## Examples of common false beliefs in mathematics. [closed]

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 2010 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. – To be cont'd May 22 2010 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. – S. Sra Sep 20 2010 at 12:39
It's a thought -- I might consider it. – gowers Oct 4 2010 at 20:13
Meta created meta.mathoverflow.net/discussion/1165/… – quid Oct 8 2011 at 14:27
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## closed as no longer relevant by Mark Sapir, Felipe Voloch, George Lowther, Mark Meckes, Ryan BudneyOct 8 2011 at 22:24

I don't think I've seen it in here:

Every vector space has a non-trivial dual space ($L^p$ for $0 < p < 1$ was a counter-example only mentioned during one of the classes in measure theory)

And of course there's the common false belief of people outside of mathematics that "mathematicians work with numbers and formulae all day long" :)

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Well, it is true that every vector space has a dual space, even $L^{1/2}$... and it is even true that every topological vector space has a continuous dual space... What you mean is that it is not true that every topological vector space has a non-trivial continuous dual space (or, that the continuous dual of a topological vector space does not necessarily separate points) – Mariano Suárez-Alvarez Jul 7 2010 at 18:54
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"It cannot be shown without some form of AC that the union (or disjoint union) of countably many countable sets is countable. I have a countably infinite set X of countably infinite sets. Therefore, the union of X cannot be shown to be countable without Choice."

The fallacy is that in many cases of interest, it is possible to exhibit an explicit counting of every element of X. In such a case a counting of X by antidiagonals is easily constructed. The usual counting of the rationals is an example of this.

I think this may even be an example of a more general phenomenon of "people think AC is necessary for a certain construction, but in fact it turns out not to be necessary for the example they have in mind". For example, AC is necessary to find a maximal ideal in an arbitrary ring ... but it isn't if you're prepared to assume the ring is Noetherian.

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If "Noetherian" is defined by the ascending chain condition or by requiring all ideals to be finitely generated, then in order to deduce the existence of maximal ideals, you still need a weak form of the axiom of choice. The usual argument uses the axiom of dependent choice. (Of course, if you define "Noetherian" to mean that every set of ideals has a maximal element, then deducing the existence of maximal ideals is a choiceless triviality.) A good reference is "Six impossible rings" by Wilfrid Hodges (J. Algebra 31 (1974) 218-244). – Andreas Blass Oct 22 2010 at 15:29
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The assumption that a cubic surface expressed as a foliation of Weierstrass curves cannot be rational, because a general Weierstrass curve is not rational.

I've seen this false assumption more than once on sci.math over the years. But there are simple counterexamples, such as:

$(x + y) (x^2 + y^2) = z^2$

On defining $u = x/y$ and $v = z/y$ one obtains $y (u + 1) (u^2 + 1) = v^2$, and hence x, y, z as rational functions of u, v.

I'd love to have a reference to a procedure for calculating the geometric genus and algebraic genus of surfaces like this, because they are rational if and only if both these quantities are zero, and for other cubic surfaces that interest me it would save a lot of fruitless hacking around trying to find a rational solution that probably doesn't exist! Are there any symbolic algebra packages that can do this?

I mean for example is $x y (x y + 1) (x + y) = z^2$ rational? I'm almost sure it isn't; but how can one be sure?

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Something I was sure about until earlier today:

Suppose $\kappa$ is an $\aleph$ number, then $AC_\kappa$ is equivalent to $W_\kappa$, namely the universe holds that the product of $\kappa$ many sets is non-empty if and only if every cardinality is either of size less than $\kappa$ or has a subset of cardinality $\kappa$.

In fact this is only true if you assume full $AC$, and $(\forall \kappa) AC_\kappa$ doesn't even imply $W_{\aleph_1}$, I was truly shocked.

Furthermore, $W_\kappa$ doesn't even imply $AC_\kappa$ in most cases.

The strongest psychological implication is that most people actually think of the well-ordering principle as a the "correct form" of choice, when it is actually Dependent Choice (limited to $\kappa$, or unbounded) which is the "proper" form, that is $DC_\kappa$ implies both $AC_\kappa$ and $W_\kappa$.

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How common is this misconception? – Thierry Zell Apr 17 2011 at 3:08
@Thierry: For the past couple of weeks I spent a lot time considering models without choice, not only I held that misconception but not once anyone corrected me about it - grad students and professors alike. – Asaf Karagila Apr 17 2011 at 6:09

Hopefully this isn't a repeat answer. False belief: a matrix is positive definite if its determinant is positive.

