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I am pretty distant from anything analytic, including analytic number theory but I decided to read the Wikipedia page on the Riemann hypothesis (current revision) and there is some pretty interesting stuff there:

Some consequences of the RH are also consequences of its negation, and are thus theorems. In their discussion of the Hecke, Deuring, Mordell, Heilbronn theorem, (Ireland & Rosen 1990, p. 359) say

The method of proof here is truly amazing. If the generalized Riemann hypothesis is true, then the theorem is true. If the generalized Riemann hypothesis is false, then the theorem is true. Thus, the theorem is true!!

What is surprising is that both a statement and its negation are useful for proving the same theorem.

Do similar situations arise with other major, notorious conjectures in mathematics? I only care about algebraic geometry and algebraic number theory for the most part but I guess it will make little sense to have such questions devoted to each area of mathematics so post whatever you've got.

To give an initial direction: are there any interesting statements one can prove assuming both some hard conjecture about motives (e.g. motivic $t$-structure, the standard conjectures, Hodge/Tate conjectures) and its negation?

Martin Sleziak
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  • Related (and closed) question: https://mathoverflow.net/questions/312439/strange-proofs-of-existence-theorems – Sam Hopkins Jun 03 '19 at 13:32
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    @SamHopkins I personally think that this question is not related to the one you have linked to and that this question is not nearly vague enough to be closed as "unclear". Opinions differ I guess. –  Jun 03 '19 at 13:34
  • Maybe I misunderstood something. For what I understand, there are many theorems that can be proven both in case RH is true and false. For example, we can prove that the series of $1/n^2$ converges. Any theorem not related to RH can be proven. I would say that any theorem which is a consequence of RH and its negation simply does not have to do with RH, although, maybe, we are not able to see it as unrelated. – Doriano Brogioli Jun 03 '19 at 15:36
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    @DorianoBrogioli well, it does have something to do in the sense that the easiest (or the only known?) proof uses "decomposition along RH". If you have many interesting examples, post many interesting answers! Though I should add that there indeed was an implicit point that the proof using "decomposition along RH" should be somehow simpler than other proofs, or be the only known proof (which in my understanding is what is happening in the example mentioned in Ireland-Rosen). –  Jun 03 '19 at 15:38
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    https://meta.mathoverflow.net/questions/4200/flood-of-similar-new-users – Todd Trimble Jun 03 '19 at 17:37
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    @ToddTrimble does this comment contribute anything to the discussion? –  Jun 03 '19 at 17:43
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    One would have to see the actual proof I guess, but in this abstract description, nothing seems even remotely "truly amazing" to me. Rather, it sounds like doing an argument by distinguishing various (in this case, two) cases. – Christian Remling Jun 03 '19 at 21:51
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    @ChristianRemling I should note that the "truly amazing" part is due to the book of Ireland & Rosen, not due to me. Possibly you are better equipped than them to judge what is or is not truly amazing, I do not know. –  Jun 03 '19 at 22:18
  • Is this so amazing to deserve an exclamation mark? It is quite common to prove a result X by cases, and the fact that one case may be empty is not so relevant in order to prove X, after all ---(Sorry I didn't note that this comment had already been made) – Pietro Majer Jun 03 '19 at 22:24
  • @PietroMajer I actually quite like such arguments, maybe I should try to join your field if such arguments are common. Anyway, all objections should be sent to Ireland & Rosen, not to me (I did not add any exclamation marks). –  Jun 03 '19 at 22:25
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    @Cutthewood My excuses, of course I like these arguments too, and I didn't mean to diminish their content. Just saying the form " A implies X and not A implies X as well" is not so strange – Pietro Majer Jun 03 '19 at 22:31
  • @Cutthewood: Yes, I know, I didn't mean to object to what you wrote, I just wanted to point out that the quote doesn't really let the reader share in the amazement. – Christian Remling Jun 03 '19 at 22:37
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    @SamHopkins: In particular, your own answer to that question is a simple example of what OP wants, I think. – Nate Eldredge Jun 04 '19 at 04:02
  • @NateEldredge indeed, the answer appears to be relevant. The question is not, in my opinion. I would upvote the answer if it is reposted here. –  Jun 04 '19 at 12:25
  • How is this situation essentially different from proving a result by considering special cases? Suppose you want to prove a result (say, quadratic reciprocity), for an arbitrary odd $p$, and you consider separately the case $p=4k+1$ and $p=4k+3$, the latter being the negation of former in the context of the problem. Both the assertion $p=4k+1$ and its negation $p=4k+3$ lead to the same conclusion, and both are essential. – Michael Jun 04 '19 at 16:06
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    @Michael: I think the difference is that, in your example, we know that the hypothesis $p = 4k+1$ is true for some $p$ and false for others, and so we really need both cases to prove the theorem for all $p$. In OP's example, the hypothesis is either universally true or universally false (there's no free variable like $p$), and hence one of the cases is completely unnecessary to treat, but we don't know which one. – Nate Eldredge Jun 04 '19 at 18:19
  • It seems that me that the italic text in this question just implies that the theory they refer to is independent of generalized RH. – user400188 Jun 05 '19 at 01:29
  • @NateEldredge, got it, thanks. – Michael Jun 05 '19 at 15:50
  • Sorry for the naive question, but has the RH been proved decidable? According to Gödel there is more than just true or false. And I wonder what would happen if a proof were based on the assumed truth value of an actually undecidable conjecture. – Manfred Weis Jun 06 '19 at 04:58
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    @ManfredWeis isn’t this essentially the case for the continuum hypothesis example in D.S.Lipham’s answer? I don’t see why a statement being undecidable should make the argument less valid. Also note that if RH is undecidable then it is also true (which means it will never be proved to be undecidable. It may be proved some day to be decidable but that hasn’t happened yet). – Dan Romik Jun 06 '19 at 07:56
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    @ManfredWeis : If RH had been proved decidable then it would have been reduced to a finite computation and we would all have heard about it. See https://mathoverflow.net/q/62144/3106 and https://mathoverflow.net/q/6250/3106 for some further discussion. – Timothy Chow Jun 06 '19 at 16:55
  • Just in case this elementary example fits the question: Proposition: Every group $G$ with at least 3 elements (and possibly infinite) has a nontrivial automorphism. Proof (a) $G$ is abelian with $G^2\neq 1$: take $x\mapsto x^{-1}$ (b) $G$ is 2-elementary abelian, thus a vector space of dimension $\ge 2$ over $Z/2$: use a nontrivial permutation of a basis to induce a nontrivial automorphism (c) $G$ is not abelian: use a nontrivial inner automorphism. Here both the statement "is abelian" and its negation are useful (to ensure that inversion is homomorphism, vs produce inner automorphism) – YCor Oct 27 '19 at 13:48

