# Examples of common false beliefs in mathematics

The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) $\sin z$ is a bounded function;
(iii) $\sin z$ is defined and analytic everywhere on $\mathbb{C}$;
(iv) $\sin z$ is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of $\sin z$ to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset $U$ of $\mathbb{R}$ must be the whole of $\mathbb{R}$. The "proof" of this statement is that every point $x$ is arbitrarily close to a point $u$ in $U$, so when you put a small neighbourhood about $u$ it must contain $x$.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

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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/… – user9072 Oct 8 '11 at 14:27

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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What is $B(a,b)$ here? – Turbo Dec 3 '13 at 10:09
Here, $B(a,b)$ is the open ball of radius $b$ centered at $a$. – Seirios Dec 3 '13 at 21:41

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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This is not a common false belief except among people who do not understand the definition of an algebra over a field – Yemon Choi Nov 12 '14 at 23:43
Moreover, surely the quaternions are a real vector space, not a complex vector space – Yemon Choi Nov 13 '14 at 1:34
@YemonChoi the field of complex number is a subring of the ring of quaternions. So quaternions is a complex vec. space.More generaly if a ring R contains a field F then R is a F-vector space,Ok? – Ali Taghavi Nov 13 '14 at 5:31
@YemonChoi I think this example is perhaps interesting unless a participant do not read it carefully. please think again to the main motivation and aim of the question of "common false..." – Ali Taghavi Nov 13 '14 at 5:44
@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 '14 at 11:58

A Banach space $X$ is reflexive if it is isomorphic to its double dual ${X^*}^*$.

(Couldn't find this is the list…)

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Even isometric fails. (Lindenstrauss & Tzafriri, in the '60s I believe.) – Hachino May 12 '15 at 8:19

"Let $E$ be a complete locally convex topological vector space (or a complete topological vector space or a complete topological group) and let $F$ be closed vector subspace (or a closed subgroup). Then the quotient $E/F$ is complete."

This just has to be true. One can almost see the proof. And in fact it is true for Banach spaces. So it has to be true for locally convex spaces as well.

Another one with completions:

"Every topological group is a dense subgroup of a complete topological group." True for abelian groups but false in general (take the homeomorphism group of $[0,1]$ with the compact open topology)

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When I was studying Banach spaces, I was confused with the following: We know that, in any Banach Space $V$, the closed unit ball is compact in the topology generated by the norm if, and only if, the dimension of $V$ is finite. But thinking about $\mathbb R$ as a vector space over $\mathbb Q$, we have an infinite-dimensional vector space which is complete in the norm (given by the modulus) but the closed unit ball is, of course, compact in topology generated by the norm.

I took some time to discover that my mistake was that I thought about $\mathbb R$ over $\mathbb Q$ as a Banach space. In fact, this vector space is a complete metric space (in the sense of Cauchy sequences), but I realized later that the word Banach space is reserved only for vector spaces defined over the fields $\mathbb R$ or $\mathbb C$.

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You can define Banach spaces over any complete field. For example, one can define p-adic Banach spaces. But Q isn't complete with respect to any of its norms. – Qiaochu Yuan May 4 '10 at 22:14
In fact the Theorem I mentioned in my answer, is based on the Riesz lemma and this lemma is not valid if the scalar fields is not complete. – Leandro May 4 '10 at 22:34
@QiaochuYuan ...unless you use the trivial norm. – Mario Carneiro Oct 20 '15 at 21:16

An elementary false belief in elementary number theory: for $a, b, c\hspace{.1cm}\varepsilon\hspace{.1cm} \mathbb{N}$

$LCM\left(a,b\right)\times GCF\left(a,b\right) = ab$ .

Thus, $LCM\left(a,b,c\right)\times GCF\left(a,b,c\right) = abc$.

In general, $\left(a_1,a_2,\ldots,a_n\right)[a_1,a_2,\ldots,a_n] = a_1a_2\ldots a_n$.

