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

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

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

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

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

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

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

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I have to say this is proving to be one of the more useful CW big-list questions on the site... – Qiaochu Yuan May 6 2010 at 0:55
The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. – To be cont'd May 22 2010 at 9:04
wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. – S. Sra Sep 20 2010 at 12:39
It's a thought -- I might consider it. – gowers Oct 4 2010 at 20:13
Meta created meta.mathoverflow.net/discussion/1165/… – quid Oct 8 2011 at 14:27

## closed as no longer relevant by Mark Sapir, Felipe Voloch, George Lowther, Mark Meckes, Ryan BudneyOct 8 2011 at 22:24

Just today I came across a mathematician who was under the impression that $\aleph_1$ is defined to be $2^{\aleph_0}$, and therefore that the continuum hypothesis says there is no cardinal between $\aleph_0$ and $\aleph_1$.

In fact, Cantor proved there are no cardinals between $\aleph_0$ and $\aleph_1$. The continuum hypothesis says there are no cardinals between $\aleph_0$ and $2^{\aleph_0}$.

$2^{\aleph_0}$ is the cardinality of the set of all functions from a set of size $\aleph_0$ into a set of size $2$. Equivalently, it is the cardinality of the set of all subsets of a set of size $\aleph_0$, and that is also the cardinality of the set of all real numbers.

$\aleph_1$, on the other hand, is the cardinality of the set of all countable ordinals. (And $\aleph_2$ is the cardinality of the set of all ordinals of cardinality $\le \aleph_1$, and so on, and $\aleph_\omega$ is the next cardinal of well-ordered sets after all $\aleph_n$ for $n$ a finite ordinal, and $\aleph_{\omega+1}$ is the cardinality of the set of all ordinals of cardinality $\le \aleph_\omega$, etc. These definitions go back to Cantor.

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I retract my above question to my suprise it indeed seems to be common. Yet, this answer is a dublicate see an answer of April 16. – quid Oct 6 2011 at 0:50
This example already appears on this very page. mathoverflow.net/questions/23478/… – Asaf Karagila Oct 6 2011 at 12:41
One of the deficiencies of mathoverflow's software is that there is no easy way to search through the answers already posted. Even knowing that the date was April 16th doesn't help. – Michael Hardy Oct 7 2011 at 20:26
@Michael Hardy: You can sort the answers by date by clicking on the "Newest" or "Oldest" tabs instead of the "Votes" tab. – Douglas Zare Oct 19 2011 at 23:03

In group theory, if $G_1 \cong G_2$ and $H_1 \cong H_2$, then

$G_1 / H_1 \cong G_2 / H_2$.

For example, $\mathbb{Z} / 2\mathbb{Z} \not \cong \mathbb{Z} / \mathbb{Z}$. The point is that the inclusion of $H_j$ into $G_j$ is needed in order to define the quocient.

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Here's one that bugged me from point set topology: "A subnet of a sequence is a subsequence".

See here for the definitions. Using this one gives a great proof that compactness implies sequential compactness in any topological space:

Let $X$ be a compact space. Let $(x_n)$ be a sequence. Since a sequence is a net and it's a basic theorem of point set topology that in a compact topological space, every net has a convergent subnet (proof in the above link), there is a convergent subnet of the sequence $(x_n)$. Using the above belief, the sequence $(x_n)$ has a convergent subsequence and hence $X$ is sequentially compact.

For a counterexample to this "theorem", consider the compact space $X= \lbrace 0,1 \rbrace ^{[0,1]}$ with $f_n(x)$ the $n$th binary digit of $x$.

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A possible false belief is that "a maximal Abelian subgroup of a compact connected Lie group is a maximal torus". Think of the $\mathbf Z_2\times\mathbf Z_2$-subgroup of $SO(3)$ given by diagonal matrices with $\pm1$ entries.

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False belief: Every commuting pair of diagonalizable elements of $PSL(2,\mathbb{C})$ are simultaneously diagonalizable. The truth: I suppose not many people have thought about it, but it surprised me. Look at $$\left(\matrix{i& 0 \cr 0 & -i\cr } \right), \left(\matrix{0& i \cr i & 0\cr } \right).$$

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A random $k$-coloring of the vertices of a graph $G$ is more likely to be proper than a random $(k-1)$-coloring of the same graph.

