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

70 Answers 70

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Theorem (false):

One can arbitrarily rearrange the terms in a convergent series without changing its value.

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    $\begingroup$ The series $\sum \frac{(-1)^i}{i}$ is a counterexample; see this Wikipedia article for a discussion. $\endgroup$
    – MJD
    Jun 4, 2014 at 18:39
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    $\begingroup$ @alexqwx Roughly, that some terms can be pushed 'infinitely far away', so that any two finite partial sums differ by arbitrarily many terms. Imagine e.g. taking the positive numbers and rearranging them as 1, 2, 3, 5, 4, 7, 9, 11, 6, 13, 15, etc. with a growing number of odd terms between each two consecutive even terms; then every number still shows up in the sequence eventually, but that doesn't mean that any concept like a 'ratio' of odd-to-even (e.g. natural density) is preserved. $\endgroup$ Jun 4, 2014 at 18:55
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    $\begingroup$ Huh? One can rearrange the terms in any series, convergent or otherwise. Did you leave something out of the statement of your theorem? $\endgroup$
    – bof
    Jun 4, 2014 at 21:25
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    $\begingroup$ Actually, you can rearrange terms in a convergent series when computing its value (or otherwise). You can even rearrange terms without changing the value, with a bit of care. What you cannot, is assume that any rearrangement will preserve convergence, or when it does preserve the value converged to. $\endgroup$ Jun 4, 2014 at 22:42
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    $\begingroup$ @kutschkem You can always make finitely many swaps of terms (using commutativity) without changing the value, but you can't make infinitely many swaps; continuity isn't preserved across an infinite number of operations. $\endgroup$ Jun 5, 2014 at 21:36
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A shape with finite volume must have finite surface area.

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    $\begingroup$ This is evidently false. Take a clay cylinder of diameter 1. Roll it in your hands so that its diameter becomes $\frac 12$. The volume is the same, but it is now four times as long as it was before, so it has twice the surface area of the original cylinder. (Half the circumference, but four times the length.) Roll it some more, and the surface area increases again. By making the snake very long and thin, you can increase the surface area to infinity while the volume remains constant. This is not only obvious, it's commonplace. $\endgroup$
    – MJD
    Jun 4, 2014 at 20:31
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    $\begingroup$ Those are good examples. I know this wouldn't confuse anyone formally trained like most of the posters here, but I was trying to suggest an example for "layman's terms, the average person" as OP requested. It seemed most of the examples so far would not even be interpretable to an untrained individual. $\endgroup$
    – DanielV
    Jun 4, 2014 at 20:41
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    $\begingroup$ @MJD: your cylinder example is nice, but it only shows that the surface area can be arbitrarily large. The OP is pointing out that the surface area can literally be infinite. $\endgroup$ Jun 5, 2014 at 0:00
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    $\begingroup$ I dunno either, I'm a physics ignoramus. Are you sure clay is made of atoms? $\endgroup$
    – MJD
    Jun 5, 2014 at 1:11
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    $\begingroup$ Mathematically this may be true, but physically impossible, as matter is discrete. The limit would be a row of atoms. $\endgroup$
    – Alex
    Jun 5, 2014 at 4:54
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I wish I'd thought of this yesterday, when the question was fresh, because it's astounding. Suppose $A$ and $B$ are playing the following game: $A$ chooses two different numbers, via some method not known to $B$, writes them on slips of paper, and puts the slips in a hat.

$B$ draws one of the slips at random and examines its number. She then predicts whether it is the larger of the two numbers.

If $B$ simply flips a coin to decide her prediction, she will be correct half the time.

Obviously, there is no method that can do better than the coin flip.

But there is such a method, described in Thomas M. CoverPick the largest numberOpen Problems in Communication and Computation Springer-Verlag, 1987, p152.

which I described briefly here, and in detail here.

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    $\begingroup$ Who said anything about a uniform distribution on $\Bbb N$? Certainly not I. $\endgroup$
    – MJD
    Jun 5, 2014 at 13:17
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    $\begingroup$ I don't think it's obvious that no such strategy exist, in fact I think most laypeople would intuitively apply it (in a heuristic form, i.e. "does that number sound big?" in the sense of "would I have been likely to choose two numbers and find this to be the larger one"). The thing that's obvious, and indeed correct, is that the strategy can easily be subverted by person A, by choosing both numbers very big and close to each other. $\endgroup$ Jun 6, 2014 at 10:59
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    $\begingroup$ This can be explained very simply. B guesses A's strategy. If she's wrong, her odds are 1/2. If she's right, her odds are better than 1/2. As long as her odds of guessing A's strategy are better than 0, her odds of winning are better than 1/2. $\endgroup$ Jun 10, 2014 at 19:54
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    $\begingroup$ @AleksandrDubinsky I think you are wrong. A's actual method is to choose from N(0,1). B guesses that A's method is to choose from N(3,1). B will lose a lot more than 50%. $\endgroup$ Jun 11, 2014 at 6:54
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    $\begingroup$ I may have been wrong. There are some guesses that can ruin B's chances. E.g., A's strategy is to make the high number even, and the low odd. B guesses A's strategy is the reverse and always loses. If we average out the chance of A deciding high=even and him choosing high=odd, B is back to 50/50. In some guessing strategies, being wrong hurts. MJD's proposed guessing strategy is great because it can never be to B's disadvantage. I wonder if it's unique, or if there are others with this property. $\endgroup$ Jun 11, 2014 at 17:28
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This is elementary compared to most of the other examples, but how about

There are more rational numbers than there are integers.

