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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

186 Answers 186

The "conditional Vitali convergence theorem": Let $X_n$ be a uniformly integrable sequence of random variables with $X_n \to X$ almost surely, and $\mathcal{G}$ a sub-$\sigma$-field. Then $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ almost surely (FALSE).

I believed this one until I read Uniformly integrable sequence such that a.s. limit and conditional expectation do not commute. It is particularly startling because the conditional versions of the monotone convergence theorem, the dominated convergence theorem, and Fatou's lemma are all true!

What is true is that $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ in $L^1$, so you do have a subsequence converging almost surely.

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Many students believe that every abelian subgroup is a normal subgroup.

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I confess that I didn't carefully comb through all the answers, although I've read through this thread a few times in the past. So maybe these are repeats.

  • "The category of compact Hausdorff spaces is complete but not cocomplete; for example, it doesn't have all coproducts."

  • "The category of torsion abelian groups is cocomplete but not complete; for example, it doesn't have all products."

One of my professors in graduate school (quite a well-known and strong mathematician actually) insisted on the first, and quite a few people here at MO have mistakenly believed the second before the error was pointed out.

The moral of the story: sometimes categorical limits/colimits aren't computed the way you might first think of, e.g., colimits of compact Hausdorff spaces aren't always computed as colimits in $\mathrm{Top}$, and limits of torsion abelian groups aren't always computed as limits in $\mathrm{Ab}$.

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Here are a few more: (Everything between quotation marks is a false belief.)

Basic logic: Among students: "If A implies B then B implies A" (or "if A implies B then not A implies not B").

Even among mature mathematicians a frequent false belief is to forget that the conclusion of a theorem need not hold once the conditions of the theorem fail. Another common frequent belief is to assume that once the conditions fail then the conclusion must fail too.

Calculus: "The derivative of a differentiable function is continuous."

"An infinite series whose general term tend to 0 is convergent."

Geometry: "The circle is the only figure which has the same width in all directions." (Feynmann regarded this mistake as one reason for the space shuttle challenger disaster).

Polytopes: Often people believe that "given a convex polytope P you can slightly move the vertices to rational positions keeping the structure of the polytope unchanged."

(From Udi de Shalit): Some people believe that "if you hold a cube along a main diagonal, the remaining vertices all lie on a plane." Some even say that their number is 4.

Algebra (Also from Udi) "I have encountered many misconceptions about solvability by radicals. Some people think that 'the solution of an irreducible equation of degree 5 and higher, say over Q, is never expressible by radicals'. Some amateur mathematicians even say that 'equations of degree 5 and higher have no solutions'."

Probability: "If you play the casino patiently and carefully you will win in the long run" (and "you do not believe that?, this is my own experience on the matter!" and "Indeed when I am calm and patient I win, but when I lose my temper I lose big time".)

"an event which may occur has positive probability": (not true for infinite probability spaces)

Various places: "If you want to prove that a certain infinite structure exists it is enough to show that there is no upper bound to the sizes of such structures."

Combinatorics: "This is a finite problem, surely you can solve it with a computer."

"Hall marriage theorem is very nice and I am surprised no combinatorialist bothered to extend it to a matching built from triples instead of pairs." (It is unlikely that a general characteriztion when a hypergraph built from triples has a perfect matching (of triangles) will be found.)

Computer science: "It is known that quantum computers can solve NP complete problems in polynomial time."

