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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 '14 at 18:08
related:… – Will Jagy Jun 4 '14 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 '14 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 '14 at 7:04
"Every true statement can be proved" – sinelaw Jun 9 '14 at 17:27

65 Answers 65

Something I used to be seduced by in my mathematical immaturity (which is sadly still existing):

Suppose that $P_n$ are a family of statements indexed by $n\in\mathbb{N}$ and we can assign meaning to $P_{\infty}$. Then if $P_n$ is true for all $n\in\mathbb{N}$, then $P_{\infty}$ is true also.

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The example I had in mind when I wrote this was $P_n$: sum of $n$ differentiable functions is differentiable; $P_\infty$: infinite sum of differentiable functions is differentiable. – Jp McCarthy Jun 6 '14 at 16:09
How about the statement that 1/n > 0 which holds for any integer n but fails in infinity? – user102184 Jun 6 '14 at 16:37
@JónÁskellÞorbjarnarson Mark would claim that $1/\infty>0$ is not meaningful because $1/\infty$ is not defined. An easier example is just a sequence of rationals converging to an irrational. – Jp McCarthy Jun 6 '14 at 16:40
@Hurkyl I agree and was trying to argue from Mark's position. – Jp McCarthy Jun 8 '14 at 22:17
This is Leibniz's Law of Continuity, right? – Tanner Swett Jun 9 '14 at 4:29

Here are some of the false statements popping into my mind that made me raise at least one eyebrow when I first realized they were not true.

Every linear function between two vector spaces is continuous.

True only as long as the domain is finite-dimensional. If it is not, then there exists a linear function that is not continuous—at any point!

The set of real numbers can in no way be (totally) ordered in such a way that every non-empty set in it has a least element.

False if choice is assumed, by the well-ordering theorem.

$\mathbb Q$ is not countable.

I am still tempted to believe it sometimes...

If the derivative of a continuous real-to-real function exists almost everywhere and (wherever it exists) vanishes almost everywhere, then the function must be constant.

False. In fact, there exists a function that satisfies the premise and it is strictly [sic!] increasing!

Any compact set is closed.

The name “compact” would suggest this, but this can be guaranteed only in Hausdorff spaces.

A set is compact if and only if every sequence in it contains a convergent subsequence.

While true in metric spaces, not only is it false in some more general topological spaces, but also neither condition implies the other!

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I totally believe that $\mathbb{Q}$ is countable. Did you mean $\mathbb{R}$? For the next, do you assume that the derivative exists everywhere, or only almost everywhere? – Daniel Fischer Jun 5 '14 at 9:14
No, I meant $\mathbb Q$. I mean, duh, $\mathbb Q$ is countable, but I always need a leap of faith to believe it. If I can enumerate the rationals in $[0,1]$, why can't I do that in an increasing order? (That is, such that $q_1<q_2<\ldots$, where $\{q_n\}_{n\in\mathbb N}=[0,1]\cap\mathbb Q$). As for the other “paradox,” I meant “exists and vanishes” is an intersection of events whose complement has measure zero. Differentiability is not assumed everywhere. I will make it clearer, thanks for making me realize the ambiguity. – triple_sec Jun 5 '14 at 10:10
I've convinced myself that $\mathbb Z^2$ is countable; obviously it is possible to devise an algorithm to count them. And I've convinced myself that this means that $\mathbb Q$ is countable, because $\mathbb Q$ is ordered pairs of elements in $\mathbb Z^2$, so there are "less" elements in $\mathbb Q$ than $\mathbb Z^2$ (thinking of graphing elements, $\mathbb Q$ leaves holes because $\frac 24 = \frac 12$), so $\mathbb Q$ is definitely not uncountably infinite. – Justin Jun 6 '14 at 2:51
@Quincunx You're perfectly right. It just rings intuitively very odd to me. For example, say that $f:\mathbb R\to\mathbb R$ is a function. Suppose tha $f$ is continuous except at a countable set of points. You may be tempted to think that this function is “almost well-behaved,” right? It turns out that there exists a function that is discontinuous precisely at all rational points. At this point, I would conclude that the function is not “almost well-behaved” at all, but rather pathological. After all, the set of discontinuities of $f$ is dense! – triple_sec Jun 6 '14 at 3:12
I guess the one feature of $\mathbb Q$ that throws me off balance is that it just somehow doesn't feel “right” to me for a countable set to be dense in an uncountable one. – triple_sec Jun 6 '14 at 3:14