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    $\begingroup$ The beauty lies in it's simplicity. The falseness of this one of those very elementary observations the part of my brain that is driven by common sense still refuses to accept. $\endgroup$
    – aRestless
    Jun 5, 2014 at 23:11
  • $\begingroup$ My brain in general refuses to accept this. :) Seems like even if you restrict $b$ to $\{1, a+1\}$, $a/b$ gives you twice as many values as $a$. $\endgroup$
    – cHao
    Jun 5, 2014 at 23:29
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    $\begingroup$ And there are half as many odd integers than (odd or even) integers. $\endgroup$
    – gnasher729
    Jun 6, 2014 at 7:30
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    $\begingroup$ @Jori: I haven’t studied the proofs and terminology in years, but I think a somewhat simple (simplistic?) way of putting it is that the existence of an injection mapping (not one-to-one) does not preclude the existence of a one-to-one mapping. P.S. Some people might consider 2∞>∞ to be obviously true. $\endgroup$
    – Scott
    Jun 7, 2014 at 20:08
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    $\begingroup$ @Jori: (2) No! As Cornelius implied in his comment, Cantor’s diagonal argument (correct link) shows that there are infinite sets (such as the set of real numbers) that cannot be put into one-to-one correspondence with the (infinite) set of natural numbers. There are (infinitely many!) different transfinite cardinals. Crudely put, not all infinities are equal. $\endgroup$
    – Scott
    Jun 9, 2014 at 14:43
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I keep harping on this, because I think it's a spectacular example of something that can be demonstrated to be completely obvious (not only because it seems so, but because it was so widely believed for so long) and yet is completely wrong:

Suppose $\Phi$ is a property that might or might not hold of some object. Then there is a collection $S_\Phi$ of all objects with property $\Phi$.

Many serious and even famous mathematicians went ahead with this intuitively obvious but utterly false principle, whose demolition shook mathematics to its foundations and marks the beginning of modern logic and set theory.

(There are many counterexamples, of which the most well known is $\Phi(x) = $ “$x$ is not a member of collection $x$”. For others, see Is the Russell paradox the only possible contradiction to the axiom schema of comprehension due to Frege (1893)? $\{x:P(x)\}$ and Paradox of General Comprehension in Set Theory, other than Russell's Paradox.)

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    $\begingroup$ +1 for a great example, but I do think the use of "properties", "objects", and "collections" takes away from the impact by being imprecise. There are definitely consistent set theories where there are collections which are not objects and the statement is true in some sense. Stating it using the word sets and in the context of traditional naive set theory and modern widely-used formalizations of set theory would make this answer better in my opinion. $\endgroup$ Jun 4, 2014 at 22:39
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    $\begingroup$ Why is this wrong? $\endgroup$
    – seldon
    Jun 5, 2014 at 6:45
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    $\begingroup$ I disagree, I think this works quite fine with undefined terms as long as you have a notion of containment, you don't need anything else. $\endgroup$ Jun 5, 2014 at 6:52
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    $\begingroup$ I really think that is precisely not the problem. The problem here has nothing to do with ZF, with axioms, or with how we exactly define "objects"; it is a fundamental problem with the intuitive notion of what it means for things to have properties. How do you feel about this formulation: “For any property $\Psi$, one can construct a catalog that lists all the books with property $\Psi$?” And the answer is, that for some properties $\Psi$, you simply cannot. Is that sufficiently concrete and not-set-theoretic? $\endgroup$
    – MJD
    Jun 5, 2014 at 15:22
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    $\begingroup$ This is Russell's Paradox, discovered in 1901, which destroyed Frege's Grundgesetze der Arithmetik (1903) and prior work. The immediate responses to the paradox include Whitehead and Russell's Principia Mathematica (1910) and Zermelo's work on axiomatic set theory starting in 1905, which eventually became ZFC, which dominates set theory to this day. $\endgroup$
    – MJD
    Jun 8, 2014 at 15:11
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In a related mathOverflow thread, Gowers pointed out the following obvious but false claim:

Let $I_1, I_2, \ldots$ be subintervals of $[0,1]$ whose total length is strictly less than 1. Then the union of the $I_i$ cannot contain $\Bbb Q\cap [0,1]$.

(Note that if $\Bbb Q$ is replaced with $\Bbb R$, the claim is true.)

I find the fact that all of $\Bbb Q$ can be covered by an arbitrarily small family of intervals to be one of the most bizarrely counterintuitive in all of mathematics.

