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It's one of my real analysis professor's favourite sayings that "being obvious does not imply that it's true".

Now, I know a fair few examples of things that are obviously true and that can be proved to be true (like the Jordan curve theorem).

But what are some theorems (preferably short ones) which, when put into layman's terms, the average person would claim to be true, but, which, actually, are false (i.e. counter-intuitively-false theorems)?

The only ones that spring to my mind are the Monty Hall problem and the divergence of $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ (counter-intuitive for me, at least, since $\frac{1}{n} \to 0$ ).

I suppose, also, that $$\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e=\sum\limits_{n=0}^{\infty}\frac{1}{n!}$$ is not obvious, since one 'expects' that $\left(1+\frac{1}{n}\right)^n \to (1+0)^n=1$.

I'm looking just for theorems and not their (dis)proof -- I'm happy to research that myself.


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The Banach-Tarski paradox is a good one. –  mjqxxxx Jun 4 at 18:08
related: mathoverflow.net/questions/23478/… –  Will Jagy Jun 4 at 18:29
This question is hard to answer, because an essential skill for a mathematician is being able to rapidly retune your intuition to match the truth. So once you know that a fact is false, very soon it no longer seems obvious. –  Nate Eldredge Jun 4 at 19:53
I strongly disagree that the Jordan Curve theorem is obviously true! I would agree that the Jordan Curve theorem for piecewise smooth curves is obviously true; but it's also pretty easy to prove, at least compared to the topological version. (Is it really obvious that the Koch snowflake doesn't have some pathological path from the inside to the outside?) –  Mike Miller Jun 6 at 7:04
"Every true statement can be proved" –  sinelaw Jun 9 at 17:27

62 Answers 62

up vote 119 down vote accepted

Theorem (false):

One can rearrange the terms in a convergent series.

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The series $\sum \frac{(-1)^i}{i}$ is a counterexample; see this Wikipedia article for a discussion. –  MJD Jun 4 at 18:39
@alexqwx Roughly, that some terms can be pushed 'infinitely far away', so that any two finite partial sums differ by arbitrarily many terms. Imagine e.g. taking the positive numbers and rearranging them as 1, 2, 3, 5, 4, 7, 9, 11, 6, 13, 15, etc. with a growing number of odd terms between each two consecutive even terms; then every number still shows up in the sequence eventually, but that doesn't mean that any concept like a 'ratio' of odd-to-even (e.g. natural density) is preserved. –  Steven Stadnicki Jun 4 at 18:55
Huh? One can rearrange the terms in any series, convergent or otherwise. Did you leave something out of the statement of your theorem? –  bof Jun 4 at 21:25
Actually, you can rearrange terms in a convergent series when computing its value (or otherwise). You can even rearrange terms without changing the value, with a bit of care. What you cannot, is assume that any rearrangement will preserve convergence, or when it does preserve the value converged to. –  Marc van Leeuwen Jun 4 at 22:42
@kutschkem You can always make finitely many swaps of terms (using commutativity) without changing the value, but you can't make infinitely many swaps; continuity isn't preserved across an infinite number of operations. –  Steven Stadnicki Jun 5 at 21:36

An 'obvious' but false theorem: There are more open sets in $\mathbb R^2$ (or $\mathbb R^n$) than there are real numbers.

And in a similar vein we have this corollary to the first statement: There are more continuous functions $\mathbb R\rightarrow\mathbb R$ than there are real numbers.

(Both statements are false.)

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The Birthday Paradox

If 30 people are randomly selected, and they have birthdays that are independently, (identically) uniformly distributed over the calendar year, then the probability that two (or more) of them have the same birthday is approximately $\frac{1}{12}$.

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Any real number can be computed somehow.

More formally:

For every real number, there exists a finite-length program that computes that number.

Since real numbers are uncountable while computable numbers are countable, that just can't be the case.

This limitation comes from the fact that we're stuck using finite-length programs. Infinite-length programs can be defined to compute any real number (trivially). So there is a sense in which all real numbers can be computed.