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"If you want to prove that a certain infinite structure exists it is enough to show that there is no upper bound to the sizes of such structures." This is not necessarily false. In some important cases this does work! One uses the compactness theorem for such proofs. –  Johannes Hahn May 7 '10 at 12:46
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Probability: There are two opposite errors. Both are common. Say we are flipping a fair coin repeatedly. (1) if there have been more heads than tails, then tails is "overdue" and thus more likely on the next flip. (2) if there have been more heads than tails, then heads is "hot" and thus more likely on the next flip. –  Gerald Edgar May 7 '10 at 15:15
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I don't know anything about polytopes, but I'm having a hard time disbelieving this false result. Are we talking about finite polytopes here? –  Tom Ellis May 9 '10 at 19:26
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Entirely finite, Tom. There are 4-dimensional polytopes with 33 vertices that cannot be presented with rational coordinates. Here is a reference arxiv.org/PS_cache/arxiv/pdf/0710/0710.4453v2.pdf –  Gil Kalai May 9 '10 at 20:46
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You list the statement "Quantum computers can solve NP complete problems in polynomial time" as a false belief, but I don't believe you actually know this belief to be false. For example, the assertion that this belief is false implies $P\neq NP$. Perhaps the false belief that you intend to mention is: "It has been proved that Quantum computers can solve NP complete problems in polynomial time." –  Joel David Hamkins May 17 '10 at 12:34

For beginner under graduate students in general topology, a false belife could be:

"If $A$ and $B$ are two homeomorphic subset of $X$ then $X-A$ is homeomorphic to $X-B$, but the knots are counter example"

However, it would be interesting to classify all locally compact Hausdorf space with this property.

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There are much simpler counterexamples, e.g., let $X$ be a closed interval, let $A$ be an endpoint, let $B$ be an interior point. –  Gerry Myerson Nov 12 at 21:50

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 at 13:31
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@quid Thank you. I did not read it already. however this one introduce an explicte counter example. –  Ali Taghavi Nov 12 at 13:59
  • The category of commutative C*-algebras is equivalent to the opposite category of locally compact Hausdorff spaces.

Let me know in case that this has been mentioned before; I haven't been able to find it. There is some discussion on math.SE.

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The Quaternions $\{x+yi+zj+wk\mid x,y,z,w\in \mathbb{R}$} is a complex banach algebra(With usual operations). Hence it is apparently a counterexample to the Gelfand=Mazur theorem

So, what is the error?

The error is the following:

However the quaternion is a vector space over the field of complex number and it is also a ring, but there is no compatibility between scalar multiplication and quternion multiplication). So it is not a complex algebra. This shows that in the definition of a complex algebra $A$, the commutative condition $\lambda (ab)=(a)(\lambda b),\;\;\lambda \in \mathbb{C},\;\;a,b\in A$, is very essential.

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Moreover, surely the quaternions are a real vector space, not a complex vector space –  Yemon Choi Nov 13 at 1:34
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@AliTaghavi You're right that $R$-multiplication induces an $F$-module ($F$-vector space) structure via the evident composite $F \times R \to R \times R \to R$. To be fair to both you and Yemon: a very common slip even among professionals is in knowing that for commutative algebras an $F$-algebra is tantamount to a homomorphism $F \to R$, but temporarily forgetting this doesn't apply in the noncommutative setting (except of course when $F$ is central in $R$) -- not rising to the level of false belief so much as a temporary slip-up. I've made that slip myself! –  Todd Trimble Nov 13 at 11:58

Let $(X,\tau)$ be a topological space. The false belief is: "Every sequence $(x_n)$ in $X$ with an accumulation point $a\in X$ has a subsequence that converges to $a$". I subscribed to this intuitively until I stumbled over a counterexample, see http://dominiczypen.wordpress.com/2014/10/13/accumulation-without-converging-subsequence/

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Here's my list of false beliefs ;-):

