes! –  Todd Trimble Mar 5 at 14:25
It's hard for me to believe it's at all common. I wasn't the downvoter, but I think it would be better if answers were rooted either in instances that can be found in the literature, or widely encountered in one's experience as an instructor. –  Todd Trimble Mar 5 at 14:52

Related to this answer: $$ \pi=\left(\frac{1}{10^5}\sum_{-\infty}^{+\infty}e^{-n^2/10^{10}}\right)^2. $$ Proof: With a computer one can verify that the first 42 billions digits of the two numbers are the same, see J. Borwein and P. Borwein, Strange series and high precision fraud, in The American Mathematical Monthly, 1992, pages 622-640.

share|improve this answer
I voted this down because I don't think it's a statement that anyone actually believes, and therefore doesn't fit the spirit of this questions, but I have to say it's pretty clever. –  Nate Eldredge Oct 19 '10 at 21:11
I would vote it down if I could. There is nothing false with believing in that Riemann sums converge to the true value of an integral (Poisson in this case). –  zhoraster Oct 20 '10 at 19:12
I must admit I'm a little bit surprised just how quickly $f(a) = (1/a \sum e^{-n^2/a^2})^2$ converges to $\pi$ as $a \to \infty$. (According to the identity given in the article, $\lim a^{-2} \log (f(a)-\pi) = -\pi^2$. This feels much faster than we have any right to expect. –  Michael Lugo Oct 26 '10 at 4:40

$\exp(\pi\sqrt{163})$ is an integer. Proof: it has a mathematically interesting definition, and the $12$ first digits after dot are zeros.

No integral power of $2$ has $7$ as first digit. Proof: compute by hands successive powers $2^n$ for an hour. You can't find one beginning with $7$. Well, if you ask a computer, it is a different tale.

share|improve this answer
This is also discussed in this MO question mathoverflow.net/questions/4775/… –  Andrey Rekalo Oct 19 '10 at 16:07
You underestimate our ability to double. 64 * (1024)^4 leads to such a power, and it does not take an hour to compute by hand. Now, if you have such a number start with 77, well, I will suggest using powers of 6 instead. Gerhard "Numbers Doubled While You Wait" Paseman, 2010.10.19 –  Gerhard Paseman Oct 19 '10 at 16:25
I'd be surprised if this actually qualifies as a "common false belief". –  Michael Hardy Dec 11 '10 at 17:37
@Gerhard. What is more your computation there gives a very easy proof by estimation rather than calculation. The existence of 1024 as an easy power of 2 means you keep adding 2.4% so will eventually get to any initial digit including 7. Starting with 64 makes it easy. –  Mark Bennet Feb 6 '11 at 11:10
I can't see how the second could possibly be a common belief among mathematicians. Since $\log_{10} 2$ is irrational and all... –  Todd Trimble Mar 31 '11 at 13:59

When I was a kid (8th grade), I solved a bunch of math problems in an exam using the ``well-known identity'' that $(x+y)^2=x^2+y^2$, which I was sure I had been taught the year before. It was of course way before I heard about characterstic two and I didn't get a good grade that day!

share|improve this answer
Quoth the question, "The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed)". –  JBL Dec 1 '10 at 23:39
Also, this is of course just a special case of the more general “law of universal linearity”, which iirc was mentioned in earlier answers… –  Peter LeFanu Lumsdaine Dec 2 '10 at 0:40

I had the false belief that recursive functions are always decidable in ZFC.

share|improve this answer

I don't know if this is what you are looking for, but I keep hearing that "a differentiable function is one that is locally linear", not one whose local variation can be approximated linearly. No one stops to think about e.g, x2, and the fact that its graph does not look like a line at any value of x.

share|improve this answer
I would say this is more a heuristic than a false statement; as such, it would be more appropriate as an answer to mathoverflow.net/questions/2358/most-harmful-heuristic (although I do not think anyone interprets it the way you apparently do). –  Qiaochu Yuan May 5 '10 at 4:53

I had a false belief in linear algebra, that a basis of a vector space could have infinitely many elements (like an orthonormal basis in Fourier analysis). That tripped me up trying to understand the definition of tensor products, and even after someone explained the issue to me I didn't believe it at first.

share|improve this answer
I don't understand. A basis can have infinitely many elements. That's no false belief, that's correct. –  Johannes Hahn Aug 22 '10 at 12:07
The false believe would be that when you define basis, you allow infinite linear combinations. If some confusion is possible, say "Hamel basis" ... Even if there is no topology defined, it still will emphasize that only finite linear combinations are considered. –  Gerald Edgar Aug 22 '10 at 12:30

protected by François G. Dorais Oct 15 '13 at 2:34

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.