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    $\begingroup$ You order the rationals $q_1, q_2, \ldots, q_n, \ldots$ and place an interval of $(\frac \epsilon 2)^n$ around each one, choosing $\epsilon$ to be less than $1$. $\endgroup$
    – dfan
    Jun 4, 2014 at 19:37
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    $\begingroup$ Of course the theorem is still true if we have a finite number of intervals; one needs the infinite construction to make this work, which is likely where the gap in intuition lies. $\endgroup$ Jun 4, 2014 at 19:48
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    $\begingroup$ @Cruncher Yes it works for any $\epsilon < 1$ since $\sum_{i=1}^\infty \left(\frac{\epsilon}{2}\right)^n < 1$. And yes he's referring to the rationals in $[0, 1]$ though if you think about the argument, you'll see that's not strictly necessary, he could cover all the rationals with finite length. Finally the statement wasn't that the intervals would be equal to $Q\cap [0, 1]$, but that they contain it. $\endgroup$
    – Polymer
    Jun 5, 2014 at 18:00
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    $\begingroup$ @Cruncher You are right, that's an issue because it does say "subitntervals" in the answer. So two things: 1) I believe that contstruction can be fixed so that our intervals go outside of $[0, 1]$; for example (this needs a little checking though, it could be impossible to do that) if some of your sub-intervals lie outside $[0, 1]$, just keep shrinking $\epsilon$ until every sub-interval is in $[0, 1]$. 2) Even if you can't do that, it's still amazing that you can cover the rationals with intervals whose sum is arbritrarily small, wether you can fit all the sub-intervals into $[0, 1]$ or not. $\endgroup$
    – Ovi
    Dec 20, 2018 at 13:44
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    $\begingroup$ @Ovi : The subintervals don't need to be centred on the rationals, so you don't need to shrink ε, just take the intersection with [0,1]. $\endgroup$ Jan 12, 2019 at 3:19
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If a function $f(x)$ has an horizontal asymptote, then $\lim f'(x) = 0$

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    $\begingroup$ +1 Love this one! A counterexample is $\frac{\sin{x^2}}{x}$. $\endgroup$ Jun 5, 2014 at 10:30
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    $\begingroup$ A sufficient condition for this to be true is for the second derivative to be bounded. $\endgroup$ Jun 5, 2014 at 12:38
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    $\begingroup$ @A.P. or even just that $\lim f'(x)$ exists. $\endgroup$
    – Ant
    Jun 5, 2014 at 17:50
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    $\begingroup$ As an undergraduate I attempted to write an $\varepsilon$-$\delta$ proof that if $f'(a)>0$ then there is some open interval containing $a$ on which $f$ is increasing. And I thought my ability to write $\varepsilon$-$\delta$ proofs was missing something because I couldn't figure out how to do it. $\endgroup$ Jun 5, 2014 at 20:41
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    $\begingroup$ Ah, we have almost-duplicate responses. The interesting thing is that you can add the following restrictions to make it even more bizarre: $f$ is $C^\infty$, bounded, and monotonic. $\endgroup$
    – Kaj Hansen
    Jun 6, 2014 at 9:28
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$ 0.\overline{9} < 1 $

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

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    $\begingroup$ The question did ask what the average person would consider true. $\endgroup$ Jun 9, 2014 at 7:56
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    $\begingroup$ This is by far the best one on the page imo. Some of the others are so complex to the average person that they wouldn't be "obviously" true to almost anyone. $\endgroup$ Jun 10, 2014 at 20:20
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    $\begingroup$ In real numbers it is false, but it is true for the hyperreals. It's only by convention that we default to using the theory of real numbers instead of some other system. $\endgroup$
    – M.M
    Jun 12, 2014 at 3:31
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    $\begingroup$ Will not considering any type of numbers 0.999... straight up is a number but this "obvious" is actually true, let me explain. 0.999... is equal to 1 - 1/∞. Which makes a weird type of infinite which is infitly close to one. $\endgroup$ Jul 23, 2014 at 20:06
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    $\begingroup$ I believed this until I was $9.\overline9$ years old. On my $10\,\text{th}$ birthday, I saw the light.    :-)    ⁠ $\endgroup$
    – Scott
    Dec 4, 2016 at 8:34
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The falseness of

Let $S$ be an infinite family of strictly positive numbers. Then $\sum S = \infty$

has been boggling people for thousands of years. It is the basis for Zeno's paradox, but if you think Zeno's paradox is old and tired, consider that it is also the basis for the Gabriel's Horn paradox (also mentioned in this thread), which still puzzles people.

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    $\begingroup$ @jpmc26 It does always get bigger as you keep adding numbers; it just doesn't necessarily do so without limit. $\endgroup$
    – Mike Scott
    Jun 5, 2014 at 7:26
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    $\begingroup$ On the other hand if the family is uncountable, it holds. $\endgroup$
    – user87690
    Jun 5, 2014 at 13:27
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    $\begingroup$ I just read this one and the comments... and it's really freaky. (I have a master's in physics, so I'm not entirely unfamiliar with these sorts of things.) $\endgroup$
    – Almo
    Jun 5, 2014 at 17:31
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    $\begingroup$ I feel like this isn't that crazy, though maybe that's because I have experience, but I mean $1.1111\cdots=1+0.1+0.01+\cdots$. $\endgroup$
    – JLA
    Jun 5, 2014 at 21:10
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    $\begingroup$ Ooooh! I read it as "strictly positive integers" which I believe would diverge. Right? (Too much programming) $\endgroup$
    – Almo
    Jun 5, 2014 at 22:30
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Every chain of subsets of $\mathbb N$ is countable.

