# Theorems that are 'obvious' but hard to prove

There are several well-known mathematical statements that are 'obvious' but false (such as the negation of the Banach--Tarski theorem). There are plenty more that are 'obvious' and true. One would naturally expect a statement in the latter category to be easy to prove -- and they usually are. I'm interested in examples of theorems that are 'obvious', and known to be true, but that lack (or appear to lack) easy proofs.

Of course, 'obvious' and 'easy' are fuzzy terms, and context-dependent. The Jordan curve theorem illustrates what I mean (and motivates this question). It seems 'obvious', as soon as one understands the definition of continuity, that it should hold; it does in fact hold; but all the known proofs are surprisingly difficult.

Can anyone suggest other such theorems, in any areas of mathematics?

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Perhaps the isoperimetric inequality. – Péter Komjáth Jan 9 '11 at 11:28
Perhaps the Kepler Conjecture: en.wikipedia.org/wiki/Kepler_conjecture – Aaron Meyerowitz Jan 9 '11 at 11:57
A former colleague of mine used to say (to students), "A theorem is obvious if a proof instantly springs to mind," a maxim I like a lot. I think what you are talking about is theorems where a plausible argument instantly springs to mind but falls short of being a proof. – gowers Jan 9 '11 at 15:12
I am tempted to vote to close as subjective and argumentative given the comments on the existing answers. Can we narrow the definition of 'obvious' being used? Something like gowers' definition is good, but depends a lot on one's training. Perhaps something like "if you asked an undergraduate if it were true, they'd bet yes." – Qiaochu Yuan Jan 9 '11 at 16:46
I disagree that the Jordan curve theorem is "obvious" but admits a surprisingly difficult proof. The proof for curves with reasonable regularity is not difficult, while the truth of the theorem for wild curves is not so obvious, I think. (At least, I think it is reasonable to argue that most people's sense of this being intuitively clear comes from imagining a rather regular curve in the plane, not a wild one.) – Emerton Jan 10 '11 at 16:46

The first of the Tait Conjectures seems intuitively obvious:

Any reduced diagram of an alternating link has the fewest possible crossings.

This 19th century conjecture is difficult to prove, with the proof coming only in 1987 by Kauffman, Murasugi, and Thistlethwaite, using the Jones Polynomial. The discovery of this proof was a huge coup for quantum topology; a quantum invariant was used to prove a difficult classical open problem.
While this is certainly hard to prove, I don't think it's unexpectedly hard to prove. Knot diagrams modulo Reidemeister moves form a rather complicated algebraic structure; and there's no reason to expect that any statement about knot diagrams should be easy to prove.

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$[0, 1]$ is connected.

But I consider this a trick answer because the real difficulty is turning this into a completely rigorous statement. Understanding the rigorous definition of connectedness (and understanding the point of making definitions like this) can be a substantial hurdle, but once this hurdle is crossed, the proof is not difficult.

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An equivalent example: The intermediate value theorem. – Lennart Meier Feb 16 '13 at 0:22

It is obvious that there is a unique point in any given affine plane in a finite-dimensional euclidean vector space which is closest to the origin.

Therefore it would seem similarly obvious that every de Rham cohomology class on a compact oriented riemannian manifold should have a unique representative with minimal $L^2$ norm: namely, its harmonic representative.

Yet it takes some effort (elliptic regularity,...) to prove that the harmonic representative does in fact exist, i.e., that it is a smooth differential form.

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Dynamic programing principle (DPP) is one of the 'obvious' and also intuitive one, in the control problem. Many papers proves its validity in various setup, and all proofs are very complicated. But, there is rarely a counter example of DPP. I wonder, if there is general framework on it. See, Dynamic programming principle (DPP)

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On page 33 of Stable Mappings and Their Singularities by Golubitsky and Guillemin (GTM 014; 1974), the following proposition is characterized as an "obvious, but surprisingly complicated result":

Proposition 1.10: Let S be a nonempty open subset of $\mathbb{R}^n$. Then S is not of measure zero.

Edit: As pointed out in the comments, Gowers already mentioned an equivalent result whose difficulty is more clear. Please vote this answer down (I can't vote down my own answers).