(A vertex coloring is proper if no two adjacent vertices are colored identically. In this case, random means uniform among all colorings, or equivalently, that each vertex is i.i.d. colored uniformly from the space of colors.)

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For some graphs $G$ and integers $k$, the opposite. The easiest example is the complete bipartite graph $K_{n,n}$ with $k=3$. The probability a $2$-coloring is proper is about $(1/4)^n$ while the same for a $3$-coloring is about $(2/9)^n$, where I've ignored minor terms like constants. The actual probabilities cross at $n=10$, so as an explicit example, a random $2$-coloring of $K_{10,10}$ is more likely to be proper than a random $3$-coloring. – aorq May 10 2011 at 0:37
This seems like a good example of a counterintuitive statement, but to call it a common false belief would mean that there are lots of people who think it's true. The question would probably never have occurred to me it I hadn't seen it here. The false belief that Euclid's proof of the infinitude of primes, on the other hand, actually gets asserted in print by mathematicians---in some cases good ones. – Michael Hardy May 10 2011 at 15:36

"Suppose that two features $[x,y]$ from a population $P$ are positively correlated, and we divide $P$ into two subclasses $P_1$, $P_2$. Then, it cannot happen that the respective features ( $[x_1,y1]$ and $[x_2,y_2]$) are negatively correlated in both subclasses

Or more succintly:

"Mixing preserves the correlation sign."

This seems very plausible - almost obvious. But it's false - see Simpon's paradox

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

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

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Here are mistakes I find surprisingly sharp people make about the weak$^{*}$ topology on the dual of $X,$ where $X$ is a Banach space.

-It is metrizable if $X$ is separable.

-It is locally compact by Banach-Alaoglu.

-The statement $X$ is weak$^{*}$ dense in the double dual of $X$ proves that the unit ball of $X$ is weak$^{*}$ dense in the unit ball of the double dual of $X.$

The first two are in fact never true if $X$ is infinite dimensional. While both statements in the third claim are true, the second one is significantly stronger, but a lot of people believe you can get it from the first by just "rescaling the elements" to have norm $\leq 1.$ (Although the proof of the statements in the third claim is not hard). The difficulty is that if $X$ is infinite dimensional then for any $\phi$ in the dual of $X,$ there exists a net $\phi_{i}$ in the dual of $X$ with $\|\phi_{i}\|\to \infty$ and $\phi_{i}\to \phi$ weak$^{*},$ so this rescaling trick cannot be uniformly applied. Really these all boil down to the following false belief:

-The dual of $X$ has a non-empty norm bounded weak$^{*}$ open set.

Again when $X$ is infinite dimensional this always fails.

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I think $M(T)$ is not metrizable in the weak$^{*}$ topology, and in fact my claim that this fails for every infinite dimensional Banach space i also think is true. The rough outline of the proof I saw was this: 1. If $X^{*}$ is weak$^{*}$ metrizable, then a first countabliity at the origin argument implies that $X^{*}$ has a translation invariant metric given the weak$^{*}$ topology. 2. One can characterize completeness topologically for translation-invariant metrics, and see directly that if $X^{*}$ had a translation-invariant metric given the weak$^{*}$ topology it would be complete. – Benjamin Hayes Oct 12 2011 at 3:42

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

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

This is more of a false philosophy than a clear mistake, but nevertheless it is very common:

A compact topological space must be "small" in some sense: it should be second countable or separable or have cardinality $\le 2^{\aleph_0}$, etc.

This is all true for compact metric spaces, but in the general case, Tychonoff's theorem gives plenty of examples of compact spaces which are "huge" in the above sense.

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"The universal cover of $SL_2(R)$ is a universal central extension" (which I believed until recently...)

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It took me a bit too long to realize that these two beliefs are contradictory:

• Period 3 $\Rightarrow$ chaos: if a continuous self-map on the interval has a period-3 orbit, then it has orbits of all periods.
• The black dots on each horizontal slice of this picture above $x=a$ show the location of the periodic points of the logistic map $f_a(y) = ay(1-y)$:

You can clearly see a 3-cycle in the light area towards the right; yet we know that if there is a 3-cycle in that slice then there must be a cycle of any period in that slice... so where are they?