  • If $U$ is a subspace of a Banach space $V$, then $U$ is a direct summand of $V$.
  • If $M/L, L/K$ are normal field extensions, then the same is true for $M/K$.
  • Submodules/groups/algebras of finitely generated modules/groups/algebras are finitely generated.
  • The Krull dimension of a subring is at most the Krull dimension of the ring.
  • The Krull dimension of a noetherian domain is finite.
  • If $A \otimes B = 0$, then either $A=0$ or $B=0$.
  • If $f$ is a smooth function with $df=0$, then $f$ is constant.
  • If $X,Y$ are sets such that $P(X), P(Y)$ are equipotent, then $X,Y$ are equipotent.
  • Every short exact sequence of the form $0 \to A \to A \oplus B \to B \to 0$ splits.
  • $R[[x,y]] = R[[x]][[y]]$ as topological rings.
  • $R[x]^* = R^*$, even if $R$ is not a domain.
  • Every presheaf on a site has an associated sheaf. (Hint: the index category of the usual colimit has to be essentially small!)
  • (Co)limits may be computed in full subcategories. For example, $Spec(\prod_i R_i) = \coprod_i Spec(R_i)$ as schemes because $Spec$ is an antiequivalence.
  • Every finite CW-complex is compact, thus every CW-complex is locally compact.
  • The smash product of pointed spaces is associative (this is even false for CW complexes when you don't use the compactly-generated product!), products commute with quotients, and so on: Topologists often assume that everything behaves well, but sometimes it does not.
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+1: you had me at "Here's my list of false beliefs". –  Pete L. Clark May 5 '10 at 23:41
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$A \to A \oplus B$ does not have to be the inclusion; likewise $A \oplus B$ does not have to be the projection. Thus the error here is: Two chain complexes, which are isomorphic "pointwise", don't have to be isomorphic. This occurs sometimes. –  Martin Brandenburg May 6 '10 at 16:12
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Your fifth example reminds me of an even more plausible false belief I once held: if $A \otimes A = 0$, then $A = 0$. –  Reid Barton May 11 '10 at 2:12
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@Regenbogen: Take the abelian group $\mathbb{Q}/\mathbb{Z}$. –  Steve D May 15 '10 at 13:44
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$f$ is just locally constant ;-) –  Martin Brandenburg May 19 '10 at 8:39

The product of two symmetric matrices is symmetric!

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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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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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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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In descriptive set theory, we study properties of Polish spaces, typically not considered as topological spaces but rather we equip them with their "Borel structure", i.e., the collection of their Borel sets. Any two uncountable standard Borel Polish spaces are isomorphic, and the isomorphism map can be taken to be Borel. In practice, this means that for most properties we study it is irrelevant what specific Polish space we use as underlying "ambient space", it may be ${\mathbb R}$, or ${\mathbb N}^{\mathbb N}$, or ${\mathcal l}^2$, etc, and we tend to think of all of them as "the reals".

In Lebesgue Sur les fonctions representables analytiquement, J. de math. pures et appl. (1905), Lebesgue makes the mistake of thinking that projections of Borel subsets of the plane ${\mathbb R}^2$ are Borel. In a sense, this mistake created descriptive set theory.

Now we know, for example, that in ${\mathbb N}^{\mathbb N}$, projections of closed sets need not be Borel. Since we usually call reals the members of ${\mathbb N}^{\mathbb N}$,

it is not uncommon to think that projections of closed subsets of ${\mathbb R}^2$ are not necessarily Borel.

This is false. Note that closed sets are countable union of compact sets, so their projections are $F_\sigma$. The actual results in ${\mathbb R}$ are as follows: Recall that the analytic sets are (the empty set and) the sets that are images of Borel subsets of $\mathbb R$ by Borel measurable functions $f:\mathbb R\to\mathbb R$.

  • A set is Borel iff it and its complement are analytic.

  • A set is analytic iff it is the projection of the complement of the projection of a closed subset of ${\mathbb R}^3$.

  • A set is analytic iff it is the projection of a $G_\delta$ subset of $\mathbb R^2$.

  • There is a continuous $g:\mathbb R\to\mathbb R$ such that a set is analytic iff it is $g(A)$ for some $G_\delta$ set $A$.

  • A set if analytic iff it is $f(\mathbb R\setminus\mathbb Q)$ for some continuous $f:\mathbb R\setminus\mathbb Q\to\mathbb R$. (Note that if $f$ is actually continuous on $\mathbb R$, then $f(\mathbb R\setminus\mathbb Q)$ is Borel.)

