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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
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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
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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
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It's a thought -- I might consider it. –  gowers Oct 4 '10 at 20:13
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Meta created tea.mathoverflow.net/discussion/1165/… –  quid Oct 8 '11 at 14:27

179 Answers 179

These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$.

Another mistake is if one i asked to solve this equation, $ \displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$.

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@Todd: No i was talking of high school students. –  Chandrasekhar May 4 '11 at 4:08

Here's another howler some people commit: If m, n are integers such that m divides n^2 then m divides n.

It's true sometimes, for example if m is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers p, q, r with p squarefree such that $ m = p q^2 $ and $ n = p q r $

The usual counterexample is that 8 divides 4^2 but not 4 ;-)

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An even more trivial counterexample is that 4 divides 2^2 but not 2 :-P –  Peter LeFanu Lumsdaine Feb 23 '11 at 9:40

If a matrix $A$ is self-adjoint/skew-self-adjoint with respect to a symmetric bilinear form, then it is diagonalizable.

True for matrices over $\mathbb{R}$, with respect to a positive definite inner product.

False over other fields. For example, over $\mathbb{C}$, $\left( \begin{smallmatrix} 1 & i \\ i & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & 1 & i \\ -1 & 0 & 0 \\ -i & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to the standard inner product.

False for other nondegenerate symmetric bilinear forms: $\left( \begin{smallmatrix} 1 & 1 \\ -1 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & -1 & -1 \\ 1 & 0 & 0 \\ -1 & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to $\left( \begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{smallmatrix} \right)$.

You can exponentiate the skew-self-adjoint matrices to get examples of matrices preserving a nondegenerate symmetric bilinear form, with Jordan blocks of the form $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$.

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You seem to have a different definition of "the standard inner product on $\mathbb{C}^n$" than I do. I think that phrase normally refers to the familiar positive definite sesquilinear form, with respect to which self-adjoint matrices are indeed diagonalizable. –  Mark Meckes Jan 28 '11 at 16:46
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Of course it's not bilinear -- an "inner product" on a complex vector space is defined to be sesquilinear, not bilinear -- I've spent a lot of time trying to get my linear algebra students to remember that. The failure of such a form to generalize to other fields is indeed sad, but I think the richness of Hilbert space theory helps to make up for that disappointment. :) –  Mark Meckes Jan 28 '11 at 21:24

If $a$ is a real zero of a cubic polynomial with rational coefficients then $a$ can be written as a combination of cube roots of rational numbers.

More generally if $a$ is a real zero of an irreducible polynomial with rational coefficients that is solvable by radicals then students expect the following:

  1. Any expression inside a radical evaluates to a real number
  2. Any sub-expression of the expression for $a$ evaluates to an algebraic number of order less than or equal to the order of $a$

Of course the problem is that from Cardan's solution to the cubic we can have negative rational numbers inside a square root. Let $c$ = $4*(-1 + \sqrt{-3})$.

$a$ = $\frac{\sqrt[3]{c}}{4} + \frac{1}{\sqrt[3]{c}}$

$f(x) = 4x^3 - 3x + \frac{1}{2}$.

So while $a$ is an algebraic number of degree three, it can not be written as combination of cube roots of rational numbers. Indeed, it is counter-intuitive that $\sqrt[3]{c}$ has degree 6 over the rational numbers yet we can use this number and simple arithmetic to produce an algebraic number of degree 3.

Also $a$ = $\sin(50^{\circ})$. For many values of $\theta$, $\sin \theta$ is a radical number. See also radical values for sine and cosine

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This has a name: en.wikipedia.org/wiki/Casus_irreducibilis –  Qiaochu Yuan Apr 6 '11 at 21:21

False Belief: "The suspension spectrum map from spaces to (edit: symmetric) spectra preserves smash-products"

The facts that one denotes the smash product of spectra and the smash product of a space with a spectrum (levelwise) with the same $\wedge$ and tends to leave away the $\Sigma^\infty$ when one embeds a space into spectra are also not helpful in getting used to the harsh reality that the above is wrong.

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I don't see that this qualifies as a false belief. In order for the question of whether it is true or false to even be meaningful, you have to first commit yourself to one of the many different notions of spectrum, not to mention smash product of spectra. –  Tom Goodwillie Oct 5 '10 at 0:35
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True. I meant symmetric spectra with the smash product coming from their description as modules over the symmetric sequence of spheres. –  Peter Arndt Oct 5 '10 at 10:52

For a bounded subset of a metric space the diameter is two times the radius!