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Does GCF mean gcd (greatest common divisor) here? – Zsbán Ambrus Nov 27 '10 at 19:33
Yes. GCF means the same: Greatest Common Factor. – Unknown Nov 29 '10 at 6:46
This kind of stuff is easy to rule out, though; it's dimensionally inconsistent. Replacing a, b, c by ka, kb, kc leads to a quick contradiction. – Qiaochu Yuan Feb 24 '11 at 21:08
@Qiaochu, that is a nice quick check. Let the RCF(remnant common factor) be the leftover factor that would make the above second equality true. There seems to be no interesting way of determining $RCF\left(a,b,c\right)$ so that $LCM\left(a,b,c\right)\times RCF\left(a,b,c\right) \times GCF\left(a,b,c\right) = abc$ – Unknown Feb 24 '11 at 22:05
@Elohemahab: actually, the correct generalization is $\gcd(a, b, c) \text{lcm}(ab, bc, ca) = \text{lcm}(a, b, c) \gcd(ab, bc, ca) = abc$. – Qiaochu Yuan May 8 '11 at 23:47

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
Disproved nicely by Reuleaux triangles. – darij grinberg Apr 12 '11 at 8:10
Disproved nicely by a two-point metric space. – Tom Goodwillie Apr 17 '11 at 1:36
An equilateral triangle in the Euclidean plane also does the job (diameter $1$ and radius $1/\sqrt{3}$): $2/\sqrt{3} > 1$. – Jean Van Schaftingen Nov 6 '13 at 15:07

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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I think you need $\kappa\leq\frak c$, no? – Asaf Karagila Jan 22 '15 at 14:44
@AsafKaragila: I believe the assumption that our measure is atomless already implies that $\kappa \leq 2^{\omega}$. – Burak Jan 22 '15 at 14:45
Take any measurable cardinal, then there is an atomless probability measure on its power set. It's just that an event is either improbable or its negation is improbable. Unless by probability measure you mean it obtains many values, not just two. – Asaf Karagila Jan 22 '15 at 14:48
Isn't that measure atomic if you are deriving it from the ultrafilter? (By an atom, I mean any $A$ of positive measure such that for any $B \subseteq A$ either $\mu(B)=0$ or $\mu(B)=\mu(A)$). I will have to catch a course now but the theorem I referred to should be in Kanomori (indeed, I checked the pdf and it is Theorem 2.5) – Burak Jan 22 '15 at 14:53
Ohhhh, right. I was thinking about atoms in the sense of Boolean algebra, as minimal positive elements. Thanks for the clarification! – Asaf Karagila Jan 22 '15 at 14:55

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

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– Michael Albanese Dec 24 '15 at 10:21

If $\alpha>0$ is not an integer, the set of functions $f:[a,b]\rightarrow\mathbb R$ such that $$\sup_{y\ne x}\frac{|f(y)-f(x)|}{|y-x|^\alpha}<+\infty$$ is ${\mathcal C}^\alpha([a,b])$.

False for $\alpha>1$, because this set contains only constant functions.

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Here's one that doesn't seem to be listed here: That ZFC isn't vulnerable to Russell's paradox because it has an axiom which implies that sets cannot be elements of themselves.

This one should be obviously false given even a moment's thought -- you cannot get rid of a paradox by adding axioms! But I've seen it repeated over and over.

The correct statement, of course, is that ZFC isn't vulnerable to Russell's paradox because it doesn't have the axiom of unrestricted comprehension; and that while ZFC does have an axiom which implies that all sets are well-founded, this is irrelevant to Russell's paradox. (It just means that the set of all sets which are not elements of themselves, if it existed, would happen to also be the set of all sets.)

Sometimes this is stated in an ambiguous way that allows for the charitable reading that modern set theory prevents Russell's paradox by preventing one from talking about a set being an element of itself, which is how Russell and Whitehead fixed the problem. But Principia Mathematica isn't exactly "modern set theory"; and I've seen this stated the blatantly wrong way often enough that I'm not too inclined to be charitable about it.

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"The set A = {a, b} has two elements..."

It's quite simple to notice that a can be the same as b, but after 5 years of university there were people still believing it...

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I'm not sure there is a false belief here, as much as awkward writing. Depending on context, I might very well write "The set $\{a,b \}$ (where $a$ and $b$ might be equal)..." if this issue mattered. – David Speyer May 6 '10 at 11:16
There are many situations where one needs to speak of a set of two numbers that may or may not be equal. E.g.: "Let x<sub>1</sub>, x<sub>2</sub> &isin; ℝ. Then among all the open intervals containing the set {x<sub>1</sub>, x<sub>2</sub>}, none of them is contained in all the others." If one is addressing mathematicians, there is no need to specify that x<sub>1</sub> might be equal to x<sub>2</sub>. – Daniel Asimov Jun 17 '10 at 23:34
Single-letter symbols are usually assumed to be variables, if the context doesn't determine otherwise, even in the absence of quantifiers. (You can put in an implicit universal quantifier to close up all sentences.) – Toby Bartels Apr 4 '11 at 9:41
In a context where one is discussing real analysis, $e$ and $\pi$ are generally taken to be the famous constants. But this is hardly universal; in other contexts, they may have very different meanings. – Toby Bartels Dec 15 '13 at 5:16
Here is a related but slightly less obvious situation. The ordered pair $(a,b)$ is generally defined in set theory to be $\{\{a\},\{a,b\}\}$. This is generally thought of as a set with two elements. But what if $a=b$? – Bruce Blackadar Sep 26 '14 at 4:12