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    $\begingroup$ So far I find this to be the most true for me. For the life of me I still cannot give myself a good metaphor that make it obviously false, even though I know mathematical counterexample. Other answers, either it never found them obvious, or if I did, I can quickly fix the misconception by an alternative way to look at it. You probably should add in counterexample for this one in the answer. $\endgroup$
    – Gina
    Jun 9, 2014 at 2:30
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    $\begingroup$ By "chain", do we mean a collection of subsets of N that is totally ordered by the subset relation? You can represent the interval $[0, 1]$ as such a chain. For each number $x$ in that interval, the subset contains (rounding down) the first $9x$ one-digit numbers, the first $90x$ two-digit numbers, the first $900x$ three-digit numbers, and so on. $\endgroup$ Jun 9, 2014 at 4:15
  • $\begingroup$ @TannerSwett That is the meaning, and yours is a good example. $\endgroup$
    – bof
    Jun 9, 2014 at 5:01
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    $\begingroup$ @TannerSwett More straightforwardly, consider Dedekind cuts of rational numbers. $\endgroup$ Jun 12, 2014 at 3:56
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    $\begingroup$ @DustanLevenstein Sure. Or, let $\mathbb Z[i]$ be the set of all Gaussian integers, and let $S_\alpha=\{z\in\mathbb Z[i]:0\lt arg(z)\lt\alpha\}$. $\endgroup$
    – bof
    Jun 12, 2014 at 4:12
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Keller's conjecture is obviously true:

Let $\Bbb R^n$ be completely covered with identical, non-overlapping $n$-cubes. There must be two cubes that share a face.

(For example, when $n=2$ we cover the plane with little square tiles, and the conjecture states that there must be two tiles that share an edge. This is true.)

However, the conjecture is false for all $n>7$.

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    $\begingroup$ I'm not sure I buy this example myself, because I'm not sure anything concerning 7-dimensional anythings can be considered intuitively obvious. $\endgroup$
    – MJD
    Jun 4, 2014 at 18:55
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    $\begingroup$ Wouldn't the fact that it's true for $n = 1, 2, 3$ (and $n = 4, 5, 6$) lead you to suspect that it might be true for all $n$? $\endgroup$ Jun 5, 2014 at 4:47
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    $\begingroup$ What does "face" even mean in 7 dimensions? Is it a 6-dimensional thing or a 2-dimensional thing? $\endgroup$
    – user541686
    Jun 5, 2014 at 7:04
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    $\begingroup$ What precisely do we mean by "share a face" in this context? $\endgroup$ Jun 5, 2014 at 10:20
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    $\begingroup$ what do you mean by "tiling"? $\endgroup$
    – electronp
    Jun 5, 2014 at 12:04
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  • If $U$ is an open subset of $\mathbb{R}^n$ that is homeomorphic to $\mathbb{R}^n$, one might think it "obvious" that it's in fact diffeomorphic to $\mathbb{R}^n$ (perhaps thinking something like "topologically it looks like $\mathbb{R}^n$, and differentiably it's locally trivial"). In fact, this is true (but by no means obvious!) for $n\neq 4$. But for $n=4$ it is false: there exist exotic $\mathbb{R}^4$'s (differentiable manifolds that are homeomorphic, but not diffeomorphic, to $\mathbb{R}^4$), including "small" ones which are diffeomorphic to an open subset of $\mathbb{R}^4$.

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

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    $\begingroup$ $20/3$? What kind of magic is that!? +1 interesting example. $\endgroup$ Jun 5, 2014 at 23:24
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$i^i$ is imaginary.$\ \ \ \ \ \ $

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    $\begingroup$ Believing $i^i$ is well-defined is illusory. $\endgroup$ Jun 6, 2014 at 13:26
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    $\begingroup$ I'm certainly not saying anything profound. One simply should not write down things that look like constant expressions but aren't. $\endgroup$ Jun 6, 2014 at 13:40
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    $\begingroup$ None of the infinitely many values for $i^i$ are imaginary :) $\endgroup$ Jun 7, 2014 at 0:10
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    $\begingroup$ @bof: Yes. And to say that "$i^i$ is imaginary" is false even when taking the multivalued nature into account. That's what I meant. $\endgroup$ Jun 7, 2014 at 3:04
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    $\begingroup$ @anorton Depends on how you define "imaginary". The usual definition is "having real part 0". and 0 certainly has real part 0. $\endgroup$
    – Snowbody
    Jun 9, 2014 at 16:50
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This part is true (Jordan-Brouwer separation theorem):

(a) Any imbedding of the $2$-sphere into $3$-dimensional Euclidean space separates the space into two disjoint regions.