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Is this really a common(!) false belief? – Martin Brandenburg Oct 3 2011 at 7:23

Fans: (related to the one of polytopes written above) all convex cones are rational, i.e. one would expect that a line would eventually hit a point in the lattice. It is obviously not true, just take the one-dimensional cone generated by $(1,\sqrt{2})$. A similar one was thinking that if I rotate the cone a bit, I can always make it rational.

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reminds me of the curious fact that some circles in the plane, too, have no points in $\mathbb Q^2$. (proven simply by cardinality!) – AndrewLMarshall Oct 4 2010 at 19:21

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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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. – thei Apr 10 2011 at 20:50
It's not a "false belief". It's a false heuristic. And it's actually here: mathoverflow.net/questions/2358/… – darij grinberg Apr 10 2011 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 2011 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 2011 at 2:28
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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 2010 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 2010 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 2010 at 15:22
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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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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 2010 at 21:11
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 2010 at 4:40
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$\exp(\pi\sqrt{163})$ 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 2010 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 2010 at 16:25
@Gerhard. You're right. Computing $2^{46}$, a number with $14$ digits, step by step requires calculating approximately $\frac12(46\times14)=322$ digits. At one digit per second, this requires less than $6$ minutes. I apologize. – Denis Serre Oct 20 2010 at 5:26
@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 2011 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 2011 at 13:59
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The sigma function

$$\sigma_{1}({p_i}^{\alpha_i}) = \displaystyle\sum_{j=0}^{\alpha_i}{{p_i}^j}$$

satisfies the inequalities

$$\sigma_{1}({p_i}^{\alpha_i}) \gt (\alpha_i + 1)(\sqrt{p_i})^{\alpha_i}$$

$$\sigma_{1}({p_i}^{\alpha_i}) \gt 1 + \alpha_i(\sqrt{p_i})^{1 + \alpha_i}$$

for prime $p_i$ and $\alpha_i \ge 1$.

The "proof" uses the Arithmetic Mean-Geometric Mean Inequality.

As a particular application of this result, Sorli's Conjecture implies the OPN Conjecture.

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Is this really a common belief? – Mariano Suárez-Alvarez Jan 28 2011 at 18:21
@Arnie Your first equation makes no sense. Presumably you want to define $\sigma_1(n)$ for every positive integer $n$, hence a product is missing on the LHS. And the sum on the RHS cannot end at $\alpha_i$. Also, I wonder what is the use of the subscript $i$ in the two inequalities. – Didier Piau Feb 6 2011 at 10:36
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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 2010 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 2010 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, 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/… (although I do not think anyone interprets it the way you apparently do). – Qiaochu Yuan May 5 2010 at 4:53
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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 2010 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 2010 at 12:30
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I had the false belief that recursive functions are always decidable in ZFC.

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$n!$ is the product of all positive integers less than or equal to $n$. In fact it should be defined in combinatorial terms.

Many assume the fact that parallel lines in Euclidean geometry do not cross is an axiom, while it can easily be proved in terms of vector space.

Many lecturers do not stress the difference between inner product and scalar product and most students think that these are different names for the same thing.

In complex numbers $i = \sqrt{-1}$. Obviously it is not correct as well.

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I wouldn't call any of these false mathematical statements except for possibly the last. You can axiomatize Euclidean geometry in any number of ways. – Qiaochu Yuan May 5 2010 at 20:52
Why should $n!$ necessarily be defined in combinatorial terms? I mean, why is defining it as $n!=\prod\limits_{k=1}^n k$ a mistake? -- As for axiomatization of Euclidean geometry, there is no definite mistake here, but the modern affine-plane-with-inner-product construction is superior to the traditional in many ways (including working over arbitrary fields of characteristic $\neq 2$), so the fifth axiom is indeed not the important thing to fuss about that it was before. Though, in hindsight, it has helped develop some highly interesting differential geometry. – darij grinberg May 5 2010 at 21:42
@Neil It's not a special case. Everyone knows the product of the empty set is 1. :-) – Dan Piponi May 6 2010 at 0:39
"The product of the first 0 positive integers" is the product of the elements of an empty set. – JBL May 6 2010 at 2:38
Am I the only one confused over what #3 (inner vs scalar) is driving at? – Thierry Zell Nov 27 2010 at 16:27
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