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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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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What I personally think it should be is... For every nontransdental, real number there exists a finite-length program that computes that number... – Binary Freak Jul 25 '14 at 3:21
@BinaryFreak I believe the non-transcendentals are all algebraic and thus computable, so that would be an obvious theorem that turns out to be true. On an interesting related note, while every non-computable number is transcendental, some transcendentals are computable ($\pi$, $e$, and some non-repeating reals like $0.123456789101112...$). – Keen Jul 25 '14 at 15:43

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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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 '14 at 2:47
depends on what field. very nice, but a trick question. I feel cheated. – Lost1 Jun 12 '14 at 11:29
@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 '14 at 15:52
@JasonDeVito ah... i was thinking take $\mathbb{Z}_2^2$ and the vector $(1,1)$... – Lost1 Jun 12 '14 at 17:46
@Lost1 According to the usual definition, the dot product must be real-valued (certainly not finite-field valued; they are not ordered fields, and without order, positive-definiteness makes no sense. And without that, this theorem is false but also has no reason to be true!) – Ryan Reich Jun 16 '14 at 4:25

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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Cauchy's theorem implies that:

if one makes a physical model of a convex polyhedron by connecting together rigid plates for each of the polyhedron faces with flexible hinges along the polyhedron edges, then this ensemble of plates and hinges will necessarily form a rigid structure.

However, there are counter-examples if you allow a general polyhedron (not convex).

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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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Your answer is essentially equivalent to this one. – Scott Jun 10 '14 at 18:08
Ahh, good catch, but I think mine is simpler to express. – CaptainCodeman Jun 10 '14 at 18:18

There are a lot of examples in (extremal) graph theory, where an obvious argument shows that a statement is true, except that there are a number of small counterexamples which are easy to overlook.

Consider the following statement: Let $G$ be a graph with $n$ vertices and the largest number of edges subject to the condition that $G$ does not contain a pair of disjoint edges (i.e. $K_2 + K_2$). Then $G$ is a star (i.e. $K_{1,n-1}$).

This is obviously true, if you think about it for a moment. But for $n=3$, a better solution is to take $G = C_3$. And for $n=4$, taking $C_3$ plus an isolated vertex is just as good as taking $K_{1,3}$.

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EDIT: The counterexample I had in mind is incorrect; I've asked another question to try and clarify the matter one way or the other. Regardless, I think it's safe to say that this is no longer obvious! But I'll go ahead and update (or delete) my answer accordingly once I've gotten a bit more clarity.

Here's a topological example that takes some thought to falsify: roughly, 'for every non-intersecting curve between two opposite corners of a square, there's a curve between the other two corners that only intersects it once'. Formally:

Let $f: [0, 1]\mapsto [0,1]^2$ be a non-self-intersecting curve with $f(0) = (0,0)$, $f(1) = (1,1)$, and $f(t)\in (0,1)^2$ for $t\in(0,1)$. Then there exists a non-self-intersecting curve $g: [0, 1]\mapsto [0,1]^2$ with $g(0) = (1,0)$, $g(1) = (0,1)$, and $g(t)\in (0,1)^2$ for $t\in(0,1)$ such that there are unique $t_0$ and $ t_1$ with $f(t_0) = g(t_1)$.

This seems obvious (at least to me) on first glance, and even on second glance, the example of the Jordan Curve theorem suggests that it should be true; after all, we get a 'left side' and a 'right side' of our curve by the JCT, and doesn't the Schoenflies theorem mean that we should be able to find an inverse mapping of our curve to the circle? But it's false; there are curves $f()$ that can't be intersected only once by any curve $g()$. Finding a counterexample makes a nice exercise...

