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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.

Thanks!

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    $\begingroup$ related: mathoverflow.net/questions/23478/… $\endgroup$
    – Will Jagy
    Commented Jun 4, 2014 at 18:29
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    $\begingroup$ another one: mathoverflow.net/questions/35468/… $\endgroup$
    – Will Jagy
    Commented Jun 4, 2014 at 18:33
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    $\begingroup$ 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. $\endgroup$ Commented Jun 4, 2014 at 19:53
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    $\begingroup$ 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?) $\endgroup$
    – user98602
    Commented Jun 6, 2014 at 7:04
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    $\begingroup$ "Every true statement can be proved" $\endgroup$
    – sinelaw
    Commented Jun 9, 2014 at 17:27

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Another example of 'obvious' being not true is Bus Waiting Time Paradox.

If the mean time between two consecutive buses arriving to a bus station is $M$, one should expect that the mean time you have to be waiting in the station before the next bus arrives is also M. But this is not true; and depending on the particular bus arrival time distribution you'll have to wait a time $M'\geqslant M$

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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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    $\begingroup$ The question asks for obvious but false statements. Are you seriously claiming that $\pi(x) < \operatorname{li}(x)$ is obvious? $\endgroup$
    – MJD
    Commented Jun 5, 2014 at 21:32
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    $\begingroup$ hard to tell whether that is supposed to be obvious... What are $\pi(x)$ and $\operatorname{li}(x)$? $\endgroup$
    – example
    Commented Jun 5, 2014 at 22:03
  • $\begingroup$ $\pi(x)$ is the prime counting function, $\mathrm{li}(x)$ is the logarithmic integral. Ramanujan asserted the inequality. Littlewood proved in 1914 the sign changed infinitely often and Skewes gave an upper bound for the first change. $\endgroup$
    – Henry
    Commented Feb 7, 2021 at 18:45
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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.

EDIT

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.

Syntax

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.

Semantics

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...

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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  • $\begingroup$ I think that the third one is neither obviously true nor obviously false. $\endgroup$
    – Asaf Karagila
    Commented Jun 4, 2014 at 22:54
  • $\begingroup$ @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. $\endgroup$
    – ex0du5
    Commented Jun 4, 2014 at 22:59
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    $\begingroup$ 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? $\endgroup$
    – Burak
    Commented Jun 4, 2014 at 23:02
  • $\begingroup$ 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". $\endgroup$
    – Burak
    Commented Jun 4, 2014 at 23:04
  • $\begingroup$ 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. $\endgroup$
    – Burak
    Commented Jun 4, 2014 at 23:14
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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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  • $\begingroup$ But it is fairly obvious and it is true. $\endgroup$
    – bjb568
    Commented Jun 9, 2014 at 1:48
  • $\begingroup$ @bjb568 I think Robert may be thinking of non-planar triangles. $\endgroup$
    – MJD
    Commented Jun 9, 2014 at 5:16
  • $\begingroup$ @MJD Well, that makes more sense, but then it isn't obvious… $\endgroup$
    – bjb568
    Commented Jun 9, 2014 at 6:05
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    $\begingroup$ It doesn't seem obvious to me that the angles of a plane triangle always have the same sum. In fact, it seems astounding, and I'm sure you could find other places on this web site where other people said it seemed astounding. $\endgroup$
    – MJD
    Commented Jun 9, 2014 at 14:14
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    $\begingroup$ For example, this highly-voted answer begins “I found it completely amazing that the angles in a triangle always added up to 180 degrees”. $\endgroup$
    – MJD
    Commented Jun 9, 2014 at 21:22
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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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  • $\begingroup$ Have you stated the false "theorem", or is this a true counter-intuitive statement? (Clearly I'm not actually intuiting anything :-) ) $\endgroup$
    – Mark Hurd
    Commented Jun 12, 2014 at 16:21
  • $\begingroup$ @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. $\endgroup$ Commented Jun 12, 2014 at 16:31
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    $\begingroup$ Good luck finding a layman whose eyes don't glaze at the term "propositional calculus", let alone one who finds something here obvious. $\endgroup$
    – rumtscho
    Commented Jun 13, 2014 at 17:46
  • $\begingroup$ @rumtscho I don't work in academia. I also only took one logic course in college, and 2 calculus courses. $\endgroup$ Commented Jun 13, 2014 at 23:21
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Perhaps this isn't quite what you were looking for, but it's still fun! How about a proof that is obviously incorrect but (to newcomers) it is difficult to figure out what is wrong.