Just not by humans. Note that, since a single infinite-length program would take up infinite memory (and we don't seem to have any infinite computers/brains), the majority of these infinite-length programs can never be known, let alone computed. So computable numbers are only those numbers computable by a finite-length program. And the set of finite-length programs is countable.

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Someone else mentioned "there are more rational numbers than integers". Along the same lines, I had a hard time accepting that

There are more integers than there are real numbers between 0 and 1

is false. I mean, I get it now, but it seemed intuitively very wrong to me before I studied transfinite numbers.

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One of the first times I got caught out being wrong about something so obvious was believing:

abs(x) is never equal to -x

Of course abs(x) is defined as -x for x < 0

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I really like "wrong proofs" as typically the insight why the proof is wrong gives you some understanding of the topic. One very simple version is this one, which I threw at my first semesters when I was a tutor:

Each binary relation which is symmetric and transitive is also reflexive and therefor an equivalence relation.


Let $\sim$ denote a symmetric and transitive relation and let $x$, $y$ be two elements with $x \sim y$. As $\sim$ is symmetric, it holds that $y \sim x$. Since $x \sim y$ and $y\sim x$ it follows by the transitivity of $\sim$ that $x \sim x$, which is the definition of reflexivity.

Edit: Since I was asked, here's why the proof is wrong (move your mouse there to show):

Take a look at the empty relation on a non-empty set $S$, so that there are no $x, y \in S$ so that $x \sim y$. This relation is symmetric and transitive, but it is not reflexive. Reflexivity needs $x \sim x$ to hold for all $x$. The proof assumes that there is a y so that x ~ y, which isn't necessarily the case for all $x$.

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+1 for set theory! –  Michael Zalla Jun 10 at 8:29

My "theorem":

The statement Everybody loves my baby, but my baby loves nobody but me is about a pair of lovers

It is so simple and so obvious, even my grandma will understand it. And no matter how much you explain the simple logic calculation which shows that we are talking about a single narcissist here, half the class of first-semester logic students will continue insisting that your proof is wrong, and they don't know what is wrong about it, but it cannot refer to a single person.

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Some people treat love as a relationship that is not defined reflexively. –  Joshua Jun 14 at 2:16

If a propositional calculus A contains all theorems of propositional calculus B under detachment and uniform substitution for propositional variables, but B does not contain all of the theorems of A, then one of the shortest single axioms of A is longer than any of the shortest single axioms of B. Or one might more haphazardly say "if propositional calculus A is bigger than propositional calculus B, then one of the shortest single axioms of A is longer than any of the shortest single axioms of B."

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Good luck finding a layman whose eyes don't glaze at the term "propositional calculus", let alone one who finds something here obvious. –  rumtscho Jun 13 at 17:46

Stein's paradox is to me the most puzzling mathematical notion I've ever known (although I'm not a mathematician), mostly because it's not a mathematical "artifact", but it's non-intuitiveness carries very tangible error consequences.

Theorem: (false)

One can do no better than ordinary decision rule for estimating the mean of a multivariate Gaussian distribution under mean squared error.

In other words, completely independent phenomena can actually be combined for a lower joint estimation error.

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Here's one of my favorites: Let's assume playing with a fair coin.

Theorem (false) In a long coin-tossing game each player will be on the winning side for about half the time, and the lead will pass not infrequently from one player to the other.

The following is from the classic of Chung & Feller Introduction to Probability Theory and It's Applications, Vol 1:

According to widespread beliefs a so-called law of averages should ensure the Theorem above. But, in fact this theorem is wrong and contrary to the usual belief the following holds:

With probability $\frac{1}{2}$ no equalization occurred in the second half of the game regardless of the length of the game. Furthermore, the probabilities near the end point are greatest.

In fact this leads to the Arc sine law for last visits (see e.g. Vol 1, ch.3, section 4, Theorem 1).

