# 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.

• 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

1. The probability that two randomly chosen positive integers are coprime is $$\frac{6}{ \pi ^2}$$.

2. There is a uniform distribution on the set of positive real numbers.

3. A sequence converges to a point $$P$$ if and only if every subsequence of it has a further subsequence which converges to $$P$$.

4. Small numbers can be found (especially by computers) very quickly.

About 1. It appears in many textbooks of Elementary Number Theory. Although the argument that most of these books give to prove it appears really elegant, but it is wrong or rather the statement is meaningless. The main thing that goes wrong is that there is no uniform distribution on the set of positive integers and hence the phrase "randomly chosen positive integers" have no meaning (with respect to the usual meaning of randomly choosing things). A few of these books also have a remark on the error in the statement (but most of the students don't read it).

About 2. Usually the main reason people think is that, since there is a uniform distribution on $$[0,1]$$ and since there is a bijective function from $$[0,1]$$ to the positive reals, so we can choose an element at random from $$[0,1]$$ and then apply this bijective function and we would get a randomly chosen element from the positive reals. But this is wrong and the error lies in the wrong interpretation of the meaning of "randomly choosing from a set". We say we can randomly choose something from a set if and only if there is a uniform distribution on the set. It is totally silent about what is a randomly chosen element of a set (or does that even make sense !). They fail to recognise that they are not justifying why the result of applying a bijective function on this element would give us a randomly chosen element (can we even speak about randomly chosen elements !).

About 3. One has to be careful about the mode of convergence. Often one can be sloppy about it and end up with some misleading results. One such example of an erroneous conclusion is "If a sequence of random variables converges in probability then it would converge almost surely" (I agree that this is really a bad example of a wrong belief as almost surely everyone knowing these modes of convergence would know that the result is not true), because every subsequence of such a sequence will also converge in probability and hence have a further subsequence which converges almost surely, so every subsequence of the parent sequence has a further subsequence which converges almost surely. So according to 3. the sequence must also converge almost surely. But everyone knows examples of sequences of random variables which converges in probability but not almost surely.

About 4. Ramsey Numbers (and particularly the example $$43 \le R(5,5) \le 48$$) are good enough to refute this belief.

Lastly, the practice of using Random Variables to frame the hypotheses in classical statistics. This is something which is not exactly a very relevant example of a false belief in mathematics, but more of a very common error among students (especially beginners) studying statistics.

• Points 1-3 are addressed in relevant undergraduate courses, and 4 is unclear/opinion based – Stanley Yao Xiao Oct 25 '18 at 17:53
• On that note, let me be more specific: I would imagine that none of 1-4 are actually 'falsely believed' by any mathematician. Perhaps some amateurs might get confused by some of them, but I don't think that is what's meant by the question. – Stanley Yao Xiao Oct 25 '18 at 18:23
• To be fair to Hardy and Wright, they let $\psi_n$ be the number of fractions $p/q$, $q>0$, $1\le p\le q\le n$, and $\chi_n$ the number in lowest terms, and write "If ... we define the probability that $p$ and $q$ are prime to each other as $\lim_{n\to\infty}(\chi_n/\psi_n)$ we obtain Theorem 332. The probability that two integers should be prime to one another is $6/\pi^2$." There is no mention of "randomly chhosen positive integers," and it's clear H & W did not hold the false belief you attribute to them. – Gerry Myerson Oct 25 '18 at 22:42
• What Gerry said. Moreover, OP did say "The properties I'd most like from examples are that they are from reasonably advanced mathematics..." and so the criticism of Stanley does apply: there is no evidence that many mathematicians don't grasp that (1) is an asymptotic statement, and that many mathematicians think what you say they think in (2). (I also don't buy the description of "random" there, constraining the meaning only to uniform distributions: the definition of random variable en.wikipedia.org/wiki/Random_variable allows for much greater latitude.) – Todd Trimble Oct 25 '18 at 23:21
• OP wrote, "I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics," so I don't think it's required that any mathematician have the false belief. – Gerry Myerson Oct 26 '18 at 1:23

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.

• 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

When people first hear of reversible computation and how reversible computation is potentially more energy efficient than conventional computation, they automatically think that reversible computation can never be practical except for very specific algorithms because one will always accumulate garbage information that one will eventually have to delete anyways which brings us back to irreversible computation. This thought is not correct. Reversible computation can compute anything that can be computed with a conventional computer with very little space and time overhead and where very little garbage information is produced.

Most people do not realize that in the future reversible algorithms will be efficient and powerful enough to replace conventional algorithms when one takes energy usage into consideration. I have heard mathematicians, computer scientists, and many otherwise knowledgable people claim that certain problems cannot be solved by reversible computers because these problems are inherently irreversible.

To give you some background, reversible computing is the type of computing where all of the computational processes are bijective.