(See also here.)

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Teaching introduction to analysis, I had students using the "fact" that if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, then $[a,b]$ can be divided to subintervals $[a,c_1],[c_1,c_2],...,[c_n,b]$ such that $f$ is monotone on every subinterval. For instance you can use this "fact" to "prove" the (true) fact that $f$ must be bounded on $[a,b]$. Also, some students used the same "fact", but with countably many subintervals. I found this mistake hard to explain to students, because constructing a counterexample (such as the Weierstrass function) is impossible at the knowledge level of an introduction course.

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Why not $x \sin(1/x)$ as example? –  quid Jan 2 at 17:33
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You can surely discuss fractal shapes without needing to go into the details of a technical counterexample. The point seems to be that it is hard to imagine that "increasing at a point" and "increasing in a neighborhood of a point" are not the same for continuous functions. You can give easy examples showing that indeed they disagree, locally, and fractals suggest that you can make the disagreement happen everywhere. You can revisit this later, once more technology has been set in place. –  Andres Caicedo Jan 2 at 23:44

Regard a reasonably nice surface in $\mathbb R^3$ that can locally be expressed by each of the functions $x(y,z)$, $y(x,z)$ and $z(x,y)$, then obviously

$\frac {dy} {dx} \cdot \frac {dz} {dy} \cdot \frac {dx} {dz} = 1$

(provided everything exists and is evaluated at the same point).

After all, this kind of reasoning works in $\mathbb R^2$ when calculating the derivative of the inverse function, it works for the chain rule and it works for separation of variables.


Note that this product is in fact $-1$ which can either be seen by just thinking about what happens to the equation $ax+by+cz=d$ of a plane / tangent plane or by looking at the expression coming out of the implicit function theorem.

I recall someone claiming that this example proves that $dx$ should be regarded as linear function rather than infinitesimal, but I cannot reconstruct the argument at the moment as this discussion was 15 years ago.

In particular, it is true under appropriate conditions in $\mathbb R^4$ that $\frac {\partial y} {\partial x} \cdot \frac {\partial z} {\partial y} \cdot \frac {\partial w} {\partial z} \cdot \frac {\partial x} {\partial w} = 1$

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This is an example of the principle that naïve reasoning with Leibniz notation works fine for total derivatives but not for partial derivatives. This is one reason why I would always write the left-hand side as $\frac{\partial{y}}{\partial{x}} \cdot \frac{\partial{z}}{\partial{y}} \cdot \frac{\partial{x}}{\partial{z}}$ if not $\left(\frac{\partial{y}}{\partial{x}}\right)_z \cdot \left(\frac{\partial{z}}{\partial{y}}\right)_x \cdot \left(\frac{\partial{x}}{\partial{z}}\right)_y$ (notation that I learnt from statistical physics, where the independent variables are otherwise not clear). –  Toby Bartels Apr 7 '11 at 12:56
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Can you help us understand it? Or is there no better way than computation? –  darij grinberg Apr 10 '11 at 18:27
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@TobyBartels, I remember an analyst colleague talking about a problem in a paper of his that he resolved by noticing (if I remember the particular example correctly) that $\partial/\partial r$ means something different in cylindrical and spherical co\"ordinates. –  L Spice Dec 12 '13 at 23:17

" Every open dense subset of $\mathbb{R}^n$ has full Lebesgue measure. "

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Common false belief: a space that is locally homeomorphic to $\mathbb{R}^n$ must be Hausdorff. More generally, many people forget that the usual definition of a manifold contains the Hausdorff and paracompact conditions.

There are of course examples that show that forgetting this assumption leads to unexpected result, and they are in fact much wilder than I knew a few weeks ago. Notably, among examples of (Hausdorff) non-paracompact "manifolds" are the well-known long line, but also the Prüfer manifold constructed from a closed half-plane by attaching to it a half plane at each boundary point.