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Are you sure? I think, I can prove that such g always exists. – studiosus Jun 4 '14 at 19:53
@studiosus I'd love to see your proof! I was fairly confident in my example, but on further thought I'm actually willing to believe that it can be falsified. I may not have been considering sufficiently degenerate $g()$ to go with my canonically degenerate $f()$... – Steven Stadnicki Jun 4 '14 at 19:57
There is no room for a proof in the comments, since it long. Ask a separate question. – studiosus Jun 4 '14 at 19:58
@mjqxxxx: No, that does not work: the straight line segment might not be contained in $\Phi([0,1]^2)$ (since you do not know what the image of the rest of the boundary of the unit square looks like). One needs more work to get a real proof. – studiosus Jun 4 '14 at 21:30
Here is a stronger (straightening) theorem: Let $S$ be a surface with some fixed triangulation (say, a triangulated plane) and $h: G\to S$ a topological embedding of a finite graph. Then $h$ is isotopic to a piecewise-linear embedding $h': G\to S$. This reduces problems of the above type to piecewise-linear ones which are easily solvable. One can give, however, a more direct solution using only Schoenflies and a bit of algebraic topology. – studiosus Jun 4 '14 at 22:03

The following is obviously false, but it is actually true, as shown in the Wikipedia article about Vitali sets.

The exists a countable collection $\left\{V_n\right\}$ of subsets of the unit circle such that:

  1. Any two distinct $V_n$ are disjoint.
  2. Any $V_n$ can be obtained from any other by a rotation.
  3. The union of all the $V_n$ is the whole circle.

All the $V_n$ must have, from property two, the same "size" (for any reasonable definition of "size"), but if the above fact was true, the sum of their (equal) sizes would be the size of the circle (positive, but finite). But if the size was zero, the the sum should be zero, and if the size was positive, the sum should be infinite.

A consequence of this is that the following is false (although we all would like it to be true):

There exists a function $\mu$ that, given a bounded subset of $\mathbb{R}$, tells you its "size". Precisely:

  1. If $A\subset\mathbb{R}$ is bounded, then $\mu\left(A\right) \in \left[0,\infty\right[$.
  2. If $\left\{A_n\right\}_{n\in\mathbb{N}}$ is a sequence of bounded disjoint subsets of $\mathbb{R}$ (that is, $A_n \cap A_m=\emptyset$ whenever $n\neq m$) with bounded union (that is, $\bigcup A_n$ is bounded), then $\sum\mu\left(A_n\right)=\mu\left(\bigcup A_n\right)$.
  3. If $A$ is bounded, $x$ is a real number, and we define $A+x=\left\{a+x:a\in A\right\}$, then $\mu\left(A+x\right) = \mu\left(A\right)$.
  4. $\mu\left(\left[0,1\right]\right) = 1$

Indeed, rewriting the first fact exchanging the circle by the half-open interval $\left[0,1\right[$ and exchanging rotations for cyclic shifts "mod $1$", we realize that if $\mu$ satisfies the first three conditions above, then $\mu\left(A\right)=0$ for all $A$.

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If you consider a finite number of sets $V_n$, and take a limit it does seem plausible, not quite obviously false. – Mark Hurd Jul 28 '15 at 7:35

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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Is there such thing as a vector space where all pairs of vectors are linearly independent? Maybe you mean a vector space basis? Or is conflating vector space and a basis for a vector space common practice ? – DanielV Jun 11 '14 at 7:49
@DanielV, not my meaning, I will reword, I meant that if every pair of vectors in $S$ is linearly independent... thanks for the comment. – James S. Cook Jun 11 '14 at 11:33
I like your example, I just wanted to make sure I wasn't misinterpreting it. If I recall correctly, I see "vector space" to refer to the span of basis vectors, so two different basis vector sets may have the same vector space. For clarity I probably would have just stated "If every pair of vectors in S is linearly independent, then S is a set of linearly independent vectors". Sorry to criticize though, I believe that even a graphical description of this claim might take some students by surprise so I think it's a good example. – DanielV Jun 11 '14 at 12:50