For example, in a reversible combinatorial circuit, all of the logic gates such as NOT gates, Toffoli gates, Fredkin gates, and CNOT gates are bijective. Quantum computing in a sense is a version of reversible computing since unitary transformations are always bijective, and reversible computing is the special case of quantum computing without superposition. On the software level, there are even some reversible programming languages where you can automatically find the inverse of a computer program.

Now, reversibility at first glance appears to be much weaker than conventional computation and practically useless. This is not true. Landauer's principle states that every bit deleted costs $$k*T*\ln(2)$$ energy ($$k=1.38\cdot 10^{-23}J/K$$ is Boltzmann's constant and $$T$$ is the temperature) and information is deleted every time one makes an irreversible computation in any way. Reversible computation is not subject to Landauer's limit, so we should expect for reversible computers to start to outperform conventional computers in very specialized reversible tasks in the near future. Now, many people who accept Landauer's principle then assume that reversible computation is inherently weak and unable to do anything interesting because either the computational overhead is too great or because one reversible computation invariably produces so much garbage information that one will need to expend the energy in reversible computation anyways after one performs the calculation. This is not true.

First of all, Charles Bennett has shown that all conventional computation can be emulated by a reversible computer with only a slight computational overhead in this paper and (see this paper by Emanuel Knill for the optimal solution to Bennett's pebble game). Any $$S$$ space and $$T$$ time computation can be calculated reversibly in $$O(T\cdot(\frac{T}{S})^{\epsilon})$$ time and $$O(S\cdot\log(\frac{S}{T}))$$ space. If $$TS(n)$$ denotes the least product of the space times the time required to perform an $$n$$ unit of time and $$1$$ unit of space computation reversibly using Bennett's pebble game, then Knill has shown that $$TS(n)=n\cdot 2^{2\sqrt{\log(n)}\cdot(1+o(n))}$$.

This computational overhead can be further managed by using partially reversible computation instead of completely reversible computation. Now, Bennett's bounds are the bounds in the worst case scenario, and in many cases, reversible computation can perform a calculation in nearly as many steps as conventional computation.

Now for some reason, people often seem to think that brute force search algorithms such as Bitcoin mining are inherently irreversible and that by Landauer's principle, Bitcoin mining on a classical computer requires one to spend a certain minimum amount of energy per hash.

I have written some code in the completely reversible programming language Janus which emulates cryptocurrency mining but which does not require one to build up any garbage information. You can find an online interpreter for Janus at http://topps.diku.dk/pirc/?id=janusB and some details about Janus in the book Introduction to Reversible Computing by Kaylan Perumalla.

The goal of the following POW problem is to find an input i where w=12321 after we perform the assignments w=i; x=((i+214)^(i+142211))+(w&1231); w-=(x&13321).

i w x

procedure proofofwork
w+=i
x^=((i+214)^(i+142211))+(w&1231)
w-=(x&13321)

procedure main
call proofofwork
from i=0 do
uncall proofofwork
i+=1
call proofofwork

until (w=12321)
uncall proofofwork


(Implementation)

The output of this program for solving a POW problem (and this output includes all possible garbage information) is

i = 20513
w = 0
x = 0


and i=20513 is a solution to this POW problem. This technique for solving POW problems reversibly holds for all POW problems such as that of finding SHA-256 hashes for Bitcoin mining and more generally for all pure brute force search problems.

You can find this erroneous claim in the following sources.

I have contacted Andrew Poelstra and he has refused to admit that yes reversible computation actually can be used to effectively mine Bitcoin.

1. Bitcoin for the Befuddled-Conrad Barski, Chris Wilmer

2. Bitcoin and Cryptocurrency Technologies-I have contacted the authors of this book, and they have refused to put a correction on their errata page or say that their errata page is no longer being updated.

Notice how in none of the above papers do any of the authors attempt to calculate nor estimate how many times $$k\cdot T$$ energy must be spent per Bitcoin hash. They do not even attempt to define an instructional mining algorithm that is much simpler to describe than Bitcoin's mining algorithm just so that it will be easy to calculate how many times $$k\cdot T$$ energy must be spent on solving these mining problems or how much information must be deleted in order to mine Bitcoin. I have contacted the authors of the above papers and books and they have so far refused to post errata or retractions. Unfortunately, unlike most misconceptions in mathematics and computer science, for some reason even so-called experts do not seem to be able to let go of this one.

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.

• 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
• I meant that $V-U$ cannot be a subspace since it doesn't contain 0. On the other hand, in any commutative ring where $1+1=0$, then the formula $(x+ y )^2=x^2+y^2$ holds. – ACL Apr 21 '16 at 10:02

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

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, $x^2$, and the fact that its graph does not look like a line at any value of $x$.

• 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

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 characteristic two and I didn't get a good grade that day!

• 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 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.

• 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

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).