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I must be completely missing the point here. Why is this hard? Can't you just show that there is a small cube inside $S$? – Deane Yang Feb 28 '11 at 2:14
I'm also missing the point. I always thought this was a trivial proposition. – Jim Conant Feb 28 '11 at 2:50
I suppose that this is trivial if one constructs Lebesgue measure by first assigning measure to rectangles and in the end declares a set to be of measure zero if...well, if it has Lebesgue measure zero. If one merely has the classical definition of a zero measure set (a set such that $\forall \varepsilon > 0$ can be covered by countably many rectangles so that the sum of their volumes is at most $\varepsilon$), this is nontrivial. Actually Gowers already mentioned an equivalent result, in which the difficulty is more clear. – Mark Feb 28 '11 at 3:34
Aha. That makes sense. – Jim Conant Feb 28 '11 at 15:39
Does this mean that it is hard to show that a rectangle does not have measure zero? – Deane Yang Feb 28 '11 at 15:54

The stability of Minkowski Spacetime

An asymptotically flat initial data set for the Einstein equations that is sufficiently "close" to the initial data for Minkowski spacetime generates a solution to the Einstein equations that approaches Minkowski spacetime asymptotically. (try saying that fast 3 times)

It is "obvious" because of our physical experience and intuition with gravity, and it is hard to prove because Einstein's equations are quite subtle and complicated.

There are other theorems in mathematical relativity that fall into this category, but this one is especially striking since it is particularly difficult to prove, while it "feels" blatantly true.

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The problem is if the near-flat initial data consists of a ton of gravitational waves all going inward, and making a black hole. This result is only obvious once you bound the waves in wavelength and energy to prevent this scenerio, and then it is just as complicated looking as the proof. – Ron Maimon Aug 27 '11 at 5:13

All isometries of the plane are affine linear.

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Is this considered hard? – Todd Trimble Jan 10 '11 at 15:20
Todd, I think something closely resembling David's statement might count as hard: the Mazur-Ulam Theorem, that every bijective isometry between real normed spaces is affine. An indication that it's not obvious is that it's false without the word "bijective": e.g. you can cook up a non-affine isometry from R to (R^2 with the infinity-norm). Reference: helsinki.fi/~jvaisala/mazurulam.pdf . I don't know whether it's appreciably easier if you stick to two dimensions. – Tom Leinster Jan 13 '11 at 18:21
But maybe you know all that, Todd... – Tom Leinster Jan 13 '11 at 18:22
Sorry Tom, didn't see your comment until now. Thanks for your comment. I had assumed "isometry" meant with respect to the usual Euclidean distance (2-norm). Here is a sketch of the proof I had in mind. First, by translation, assume WLOG that the origin is taken to itself. Then |f(v)|^2 = |v|^2 for all v. Since d(fv, fw) = d(v, w) for all v, w, it follows that |fv - fw|^2 = |v - w|^2. Since the inner product of v and w can be defined in terms of |v|^2, |w|^2, and |v-w|^2, it follows that f preserves inner products. So it takes an orthonormal basis to an orthonormal basis. Finally, [cont'd] – Todd Trimble Mar 5 '11 at 15:34
Since the inner product is linear in each argument, it quickly follows that if e_i is orthonormal, then the inner product of f(av + bw) against all the f(e_i) matches the inner product of af(v) + bf(w) against all the f(e_i). Since f(e_i) is an orthonormal basis, this shows f preserves linear combinations av + bw. – Todd Trimble Mar 5 '11 at 15:39

In the same genre, if not the same type: The Fundamental Theorem of Algebra. Easily understood by high schoolers, plausible, beautifully simple to state. As far as I know, there are no nice proofs understandable by a good (not brilliant) high school student.