Let $x = y$. Then $$x^2 = xy$$ $$x^2-y^2=xy-y^2$$ $$(x+y)(x-y)=y(x-y)$$ $$x+y = y$$ $$2y=y$$ $$2=1$$

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    $\begingroup$ yeah, you can "prove" all sort of things when you divide by 0 ... ;) $\endgroup$
    – scibuff
    Commented Feb 7, 2021 at 22:13
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An "obviously true" theorem:

If take a 3d object $U$ and chop it into finitely many pieces, any object $V$ I rearrange those pieces into will have the same volume as the object $U$ I started with.

But in fact, the Banach-Tarski paradox tells us this isn't true -- if we construct our finitely-many subsets of $U$ "weirdly" enough, we can actually build a $V$ with any volume we'd like.

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Any continuous function is differentiable at least somewhere, right?

False, the Wierstrass function is a famous counterexample

https://en.wikipedia.org/wiki/Weierstrass_function

It is continous everywhere, yet it is differentiable nowhere.

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The Freshman's dream: $$(x+y)^p=x^p+y^p$$.

Obviously false. But true in characteristic $p$.

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An analytic function with compact support vanishes identically.

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    $\begingroup$ That is actually true, as long as we stay in $\mathbb{C}^n$ resp. $\mathbb{R}^n$. It becomes false when one takes a compact analytic manifold, but then it's obvious that it's false. $\endgroup$ Commented Jun 4, 2014 at 19:45
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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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    $\begingroup$ 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. $\endgroup$ Commented Jun 6, 2014 at 2:54
  • $\begingroup$ That depends on the topology, but it might be that the argument I'm thinking of used the space of all sequences (both finite and infinite) of natural numbers or developed the homeomorphism with the exclusion of $\mathbb Q$. $\endgroup$
    – jjfunk
    Commented Jun 6, 2014 at 4:04
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    $\begingroup$ "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. $\endgroup$ Commented Jun 6, 2014 at 4:51
  • $\begingroup$ @jjfunk: Is it possible that the map you are talking about is the homeomorphism between the Baire space and the irrationals given by infinite continued fractions? $\endgroup$
    – Burak
    Commented Apr 9, 2016 at 0:58
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Let $n_1$ and $n_2$ be positive integers. Suppose $A\subset \mathbb{N}$ with $\vert A\vert = n_1 n_2,$ such that,

for each $0\leq j\leq n_1-1,\quad \vert\ \{\ a \in A: a\equiv j \pmod {n_1}\ \}\ \vert = n_2,\ $ and

for each $0\leq k\leq n_2-1,\quad \vert\ \{\ a \in A: a\equiv k \pmod {n_2}\ \}\ \vert = n_1.$

Then, it seems "obvious/trivial" that $\ \{\ a \pmod {n_1 n_2}: a\in A \} = [ n_1 n_2]:= \{ 0, 1, \ldots, n_1 n_2 - 1\},\ $ or at least trivial if $n_1$ and $n_2$ are coprime.

For example, take $n_1 = 2, n_2 = 3.$ Suppose that $a_1, a_2,\ldots, a_6$ are integers such that :

For each $i=0,1,2,\ $ two of the $a_k,\ 1\leq k\leq 6$ are congruent to $i \pmod 3.$

And for each $i=0,1,\ $ three of the $a_k,\ 1\leq k\leq 6$ are congruent to $i \pmod 2.$

The proposition is that this is enough to imply that $\ \{\ a \pmod 6: a\in A \} = \{ 0, 1, \ldots, 5\}. $

However, $\{ 0,1,2,3,7,8\}\ $ is a counterexample.

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When I first learnt elementary calculus, I found it surprising that the harmonic series $1+1/2+1/3+\cdots$ diverges, since previously I believed that

If $(a_n)$ in a sequence of positive real numbers such that $a_n\to0$ as $n\to\infty$, then $\sum_{n=1}^{\infty}a_n$ is convergent.

The condition that $a_n\to0$ is a necessary, but not sufficient, condition for the series to converge.

A similar false belief is that

If $f$ is an increasing differentiable function such that $f'(x)\to0$ as $x\to\infty$, then $f$ is bounded above.

The counterexample $f(x)=\log x$ is notable because it can be used to be used to prove the divergence of the harmonic series, and vice versa.

In fact, the converse of the second statement is also not true: that is, if $f$ is an increasing differentiable function, and $f$ is bounded above, then it does not follow that $f'(x)\to0$ as $x\to\infty$. See here.

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