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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What "people"? Non-mathematicians? – Todd Trimble May 4 '11 at 0:03
@Todd: No i was talking of high school students. – S.C. 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

True: Given a graded algebra $A$, there is a notion of a "homogeneous" ideal of $A$. It is a property that connects an ideal of $I$ with the grading and is often necessary to require. For example, if $I$ is a homogeneous ideal of $A$, then the algebra $A / I$ is graded again. If $I$ is not homogeneous, then it is not graded in general (since the projections of different graded components of $A$ onto $A / I$ might have nonzero intersection).

False: Given a filtered algebra $A$, there is a notion of a "filtered" ideal of $A$.

There is no such notion. We can require $I$ to be generated by $I\cap A_n$ for some $n$, or actually to lie inside $A_n$ for some $n$, but in most cases none of these is actually needed. (Correct me if I am wrong.) Formulations like "Let $I$ be an ideal compatible with (or respecting) the filtration" are cargo cult.

But: Given a filtered algebra $A$ and a generating set $G$ of an ideal $I$ of $A$, it is an important question whether $I\cap A_n$ is generated by $G\cap A_n$ for every $n\in \mathbb N$. This is not always satisfied, often nontrivial (in many cases it can be proved by using the diamond lemma to show that every element of $A_n$ has a unique "remainder" modulo $I$ in a certain sense, and this remainder can be obtained by repeated subtraction multiples of elements of $G\cap A_n$) and used tacitly in various texts.

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Good point, but "cargo cult"? – Tom Goodwillie Mar 15 '11 at 14:32
What I mean is: People use these formulations as a protective charm against a danger they don't see but intuitively feel is there, although closer inspection shows that it is pure superstition. – darij grinberg Mar 15 '11 at 17:26

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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Multiplication of differential forms is inherently anti-commutative. Thus, if $x$ and $y$ are coordinates on a surface, then $dx \wedge dy$ makes sense but $(dx)^2+(dy)^2$ is either nonsense or, if it means anything, is $0$.

I'm not sure why I believed this, but I did for several years. I tried my best to avoid creating this impression in my students, but I think it still happened in some of them, simply because the curriculum spends a lot of time on integration and Stokes theorem and very little time on metrics, curvature, etc.

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Probably my fault for not paying enough attention in analysis, but:

Any continuous function on the interval that has derivative equal to zero almost everywhere is constant.

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As a sequel of this famous answer on $\dim(U+V+W)$, the following inequality is not true $\forall n \ge 4$:
$$\dim(\sum_{i = 1}^{n} U_i) \le \sum_{r=1}^{n} (-1)^{r+1} \sum_{i_1 < i_2 < \dots < i_r} \dim(\bigcap_{s=1}^{r}U_{i_s}) = \alpha$$
Darij Grinberg has found a counter-example (see this post).

Same flavor: for $n \le 5$, it is true that $\alpha \ge 0$ (see this proof), but it's false for $n>5$ (see this comment).

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False belief: Any orthonormal basis of a subvectorspace $W\subset V$ of an inner product space $V$ can always be extended to an ONB of $V$.

Counterexample: Let $V$ be $\bigoplus_{i\ge 1} \mathbb{R}$ with the inner product given by $\langle a_*,b_*\rangle =\sum_{i\ge 1} a_ib_i$ and let $W$ be the subvectorspace of $V$ spanned by $e_1+e_i$ for $i\ge 2$. The given set is basis and we can apply Gram-Schmidt to obtain an ONB.

However $W^\perp = 0$ so there is no way to complete it. Related false belief: $(W^\perp)^\perp=W$. These beliefs are all true in finite dimensions, but false in general.

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That's why we like Hilbert spaces (inner product spaces that are complete w.r.t. the inner-product norm) much better than arbitrary inner-product spaces. – Noam D. Elkies Mar 4 at 4:08

If $H$ and $K$ are subgroups of $G$, then $HK$ is a subgroup of $G$.

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Hm, wonder how common that false belief actually is. It seems obviously implausible in the nonabelian case. – Todd Trimble Sep 6 '15 at 15:39
Its common, specially when undergraduates use product formula : $|HK|=\frac{|H||K|}{| H \cap K | }$ Because all of them are subgroup, except $HK$ probably. – user68208 Apr 10 at 17:30

Many people believe that Cantor proved the uncountability of the real line using a diagonal argument. This paper does not that provide that proof; Cantor's stated purpose was to prove the existence of `uncountable infinities' without using the theory of irrational numbers.