Added: Let me give a particular case of this false belief to illustrate what kind of weird things can happen that most people would not realize when they are sloppy with the paracompact hypothesis: there exists a path-connected, locally contractible, simply-connected space that admits non-trivial locally trivial bundles with fiber $[0,1]$. Indeed, the first octant in the product of two long line is not homeomorphic to a product a long ray with an interval, but has a natural bundle structure over a long ray.

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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

It always confused me as an undergraduate that $\mathbb{Q}\subset\mathbb{R}$ has an open neighborhood $N_\epsilon \supset\mathbb{Q}$ of arbitrarily small measure $\epsilon$, because $\mathbb{Q}$ is dense in $\mathbb{R}$.

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Let $B(r_1) \subset B(r_2)$ be two open balls of radius $r_1$ and $r_2$ respectively. Then $r_1 \leq r_2$.

Bounded metric spaces give trivial counterexamples. Also, $B \left( \frac{1}{6}, \frac{2}{3} \right) \subsetneq B \left( \frac{1}{2}, \frac{1}{2} \right)$ in $(0,+ \infty)$.

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I guess you don't want commonly held beliefs of students that for every real number there is a next real number, or that convergent sequences are eventually constant. A version I saw in a book asked whether points on a line "touch." Understanding the topology of a line is a challenge for many students, although presumably not for most mathematicians.

Here is a more esoteric belief that I have even seen in some books:

"The Banach-Tarski Paradox says that a ball the size of a pea can be cut into 5 pieces and reassembled to make a ball the size of the sun."

As a consequence of the Banach-Tarski paradox, a ball the size of a pea can be partitioned (not really "cut") into a finite number of pieces which can be reassembled into a ball the size of the sun, but a simple outer measure argument implies that the number of pieces must be very large (I roughly estimate at least $10^{30}$). The number 5 probably comes from the fact that the basic Banach-Tarski paradox is that a ball of radius 1 can be partitioned into 5 pieces which can be reassembled into two disjoint balls of radius 1. (It can almost, but not quite, be done with four pieces; one of the five pieces can be taken to be a single point.)

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Since points do not touch, this was an objection to the set theoretic view of the geometric continuum as a set of points, for example by Veronese. A decent account of this can be found in Debates about infinity in mathematics around 1890: The Cantor-Veronese controversy, its origins and its outcome, by Detlef Laugwitz. –  Andres Caicedo Dec 2 '13 at 23:22

A common trap which sometimes I see people fall is that a Hermitian matrix $M$ is negative definite if and only if its leading principal minors are negative.

What is true is the Sylvester's criterion, which says that $M$ is positive definite if and only if its principal minors are positive. Thus, the true statement is that $M$ is negative definite if and only if the principal minors of $-M$ are positive.

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False belief: A function being continuous in some open interval implies that it is also differentiable on some point in that interval:

Counterexample:

The Weierstrass function is an example of a function that is continuous everywhere but differentiable nowhere:

$f(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x)$

Where $a \in (0, 1)$, $b$ is a positive odd integer, and $ab > 1 + \frac{3\pi}{2}$. The function has fractal-like behavior, which leads to it not being differentiable. This notion is rather disheartening to most calculus students, though.

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Related: if f is continuous on the interval I, there must be an interval J in I on which f is monotone. Easily believed by the beginner. –  Thierry Zell Aug 31 '10 at 2:34

Here are two group theory errors I've seen professionals make in public.

1) Believing that if $G_1 \subset G_2 \subset \cdots$ is an ascending union of groups such that $G_i$ is free, then $\bigcup_{i=1}^{\infty} G_i$ is free. Probably the vague idea they have is that any relation has to live in some $G_i$, so there are no nontrivial relations.

