# 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

## 250 Answers

The derived subgroup of a finite group equals to the set of all its commutators

or equivalently

A product of two commutators in a finite group is always a commutator

This mistake is very widespread, probably because counterexamples to it tend to be quite large. The smallest group, for which it is not true has order $$96$$.

## From Keith Devlin

"Multiplication is not the same as repeated addition", as put forward in Devlin's MAA column.

I'm not really sure how I feel about this one; I might be one of the unfortunate souls who are still prey to that delusion.

## Caution

In case you missed it, the column ended up spilling a lot of electronic ink (as evidenced in this follow-up column), so I don't believe it would be wise to start yet a new one on MO. Thanks in advance!

• I followed your link, and I cannot even tell what is wrong about attaching helium balloons to both sides of a balance to model substraction on both sides of an equation. – user11235 Apr 10 '11 at 20:32
• The more I think about this "error", the less I am convinced. It's like saying that you cannot say that $\binom n k$ is the number of $k$-element sets in an $n$-element set because then you will be unable to generalize to complex values of $n$. Or you cannot define the chromatic polynomial as the function counting the colourings and then plug in $-1$ to get the acyclic orientations of the graph. Also, I think it is perfectly understandable what it means to add something halfways. – user11235 Apr 10 '11 at 20:50
• It's not a "false belief". It's a false heuristic. And it's actually here: mathoverflow.net/questions/2358/most-harmful-heuristic – darij grinberg Apr 10 '11 at 21:17
• When I taught elementary teachers the course on arithmetic, they all had been taught that multiplication is repeated addition, but I myself thought it was the cardinality of the cartesian product. We enjoyed discussing this difference in point of view. – roy smith May 9 '11 at 3:06
• The "repeated addition" characterization has an advantage over the "cardinality of the Cartesian product" characterization (which possibly in some contexts could be considered a disadvantage). That is that it's not self-evident that it's commutative, and so one has a useful exercise for certain kinds of students: figure out why it's commutative. – Michael Hardy May 20 '11 at 2:28

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 $\mathbb{Z}(p^{\infty})$.

• Is that actually a common false belief? After all, $\mathbb{Z}_p$ is uncountable, as everyone realizes! – Todd Trimble Mar 5 '15 at 14:25
• "$\mathbb{Z}_p$ is countable" is also a false belief for people who didn't really read the definition of $\mathbb{Z}_p$, but I don't know how much it is common. – Sebastien Palcoux Mar 5 '15 at 14:34
• 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 '15 at 14:52

In algebraic topology, I thought for a while:

• "For $$k \geq 2$$, $$H_k$$ is the abelianization of $$\pi_k$$." False. True for $$k = 1$$. Also for all $$k$$ up to $$n-1$$ if the space is $$(n-1)$$-connected for $$n \geq 2$$ (vacuously, since this says the first $$n-1$$ homotopy groups are trivial and for these, the Hurewicz homomorphism is the isomorphism, $$\pi_k \cong H_k$$). See the Hurewicz theorem for more.
• "Generically, all the $$\pi_k$$ are nonabelian." False. For $$k \geq 2$$, $$\pi_k$$ is abelian.

Edit: I had a third error in thinking when I first posted this, mangling the above into something further from true. Which I suppose makes the first version of this post meta-appropriate for this thread (but I've fixed it anyway). Thankfully, user Michael gently pointed out my mangling.

• First bullet: did you mean "True for $n=1$"? – Michael Jan 15 at 23:06
• @Michael : It's not always true for $n=1$, $\pi_1$ can be abelian, e.g. the fundamental group of the circle. For $n > 1$, $H_n \cong \pi_n$. It's easy to imagine "$\pi_n$s are (usually) nonabelian monsters and their associated homology groups are friendly abelian groups", but this difference *only* happens for $n=1$. – Eric Towers Jan 15 at 23:31
• I think you are confusing a few things here. Compare $H_2$ of the 2-dimensional torus with its $\pi_2$, for example. – Michael Jan 15 at 23:34
• @Michael : After actually looking up what I was talking about, I find that I have mashed together (at least) two errors to make another. Yay? – Eric Towers Jan 16 at 4:42
• @Michael : I think I've disentangled my mangling. I may still have a fumble-thought in the first bullet that I'm just not seeing. – Eric Towers Jan 16 at 5:11

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

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 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

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