(The other cycles are there of course, but they are repelling and hence are not visible. You can see artifacts from these repelling cycles near the period-doubling bifurcations in this picture)

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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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$$2^{\aleph_0} = \aleph_1$$

This is a pet peeve of mine, I'm always surprised at the number of people who think that $\aleph_1$ is defined as $2^{\aleph_0}$ or $|\mathbb{R}|$.

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Something I was sure about until earlier today:

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

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

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

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

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

I have heard the following a few times :

"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."

$f(z)=e^z$ of course is a counterexample.

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No true. Take $f(z)=z^3-3z$ and restrict it to the complement of $\lbrace 1,-1\rbrace$ so that $f'(z)$ is never $0$. It maps this domain onto $\mathbb C$. – Tom Goodwillie May 4 2011 at 0:16
• Many students have the false belief that if a topological space is totally disconnected, then it must be discrete (related to examples already given). The rationals are a simple counter-example of course.

• It is common to imagine rotation in an n-dimensional space, as a rotation through an "axis". this is of course true only in 3D, In higher dimensions there is no "axis".

• In calculus, I had some troubles with the following wrong idea. A curve in a plane parametrized by a smooth function is "smooth" in the intuitive sense (having no corners). the curve that is defined by $(t^2,t^2)$ for $t\ge0$ and $(-t^2,t^2)$ for $t<0$ is the graph of the absolute value function with a "corner" at the origin, though the coordinate functions are smooth. the "non-regularity" of the parametrization resolves the conflict.

• When first encountering the concept of a spectrum of a ring, the belief that a continuous function between the spectra of two rings must come from a ring homomorphism between the rings.

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Unfortunately, "smooth" is a word which means whatever its utterer does not want to specify. Differentiable, C^infty, continuous, everything is mixed. – darij grinberg Apr 14 2011 at 15:12
I don't think the curve (-t^2,t^2) is the graph of the absolute value function. – Zsbán Ambrus May 2 2011 at 16:36
+1 for the discrete $\neq$ totally disconnected example. – Jim Conant May 4 2011 at 15:12
Discrete $\ne$ totally disconnected is a good one that I thought of today and just had to check to see if it was posted already. It adds to the confusion that every finite subset of a totally disconnected space must have the discrete topology, and that in most topological spaces encountered "in nature," the connected components are open sets. – Timothy Chow Oct 20 2011 at 14:30

A degree $k$ map $S^n\to S^n$ induces multiplication by $k$ on all the homotopy groups $\pi_m(S^n)$.

(Not sure if this is a common error, but I believed it implicitly for a while and it confused me about some things. If you unravel what degree $k$ means and what multiplication by $k$ in $\pi_m$ means, there's no reason at all to expect this to be true, and indeed it is false in general. It is true in the stable range, since $S^n$ looks like $\Omega S^{n+1}$ in the stable range, "degree k" can be defined in terms of the H-space structure on $\Omega S^{n+1}$, and an Eckmann-Hilton argument applies.)

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If $n$ is even and $x \in \pi_{2n-1}(S^n)$ and $f$ a degree $k$ map and $H$ the Hopf invariant, then $H(f_* (x)) = k^2 H(x)$. A related misbelief: if $M$ is a framed manifold and $N\to$M a finite cover, of degree $d$. Then the framed bordism classes satisfy $[N]=d [M]$. Completely wrong. – Johannes Ebert Apr 14 2011 at 9:04

I saw many students using the "fact" that for a subset $S$ of a group one has $SS^{-1}=\{e\}$

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This is an interesting example, because it addresses the mistakes that come from the all-too frequent confusion with notations. But we need our shortcuts, our $f^{-1}(x)$ versus $x^{-1}$, etc. Obtaining concise notations while avoiding confusion: a tricky proposition! – Thierry Zell Apr 14 2011 at 15:50

If every collection of disjoint open sets in a topological space is at most countable, then the space is separable

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(*) "Let $(I,\leq)$ be a directed ordered set, and $E=(f_{ij}:E_i\to E_j)_{i\geq j}$ be an inverse system of nonempty sets with surjective transition maps. Then the inverse limit $\varprojlim_I\,E$ is nonempty."

This is true if $I=\mathbb{N}$ ("dependent choices"), and hence more generally if $I$ has a countable cofinal subset. But surprisingly (to me), those are the only sets $I$ for which (*) holds for every system $E$. (This is proved somewhere in Bourbaki's exercises, for instance).