Note: Please note their remarkable statements cited from Chapter III: Fluctuations in Coin Tossing and Random Walks:

(Chung & Feller): For example, in various applications it is assumed, that observations on an individual coin-tossing game during a long time interval will yield the same statistical characteristics as the observation of the results of a huge number of independent games at one given instant. This is not so.

and later on:

(Chung & Feller): Anyhow, it stands to reason that if even the simple coin-tossing game leads to paradoxical results that contradict our intuition, the latter cannot serve as a reliable guide in more complicated situations.

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I really want the following to be true:

Theorem: Let $S$ a subset of a vector space. If $S$ is pairwise linearly independent (meaning each $\{v,w \} \subseteq S$ is linearly independent) then $S$ is linearly independent.

And yet, it is false. For example, $$ \{ v,w,v+w \} $$ If $S$ only had two elements then we win by default. In any event, students tend to believe this. I mean, it's linear algebra, the principle of superposition ought to apply right? Something is the sum of its parts, linear independence begets linear independence... very seductive, very wrong.

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Infinite terms always have a sum equal to infinity.

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Do you mean infinitely many terms? If not, what is an "infinite term"? –  WChargin Jun 15 at 0:07

The following statement is wrong:

The inner angles of a triangle always sum to 180 degrees.

While it sounds plausible that the sum of the angles is a constant, it is actually a property of the space. In Euclidean space the inner angles of a triangle always sum to 180 degrees.

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For example, this highly-voted answer begins “I found it completely amazing that the angles in a triangle always added up to 180 degrees”. –  MJD Jun 9 at 21:22

What about this:

$\mathbb{R}$ and $\mathbb{R}^2$ are not isomorphic (as Abelian groups with addition).

It falls under the category of "Let's take the Hamel basis of $\mathbb{R}$...", but I like it a lot.

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Geometry proofs done informally by drawing figures on the blackboard. You then bypass the axioms of Euclidian geometry, you pretend that you don't need to invoke them as the figures drawn seem sufficient. However, in Earth's gravity Euclidean geometry is only an approximation.

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"A sequence of numbers in which every number is larger than the previous, will always eventually go above a given value L."

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The following statement I once believed to be "obvious":

If $f:\mathbb{R} \rightarrow [0,\infty)$ continuous is such that $\int_{-\infty }^{\infty }f(x)\text{d}x<{\infty } $, then $\lim \limits_{x \to \pm\infty} f(x) = 0$

which is actually false.

(Note: It is true if $f$ is uniformly continuous!)

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Can you provide a counterexample? –  hallaplay835 Jun 11 at 22:26
This is pretty easy to give a counterexample to: let $f$ have bumps as far down as you like, just with sufficiently small widths so that the sum of their integrals goes to zero. You can make $\limsup_{x \to \infty} f(x) = \infty$ this way. –  Dustan Levenstein Jun 12 at 3:43

I'm surprised noone gave this answer already, so here it is:

There are more integers than there are natural numbers.

It's obvious, isn't it?

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I hesitate with this one, because it depends on (somewhat unsound) definition of "more". –  DanielV Jun 7 at 16:24
There absolutely are more integers than natural numbers (by inclusion). And cardinality is a slightly contrived notion if you think about it for a moment. –  Tibor Jun 8 at 9:51
@Tibor: No, cardinality is not a contrived notion if you think about it for a long moment. –  Asaf Karagila Jun 8 at 21:09
All of mathematics is contrived, by definition. :) –  Ryan Jun 9 at 1:32

Image of a measure zero set under a continuous map has measure zero!

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If the dot product of two vectors is 0, then they are linearly independent.

My prof threw this question at me today and I fell for it.

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haha, took me a second. –  Bennett Gardiner Jun 11 at 2:47
@Lost1: It actually doesn't depend on the field: $\vec{v}\cdot \vec{0} = 0$ and $\{\vec{v}, \vec{0}\}$ is never independent. –  Jason DeVito Jun 12 at 15:52
@JasonDeVito ah... i was thinking take $\mathbb{Z}_2^2$ and the vector $(1,1)$... –  Lost1 Jun 12 at 17:46

Theorem: Let $f_1(x,y)$ and $f_2(x,y)$ be two joint probability densities, each having its $x,y$ components positively correlated ($Cov_1(x,y)>0$, $Cov_2(x,y)>0$). Let $f_3=\alpha f_1 + (1-\alpha) f_2$ be the mixing density, for some $0\le \alpha\le 1$. Then $Cov_3(x,y)>0$.