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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Here are some "obvious" statements to which Richard's paradox can apply to:

  1. For a given predicate $P$, there exists a set $S$ of $x$ for which $P(x)$ is true. (Russell's paradox)
  2. The set of integers and the set of real numbers are the same infinite size. (Cantor's diagonal argument.)
  3. There exists a formalization of arithmetic in which all true statements are exactly those which are provable. (Gödel's theorem(s))
  4. There exists a computer program (Turing machine) that can effectively determine if any other computer program doesn't halt. (Halting problem)
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I already mentioned #1. – MJD Jun 4 '14 at 19:53
I think #4 is not merely not obviously true, but obviously false, because if it were true, then we wouldn't have to actually run our computer programs to find out what they did, instead we could use this hypothetical procedure to tell us, without running them. But then what would be the point of running the programs at all? The universe just doesn't work that way; if you want to find out what happens, then, at least some of the time, you just have to try it and see. – MJD Jun 4 '14 at 19:55
It doesn't matter; my comment is the same either way. I have a blog post drafted that claims that Rice's theorem is the most obvious theorem in mathematics, and the undecidability of the halting problem is a special case of that. – MJD Jun 4 '14 at 23:59
This is not about general predictions of the future; this has to do with time only insofar as countable infinity can be mapped to discrete Turing machine steps that take place in time. E.g. A Turing machine that halts when it finds $a,b,c,n\in\mathbb{Z}_+:a^{n+2}+b^{n+2}=c^{n+2}$ we know doesn't halt thanks to Wiles's proof. If the halting theorem were false, then a "theorem-proving machine" based on a formalization of any area of mathematics would spit out all true statements eventually. This was "obvious" to Russell, Whitehead, and even Hilbert until the 1930's thanks to Gödel, Turing et al. – Matt Jun 6 '14 at 19:42
@MJD If the halting theorem were false, it doesn't imply that we would be able to tell the outcome of any computation. It merely would imply that an algorithm exists. There is still a substantial gap between the existence of such an algorithm $U$, and both (1) our physical ability to run $U$ (it may require more memory than the number of atoms known to exist) and (2) more importantly our knowledge that a given algorithm is in fact $U$. (1) and (2) do not automatically follow from the halting theorem's falsity. The reasons for it being true are more subtle than you are framing it. – Matt Jun 6 '14 at 19:53

If you start thinking about the rigidity of thin shells in $\mathbb{R}^3$, you quickly encounter a slew of counterintuitive results.

For instance, it is obvious that a spherical shell is ($C^2$) rigid, and this is in fact true. A smooth, closed, compact surface with everywhere positive Gaussian curvature is likewise rigid. One might imagine these results generalize to

  • Any closed surface;
  • Any closed surface with positive Gaussian curvature everywhere but at finitely many points;
  • Any surface with boundary with everywhere positive Gaussian curvature;

and all of these are false.

Moreover, after thinking about reflections, or poking and prodding at a ping-pong ball, it is intuitively obvious that a spherical shell is not $C^0$ rigid. But you can't really "see" any difference between a $C^1$ and a $C^2$ deformation of the sphere, so surely the sphere is $C^1$ rigid? Far from it -- given any arbitrary closed surface that is topologically a sphere, and distance $\epsilon$, it is possible to $C^1$-isometrically embed a sphere $\epsilon$-close to the target surface!

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Here is a proposition, when put in layman's terms, an average person would claim to be true.

Every subset of real numbers has a measure.

How can this be false, when you mark a region, say in two dimensions, of course, it has an area?! Unless you constructed a Vitali set at some point, we tend to think that the concept of length/area/volume should extend to all possible subsets.

Here is another such false proposition.

Axiom of Determinacy

If we are playing a two-player infinite game where we create a real number in $[0,1]$ by choosing decimal digits in turns and one of us tries to land the resulting number in a pre-determined payoff set that is known to both of us and the other tries to avoid it, how could it be that there is a game where neither of us have a winning strategy? We both have complete information about the payoff set what numbers to avoid and what numbers to hit, one of us should be able to come up with a strategy. Well, unfortunately no.