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More to the point, I think, is that the paper proves that the power set of any set has greater cardinality than the set itself. This is the first proof that there is no greatest cardinality. (The uncountability of the real line easily follows, even if Cantor does not mention it because he has bigger fish to fry.) – John Stillwell May 31 '10 at 5:12
Just to fill in some history here: if I remember right, Cantor first proved the uncountability of the reals by other arguments, then later (as you reference) found the diagonal argument, as a proof of the more general statement about power sets. – Peter LeFanu Lumsdaine Sep 27 '10 at 3:01
The link in the answer goes to the wrong page - it should go to page 75, not page 72. – David Roberts Jun 13 '12 at 6:41
And it looks like a diagonal argument to me. – David Roberts Jun 13 '12 at 6:43

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Belief 2. The most naive statement which would, if true, generalize the Hodge Theorem to noncompact manifolds is this.

The inclusion of the complex of coclosed harmonic forms into the de Rham complex of a riemannian manifold is a quasi-isomorphism.

This statement happens to be true.

Here is a reference:

The simplest example is that of the real line with its standard metric. In degree zero the complex of coclosed harmonic forms is $\mathbb C\oplus\mathbb Cx$, and in degree one it is $\mathbb Cdx$, which gives the right cohomology.

Here is the (trivial) algebra background.

Let $A$ be a module over some unnamed ring, and let $d,\delta$ be two endomorphisms of $A$ satisfying $d^2=0=\delta^2$. Put $\Delta:=d\delta+\delta d$. Assume $A=\Delta A+A_{d,\delta}$ where $A_{d,\delta}$ stands for $\ker d\cap\ker\delta$. Write $A_{\delta,\Delta}$ for $\ker\Delta\cap\ker\delta$.

We claim that the natural map $$H(A_{\delta,\Delta},d)\to H(A,d)$$ between homology modules is bijective.

Injectivity. Assume $\delta da=0$ form some $a$ in $A$. We must find an $x$ in $A_{\delta,\Delta}$ such that $dx=da$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta db+c$ does the trick.

Surjectivity. Let $a$ be in $\ker d$. We must find $x\in A$, $y\in A_{d,\delta}$ such that $a=dx+y$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta b$, $y:=\delta db+c$ works.

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Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows:

There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\\ E_s\\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.

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"$\emptyset$ is a basis for {0}" is an immediate consequence of the definitions. There is no false belief in this point. – Johannes Hahn May 12 '10 at 15:14
Dear Johannes, please reread my post. – Pierre-Yves Gaillard May 12 '10 at 15:18
Even a cursory examination of Nick Higham's book amazon.co.uk/Functions-Matrices-Computation-Nicholas-Higham/dp/… will show that both these opinions on the evaluation of matrix exponentials are hopelessly naive. – Robin Chapman May 15 '10 at 9:17
Your opinions are normative statements: "one should" and "it is better". It is naive to suppose that there is one best method that one should use to compute the matrix exponential. – Robin Chapman May 15 '10 at 14:07
I don't think the OP wants examples of normative statements. As I read it, the question is about conceptual errors regarding non-normative mathematical statements. – Qiaochu Yuan May 17 '10 at 6:19

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
@ Todd Trimble: true, but then also the belief about $sin$ suggested by the OP is only common among people who have not completed a course in complex analysis. – Delio Mugnolo Dec 13 '13 at 8:34
I thought "bounded" is only defined on metric spaces, and this is true on metric spaces. Is that wrong? – Akiva Weinberger Sep 1 '15 at 2:48
I have seen analysis textbooks take this as a definition. I hope they realize that they are contributing to future confusion in their readers once they move on to topology or even metric spaces. @AkivaWeinberger, The Heine-Borel theorem stated in this way makes sense for arbitrary metric spaces, but it is only true for complete metric spaces for which balls are totally bounded. The correct statement of H-B for general metric spaces is "a metric space is compact iff it is complete and totally bounded". – Mario Carneiro Oct 20 '15 at 21:27
@AkivaWeinberger: Yes, it is wrong. The closed unit ball of an normed vector space is compact if and only if the space is finite dimensional. – ACL Apr 21 at 13:43

I have checked the existing answers, but I think this one is not given yet. Sorry, if I missed it.

