Examples of common false beliefs in mathematics

Question

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.

Answer 1

Let $R$ be a ring with identity $e$, $A, B\in R$, $A\neq 0$, $B$ is invertible element. If $A\cdot B = A$ then $B = e$.

Answer 2

Assume that $a,b\in \mathbb{R}\setminus \{0\}$ which satisfy $a^{3}= 2b^{3}$.

Then $a-2b$ is a non zero nilpotent element of group ring $\mathbb{Z}_{3} \mathbb{R}$, that is $(a-2b)^{3}=0$.

This would be a counterexample to the zero divisor Kaplansky conjecture

The false lies in an obvious abuse in the definition of the group ring multiplication.

Answer 3

Let $M \subset B(H)$ be a von Neumann algebra, $p \in B(H)$ a projection and $q=I-p$.

False belief: If $pM=Mp$ then $M=pMp \oplus qMq$.
(I think it is a quite common careless mistake)

Counter-example: diagonal embedding of $\mathbb{C}$ into $M_2(\mathbb{C})$.

Answer 4

A common false assumption is that that two non-orthogonal pure states of a quantum mechanical system may never be unambiguously distinguished by a measurement. (See https://arxiv.org/pdf/quant-ph/9807022.pdf)

Another false belief is that a quantum computer is similar to an analogue computer, in that large computations will necessarily fail because of accumulated error. (See, for example, https://arxiv.org/abs/quant-ph/9712048)

For that matter, another common false believe is that Bell Inequalities aren't violated, although it is mostly held by people who have never heard of Bell Inequalities.

Answer 5

Consider an $n\times n$-matrix $A$, which represents a linear map $f : \mathbf{k}^n \to \mathbf{k}^n$ (where $\mathbf{k}$ is the base field) with respect to the standard basis $\left(e_1,e_2,\ldots,e_n\right)$ of $\mathbf{k}^n$. Let's do some multilinear algebra.

The $d$-th exterior power of $A$ is the matrix $\Lambda^d A$ that represents the map $\Lambda^d f : \Lambda^d \mathbf{k}^n \to \Lambda^d \mathbf{k}^n$ with respect to the standard basis $\left(e_{i_1} \wedge e_{i_2} \wedge \cdots \wedge e_{i_d}\right)_{i_1 < i_2 < \cdots < i_d}$ of $\Lambda^d \mathbf{k}^n$. Exterior powers of matrices have lots of good properties, such as the functoriality $\Lambda^d\left(AB\right) = \left(\Lambda^d A\right) \left(\Lambda^d B\right)$ and the determinant formula $\det\left(\Lambda^d A\right) = \left(\det A\right)^{\dbinom{n-1}{d-1}}$ and the fact that taking the $d$-th exterior power commutes with transposing the matrix: \begin{align} \Lambda^d\left(A^T\right) = \Lambda^d\left(A\right)^T. \label{eq.darij5.1} \tag{1} \end{align} In particular, if the matrix $A$ is symmetric, then so is $\Lambda^d\left(A\right)$.

Exterior powers are the fermionic twin (or, depending on one's taste, the Koszul dual, or the partner under combinatorial reciprocity) of symmetric powers, and normally one expects all properties of one notion to work similarly for others. And indeed, much of the time this is true. The $d$-th symmetric power of $A$ is the matrix $\operatorname{Sym}^d A$ that represents the map $\operatorname{Sym}^d f : \operatorname{Sym}^d \mathbf{k}^n \to \operatorname{Sym}^d \mathbf{k}^n$ with respect to the standard basis $\left(e_{i_1} e_{i_2} \cdots e_{i_d}\right)_{i_1 \leq i_2 \leq \cdots \leq i_d}$ of $\operatorname{Sym}^d \mathbf{k}^n$. Again, there is a functoriality $\operatorname{Sym}^d\left(AB\right) = \left(\operatorname{Sym}^d A\right) \left(\operatorname{Sym}^d B\right)$ and the determinant formula $\det\left(\operatorname{Sym}^d A\right) = \left(\det A\right)^{\dbinom{n+d-1}{d-1}}$... but the analogue of \eqref{eq.darij5.1} does not hold! The $d$-th symmetric power $\operatorname{Sym}^2 A$ of a symmetric matrix $A$ won't usually be symmetric. For instance, \begin{align} \operatorname{Sym}^2 \begin{pmatrix} a&b \\ c&d \end{pmatrix} = \begin{pmatrix} a^2 & ab & b^2 \\ 2ac & ad+bc & 2bd \\ c^2 & cd & d^2 \end{pmatrix}, \end{align} and this does not become symmetric just by requiring $b=c$.

