# Examples of common false beliefs in mathematics

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 $\mathbb{C}$;
(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 $\mathbb{R}$ must be the whole of $\mathbb{R}$. 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.

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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/… – user9072 Oct 8 '11 at 14:27

Assume that $a,b\in \mathbb{R}\setminus \{0\}$ which satisfy $a^{3}= 2b^{3}$.

Then $a-2b$ is a non zero nilpotent element of group ring $\mathbb{Z}_{3} \mathbb{R}$, that is $(a-2b)^{3}=0$.

This would be a counterexample to the zero divisor Kaplansky conjecture

The false lies in an obvious abuse in the definition of the group ring multiplication.

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This does not seem like a common false belief. – Yemon Choi May 15 at 11:47

I'm not sure how common it is but I've certainly been able to trick a few people into answering the following question wrong:

Given $n$ identical and independently distributed random variables, $X_k$, what is the limiting distribution of their sum, $S_n = \sum_{k=0}^{n-1} X_k$, as $n \to \infty$?

Most (?) people's answer is the Normal distribution when in actuality the sum is drawn from a Levy-stable distribution. I've cheated a little by making some extra assumptions on the random variables but I think the question is still valid.

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I don't understand your third paragraph. Are you saying that under the assumptions in the 2nd paragraph, the limiting distribution (rescaling if necessary) is always Levy-stable? – Yemon Choi Apr 12 '11 at 1:28
@Yemon, Yes, this is what I was implying. Perhaps I was a little too cavalier? Certainly the sum of (well enough behaved) i.i.d. r.v.'s with power law tails converge to a Levy-Stable distribution... – dorkusmonkey Apr 12 '11 at 23:53
Generally such a limiting distribution doesn't exist. Perhaps you need to divide your sum by the square root of $n$? – John Bentin Dec 29 '11 at 13:56

I once very briefly thought that:

Given a vector space $V$ and a sub-space $U \subset V$ that $V-U$ is also a subspace.

I've heard this several times as a TA also.

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Why the downvote! I heard this from more than one student in introductory linear algebra classes and when marking. – Benjamin May 12 '15 at 22:21
I think this falls under $(x+y)^2=x^2+y^2$, – Thomas Rot Aug 10 '15 at 12:48
I never said it always fails, just that it's not generally true and I thought it was for about 1 min once. – Benjamin Aug 10 '15 at 19:12
It always fails... But I don't think this is a common held belief. – Thomas Rot Aug 10 '15 at 21:40
@ThomasRot But it always fails, while $(x+y)^2=x^2+y^2$ sometimes holds, especially in characteristic 2. – ACL Apr 21 at 6:37

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.

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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

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!

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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.

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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.

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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
Yes, I did not read the question very carefully. I realize it is not a good comment, and, yes, it is more of a abd heuristic than anything else. – Herb May 25 '10 at 23:59
it is also a comment on the imprecision of the words locally, infinitesimally,.... This once led Oort-Steenbrink to give some careful restatements of results previously called as "local Torelli theorems"... – roy smith Apr 14 '11 at 19:02

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.

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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
+1 for Gerald's corrected version. – darij grinberg Mar 6 '11 at 0:16

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

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