In words: mixing populations preserves the correlation sign. In other words: if the average MSE male user is brighter than the mean, and if the average MSE female user is brighter than the mean, then the average MSE user is brigther than the mean. Obviously true.

False. See Simpson's paradox.

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Perhaps the following example will make the obviousness more obvious. Baseball team $A$ has a better win-lose ratio than team $B$ in the first part of the season. Then there is a strike, and some games are missed. When the season resumes, team $A$ also has a better win-lose ratio than team $B$ in the second portion of the season. Therefore, team $A$ has a better win-lose ratio than team $B$ overall. (Wrong!) Baseball did play such a split season in 1981, but I don't know if the paradox actually occurred then. Probably not, as it requires $A$ and $B$ to play very different numbers of games. –  MJD Jun 6 at 0:43

$ 0.\overline{9} < 1 $

Probably the most famous of the "obvious" but false.

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The question did ask what the average person would consider true. –  MichaelRushton Jun 9 at 7:56
This is by far the best one on the page imo. Some of the others are so complex to the average person that they wouldn't be "obviously" true to almost anyone. –  Sam Creamer Jun 10 at 20:20

$\textbf{Counter-Intuitive Example}$

$$\ \ \textbf{D}_v f(\textbf{a}) = 0, \forall \textbf{v},a \not \Rightarrow f \ \ \text{continuous}.$$

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Here is one thing commonly thought to be true but is quite horribly wrong on many accounts:

There is a notion of mathematics where we can say things are 
"actually true" or "actually false".  

An example of making this error: the OP. Other examples: the many responses.

There are several reasons why this is wrong. First, in the system most mathematicians assume when not being explicit, we have no standard model (we have no models within that system because any model would show the system consistent which we know we cannot show in that system through blah blah Godel blah.. I know you don't want details, just explaining what I'm getting at). Truth and falsity are semantic - they exist in models and so without one, we don't make claims of truth or falsity.

But also, mathematics is not "the system most mathematicians assume when not being explicit" - it is formalisation in general. There are many systems seriously investigated by mathematicians that make numerous "counterintuitive" derivations. For instance, these are all obvious and wrong in different systems:

  • A statement cannot be both true and false. (In paraconsistent logics, statements can be both true and false and the system does not collapse to trivial - in fact, a number of dialetheists argue this is a much more accurate logical system for real world reasoning).
  • There are discontinuous total functions. (In a number of constructive systems it is not possible to prove the existence of discontinuous total functions. Some are even strong enough to prove that all total functions are continuous.)
  • Every infinite set A has the same cardinality as AxA. (This is not necessarily true in systems without the axiom of choice. Famously, Tarski tried to publish his result on this implication and was rejected by both Frechet and Lebesgue. Frechet thought the paper was obvious and well-known and had no mathematical merit. Lebesgue thought both the axiom of choice and the implication from it were both wrong, so the paper had no mathematical merit.)

I only bring these examples up not as answers to the OP but simply to illustrate my real answer that the question itself demonstrates an extremely common assumption in mathematics that is in fact wrong.


This is an area that I think is often a place of common misunderstanding, and discussion in the comments makes it clear I should elaborate. Modern mathematics separates out the domains we make statements on into syntax and semantics.


The syntax is the theory - the formal language, axioms specified as sentences in the formal language, and some metalogical rules of inference. In the syntax, we talk about sentences, propositions, terms, derivations, and proofs. It is a place of symbol manipulation.


The semantics is the model - it is the meaning we ascribe to the statements of the theory. An interpretation of a theory is a model that assigns to each formula of the theory a meaning value - typically truth. Truth is semantic and is specific to a model.