But this part, which would seem to be a natural generalization of the Jordan-Schönflies Curve Theorem, is not true:

(b) The regions are homeomorphic to the inside and outside of the unit sphere.

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    $\begingroup$ I was thinking of this example (the horned sphere) myself. My answer can be viewed as a sort of 2d analog in some ways... $\endgroup$ Jun 4, 2014 at 18:37
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    $\begingroup$ This is the first result in this thread I actually find unintuitive. What the heck? $\endgroup$ May 22, 2019 at 17:44
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I really like "wrong proofs" as typically the insight why the proof is wrong gives you some understanding of the topic. One very simple version is this one, which I threw at my first semesters when I was a tutor:

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

"Proof":

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

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

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

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  • $\begingroup$ Where is the flaw in this argument? $\endgroup$
    – beep-boop
    Jun 5, 2014 at 22:33
  • $\begingroup$ @user21820 Sorry, for formulating too ambiguous. I edited and it should be clear now what's meant. $\endgroup$
    – aRestless
    Jun 6, 2014 at 10:39
  • $\begingroup$ Sorry I misread. $\endgroup$
    – user21820
    Jun 6, 2014 at 10:58
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    $\begingroup$ A relation is not simply "reflexive"; rather it is reflexive on some set. The relation $\sim$ on $\{1,2,3,4\}$ where by $3\sim3$, $4\sim4$, $3\sim4$, and $4\sim3$, and nothing else is related, is not reflexive on the set $\{1,2,3,4\}$, but there is indeed a set on which it is reflexive. $\endgroup$ Jun 13, 2014 at 18:18
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    $\begingroup$ The symmetric transitive relations comprise a very important class of relations, sometimes called the partial equivalence relations. Just as you can mod out by an equivalence relation to get a quotient, so you can restrict to and mod out by a partial equivalence relation to get a subquotient, that is a subset of a quotient (or equivalently a quotient of a subset). … $\endgroup$ Jan 12, 2019 at 3:54
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Theorems that are intuitively true, but actually flawed:

  • There is no continuous, nowhere-differentiable real function.

  • There is no real function that is differentiable and not monotonic on any non-trivial interval.

  • If a real function satisfies $\forall x, y, f(x+y) =f(x) +f(y) $, it is of the form $x\to ax$.

  • Infinite sums and integrals can be swapped anytime.

  • A connected metric space is path-connected.

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

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

The following is from W. Fellers classic of Introduction to Probability Theory and It's Applications, Vol 1:

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

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

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

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

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

and later on:

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

[2015-07-16] According to a comment from @HenningMakholm some examples exposing striking aspects.

  • Suppose that a great many coin-tossing games are conducted simultaneously at the rate of one per second, day and night, for a whole year. On the average, in one out of ten games the last equalization will occur before $9$ days have passed, and the lead will not change during the following 356 days. In one out of twenty cases the last equalization takes place within $2\frac{1}{2}$ days, and in one out of a hundred cases it occurs within the first $2$ hours and $10$ minutes.

  • Suppose that in a learning experiment lasting one year a child was consistently lagging except, perhaps, during the initial week. Another child was consistently ahead except, perhaps, during the last week. Would the two children be judged equal? Yet, let a group of $11$ children be exposed to a similar learning experiment involving no intelligence but only chance. One among the $11$ would appear as leader for all but one week, another as laggard for all but one week.

The examples above are in fact a consequence of the Arc sine law for last visits.

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  • $\begingroup$ Could you add some explanation to the third quote about how it can coexist with the usual shorthand assumption that a "coin toss" is independent of every other coin toss anywhere in spacetime? How would the coins observed in the long game "know" they are being used in a long game rather than carried out simultaneously in different games? $\endgroup$ Jul 15, 2015 at 20:49
  • $\begingroup$ @HenningMakholm: I will add some information. In fact behind the scene it reduces to a purely combinatorial aspect. There are more possibilities to preserve or increase the height within $k$ steps than to go down. Maybe we could informally say, that it's not the coin, which knows anything about its path, but instead the surrounding space with the specific lattice structure, which has a certain knowledge of the embedded coin and its current height. :-) $\endgroup$
    – epi163sqrt
    Jul 15, 2015 at 22:10
  • $\begingroup$ Probably the most needed information is what the "observations" they speak about are. As quoted it sounds like they claim you can hand them a list of 1000 coin tosses that either originates from one long game or from 1000 different games, and they'd be able to guess which with better than 1/2 chance, by some kind of statistical analysis. $\endgroup$ Jul 15, 2015 at 22:28
  • $\begingroup$ @HenningMakholm: I've added two examples which could be helpful. Best regards, $\endgroup$
    – epi163sqrt
    Jul 16, 2015 at 20:41
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    $\begingroup$ @ApoorvPotnis: I'm sorry, this was a mistake. I'm also referring to W. Fellers book. The arc sine law was first proved by K. L. Chung and W. Feller and it seems I mixed it up. I've updated the post. Best regards, $\endgroup$
    – epi163sqrt
    Jan 21, 2020 at 21:04
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$\begingroup$

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

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

which is actually false.