Of course, other useful cases where (*) holds are when the $E_i$'s are finite, or more generally compact spaces with continuous transition maps.

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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 Schwarzmann Apr 11 2011 at 15:34
Disproved nicely by Reuleaux triangles. – darij grinberg Apr 12 2011 at 8:10
Disproved nicely by a two-point metric space. – Tom Goodwillie Apr 17 2011 at 1:36

## From Keith Devlin

"Multiplication is not the same as repeated addition", as put forward in Devlin's MAA column.

I'm not really sure how I feel about this one; I might be one of the unfortunate souls who are still prey to that delusion.

## Caution

In case you missed it, the column ended up spilling a lot of electronic ink (as evidenced in this follow-up column), so I don't believe it would be wise to start yet a new one on MO. Thanks in advance!

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The more I think about this "error", the less I am convinced. It's like saying that you cannot say that $\binom n k$ is the number of $k$-element sets in an $n$-element set because then you will be unable to generalize to complex values of $n$. Or you cannot define the chromatic polynomial as the function counting the colourings and then plug in $-1$ to get the acyclic orientations of the graph. Also, I think it is perfectly understandable what it means to add something halfways. – thei Apr 10 2011 at 20:50
It's not a "false belief". It's a false heuristic. And it's actually here: mathoverflow.net/questions/2358/… – darij grinberg Apr 10 2011 at 21:17
When I taught elementary teachers the course on arithmetic, they all had been taught that multiplication is repeated addition, but I myself thought it was the cardinality of the cartesian product. We enjoyed discussing this difference in point of view. – roy smith May 9 2011 at 3:06
The "repeated addition" characterization has an advantage over the "cardinality of the Cartesian product" characterization (which possibly in some contexts could be considered a disadvantage). That is that it's not self-evident that it's commutative, and so one has a useful exercise for certain kinds of students: figure out why it's commutative. – Michael Hardy May 20 2011 at 2:28

Real projective space ${\mathbb{RP}}^3 = (\mathbb R^4 - 0)/\mathbb R^*$ is non-orientable.

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"Non-orientable surfaces do not embed in orientable three-manifolds." is also a classic. – Sam Nead Apr 10 2011 at 19:33
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I'm not sure that anyone holds this as a conscious belief but I have seen a number of students, asked to check that a linear map $\mathbb{R}^k \to \mathbb{R}^{\ell}$ is injective, just check that each of the $k$ basis elements has nonzero image.

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Higher-level version: $n$ vectors are linearly independent iff no two are proportional. I've seen applied mathematicians do that. – darij grinberg Apr 10 2011 at 18:45

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 2011 at 18:28
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 2011 at 19:25
At least it's better than people thinking multiplication is constant-time. :P – Harry Altman Apr 10 2011 at 19:35

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 $z(x,z)$, 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 2011 at 12:56
Can you help us understand it? Or is there no better way than computation? – darij grinberg Apr 10 2011 at 18:27
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This might not be common, but I once believed the following.

Let $A, B$ be integers, and define a sequence by the linear recurrence $s_n = A s_{n-1} + B s_{n-2}$ with the base case $s_0 = 0$, $s_1 = 1$. Two important special cases are the Fibonacci sequence ($A = B = 1$) and the sequence $s_n = 2^n - 1$ (where $A = 3$, $B = -2$). Then, for any integers $n$ and $k$, $\gcd(s_n, s_k) = s_{\gcd(n,k)}$.

This is true in the two mentioned special cases, so it's tempting to believe it's true in general. But there's a counterexample: $A = B = k = 2$, $n = 3$.

Update: corrected the powers of two minus one example from B = 2 to B = -2. Thanks to Harry Altman.

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I'm not sure how common it is but I've certainly been able to trick a few people into answering the following question wrong:

Given $n$ identical and independently distributed random variables, $X_k$, what is the limiting distribution of their sum, $S_n = \sum_{k=0}^{n-1} X_k$, as $n \to \infty$?

Most (?) people's answer is the Normal distribution when in actuality the sum is drawn from a Levy-stable distribution. I've cheated a little by making some extra assumptions on the random variables but I think the question is still valid.

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