Although the incompleteness theory of Gödel is generally correctly understood, the consequence of it has multiple false beliefs:

• Due to the incompleteness theory it is not possible to make an AI. Humans will always be be superior to the AI. This assumes that human thinking is complete and will eventually find the answer on any question.

• Due to the incompleteness theory, it is not possible to formalize mathematics. This is refuted by many proof systems, which can formalize almost all mathematics.

As side note, I think this is partly fueled how logic is taught. It puts more emphasis on impossibilities (incompleteness theory), than possibilities (a proof system).

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+1. I always found it wrong that classes in logic put so much emphasis on negative results. (And I wish they had prepared me better for proof assistants... though I guess one semester does not suffice for the ones that exist today.) – darij grinberg Sep 5 '15 at 22:17
There's arguably too much fascination with incompleteness and not enough with completeness, which is more of a cornerstone of model theory. – Todd Trimble Sep 6 '15 at 1:50

Here is a short list of some false beliefs I had when I was studying mathematics, I suppose they may be common but I have never checked:

• I was in the last year of high school and studying university-level math in advance. I remember trying for a week to prove that a continous injective map from an open subset of $\mathbb{R}^2$ to $\mathbb{R}^2$ that preserves "being aligned" (I mean that maps aligned triples to aligned triples) must be the restriction of an affine map (over $\mathbb{R}$). That is disproved by restrictions of projective transformations... Which I knew of but I was not able to see they contradicted my belief. When my teacher told me "What about projective transformations?"... I felt dumb.
• I was in the 1st year of PhD studies. My advisor, Adrien Douady, had an idea to build polynomial Julia sets with positive Lebesgue measure. Julia sets are fractals, often with complicated topological structures at every scale. Surely that must be the source of measure? So as an exercise, I tried for a week to prove that Jordan curves are necessarily of Lebesgue measure 0. I told Adrien about my attempts. He gave me a counter-example. I felt dumb.
• Learning that there are closed subsets of the interval with positive Lebesgue measure but no interior did not surprise me as much, as the construction is very simple, but still that's a bit counterintuitive.
• When you zoom on the Mandelbrot set, you see all that round components with smooth boundaries. They look so round. Surely they must be circles, for otherwise the difference would be visible. Well... they are not (except one). Guess how I felt when I learned.
• Frankly, when learning the first time about complex numbers, did anybody here expect that, adding the square root of -1 to the reals would add the roots of all other polynomials?
• I was giving a lecture to math teachers about sensitivity to initial condition (call it chaos) and showing strange attractors on the computer, one told me that by the very presence of chaos, what we see may be quite far from the actual behaviour of the equation, save reality. It turns out hyperbolic systems are stable, so I believe this is still representative (it does not prove it but it is an encouraging hint).
• ... Chaos in deterministic systems. I won't develop on that.
• Surely before hearing of set theory and Cantor's argument, you will believe that all sets are countable. Then after learning that this is not the case, you will think that $\mathbb{R}^2$ must be bigger than $\mathbb{R}$, right?
• You have a $C^\infty$ function on the right half plane, all of whose derivatives have a continuous extension to the boundary line. Surely, it must be easy to extend it to a $C^\infty$ function of the whole plane, isn't it? Well... You can but I would not call it easy.
• Short statements have short proofs. Disproved by Fermat's last theorem (among others).
• I was quite disapointed to learn that there cannot be a finite non-commutative field (division algebra).

I have a few other examples, that I would not term "common false beliefs" but rather "fun and surprising math facts". Is there already a MO question about that?

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Every matrix is the sum of a symmetric and an antisymmetric matrix. Hence:

If $V$ is a vector space and $k$ is a number, then the $k$-th tensor product of $V$ with itself decomposes as a direct sum into symmetric and antisymmetric tensors: $$\underbrace{V \otimes ... \otimes V}_{k\text{ times}} = \Lambda^kV \oplus \mathrm{Sym}^kV$$

Recall (in the finite-dimensional case) the dimensions: $$\dim \Lambda^k V = \binom{n}{k} \quad\text{ and }\quad \dim\mathrm{Sym}^kV = \binom{n+k-1}{k}$$

Looking at $k=1$ shows that we have non-trivial intersection.

Looking at $n=k=3$ shows that the sum is not exhausting.

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Is this a common false belief? – Jim Conant Oct 18 '15 at 2:52
@JimConant: I believed this once. Of course, if you count dimensions it's obviously false. But if $V$ is an infinite-dimensional Hilbert space it sure seems natural to decompose the full tensor product into its Bosonic and Fermionic parts, and you might not think right away to ask whether it works in finite dimensions. That's my excuse, anyway! – Nik Weaver Mar 4 at 4:22

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

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

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

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