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

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    $\begingroup$ Can you provide a counterexample? $\endgroup$
    – dazedviper
    Jun 11, 2014 at 22:26
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    $\begingroup$ This is pretty easy to give a counterexample to: let $f$ have bumps as far down as you like, just with sufficiently small widths so that the sum of their integrals goes to zero. You can make $\limsup_{x \to \infty} f(x) = \infty$ this way. $\endgroup$ Jun 12, 2014 at 3:43
  • $\begingroup$ @hallaplay835 c.f. Stein&Shakarchi's Real Analysis book, the chapter on $L^1$ functions. It was an excercise. $\endgroup$ Jun 12, 2014 at 4:17
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    $\begingroup$ @hallaplay835 If $f(x) = \cos(x^2)$, then clearly $f$ doesn't even have a limit as $x$ approaches infinity. But the integral of $f$ does exist as an improper Riemann integral (Hint: Make a change of variable $y=x^2$ and $dx=\frac{1}{2\sqrt(y)}$ and Abel's test does the rest!). $\endgroup$
    – Mark Viola
    Mar 9, 2015 at 19:09
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    $\begingroup$ You can assume $f$ to be positive. That makes the condition a bit more misleading. In fact, this is exactly what is claimed in many quantum mechanics books in attempts to prove the decay of distribution function at infinity. $\endgroup$
    – Hans
    Jun 7, 2016 at 19:14
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There are a good number of counterintuitive probability situations out there. One of my favorites is nontransitive dice:

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

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

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    $\begingroup$ There are a number of similarly counterintuitive statements in multiway voting, of which this could be seen as a special case. For example, in figure skating competitions at the championship level there are nine judges. It would sometimes happen that a majority of judges would rank skater $A$ above skater $B$, a majority would rank $B$ above $C$, and a majority would rank $C$ above $A$. The scoring rules had to have conditions in place to deal with this kind of situation. $\endgroup$
    – MJD
    Jun 5, 2014 at 21:35
  • $\begingroup$ In fact, you can come up with any crazy directed graph among the candidates that you want, and there is a population of apparently rational linear-order voters whose voting behavior will result in that outcome. It turns out that the ONLY behavior for voters that results in consistent majority decisions is blindly following a party line. $\endgroup$ Jun 7, 2014 at 17:50
  • $\begingroup$ Is nine the minimum number of sides/judges requited that non-transitivity phenomenon can be achieved? $\endgroup$
    – Akavall
    Jun 15, 2014 at 2:20
  • $\begingroup$ Definitely not. Most non transitive dice are described as being 6-sided. It's just that the numbers on them are from 1-9 (not necessarily all used). $\endgroup$
    – Duncan
    Jun 15, 2014 at 5:28
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    $\begingroup$ @duncan Suppose you have three skaters, X Y and Z. Judge 1 prefers X to Y and Y to Z; judge 2 prefers Y to Z and Z to X; judge 3 prefers Z to X and X to Y. Now two of three judges prefer X to Y; two prefer Y to Z, and two prefer Z to X. $\endgroup$
    – MJD
    Mar 14, 2015 at 17:33
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The real numbers/Cantor set are countable.

There are several false "obvious" proofs:

  1. "Proof". Consider the tree $\{0,\ldots,9\}^\Bbb N$, then every real number corresponds to a node in the tree. Since there are only countably many levels and each is finite, it follows that the real numbers are finite.

    Why does it fail? This set is actually not a tree. You can order it so it looks like a tree, but in fact the tree would be composed of initial segments of each functions ordered by continuation. This tree, then, would have a last level (namely a level that no point there has a successor), and it would be exactly the level of the functions themselves (the previous levels would be proper initial segments of the functions).

    If we remove that last level, then the tree is indeed countable, but now each real number corresponds to a branch in the tree rather than a node. (It's the unique branch whose limit equals to the function, which previously appeared on that final level.)

  2. "Proof". The rational numbers are countable, and between every two real numbers there is a rational number. Therefore this defines a bijection between pairs of real numbers and the rational numbers.

    Why does it fail? Because there are many, many, many pairs being mapped to the same rational number, this is not actually a bijection.

  3. "Proof". The Cantor set is closed, its complement is open, so it is a countable union of intervals, so the Cantor set is countable.

    Why does it fail? Because not every point in the Cantor set is an endpoint of such interval. For example $\frac14$. It is true that the endpoints of these intervals form a countable dense subset, though.

  4. BONUS!, $\mathcal P(\Bbb N)$ is countable.

    "Proof". For every finite $n$, $\mathcal P(n)$ is finite, and $\mathcal P(\Bbb N)=\mathcal P(\bigcup n)=\bigcup\mathcal P(n)$, is a countable union of finite sets, which is countable.

    Why does it fail? Because the union only includes finite subsets of $\Bbb N$, but none of its infinite subsets.