5 Answers5

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Theorem. The Stone-Čech compactification of the real line contains $2^\mathfrak c$ topologically distinct continua.

Here a continuum is defined to be a compact connected Hausdorff space. $2^\mathfrak c$ is easily seen to be an upper bound in the problem.

The proof was divided into two parts:

Case 1: The Continuum Hypothesis fails ($\mathfrak c>\omega_1$).

Dow, Alan, Some set-theory, Stone-Čech, and $F$-spaces, Topology Appl. 158, No. 14, 1749-1755 (2011).

Case 2: The Continuum Hypothesis holds ($\mathfrak c\leq\omega_1$).

Dow, Alan; Hart, Klaas Pieter, On subcontinua and continuous images of $\beta \mathbb{R} \setminus \mathbb{R}$, Topology Appl. 195, 93-106 (2015).

The assumptions $\neg$CH and CH are critical to the constructions. Also the types of continua constructed are very different in Case 1 vs Case 2. In fact, all continua of the type constructed for Case 1 are homeomorphic under CH.

This is the only theorem I know of which was proved using CH in this way. What's especially interesting is that both cases are necessary to this proof because CH and $\neg$CH are each consistent with ZFC. This may be different from the use of RH and $\neg$RH, or some other conjecture and its negation. If the conjecture is eventually proved, for instance, then you could throw away the other half of your proof.

D.S. Lipham
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This paper proves the existence of an essentially-deterministic algorithm for the following problem:

Input: an integer $n$

Output: An $n$-bit prime.