2) Consider a group $G$ acting on a vector space $V$ (over $\mathbb{C}$, say). Assume that $G$ acts as the identity on a subspace $W$ and that the induced action of $G$ on $V/W$ is trivial. Then I've seen people conclude that the action of $G$ on $V$ is trivial. Of course, this is true if $G$ is finite since then all short exact sequences of $G$-modules split, but it is trivial to construct counterexamples for infinite $G$.

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So people think $\mathbb{Q}$ is a free group? Curious. –  Pete L. Clark May 5 '10 at 23:39
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Sadly enough, I suspect that many people who care about geometric/combinatorial group theory do not think of Q as a group... –  Andy Putman May 6 '10 at 0:28
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Pete, they don't think... they BELIEVE! –  Victor Protsak May 14 '10 at 6:56
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Come on Steve -- aren't all abelian groups finitely generated? =) –  Andy Putman May 15 '10 at 15:04
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@yatima You probably mean $G_i=\{n/i!\}$. –  The User May 11 '13 at 23:58

Many students believe that 1 plus the product of the first $n$ primes is always a prime number. They have misunderstood the contradiction in Euclid's proof that there are infinitely many primes. (By the way, 2 * 3 * 5 * 7 * 11 * 13 + 1 is not prime.)

Much later edit: As pointed out elsewhere in this thread, Euclid's proof is not by contradiction; that is another widespread false belief.

Much much later edit: Euclid's proof is not not by contradiction. This is another very widespread false belief. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. In fact, if the derivation of an absurdity or the contradiction of an assumption is a proof by contradiction, then Euclid's proof is a proof by contradiction. Euclid says (Elements Book 9 Proposition 20): The very thing (is) absurd. Thus, G is not the same as one of A, B, C. And it was assumed (to be) prime.


Nb. The above edits were not added by the OP of this answer.

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When I was 11 y.o. I was screamed at by a teacher and thrown out of class for pointing this out when he claimed the false belief stated (it wasn't class material, but the teacher wanted to show he was smart). I found the counterexample later at home. I didn't let the matter drop either... I knew I was right and he was wrong, and really had a major fallout with that math teacher and the school; and flunked math that year. –  Daniel Moskovich May 5 '10 at 1:19
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@Daniel: Sorry to hear that. When my daughter Meena was the same age (11), her teacher asserted that 0.999... was not equal to 1. Meena supplied one or two proofs that they were equal, but her teacher would not budge. Maybe this is another example of a common false belief. –  Ravi Boppana May 5 '10 at 2:59
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@Daniel: I've heard a worse story. A college instructor claimed in Number Theory class that there are only finitely many primes. When confronted by a student, her reply was: "If you think there are infinitely many, write them all down". She was on tenure track, but need I add, didn't get tenure. –  Victor Protsak May 5 '10 at 5:38
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This false belief leads to a proof of the Twin Prime conjecture: For every $n$, $(p_1 p_2 \cdots p_n -1, p_1 p_2 \cdots p_n +1)$ are twin primes, right? –  David Speyer May 6 '10 at 15:50
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Daniel, about the same age, I was asked to leave class for claiming that pi is not 22/7. The math teacher said that 3.14 is an approximation and while some people falsly believe that pi=3.14 but the true answer is 22/7. Years later an Israeli newspaper published a story about a person who can memorize the first 2000 digits of pi and the article contained the first 200 digits. A week later the newspaper published a correction: "Some of our readers pointed out that pi=22/7". Then the "corrected" (periodic) 200 digits were included. Memorizing digits of pi is a whole different matter if pi=22/7. –  Gil Kalai May 11 '10 at 5:45

By definition, an asymptote is a line that a curve keeps getting closer to but never touches. The teaching of this false belief at an elementary level is standard and nearly universal. Everybody "knows" that it is true. A tee-shirt has a clever joke about it. In the course of describing the function $f(x) = \dfrac{5x}{36 + x^2}$, I mentioned about an hour ago before a class of about 10 students that its value at 0 is 0 and that it has a horizontal asymptote at 0. One of them accused me of contradicting myself. What of $y = \dfrac{\sin x}{x}$? And even with simple rational functions there are exceptions, although there the curve can touch or cross the asymptote only finitely many times. And $3 - \dfrac{1}{x}$ gets closer to 5 as $x$ grows, and never reaches 5, so by the widespread false belief there would be a horizontal asymptote at 5.