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I do not know why this is plausible. What would be an "obvious" reason to expect that a degree 6 polynomial with real coefficients has a complex root? – Andrés E. Caicedo Jan 10 '11 at 23:30
The division theorem implies a non constant polynomial defines an open map from the sphere to itself. Since any such map is also continuous, hence closed, it is surjective. that's the best I can do. It seems conceivable to make that seem plausible to someone with a little intuition, if not obvious. – roy smith Jan 11 '11 at 0:43
Here is a proof whose main idea is understandable by many high school students. The winding number of the image of the circle of radius $r$ changes from $0$ at $r=0$ to the degree of the polynomial for $r$ large, and it can only change when there is a $0$ of the polynomial. – Douglas Zare Jan 11 '11 at 1:11
If I'm following this correctly, some comments say the example is unsuitable because proving the theorem is actually easy, while the oldest comment says it's unsuitable because it's not obvious. What a mess! Keeping in mind how long it took between the result being conjectured and the first actual proof, I think we are too far removed in time from the result to truly appreciate it from a historical perspective, and the FTA is too fundamentally rooted in students educations to imagine how hard it would be for someone who was a blank slate. – Thierry Zell Jan 11 '11 at 14:43
@Thierry: agreed. Anything Euler attempted unsuccessfully to prove can't be all that easy. I think your comments apply to comments on several other answers as well in relation to how rooted algebraic topology is in many mathematicians' educations these days. – Qiaochu Yuan Jan 12 '11 at 14:34

On an elementary level, the intermediate value theorem is surprisingly deep.

On a less elementary level, the prime number theorem is "obvious" from $\sum_{p\leq x}1/p\sim \log\log x$ (that was noticed by Euler) and Dirichlet's theorem on primes in arithmetic progressions is "obvious" if you use the sieve of Erathostenes.

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But historically: The fifth postulate of Euclid (to a point outside a straight line passes exactly one line parallel to this line). At first it seems an obvious fact, and tried to prove it....

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A differentiable manifold M that is homeomorphic to the n-sphere is also diffeomorphic to the n-sphere . Obvious, but wrong ! (But right for 1-, 2-, 3-, 5-, 6- and 12-spheres).

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And unknown for $4$-spheres – Jim Conant Aug 27 '11 at 11:16

P is not equal to NP. This is "obvious", is a straightforward arithmetic proposition doesn't involve any fancy set theory or spacefilling curves, and yet it's so hard that there have whole workshops ("Barriers") and important papers (BGS, natural proofs etc.) devoted to the question of what makes it so hard. Scott Aaronson describes "a would-be P≠NP prover who hasn’t yet grasped the sheer number of mangled skeletons and severed heads that line his path." P≠PSPACE is even more obvious and yet there is a comparable lack of progress.

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I think a very good example is Kepler Conjecture:

http://en.wikipedia.org/wiki/Kepler%27s_conjecture

This conjecture stated that the most "tight" stack of same balls have only two kinds of arrangement with a fixed density.

Every physicist knows that it's true, no mathmaticans ever proved it.

Fortunately, Hales used computers to step forward a little little bit.

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Since this was resurrected, here is the statement that at this time seems to me to have the greatest gap between obviousness of truth and obviousness of proof:

• There exists a natural set theoretic universe in which every subset of [0,1] is Lebesgue measurable, so that the reals admit no well order and do not inject into aleph-1.

Here are a different class of obvious theorems, these are only obvious in the sense of physical intuition. They took a long time to prove:

• The existence of solid matter occupying space (in the lowest energy state, the electron-nucleus system occupies a volume proportional to the number of nuclei)
• The positive energy theorem--- every asymptotically flat solution of GR obeying the appropriate energy condition has a positive mass at infinity, with zero mass only for Minkowski space.
• Hard sphere collisions on a negatively curved space are ergodic.

Here is a physically obviously true statement, which can be seen from physical intuition, but which is not proven (as far as I know):

• The asymptotic fate of GR with a positive cosmological constant is within any causal patch, and except for a set of initial conditions of measure zero, a deSitter space.

The reason this is obvious is because the deSitter space maximizes the horizon area, which is a measure of the entropy.

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Very good candidates for this question are theorems that amount to saying that some sequence behaves randomly in a way. Both the fact that the statements are obvious and the fact that they are usually hard to prove, are explained by the fact that there is just no reason for the sequence to nót behave randomly. The primes are of course notorious for this. Easy example that springs to mind: there are about as many primes whose last (decimal) digit is $1$ as there are primes whose last digit is $3$.