The deeper (more functorial) reason for this weirdness is that the exterior power functor commutes with duality (i.e., we have $\Lambda^d\left(V^*\right) \cong \left(\Lambda^d V\right)^*$ canonically when $V$ is finite-dimensional), but the symmetric power functor does not. Instead, we have $\operatorname{Sym}^d\left(V^*\right) \cong \left(\Gamma^d V\right)^*$, where $\Gamma^d$ is the "$d$-th symmetric tensors" functor (sending $V$ to the subspace of $V^{\otimes d}$ consisting of the tensors invariant under permuting the tensorands). When $\mathbf{k}$ has characteristic $0$, it is possible to identify $\Gamma^d V$ with $\operatorname{Sym}^d V$ by an isomorphism, but this will scale the standard basis by some factorials.

If I am not mistaken, we can in fact scale the standard basis of $\operatorname{Sym}^d V$ by square roots of factorials to salvage the equality $\operatorname{Sym}^d\left(A^T\right) = \operatorname{Sym}^d\left(A\right)^T$ (for example, replacing $e_i e_j$ by $\sqrt2 e_i e_j$ when $i\neq j$), but whether this is a good idea is not so clear. (For starters, this requires the relevant square roots to be defined in $\mathbf{k}$.)

This gave me a little jump scare yesterday when I was surprised my formulas for symmetric powers were missing some obvious symmetry. In hindsight, I should not have been expecting this much symmetry -- after all, determinants are much better behaved than permanents...

Answer 6

Surely, the densest packed set of given size in a lattice is a ball under the graph metric?

Not so fast... in the taxicab metric, a square is strictly denser as we can see explicitly for $5^2=1+4+8+12$ vertices:

Answer 7

I don't know how common a belief this is, but I just realised yesterday that I was thinking: if I take the solution set in $k^n$ to a system of homogeneous linear equations, then I obtain a subspace of $k$; and then tensoring that subspace with an extension field $E$ of $k$ gives me the solution set of the same system of equations in $E^n$.

Of course, that's true. But now imagine that $k$ has characteristic $p$. Then the solution set in $k^n$ to a system of homogeneous linear equations in $p$th powers of the variables (i.e., $\sum_i a_i x_i^p = 0$) is also a subspace of $k$; but tensoring that subspace with $E$ does not necessarily give the full solution set of the same system of equations in $E^n$!

The example that forced me to confront that this was false was slightly more complicated, but, now that I know the problem's there, it's easy to manufacture examples. For instance, if we put $k = \mathbb F_p(t)$ and $E = \mathbb F_p(t^{1/p})$, then the solution sets to $x_1^p = t x_2^p$ in $k^2$ and $E^2$ exhibit this phenomenon.

Answer 8

Another common mistake. If $W = _P(e_1,\ldots, e_{n})$ is a vector space and $V$ is a subspace of $W$ of dimension $k$, then $V = _P(e_{i_1},\ldots, e_{i_k})$.

Answer 9

I don't know how common this is, but it occurs as a corollary of a theorem in the fine, and widely used, text by Shafarevich on algebraic geometry: namely, if $f \colon X \longrightarrow Y$ is a surjective algebraic map of varieties, then 1) for all $y \in Y$, the fiber over $y$ has dimension $≥ \dim(X)-\dim(Y)$; 2) on some non empty open set in $Y$ the dimension of the fibers equals $\dim(X)-\dim(Y)$; 3) for all $r$, the set of $y \in Y$ such that the fiber over $y$ has dimension $≥ r$, is closed in $Y$.

The first two are true, but the third is false. Upper semicontinuity of fiber dimension is true on the source, not the target. For the conclusion as stated to hold, one can add properness to the hypothesis on the map. I think this is not at all widely believed by experts, but for some reason it persists in the text, hence may be believed by students.

Since I have myself written notes in which blatantly false statements occur, I do not think for a moment that Shafarevich himself believed this false statement. But such things do slip by, and may mislead beginners. In fact I believed it for some time until enlightened by a friend.