Let $S\subset X$ be bounded. The definitions are:

$\mathrm{diameter}(S):=\sup\{d(x,y)\,|\,x,y\in S\}$

$\mathrm{radius}(S):=\inf\{r>0\,|\,\exists x\in X:\,S\subset B(x,r)\}$

where $B(x,r)$ denotes the open ball of radius $r$ around $x$.

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Hםw do you define the radius of an arbitrary bounded subset? –  Mark Apr 11 '11 at 15:34
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Disproved nicely by Reuleaux triangles. –  darij grinberg Apr 12 '11 at 8:10
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Disproved nicely by a two-point metric space. –  Tom Goodwillie Apr 17 '11 at 1:36

Coordinates on a manifold do not have an immediate metric meaning. Until becoming familiar with differential geometry one tends to think they do. (Einstein wrote that he took seven years to free himself from this idea.)

For example, linear control theory is for the most part metric with variables in $R^n$. When moving away from linear control theory, variables are represented as coordinates on a manifold. Nevertheless, much of the literature tends to either abandon metric notions altogether, or to keep using an Euclidean metric though it is no longer very useful.

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A set is compact iff it is closed and bounded.

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This is perhaps a common false belief among undergraduates, but one that is dispelled by just a superficial acquaintance with functional analysis. –  Todd Trimble Dec 9 '13 at 2:45

That Darboux functions are continuous is certainly a widely held belief among students, at least in France where it is induced by the way continuity is taught in high school.

I remember having gone through all the five "stages of grief" when shaken from this false belief with the $sin(1/x)$ example : denial, anger ( "then the definition of continuity must be wrong ! Let's change it !), bargaining ("Ok, but a Darboux function must surely be continuous except at exceptional points. Let's prove that..."), depression (when shown a nowhere continuous Darboux function), acceptance ("Hey guys, you really think the intermediate value theoem has a converse ? C'mon, you're smarter than that...")

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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 '10 at 18:54

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 '10 at 19:21

"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 '10 at 15:29

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 '11 at 3:08
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@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 '11 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 '11 at 7:23

Here are some various examples (I hope that some of them weren't already mentioned):
1. If a space $X$ have two different norms $\| \cdot \|_i, i=1,2$ such that $\| \cdot \|_1 \leq \| \cdot \|_2$ then the completion with respect to $\| \cdot \|_1$ is contained in the completion with respect to $\| \cdot \|_2$.
2. If $M_1,M_2$ are isomorphic modules and $N_1,N_2$ are isomorphic submodules then $M_1/N_1$ and $M_2/N_2$ are isomorphic.
3. If $A,B$ are subsets of topological spaces $X,Y$ (resp.) and $A,B$ are homeomorphic then the closures $\overline{A}$ and $\overline{B}$ are also homeomorphic.
4. The standard construction of adjoining unit to the Banach algebra $A$ yields nothing new if $A$ already was unital.
5. The phrase "a function is almost everywhere continuous" means the same as: "the function is almost everywhere equal to the continuous function".
6. Suppose you are trying to prove that some function space $F$ is complete (say that functions are defined on $X$ and real valued): you take a Cauchy sequence $\{f_n\}_n$ and prove that for each point $x \in X$ the sequence $\{f_n(x)\}_n$ is Cauchy. Then form the completeness of $\mathbb{R}$ you obtain a function $f$. The false belief is that it is now enough to show that $f$ belong to $F$.
7. If you have an ascending family $\{A_i\}_i$ then to obtain it's union $\bigcup_{i}A_i$ it is enough to take some countable subfamily
8. A convergent net $\{x_i\}_i$ in a metric space is bounded and the set $\{x_i\}_i \cup \{x\}$ is compact (where $x$ is the limit).
9. If $D$ is an open dense subset of a topological space $X$ then $card \; D= card \; X$

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I don't know how common this mistake is, but I think it's worth mentioning. I used to think that existence of non-measurable sets is guaranteed by the axiom of choice only.

In the presence of AC, there cannot be a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ that extends the usual Lebesgue measure.

It is true that we cannot extend the Lebesgue measure in a translation-invariant way by various Vitali set constructions. On the other hand, if you do not insist that the extension is translation-invariant, it might be possible to do this relative to a real-valued measurable cardinal assumption.

Theorem (Ulam): If there exists a cardinal $\kappa$ such that there exists an atomless $\kappa$-additive probability measure on $\mathcal{P}(\kappa)$, then there exists a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ extending the Lebesgue measure.

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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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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
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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
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$\mathrm{polymod}$ 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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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
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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
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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
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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

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
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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
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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
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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
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@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 '10 at 5:26
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@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
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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
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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

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 '11 at 18:21
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@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. –  Did Feb 6 '11 at 10:36

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

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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 '10 at 12:07
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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

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