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Here is an example proof of which is not conceptually difficult, but tedious to write down in full detail: There exists a universal Turing machine.

The reason I find this obvious is that, after learning the definition of what a TM is, a moment's thought will suggest that we should be able to encode information of any TM to a natural number since it is a finitary object. Then, it is conceivable that there is an algorithm which decodes any natural number and carry out the instructions.

Of course, constructing the universal machine and proving that it can actually simulate all the other machines involve some unpleasant details.

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I don't think this is a good example for what the OP had in mind. The proof is maybe not nice to write down, but it is conceptually very easy. The very reason it is "obvious" is just because the proof is immediately obvious. The reason the proof isn't nice is rooted in the fact that Turing machines are not nice because of their minimalistic definition. If one would choose to define a theoretical computation machine with abilities more akin to a real world computer, then writing down the proof would reduce to some easy programming. – Johannes Hahn May 4 '13 at 14:10

I think Godel's completeness theorem is very intuitive. For example, can you imagine a first order theorem that would be true for all groups, that you wouldn't be able to prove (by Godel's definition of `prove'). Of course not! But the proof of the completeness theorem is hard.

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among people trained on the incompleteness theorem, this is not obvious. – Will Sawin Sep 30 '11 at 3:59
The completeness theorem is not really hard to prove. The propositional part may be a bit messy (which can be alleviated by a wise choice of a proof system), but the main Henkin construction is quite straightforward. – Emil Jeřábek Sep 30 '11 at 11:36
Will, I disagree. The incompleteness theorem talks about something quite different. About whether there is a small set of axioms that imply every true statement about the integers (or other models). I dislike the comparison people make between Godel's completeness theorem and his incompleteness theorem. They talk about very different things. Emil, you are correct. The proof is not very hard. But it is much easier to state the theorem and believe it than to actually prove it. – James D. Taylor Oct 1 '11 at 15:24
I agree that the completeness theorem sounds very intuitive, but I think this is misleading. It takes some serious thought to convince yourself that a particular definition of formal proofs captures mathematical practice (even if you believe intuitively that some definition should work, it's much less obvious that a given deductive system really is complete). Furthermore, I'd bet that many mathematicians would find it equally intuitive that there should be a complete proof system for second-order logic, and of course incompleteness tells us there isn't. So completeness is pretty subtle... – Henry Cohn Sep 8 '12 at 18:17

I'm not sure if it is considered obvious, but I think the Collatz conjecture is also a good example (not just hard to prove, but still unproven!). I've always found it very frustrating that such a simple conjecture does not lend itself to a simple proof.

'Obvious' here sort of means: Try a lot of values, and you'll get a feeling THAT it indeed holds, but also WHY it holds.

http://en.wikipedia.org/wiki/Collatz_conjecture

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How can you tell that there is no counterexample at $2^{3013}+3^{2012}+5^{1003}+7^{691718}+1$? Or perhaps the number after that? Or after that? Etc.? – Asaf Karagila Aug 26 '11 at 9:53
I don't think that being easy to check for the first few trillion values makes it 'obvious'. You say that when you try a lot of values you get a feeling 'why' it holds. Can you say any more about this? Are you sure it isn't really a feeling why it should hold for almost all values (for which there is a rather weak heuristic argument)? – Alec Edgington Aug 27 '11 at 7:11
In Pauls defense, I think the conjecture that there is no positive integer whose Collatz-iterations escape to infinity is at least somewhat obvious – Woett Sep 30 '11 at 22:23

No one has mentioned the 4-color map theorem yet. it would seem to be a classic example.

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No one has mentioned it because it isn't obvious. – Johannes Hahn Feb 28 '11 at 20:05
It appears to be obvious. Combine the obvious fact that more than 3 is necessary with the fact you can't construct a 5 region map that requires 5 colors (follows directly from the fact that the complete graph of 5 node is not plainer) leads most people who look at it casually to mistakenly conclude that 4 is sufficient. – Mark Biggar Feb 28 '11 at 21:42
Sorry, "mistakenly" is the wrong word as the conclusion is correct it's just not justified directly by the premises given. – Mark Biggar Mar 1 '11 at 16:24

Maybe on the boundary of what's allowed, but I would say most basic geometric things like Pythagoras' Theorem, trigonometry with sine/cosine, the area of a circle, etc. etc. etc.; here of course the difficulty is in defining what we mean by length, area, angle, etc. - in which case some of these become axiomatic, but then the difficulty is shifted onto proving that things do work correctly.