The "problem"

A model is a consistent interpretation of the truth meaning of a theory. If a theory has a model, it has almost trivially been shown to be consistent. But... it is well known that a theory strong enough to express the Gödel diagonalisation can never prove it's own consistency. For these theories, we will never have a model and cannot make statements about the meaning of any formula.

In these theories, it is wrong to talk about truth or falsity. We don't have a model giving meaning to that. We will never have a model.

That's not really a problem. For centuries, mathematicians had loosely combined derivation and truth and had mostly discussed them as one thing. Derivation and proof were seen as the important part of mathematics and formalization. You still have that.

Also, it is perfectly meaningful to derive results that say "if this theory is consistent and has a model, then...". Model theory has been doing that for nearly a century.

What about truth predicates?

But people seem to want more. They want to talk about truth, as that is a form of meaning that holds a special place. They often go to great lengths to try to continue to assign truth and falsity. One common approach is to form truth predicates - predicates in the syntax that have the property that asserting the predicate on a formula corresponds to asserting the validity of the statement (that it is true in all models).

Note the switch - a truth predicate is syntactical. We still aren't talking about true or false here - the context of their use is still whether statements including the predicate "are derivable" or "obtain". Theories may have multiple models - most theories are not categorical just from things like Löwenheim–Skolem, so predicates cannot talk about truth. They can talk about validity - and that's really what is going on here - but even that is extremely problematic.

Incomplete theories cannot actually derive anything about validity on the total theory. And actually, this is where Tarski's theorem on nondefinability comes in and it is shown that such a predicate doesn't actually exist. So others keep at it with a hierarchy and reflection extensions of the base theory, seeking out some approximation of a fixed point for validity.

But this doesn't actually buy anything to do with insight into truth. It cannot. There is nothing you can do to reach truth because you cannot know if the theory is consistent or not and whether truth exists. And no attempts to reach beyond derivability actually give a predicate that can be used and say "this is true". The predicate is only useful to say "this is provable".

But there are already provability predicates, and that investigation is much more profitable. Truth predicates are a voiceless oracles. They do not help anyone make assertions on truth. They are simply reformulations of "if we knew that X was consistent, and we had some platonic sight that could see the truth values in all models, and we could collate the infinite possibilities and see the validities forever hidden, then this predicate applied to this class of statements would agree with those assertions that are valid". But if we had that supernatural sight, we could more easily just say "hey, that's true in that model - and that's false over there." Without that, we can use the predicate to say "truth is preserved in this derivation". Which doesn't add anything.

A truth predicate doesn't talk about truth. It is irrelevant to the point.


So.. life goes on. My whole point in posting this answer was to illustrate that the initial question was making a common obvious assumption that is actually wrong. You should not talk about truth in the commonly used ambient theory - just talk about what is provable and you are fine. If you want to talk about truth, ensure you specify the ambient theory and it is one where such discussions are meaningful. Or talk about conditional models as model theorists do.

It may not be intellectually satisfying to some people. Clearly, as of writing this, my answer has received 3 downvotes and two upvotes, so it doesn't sit right with some anonymous readers of a math web site. But there is nothing controversial about the point. It has been known for almost 100 years and it is still a common mistake.

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Your third paragraph, in its most precise form, is Tarski's undefinability theorem. But being aware of what it is, I still disagree with the preceding two paragraphs. Relative to another system, you can talk about truth, for example arithmetical truth is definable in ZFC (but not in PA). Truth in the set theoretic universe is not definable in ZFC, but so what? –  Burak Jun 4 at 23:02

Lebesgue once stated that the projections of Borel sets in $\mathbb R ^2$ on to one of its axes are also Borel sets. This fact is actually false, the realization of which is attributed to the short-lived mathematician Mikhail Yakovlevich Suslin.

Unfortunately it is very difficult to find a counterexample. The only one I've ever seen takes a result in descriptive set theory about $\mathbb N ^ {\mathbb N}$ and uses the fact that it's homeomorphic to $\mathbb R$.