In keeping with the OP's desire to know the psychological reason for the error, it seems for some reason common in my experience for people to assume unconsciously that maps are proper.

Answer 10

False belief: a subgroup isomorphic to a quotient is a retract.

Formally: Let $H,N$ be subgroups of $G$ with $N$ normal and $H \simeq G/N$, then $H$ is a retract of $G$.

It is false, because otherwise $C_2$ would be a retract of $C_4$, but it is not.

In fact, $H$ is a retract of $G$ if and only if $G$ is isomorphic to $H \ltimes N$ (semidirect product).

This false belief caused this post.

Answer 11

I cannot believe this example was not yet given (but, if my belief is false, I will happily delete this answer):

It is very common among "lay people" (who do not understand what it means for lines to be parallel) to believe that "in some kind of geometry" (frequently described as non-euclidean) parallel lines can intersect.

One finds many instances (my guess, the count is in hundreds of thousands) of this false believe just by searching the internet. Here is a random example, the article "How Looking At A Basketball Disproves Something Everybody Learns In High School Geometry" from "Business Insider", 2014. The article concludes with

And voila! We’ve successfully disproven “parallel lines never intersect” using just a basketball.

Answer 12

If an Abelian category $\mathcal{A}$ is a full subcategory of an Abelian category $\mathcal{B}$, then for all objects $M,N$ of $\mathcal{A}$, we have an injection $$\operatorname{Ext}^i_{\mathcal{A}}(M,N) \hookrightarrow \operatorname{Ext}^i_{\mathcal{B}}(M,N).$$

As an example, let $G$ be the free group on $2$ letters, $A$ its abelianization, $\mathcal{B} = G-mod$, $\mathcal{A}=A-mod$, and $M=N=\mathbb{Z}$ with the trivial action. Then $\operatorname{Ext}^i_{\mathcal{A}}(M,N) \cong \mathbb{Z}$, while $\operatorname{Ext}^i_{\mathcal{B}}(M,N) \cong 0$.

(This example comes from the topological fact that a torus has nontrivial $H^2$, while a punctured torus has trivial $H^2$. In algebra, it's related to the idea that group homology $H_1$ is space of generators for a group while $H_2$ is a space of relations.)

This false belief came up a context where $\mathcal{B}$ was the category of all Galois representations while $\mathcal{A}$ was a certain subcategory. See the comments to Status of the conjectured vanishing of Bloch-Kato H^2.

Answer 13

Initially when I started studying sequences I believed that:

Consider $(x_n)$ and $(y_n)$ are two convergent real sequences having limits $x$ and $y$ as limits respectively. If $ x_n < y_n \quad \forall n\in \mathbb{N}$ then $x < y$

which turned out to be false.

Answer 14

Common false belief:

Consider a non-linear system $\dot{x} = f(x)$ such that $f(0) = 0$. Suppose that $V : D \rightarrow \mathbb{R}$ is a Lyapunov function, i.e. $V$ is positive-definite in $D$ and $\dot{V}$ is negative-definite in $D$. Then $D$ is contained in the basin attraction of the equilibrium point $0$.

Here is a example of a system with a Lyapunov function that is positive definite everywhere, its time-derivative is negative definite everywhere and there are trajectories that does not converge to $0$.

This misconception is common among students in some engineering courses where the proof of the Lyapunov stability theorem is sometimes omitted. Many students rely on their intuition from simple examples and skip or forget the proof, leading to this false belief.

Answer 15

“There are no trigonometric proofs, because all the fundamental formulae of trigonometry are themselves based upon the truth of the Pythagorean Theorem.”

that quote is from Elisha Scott Loomis' book, "The Pythagorean Proposition: Its Demonstrations Analyzed and Classified, and Bibliography of Sources for Data of the Four Kinds of Proofs" and is cited in the abstract of Ne'Kiya D Jackson and Calcea Rujean Johnson's latest American Mathematical Society paper, "An Impossible Proof of Pythagoras":

In the 2000 years since trigonometry was discovered it's always been assumed that any alleged proof of Pythagoras’s Theorem based on trigonometry must be circular. In fact, in the book containing the largest known collection of proofs (The Pythagorean Proposition by Elisha Loomis) the author flatly states that “There are no trigonometric proofs, because all the fundamental formulae of trigonometry are themselves based upon the truth of the Pythagorean Theorem.” But that isn’t quite true: in our lecture we present a new proof of Pythagoras’s Theorem which is based on a fundamental result in trigonometry—the Law of Sines—and we show that the proof is independent of the Pythagorean trig identity $\sin^2x + \cos^2x = 1$.