Which runs in time sub-exponential in $n$ for infinitely many inputs.

On a high level the argument works as follows:

  1. Suppose a certain kind of pseudo-random generator exists. Then we can derive an algorithm to compute the above.
  2. Suppose that kind of pseudo-random generator does not exist. Then we have a collapse of complexity classes $\mathsf{PSPACE} = \mathsf{ZPP}$ which then implies some circuit lower bounds, which in turn yields a pseudo-random generator with the desired properties.
Izaak Meckler
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  • is the existence of a pseudo-random generator hard? –  Jun 03 '19 at 19:56
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    Yes -- no one knows how to unconditionally prove the existence of such things. For most kinds of pseudo-random generator, their existence is stronger than $P \neq NP$. – Izaak Meckler Jun 03 '19 at 19:59
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    This is indeed an intriguing result, however, the algorithm only runs in time subexponential in $n$ rather than polynomial, and it does not work on all inputs $n$, but only for some infinite subset. – Emil Jeřábek Jun 04 '19 at 10:05
  • Thanks Emil -- amended. – Izaak Meckler Jun 06 '19 at 05:31
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    Please, what does essentially-deterministic mean? – Hilder Vitor Lima Pereira Jun 06 '19 at 12:56
  • @HilderVítorLimaPereira : It means that the algorithm is randomized, but produces the same output most of the time. This is to distinguish it from randomized algorithms for producing primes that give a different prime (almost) every time you run the algorithm. – Timothy Chow Jun 06 '19 at 16:32
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I think this is much less surprising today. It's not uncommon to argue that a statement about some structures is true because one can decompose structures into random-like and structured parts ('structure and randomness'), and in either case get to the desired conclusion. Usually one is trying to prove that a certain statement is true for a whole class of structures by this method; there is no one special structure for which we really want to prove it. Take Roth's theorem as an example: we want to prove any positive-density set of integers in $N$ (for some large $N$) contains a three-term AP. We can do this if the set looks random (small Fourier coefficients) easily enough, but if the set does not look random, then there is a large Fourier coefficient, so the set looks structured, namely it has a noticeably larger density on some long AP than on $[N]$. But then we can pass to the AP and iterate this argument.

It just happens that when we try to prove things about the primes, we really only want to know about the one structure, contained in a class of structures with prime-like properties (which might not have more than one member, depending on the properties used in the proof). But the proof method is still to show that the whole class of structures satisfies the desired conclusion.

In this case (G)RH is a statement that the primes behave in some random-like way, in a fairly strong quantitative sense. And so its negation is the assertion of some structure beyond the obvious - which of course can be useful.

Of course, if you want to find other examples of theorems which you can prove by appealing to some major open problem being either true or false, you should probably have some reason why either outcome helps. Quite a few major open problems (or major theorems) can be read as some kind of quasirandomness statement, and for that there is a reason why the conjecture failing might be useful, so in principle there should be a decent list of possible candidates.

user36212
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  • I guess the Hodge conjecture could be interpreted as saying "there are many algebraic cycles" or maybe "algebra controls topology" (feel free to bash me for such a formulation). Would be pretty fun is there was a similar example in this setting. –  Jun 03 '19 at 15:54
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Here is a baby answer.

Theorem: There exist two irrational numbers $p,q$ such that $p^q$ is rational.

Proof: In case $\sqrt 2^{\sqrt 2}$ is rational we can take $p=q=\sqrt 2$.

Otherwise take $p= \sqrt 2^{\sqrt 2}$ and $q=\sqrt 2$. We have $p^q = \big(\sqrt 2^{\sqrt 2}\big)^{\sqrt 2} = \sqrt 2^{\sqrt 2 \cdot \sqrt 2} = \sqrt 2^2=2$ is rational.