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For this to be a false definition, it would have to be a definition in the first place. And this means you have to define a "curve" first, and then define "get closer" and "touch". –  Laurent Moret-Bailly Mar 6 '11 at 16:01
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@Laurent: It's hard to imagine a comment more irrelevant to what happens in classrooms than yours. –  Michael Hardy Mar 7 '11 at 4:38
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It happens to be the literal meaning of the word asymptote "not together falling". You could say that it is a bad choice of name, but for hyperbolas it worked just fine and then it was mercilessly generalized. –  user11235 Apr 8 '11 at 14:35

The cost of multiplying two $n$-digit numbers is of order $n^2$ (because each digit of the first number has to be multiplied with each digit of the second number).


A lot of information is found on http://en.wikipedia.org/wiki/Multiplication_algorithm .

The first faster (and easily understandable) algorithm was http://en.wikipedia.org/wiki/Karatsuba_algorithm with complexity $n^{log_2 3} \sim n^{1.585}$.

Basic idea: To multiply $x_1x_2$ and $y_1y_2$ where all letters refer to $n/2$-digit parts of $n$-digit numbers, calculate $x_1 \cdot y_1$, $x_2\cdot y_2$ and $(x_1+x_2)\cdot(y_1+y_2)$ and note that this is sufficient to calculate the result with three such products instead of four.

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It would be better if these misconceptions would come with explanations how things really are... –  darij grinberg Apr 10 '11 at 18:28
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Along these lines: there is a widespread misapprehension that multiplication is the same thing as a multiplication algorithm (whichever one the speaker learned in elementary school). –  Thierry Zell Apr 10 '11 at 19:25
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At least it's better than people thinking multiplication is constant-time. :P –  Harry Altman Apr 10 '11 at 19:35

I used to believe that a continuous algebra homomorphism from $k[[x_1,\dots, x_m]]$ to $k[[y_1,\dots,y_n]]$, with $m > n$, could not be injective. Konstantin Ardakov set me straight on this.

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Whoa, that's not true?! What's the counter-example? –  David Speyer May 21 '10 at 19:44
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Not true and deeply frustrating. Take a map from $k[[x,y,z]]$ to $k[[u,v]]$ that sends $x$ to $u$, $y$ to $uv$ and $z$ to $uf(v)$ for some $f\in k[[v]]$ and think about what the kernel is. It isn't hard to see that only for countably many choices of $f$ can it possibly be zero. –  Simon Wadsley Jun 7 '10 at 19:45
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@Simon: Were you thinking of $k$ a countable field, and did you mean "can it possibly be non-zero"? (It seems the condition on $f$ is that it should not be a solution of any monic poly. whose coefficients are in $k(v)$.) –  fherzig Jun 7 '10 at 22:00
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Yes. I had just come back to say that I was assuming that $k$ is countable (in fact in my mind I was thinking of $k$ as finite because that is the case that I think about most) and as you say I should have said 'can it possibly be non-zero'. Of course the countability of $k$ is not necessary for such an example it just makes it easy to see that a suitable $f$ exists. –  Simon Wadsley Jun 8 '10 at 8:25
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Thanks, Simon, I had meant to get around to coming back and answering David! It's disturbing, isn't it? I had a paper ready to go apart from one lemma I was confused about that used needed this (false) statement, and indeed, not only is the statement wrong, the thing I thought I had almost proved turned out to be false! –  JSE Jun 8 '10 at 22:15

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

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