Both of these propositions are inconsistent with the axiom of choice, with which you can construct the counterexamples that will not "nicely behave".

Fact: The latter proposition implies the former. (in ZF, with which AD is believed to be consistent).

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It seems a bit of a stretch to call AD "false." AC is not on any higher epistemological plane. – Kevin Carlson Jun 5 '14 at 20:51
How would you put "every set of real numbers has a measure" in layman's terms, so that an average person would claim it to be true? Be sure to include countable additivity in your layman's formulation, since that's needed to make the proposition false. – bof Jun 5 '14 at 22:23
@bof: The concept of measure originates from our intuitive notion of length/area/volume. I doubt that an average person would object to that. Pick a unit square and color a subset of it, it should have some "area" comparable to areas of basic shapes. Or, ask this, if I throw a dart to this square, do you think that there is a well-defined probability that I hit the colored region? As for the countable additivity, doesn't our intuition about "area" include this? If I give you disjoint pieces that you can label by naturals, of course the area of the union is the sum of the areas? – Burak Jun 5 '14 at 22:48
@KevinCarlson No stretch, but common usage. "False" (without qualifiers, as in "false in this or that model") is shorthand for "false under the standard interpretation", which in this case means under the standard set of axioms, namely $\mathsf{ZFC}$, including choice. – Andrés E. Caicedo Jun 6 '14 at 2:58

I think this is not covered in any of the other answers (although, to be sure, there are a lot of them). The Simpson's paradox one is close, but I think this is different and somewhat easier to understand:

If $X$ is positively correlated with $Y$, and $Y$ is positively correlated with $Z$, then $X$ is positively correlated with $Z$.

In other words, positive correlation is transitive. I think it's fairly intuitive, yet false.

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Before 1955 everyone “knew” that to know the nth decimal digit of $\pi$ (and for any other irrational) it was necessary to know the previous digits. A genius like Archimedes ("There was more imagination in the head of Archimedes than in that of Homer": Voltaire) "knew" very well this as History shows. However, The Bailey–Borwein–Plouffe formula (BBP formula) finished with this sacred for centuries “knowledge” and now it is possible to know, for example, the 33-th digit without to know the precedent ones.

Concerning the intuitive perception, it is false that a continuous numerical function must be derivable at least in one point; it is false too that a little square cannot contains a curve of infinite length.

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The BPP formula for pi doesn't give decimal digits, but hexadecimal digits. – Gerry Myerson Dec 5 '15 at 11:30

A subgroup of a finitely generated group may not be finitely generated and there are up to isomorphism at most two $pq$ groups, where $p$ and $q$ are prime.

To strengthen your example, $\sum_{p \ prime} 1/p$ diverges.

The continuum hypothesis also seems like it has to have in answer in ZFC, which it doesn't.

On another page, mathematicians thought that cyclotomic fields "obviously satisfy the unique factorization theorem" leading to some false proof attempts to Fermat's last theorem.

Next, one might think that the "angle trisection" is possible or that any set of analytic functions $\{f_\alpha\}$, such that for every $z \in \mathbb C$, the set $\{f_\alpha(z)\}$ is countable, itself has to be countable.

These are just some random examples that came to mind and since the term "obvious" is subjective, you may very well disagree with items on my list. I guess it heavily depends on your mathematical background.

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I disagree that there is anything intuitively obvious about the continuum hypothesis, and I find even less that is intuitively obvious about what can or cannot be proved in ZFC, which is so complex that you have to study really hard just to understand what the axioms are saying. – MJD Jun 4 '14 at 18:31

The following is a very well-known example, though probably slightly outside of the world of mathematics, rather physics. A great many people would 'intuitively' consider the following to be true:

The heavier the object, the faster it falls down.

In fact, the story goes that this was supposed to be common knowledge until Galileo Galilei disproved it (as the story goes, by dropping two balls of the tower in Pisa, which never happened though).