Answer 16

Characters of a Hopf algebra form a group under convolution .-

It is, of course, the set of non-commutative (or vector valued) characters as it is well-known that, a Hopf algebra $(\mathcal{H},\mu_\mathcal{H},1_\mathcal{H},\Delta,\epsilon,S)$ [1] and a commutative algebra $(\mathcal{A},\mu_\mathcal{A},1_\mathcal{A})$ being given (all over the same commutative ring $k$), the set $Hom_{k_AAU}(\mathcal{H},\mathcal{A})$ is a group under convolution (the inverse being performed through precomposition with $S$). The story is the following.

Until yesterday and based on hasty readings (or rather non-revisited readings of hasty writers :), I believed that the above was true even if the algebra $\mathcal{A}$ was non-commutative. Which is false as shows the counterexample below.

Take any commutative ring $k$ and two letters (or non-commuting variables), $\{a,b\}$, then, with the Hopf algebra $\mathcal{H}=(k<a,b>,conc,1_{\{a,b\}^*},\Delta,\epsilon,S)$ and the (non-commutative) algebra $\mathcal{A}=(k<a,b>,conc,1_{\{a,b\}^*})$, we get a $\mathcal{A}$-valued character $Id:\ \mathcal{H}\to \mathcal{H}$ such that its convolutional square $Id^{*\,2}$ is NOT a character.

If one prefers a more standard world and matrix-valued characters, one can consider the following (and essentially the same counterexample)

1. $k$, a field
2. a two-dimensional vector space $V=k.a\oplus k.b$
3. The Hopf algebra $T(V)$
4. The algebra of 2x2 matrices over $k$
5. The character on $T(V)$, given by $a\to \Big(\begin{matrix}0 & 1\\ 0 & 0\end{matrix}\Big)$ and $b\to \Big(\begin{matrix}0 & 0\\ 1 & 0\end{matrix}\Big)$.

[1] The order is always (space, product,unit,coproduct,counit,antipode).

Answer 17

Way late to the party...

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

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

Answer 18

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!

Answer 19

In algebraic topology, I thought for a while:

• "For $k \geq 2$, $H_k$ is the abelianization of $\pi_k$." False. True for $k = 1$. Also for all $k$ up to $n-1$ if the space is $(n-1)$-connected for $n \geq 2$ (vacuously, since this says the first $n-1$ homotopy groups are trivial and for these, the Hurewicz homomorphism is the isomorphism, $\pi_k \cong H_k$). See the Hurewicz theorem for more.
• "Generically, all the $\pi_k$ are nonabelian." False. For $k \geq 2$, $\pi_k$ is abelian.

---

Edit: I had a third error in thinking when I first posted this, mangling the above into something further from true. Which I suppose makes the first version of this post meta-appropriate for this thread (but I've fixed it anyway). Thankfully, user Michael gently pointed out my mangling.

Answer 20

For $p$ prime and the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \cdots$ given by multiplication by $p$, then $\bigcup_n \mathbb{Z}/p^n\mathbb{Z}$ is not the group of $p$-adic integers $\mathbb{Z}_p$, but its Pontryagin dual, the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$.

Answer 21

I once very briefly thought that:

Given a vector space $V$ and a sub-space $U \subset V$ that $V-U$ is also a subspace.

I've heard this several times as a TA also.

Answer 22

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.

Answer 23

One common misconception is that Category Theory is abstract. It is in fact, no more abstract than set theory or calculus. This common complaint is the usual complaint that people have when they first meet set theory or category theory. Since people interested in category theory are usually already interested in mathematics, it is strange - but perhaps not that strange - that they voice this complaint. It's underlying reason is that it's notions are unfamiliar, rather than abstract. And it doesn't help that there isn't a bestiary of easy examples that we would have for algebra and calculus.

This needs to be developed.

Answer 24

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.

Answer 25

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 characteristic two and I didn't get a good grade that day!

Answer 26

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, $x^2$, and the fact that its graph does not look like a line at any value of $x$.