# 'Obvious' theorems that are actually false

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.

Thanks!

• – Will Jagy Jun 4 '14 at 18:29
• another one: mathoverflow.net/questions/35468/… – Will Jagy Jun 4 '14 at 18:33
• 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?) – user98602 Jun 6 '14 at 7:04
• "Every true statement can be proved" – sinelaw Jun 9 '14 at 17:27

## 70 Answers

In my opinion, the most interesting (but also sometimes not intuitive) results in mathematics are those that state a theorem that ends up being false because it actually holds in many cases, except for very few or very strange cases. In other words, the most "obvious" false theorems to me are those that have very difficult counterexamples.

Some examples:

• Banach-Tarski: There exists a strict subset $A$ of the Euclidean $n$-ball $B$ such that one can partition $A$ and $B$ into an equal number of further subsets that can be mapped to each other by isometries. This shows that not all sets are measurable, and that it's possible to perform partitions that do not preserve measure.

• Non-finiteness of differentiable structures: For $\mathbb{R}^n$ with $n = 4$, there are an uncountable number of distinct differentiable structures.

• Divergence of Fourier series: There exists an integrable function on $[-\pi, \pi]$ whose Fourier series diverges everywhere. This is extremely unusual because for any typical function we might write down, usually its Fourier series might diverge at one or a finite number of points, but will probably converge everywhere else.

• This makes me think of: a subset of $\mathbb{R}^n$ is either closed or open (but not both). Which of course is not true because the empty set is open and closed. (But also, I don't think $\mathbb{Q}$ is either closed or open in $\mathbb{R}$). – Ruben Jun 4 '14 at 19:36
• You don't need a weird example to find a subset of $\Bbb R^n$ that is neither open nor closed; just take a closed ball, and remove a boundary point; or take a sequence of points that converges to a limit, but omit the limit. The terminology may lead one to think that sets are either open or closed, but there is nothing about the actual concepts that would make it appear so. – MJD Jun 4 '14 at 20:07
• What is an example of function with everywhere divergent Fourier series? – Ruslan Jun 5 '14 at 11:59
• @Ruslan, The example is nontrivial and is due to Kolmogorov. An earlier, almost-everywhere divergent construction was one of Kolmogorov's first claims to fame. You'll have to consult Google on this one, I do not know the example. – Christopher A. Wong Jun 5 '14 at 19:14
• @Ruslan Google leads us back here. – Mark Hurd Jun 6 '14 at 9:44

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.)

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!

• 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

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.

• For what sorts of manifolds does it fail? – MJD Jun 5 '14 at 20:39
• @MJD As I recall, there is an invariant of the fundamental group (called the Whitehead group) which is defined in terms of matrices over the group ring, and this needs to be nonzero for the manifold to stand a chance of being a counterexample. This rules out a lot of fundamental groups, including the trivial group and $\mathbb Z$. It's been a while since I've thought about this. – Cheerful Parsnip Jun 5 '14 at 20:46

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 its 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.

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.

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

• 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

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.

• It is true, however, that if both $X$ and $Z$ are positively and perfectly correlated with $Y$, then $X$ is positively and perfectly correlated with $Z$. – user1551 Aug 10 '17 at 17:00

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

• Someone needs to tell this to the authors of college-level remedial elementary-algebra textbooks. These consistently state that as a theorem when x is real and y and z are rational. – Toby Bartels Jan 12 '19 at 11:29

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.

• 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

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.

• 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

• A related non-theorem that my students want to believe is that $\sqrt{a^2}=a$. (Then again, I have students who want to believe that $\sqrt{a+b}=\sqrt a+\sqrt b$.) – Toby Bartels Jan 12 '19 at 5:09

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).

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...

• Are you sure? I think, I can prove that such g always exists. – Moishe Kohan 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
• I'd agree w/ @Alexey. Add a curve from $(1,1)$ to $(0,0)$ to extend $f$ to a closed curve $\tilde{f}$ that $(0,1)$ is inside and $(1,0)$ is outside. Find a homeomorphism $\Phi:\mathbb{R}^2\rightarrow\mathbb{R}^2$ of the plane that maps $\tilde{f}$ to the unit circle (by the Jordan-Schoenflies theorem). Draw a straight segment connecting the images of $(0,1)$ and $(1,0)$ (this crosses the unit circle exactly once). The pre-image of this segment (i.e. its image under $\Phi^{-1}$) is the desired curve $g$. (Right?) – mjqxxxx Jun 4 '14 at 20:51
• @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. – Moishe Kohan 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. – Moishe Kohan 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 $A$.

• 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

### 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 $\frac{1}{12}$.

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.

• The BPP formula for pi doesn't give decimal digits, but hexadecimal digits. – Gerry Myerson Dec 5 '15 at 11:30
• I think "1955" here must be an error. – MJD Nov 8 '19 at 10:14
• The idea that "a little square cannot contain a curve of infinite length" becomes less "obvious" when people realize that each tiny cell in their bodies contains very long strands of DNA. (Of course the DNA isn't quite infinitely long and the cell is 3-dimensional, so this isn't a disproof of the "obvious" idea, but it seems to ruin the obviousness.) – Andreas Blass Sep 5 '20 at 15:00

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 $K_{1,3}$.

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.

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.

• 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

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)
• – 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
• @MJD I don't see the essence of your point. Part of the halting problem is if a program doesn't halt, how will we know that? We can't just "try it and see"--how long do we wait before deciding it doesn't halt? This is the essential problem. This kind of statement may have been more "obvious" at the beginning of the 20th Century when automation seemed to be the name of the game, also reflected in the goals of Russell and Whitehead in formalizing mathematics. It's great that it is obviously false to you now, but I would say it's because of results like this that we are now so enlightened. – Matt Jun 5 '14 at 17:15
• 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

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!

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.

• 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

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...

• 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

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).

• It seems a bit of a stretch to call AD "false." AC is not on any higher epistemological plane. – Kevin Arlin 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

$\textbf{Counter-Intuitive Example}$

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

• By $\mathbf{D}_v$ do you mean $f^{(v)}$, as in the $v$-th derivative? – Mr Pie Feb 29 '20 at 1:26

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.

All infinities are of the same size.

But Cantor's theorem shows otherwise.

Infinitely many terms always have a sum equal to infinity.

• 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

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.