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-1 (sorry for the downvote): The difficulty is in statement of the theorems, not the proof. – Daniel Moskovich Jan 9 '11 at 14:31
Same kind: the surface of the sphere = 4π (Archimedes). Well, not really obvious and not really easy to prove. I'm down :( – Patrick I-Z Jan 9 '11 at 16:29

I think this answers your question in a perverse way: All statements in the theory of Natural Numbers provable from the ZFC axioms of set theory. They are obviously true by definition.

EDIT: Looking at this objectively, it probably sounds like I'm saying if a statement is true, then it's obviously true. However, that was not my intent, and I apologize for what may have sounded like a thoughtless response. This is how I see it:

All statements expressible in the language of arithmetic can be represented by formulas in the language of set theory that are only $\Delta_1$ in the Levy hierarchy. In particular, all transitive models will agree on whether they are true. If we further restrict ourselves to only consider the true statements in $\mathbb{N}$ that are ZFC theorems, then all ZFC models will agree that these statements must be true so they are about as obvious to ZFC models as possible. Now if you are an oracle having knowledge of all such true statements, then you will probably develop an intuition that makes them all seem "intuitively obvious." This reflects the answers suggesting that a theorem is obvious after you prove it.

To add one more related point here, when addressing G$\ddot{\textrm{o}}$del's Incompleteness Theorem, one can naively ask about completing PA in the "obvious" way, i.e., by extending it to be the theory consisting of all true statements in $\mathbb{N}$. But of course such a completion is not computable.

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What is this definition of "obviously true" that makes these obviously true? – Chris Eagle Jan 10 '11 at 4:00
By definition, the theory of the Natural numbers is the collection of all statements in the language of arithmetic that are true in $\mathbb{N}$. When we say that a statement in the language of arithmetic is true, we mean that is true in $\mathbb{N}$. Consequently, a statement in the language of arithmetic is true if and only if it is in the theory of the Natural numbers. The argument of whether the statement is in the theory of the Natural numbers takes place in the set-theoretic universe $V$, which models all of the theorems of ZFC. – Jason Jan 10 '11 at 4:29
My point was that given you know a statement is in the theory of the Natural numbers, if you were asked whether it were true or not, you'd be able to respond without hesitation with an affirmative answer. – Jason Jan 10 '11 at 5:03

Fermat's Last Theorem. (Should I state it?)

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Why is Fermat's last theorem 'obvious' ? – Denis Serre Jan 9 '11 at 13:40
It was already obvious to Fermat! ;-) There are different ways to interpret the word "obvious". Not so many people doubt it's true. And just a few understand its proof. – Wadim Zudilin Jan 9 '11 at 13:48
Wadim, sorry for the downvote; but I think you're misunderstanding the English language here. Although the meaning of the statement is obvious, its truth is definitely not. Suppose you'd never heard of it before, and someone asked you to guess the truth or falsity of the statement, in less than an hour, say for a bet of 100 dollars or similar. I cannot believe that any human being could possibly guess "true" and have enough "reasonable" confidence to want to make the bet. – Zen Harper Jan 9 '11 at 14:06
Zen, I guess you are not the one who tells me who could tell me that the Jordan curve theorem is obvious (how many students in your class would agree?). The word "obvious" is too subjective. The things which are obvious to you may be not obvious to others. Therefore, we have a perfect voting system. Everybody feel free to vote down! I have to go to bed to get rid of this nightmare. English language lessons are too much for Sunday. G'night. – Wadim Zudilin Jan 9 '11 at 14:16
Wadim, for what it's worth I also disagree that the Jordan curve theorem is obvious. It's certainly not any more obvious than the nonexistence of space-filling curves. – Qiaochu Yuan Jan 9 '11 at 15:04

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