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I hope you are misremembering or misunderstanding the arguments you have seen, because $\mathbb N^{\mathbb N}$ is not homeomorphic to $\mathbb R$. For example, one is connected and the other one is not. One is $\sigma$-compact and the other one is not. –  Andres Caicedo Jun 6 at 2:54
"That depends on the topology". Obviously. "The space of all sequences (both finite and infinite) of natural numbers." Perhaps, but that is not $\mathbb N^{\mathbb N}$. "With the exclusion of $\mathbb Q$." But then that is not $\mathbb R$. Anyway, a word or two clarifying the imprecision would not be superfluous. –  Andres Caicedo Jun 6 at 4:51

For me a nice example of all of "evidence" suggesting it was true is

$$\pi(x) < \operatorname{li}(x)$$

until Skewes showed that $\pi(x) - \operatorname{li}(x)$ changes sign infinitely often

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The question asks for obvious but false statements. Are you seriously claiming that $\pi(x) < \operatorname{li}(x)$ is obvious? –  MJD Jun 5 at 21:32
hard to tell whether that is supposed to be obvious... What are $\pi(x)$ and $\operatorname{li}(x)$? –  example Jun 5 at 22:03
  • If $U$ is an open subset of $\mathbb{R}^n$ that is homeomorphic to $\mathbb{R}^n$, one might think it "obvious" that it's in fact diffeomorphic to $\mathbb{R}^n$ (perhaps thinking something like "topologically it looks like $\mathbb{R}^n$, and differentiably it's locally trivial"). In fact, this is true (but by no means obvious!) for $n\neq 4$. But for $n=4$ it is false: there exist exotic $\mathbb{R}^4$'s (differentiable manifolds that are homeomorphic, but not diffeomorphic, to $\mathbb{R}^4$), including "small" ones which are diffeomorphic to an open subset of $\mathbb{R}^4$.

  • Much less profound, but still fun: it's "obvious" that the sum of two convex open sets in the plane whose border is $C^\infty$ also has a $C^\infty$ border (perhaps thinking something like "the border of the sum is parametrized by a smooth function of the borders of the summands"). But this is false: in fact, the border of the sum is always $C^{20/3}$ (meaning six times differentiable and with a sixth derivative which is appropriately Hölder) and no more in general. A simple counterexample is given by the epigraphs of $x^4/4$ and $x^6/6$. For details, see Kiselman, "Smoothness of Vector Sums of Plane Convex Sets", Math. Scand. 60 (1987), 239–252.

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$20/3$? What kind of magic is that!? +1 interesting example. –  Olivier Bégassat Jun 5 at 23:24

There are a good number of counterintuitive probability situations out there. One of my favorites is nontransitive dice:

There are 3 dice, A, B and C. The dice have numbers from 1-9 on their sides (repeats possible). If die B beats (higher number) die A more than half the time and die C beats die B more than half the time, then die C will beat die A more than half the time.

This is not necessarily a true statement. Dice can be designed such that the "x beats y" property is not transitive. A beats B, which beats C, which beats A.

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There are a number of similarly counterintuitive statements in multiway voting, of which this could be seen as a special case. For example, in figure skating competitions at the championship level there are nine judges. It would sometimes happen that a majority of judges would rank skater $A$ above skater $B$, a majority would rank $B$ above $C$, and a majority would rank $C$ above $A$. The scoring rules had to have conditions in place to deal with this kind of situation. –  MJD Jun 5 at 21:35

The Hauptvermutung states that there is essentially only one PL structure on a manifold. More precisely, it states that any two triangulations have a common subdivision. The reason why this seems "obviously true" is that you can take both triangulations and superimpose them one on top of the other, subdividing the manifold into a bunch of cells, and then taking the barycentric subdivision to get a triangulation. It turns out that this is false and one needs some pretty subtle invariants to detect it. The problem with the argument that I gave is that one triangulation could be very wild with respect to the other (fractally wiggly) so that their union does not subdivide the manifold into a nice collection of cells.

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