One of the first physics classes many people have (I'm talking primary school here) aims at showing this theorem is false, and actually everything falls with the same acceleration (ignoring resistance by air) regardless of weight.

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Everything falls with the same acceleration not at the same speed... big difference. – Dan Jun 5 '14 at 8:29
Aristotle's claim was actually that objects fall with speed proportional to their weight, which is much more specific, and so much less obvious-seeming. It is incredible that Western civilization swallowed this whopper for as long as it did. – MJD Jun 5 '14 at 11:47
@MJD Universally people tend to believe things according to popularity rather than evidence or experience. This is as true today as it was in the past, and studies show that (counterintuitively) well-educated people are better at doing it than poorly-educated people. – AndrewC Jun 5 '14 at 16:22
I make fun of this idea in my lessons when I've checked several students answers and pick the mode answer, and announce "This one must be right, because as we all know, democracy is the route to all truth!" – AndrewC Jun 5 '14 at 16:22
But a gas balloon is lighter (relative to its volume) than eg a rock and it even rises instead of falling down. – Robert Jun 8 '14 at 20:57

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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I think that the third one is neither obviously true nor obviously false. – Asaf Karagila Jun 4 '14 at 22:54
@AsafKaragila: I do too. I put it there because there is a famous anecdote about two mathematicians who believed it obvious, but on both sides. It was more to illustrate the point that many make the category mistake of thinking theorems "true or false" instead of "obtains in ambient system S". Truth or falsity, indeed even validity, are not really applicable when your ambient system cannot provide a semantics on which to make those claims. – ex0du5 Jun 4 '14 at 22:59
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 '14 at 23:02
If you believe that there is a set theoretic universe and first order logic captures its truth (meaning that axioms are true [whatever this means]), then we can interpret being provable/disprovable as being true/false. Why do we have to be able to define (formal) truth in the universe which we presume to exist? If talking about truth and falsity bothers you, you can re-read the question and all the answers (that you seem to claim are faulty) as "provably true" and "provably false". – Burak Jun 4 '14 at 23:04
One last quick remark: Formal truth in the set-theoretic universe is actually definable, but not uniformly. What this means is that if you put a bound on the complexity of your formulas (say in the Levy hierarchy), then you can define the truth for those formulas (using a single formula having greater complexity). Not being able to do this at a single stroke (with a single formula) should not make you think that mathematical truth does not make sense nor actually true/false are meaningless. – Burak Jun 4 '14 at 23:14

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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"Obviously" $$(x^y)^z = x^{y\cdot z}$$ for $x,y,z \in \mathbb{C}$ such that given expressions are defined.

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A linear order can be uniquely (up to isomorphism) reconstructed from the set of order types of its proper initial segments.

Update: Even if we know the cardinality of the linear order, and know that it does not have a maximal element, this "theorem" still does not hold.

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$\textbf{Counter-Intuitive Example}$

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

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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 '14 at 0:07
That's infinitely many terms. :) – Harshal Gajjar Jun 15 '14 at 4:05
Hey I don't think that's true because a geometric progression has infinite terms it still converges. – Jasser Oct 3 '14 at 13:55

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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It's not exactly a theorem, but it fools every math newcomer:

$e = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n$

$(1 + 1/\infty)$ is $1$, obviously. And 1 to the power of $\infty$ is obviously still 1.

Nope, it's 2.718...

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I believe the poster gave this example in the question, but it is a good one. – DanielV Jun 5 '14 at 13:26
Nope, $1^\infty$ is an indeterminate form. – Hakim Jun 5 '14 at 20:25

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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Have you stated the false "theorem", or is this a true counter-intuitive statement? (Clearly I'm not actually intuiting anything :-) ) – Mark Hurd Jun 12 '14 at 16:21
@MarkHurd It's a "false theorem". The pure implicational calculus has a 13 letter single axiom, but the single axioms for say BCI are longer. – Doug Spoonwood Jun 12 '14 at 16:31
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 '14 at 17:46
@rumtscho I don't work in academia. I also only took one logic course in college, and 2 calculus courses. – Doug Spoonwood Jun 13 '14 at 23:21

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