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    $\begingroup$ I think the non-existence of irrational numbers may itself be in this category (obviously true, but false), but was discovered to be false so long ago that we no longer remember how bizarre it must have seemed. $\endgroup$
    – MJD
    Jun 4, 2014 at 19:29
  • $\begingroup$ @MJD: True, and in two centuries, they will find it strange that at some point people thought that $\Bbb R$ can be countable. Or at least that what I hope! $\endgroup$
    – Asaf Karagila
    Jun 4, 2014 at 19:33
  • $\begingroup$ Now it strikes me as odd that nobody before Cantor observed that if $\Bbb R$ were countable, it could be covered by a family of disjoint intervals whose total length was less than any given $\epsilon$. $\endgroup$
    – MJD
    Jun 4, 2014 at 19:39
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    $\begingroup$ @Christopher: No, the tree has $\omega+1$ levels. The last is the infinite sequence $0.333\ldots$. $\endgroup$
    – Asaf Karagila
    Jun 5, 2014 at 6:27
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    $\begingroup$ @Cory: Or $\frac13$, yes. There are many examples. But this doesn't quite explain why the proof fails. It just shows that it does. I was trying to actually explain the reason for the failure. $\endgroup$
    – Asaf Karagila
    Sep 30, 2014 at 19:32
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Hypothesis: Every infinitely-differentiable function is real-analytic somewhere.

This is false, as shown by (for example) the Fabius function.

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    $\begingroup$ See this interesting answer by Dave L Renfro, and the posts it links to. $\endgroup$ Jun 6, 2014 at 3:02
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$\begingroup$

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

There are more integers than there are natural numbers.

It's obvious, isn't it?

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    $\begingroup$ I hesitate with this one, because it depends on (somewhat unsound) definition of "more". $\endgroup$
    – DanielV
    Jun 7, 2014 at 16:24
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    $\begingroup$ There absolutely are more integers than natural numbers (by inclusion). And cardinality is a slightly contrived notion if you think about it for a moment. $\endgroup$
    – user10904
    Jun 8, 2014 at 9:51
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    $\begingroup$ @Tibor: No, cardinality is not a contrived notion if you think about it for a long moment. $\endgroup$
    – Asaf Karagila
    Jun 8, 2014 at 21:09
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    $\begingroup$ All of mathematics is contrived, by definition. :) $\endgroup$
    – Ryan
    Jun 9, 2014 at 1:32
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    $\begingroup$ “I'm surprised no one gave this answer already”. Actually, the fourth subparagraph of this answer, “There are just as many even numbers as natural numbers.”, was posted Jun 5 at 13:58 (implicitly, it is listing things that are true but non-obvious), and this comment, “… there are half as many odd integers” [as integers], was posted Jun 6 at 7:30. $\endgroup$
    – Scott
    Jun 9, 2014 at 21:40
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Image of a measure zero set under a continuous map has measure zero!

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  • $\begingroup$ Isn't is true?!?! $\endgroup$
    – MR_BD
    Feb 12, 2017 at 7:07
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    $\begingroup$ @LeilaHatami No, see here. Essentially, we can bijectively (and continuously) map the cantor set to [0,1], which maps a measure 0 set to a measure 1 set. $\endgroup$
    – Mark
    Feb 22, 2017 at 7:30
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The probability that you hit any single point on a dart board is $0$ but the probability that you hit the dart board is $1$ (as long as you're not as bad as I am at throwing darts ;D).

EDIT:

As @JpM pointed out I didn't follow the format of these posts albeit the idea can (easily in my opinion) be understood from what I've said above.

Pseudo-Claim: The probability of hitting a single point on a dart board is greater than $0$ since the probability of hitting it at all (assuming that you will hit the dart board) is $1$.

Seems obvious in the sense that a bunch of $0$ can't add up to be $1$ so each point must have some probability. Actually false because of some properties of measures.

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  • $\begingroup$ ...is this not an unobvious theorem that is actually true?? $\endgroup$ Jun 5, 2014 at 7:41
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    $\begingroup$ In particular, there's an "obvious" intuition that the point you actually did hit must have had probability greater than $0$, "since probability $0$ events are impossible". I guess that this example is so obvious that it has to be made axiomatically false in the theory in order to get it out of the way and do some work :-) $\endgroup$ Jun 5, 2014 at 8:27
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    $\begingroup$ @DanielV I would say one of two things based on my mood: a) A lay-person would not claim such a thing, i.e. they wouldn't say the probability of hitting a single point is an infinitesimal. b) Generally (in fact I have not seen otherwise, then again I'm young and naive) measures are defined with values in the extended real numbers, meaning without infinitesimals. $\endgroup$
    – DanZimm
    Jun 5, 2014 at 12:23
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    $\begingroup$ @JpMcCarthy ok I changed the format of my post to follow exactly how the question was asked. $\endgroup$
    – DanZimm
    Jun 5, 2014 at 13:03
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    $\begingroup$ I think the essential thing some people simply can't get their head around is that an event with probability 0 can still happen. $\endgroup$
    – Neil W
    Jun 11, 2014 at 2:43
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My "theorem":