Daron
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    This was the first example that came to my mind. I wonder if the commenters sceptical about the "truly amazing"ness of the example in the original post really find this one (which is accessible to any mathematician) to be just an ordinary proof, with nothing special about it (in a psychological, not rigorous, sense). – LSpice Jun 05 '19 at 19:17
  • @LSpice It is not quite as "amazing" (in my opinion) because one can actually just prove $\sqrt{2}^\sqrt{2}$ is irrational. In other words, this proof is inessential. The rationality/irrationality of $\sqrt{2}^\sqrt{2}$ is neither an open problem nor an independent axiom. But I still like the argument :) – D.S. Lipham Jun 05 '19 at 22:43
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    @LSpice I find it to be disappointing for an almost opposite reason as D.S. Lipham. A simple cardinality argument gives the theorem. It's like this "Thm: There is a room in my house without my dog in it. Prf: If my dog is not in the kitchen, we are done. If my dog is in the kitchen, then my dog is begging for food scraps in the kitchen, which means my dog is not in the bedroom, so we are done." Better to say I have four rooms in my house and one dog. – Will Sawin Jun 06 '19 at 05:46
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    @D.S.Lipham the proof above is much easier than the proof that $\sqrt{2}^\sqrt{2}$ is irrational, so there is still something quite nice and elegant about it. I do agree that knowing about this more explicit result spoils the magic a bit. It’s also worth pointing out that p=e, q=log(2) gives another explicit way of proving the theorem being claimed (and I believe that the irrationality of log(2) is much easier to prove than irrationality of $\sqrt{2}^\sqrt{2}$. The irrationality of e is a trivial result of course.) – Dan Romik Jun 06 '19 at 07:36
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    @WillSawin what cardinality argument did you have in mind? – D.S. Lipham Jun 06 '19 at 08:22
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    @D.S.Lipham Fix any positive rational number $a$ and take $p=a^{1/q}$ for $q$ in $\mathbb R$. We have only countably many rational $q$, and only countably many $q$ with $p$ rational, because $a^{1/q}$ is an invertible function. So there must be (uncountably many!) irrational pairs with this property. – Will Sawin Jun 06 '19 at 14:01
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    @WillSawin : You are of course correct. What I like about this example is not the theorem (as stated above) per se, but the light it sheds on the meaning of the word explicit. Suppose I want an explicit example of $a$ and $b$ with $a$ irrational, $b$ algebraic irrational, and $a^b$ rational. All the numbers appearing in this proof are explicit, but somehow the example fails to be explicit because of the use of the law of the excluded middle. – Timothy Chow Jun 06 '19 at 14:23
  • @TimothyChow Conversely the cardinality-based proof doesn't have any explicit numbers at all but can easily be mined to give an explicit procedure to compute the digits of the irrational numbers. – Will Sawin Jun 06 '19 at 14:52
  • @WillSawin I see. So each positive rational number $a\neq 1$ is equal to $p^q$ for some irrationals $p$ and $q$. – D.S. Lipham Jun 06 '19 at 22:24
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Every proof by contradiction can be seen as following the template identified in the theorem.

Namely, when we've proved a statement $S$ by contradiction, then $S$ follows from $S$ and also from $\neg S$.

Joel David Hamkins
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    I do not see how this addresses the question. The question specifically asked about "major, notorious conjectures in mathematics". A somewhat subjective definition but you probably agree that not every statement $S$ belongs to that category. Actually, now that I think of it $S$ would simultaneously have to be proven and to be a conjecture so I think this answer deserves some further clarification. –  Jun 03 '19 at 15:25
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    I took the central phenomenon of your question to be the situation where one statement follows from another and also from its negation. While that might seem unusual or even amazing, as you say, the point of my answer is that actually this happens quite frequently in mathematics. – Joel David Hamkins Jun 03 '19 at 15:28
  • I appreciate your shift of perspective. It is actually pretty interesting. Still, you addressed a very different question. If I could downvote, I would (but I also thank you for this insight, even if it was not presented in the appropriate place!). –  Jun 03 '19 at 15:35
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    I do not think that the proof structure being discussed by the question asker follows your pattern. Their pattern is [(P) --> (Q)] and [(not P) --> (Q)] therefore, (Q). That is standard proof by cases. You did something different. – Toothpick Anemone Jun 04 '19 at 20:15
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    @ToothpickAnemone - The answer was following (a special case of) the pattern posed in the question, not vice versa. Joel has explored the special case of P = Q. – Jirka Hanika Jun 04 '19 at 21:43