# 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 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.

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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/… –  quid Oct 8 '11 at 14:27

$\mathbb{R}^2$ has a unique complex structure; it's just $\mathbb{C}$ right?

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When $H$ and $K$ are subgroups of $G$ then $HK$ is subgroup of $G$...

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If every collection of disjoint open sets in a topological space is at most countable, then the space is separable

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Way late to the party...

"$\mathrm{polymod}\ p$ and $\mathrm{mod}\ p$ are the same thing."

And it's cousin: "$\forall{x}, f(x) \cong g(x) \pmod{q} \implies f(x) = g(x)$"

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What does polymod mean? –  darij grinberg Oct 20 '10 at 11:47
Probably I understand what this means: if $f(x)=0\pmod 2$ for all $x$, then $f=0$ over $\mathbb F_2$. This is similar to my second example: mathoverflow.net/questions/23478/… –  zhoraster Oct 20 '10 at 18:33
$\mathrm{polymod}$ is "polynomial mod". Two polynomials are congruent $\mathrm{polymod} p$ iff the coefficients each power of the variable are congruent $\pmod{p}$. The equivalence classes are sets of polynomials where each coefficient ranges over an equivalence class $\pmod{p}$. For the cousin, there are many local/globals but they all seem to require additional conditions (q.v. Hensel lifting). I think the set from which $x$ was chosen was left unspecified because this "imprecise mental abbreviation" pops up at various levels of sophistication each with a different such set. –  Anonymous Oct 23 '10 at 15:22

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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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

This might not be common, but I once believed the following.

Let $A, B$ be integers, and define a sequence by the linear recurrence $s_n = A s_{n-1} + B s_{n-2}$ with the base case $s_0 = 0$, $s_1 = 1$. Two important special cases are the Fibonacci sequence ($A = B = 1$) and the sequence $s_n = 2^n - 1$ (where $A = 3$, $B = -2$). Then, for any integers $n$ and $k$, $\gcd(s_n, s_k) = s_{\gcd(n,k)}$.

This is true in the two mentioned special cases, so it's tempting to believe it's true in general. But there's a counterexample: $A = B = k = 2$, $n = 3$.

Update: corrected the powers of two minus one example from B = 2 to B = -2. Thanks to Harry Altman.

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## 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!

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

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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

This might not be common, but it gave me a headache once. I'll delete if it gets heavily downvoted.

I once had to think really hard about a contradiction in the great scheme of things that followed from my unwitting assumption that if $f$ was a function from a semigroup to a semigroup, then if its kernel was a congruence, $f$ had to be a homomorphism. I encountered a function whose kernel clearly was a congruence but which clearly wasn't a homomorphism, and it took about an hour's walk in a park for my vague notions and incoherent thought to produce the necessary realization.

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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.

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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

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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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

$\exp(\pi\sqrt{163})$ 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.

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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!

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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

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

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

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