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

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

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    $\begingroup$ Some people treat love as a relationship that is not defined reflexively. $\endgroup$
    – Joshua
    Jun 14, 2014 at 2:16
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    $\begingroup$ @Joshua I have to yet meet a student who notices that this is the root of the problem, and explain how to avoid it in the logical "proof". You are right, of course, once we define the set of "everybody" not as "all humans" but "all humans who are capable of loving my baby", and decide that my baby is not in the set, the result is different. $\endgroup$
    – rumtscho
    Jun 15, 2014 at 17:19
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    $\begingroup$ I have no idea what this answer is trying to say. $\endgroup$ Sep 17, 2015 at 0:45
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    $\begingroup$ @goblin As can be seen from this proof in logic the only possible meaning of the statement is that my baby is me. $\endgroup$
    – Mark Hurd
    Jun 4, 2016 at 13:44
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Consider a function $f:(0, \infty) \rightarrow \mathbb{R}$ that is $\mathcal{C}^\infty$ on that interval. At first glance, one might think that, if $\lim(f) = 0$ as $x \rightarrow \infty$, then $\lim(f') = 0$ as $x \rightarrow \infty$. However, this is false. Here is but one counterexample:

$$f(x) = \frac{1}{x}\sin(x^2)$$

Further, if we add the stipulation that $f$ also be monotonic, counterexamples can still be found (though they are quite pathological).

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    $\begingroup$ Your "another" was already mentioned......... $\endgroup$
    – user55315
    Jun 4, 2014 at 19:34
  • $\begingroup$ Oops, didn't see it there. Thanks. $\endgroup$
    – Kaj Hansen
    Jun 4, 2014 at 21:15
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    $\begingroup$ Caould you give a $ \mathcal{C}^\infty$ monotonic, counterexamples? Thank you very much! $\endgroup$
    – 2016
    Jun 5, 2014 at 13:49
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    $\begingroup$ Yes, you can get one by tinkering with the responses here a bit: math.stackexchange.com/questions/788813/… $\endgroup$
    – Kaj Hansen
    Jun 6, 2014 at 2:35
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A simple arc (homeomorphic image of the closed unit interval) in the plane has $2$-dimensional Lebesgue measure zero.

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  • $\begingroup$ this is obvious right -- isn't it? $\endgroup$
    – athos
    Jul 7, 2014 at 2:32
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    $\begingroup$ @athos It should be obvious, otherwise it wouldn't be a valid answer to this question. Are you asking why it's actually false? There is probably a simpler direct construction of a "fat arc" in the plane, but it follows from a more general theorem of R. L. Moore and J. R. Kline (On the most general plane closed point-set through which it is possible to pass a simple continuous arc, Ann. of Math. (2) 20 (1919), 218-223) that every (homeomorph of the standard) Cantor set in the plane is contained in an arc; apply this to a fat Cantor set in the plane. $\endgroup$
    – bof
    Jul 7, 2014 at 3:28
  • $\begingroup$ thanks for the link... a bit tough for me, will bookmark it for later digestion :) $\endgroup$
    – athos
    Jul 7, 2014 at 3:52
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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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  • $\begingroup$ What I personally think it should be is... For every nontransdental, real number there exists a finite-length program that computes that number... $\endgroup$ Jul 25, 2014 at 3:21
  • $\begingroup$ @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 $0.123456789101112...$). $\endgroup$
    – Keen
    Jul 25, 2014 at 15:43
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Theorem: Let $f_1(x,y)$ and $f_2(x,y)$ be two joint probability densities, each having its $x,y$ components positively correlated ($Cov_1(x,y)>0$, $Cov_2(x,y)>0$). Let $f_3=\alpha f_1 + (1-\alpha) f_2$ be the mixing density, for some $0\le \alpha\le 1$. Then $Cov_3(x,y)>0$.

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

False. See Simpson's paradox.

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    $\begingroup$ Perhaps the following example will make the obviousness more obvious. Baseball team $A$ has a better win-lose ratio than team $B$ in the first part of the season. Then there is a strike, and some games are missed. When the season resumes, team $A$ also has a better win-lose ratio than team $B$ in the second portion of the season. Therefore, team $A$ has a better win-lose ratio than team $B$ overall. (Wrong!) Baseball did play such a split season in 1981, but I don't know if the paradox actually occurred then. Probably not, as it requires $A$ and $B$ to play very different numbers of games. $\endgroup$
    – MJD
    Jun 6, 2014 at 0:43
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Claim:

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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    $\begingroup$ haha, took me a second. $\endgroup$ Jun 11, 2014 at 2:47
  • $\begingroup$ depends on what field. very nice, but a trick question. I feel cheated. $\endgroup$
    – Lost1
    Jun 12, 2014 at 11:29
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    $\begingroup$ @Lost1: It actually doesn't depend on the field: $\vec{v}\cdot \vec{0} = 0$ and $\{\vec{v}, \vec{0}\}$ is never independent. $\endgroup$ Jun 12, 2014 at 15:52
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    $\begingroup$ @JasonDeVito ah... i was thinking take $\mathbb{Z}_2^2$ and the vector $(1,1)$... $\endgroup$
    – Lost1
    Jun 12, 2014 at 17:46
  • $\begingroup$ @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!) $\endgroup$
    – Ryan Reich
    Jun 16, 2014 at 4:25

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