Arriving at the same result with the opposite hypotheses“Strange” proofs of existence theorems



Arriving at the same result with the opposite hypotheses


“Strange” proofs of existence theorems













10












$begingroup$


I am pretty distant from anything analytic, including analytic number theory but I decided to read the Wikipedia page on the Riemann hypothesis 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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Related (and closed) question: mathoverflow.net/questions/312439/…
    $endgroup$
    – Sam Hopkins
    11 hours ago






  • 3




    $begingroup$
    @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.
    $endgroup$
    – Cut the wood
    11 hours ago






  • 1




    $begingroup$
    meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
    $endgroup$
    – Todd Trimble
    6 hours ago






  • 1




    $begingroup$
    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.
    $endgroup$
    – Christian Remling
    2 hours ago






  • 1




    $begingroup$
    @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
    $endgroup$
    – Pietro Majer
    2 hours ago
















10












$begingroup$


I am pretty distant from anything analytic, including analytic number theory but I decided to read the Wikipedia page on the Riemann hypothesis 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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Related (and closed) question: mathoverflow.net/questions/312439/…
    $endgroup$
    – Sam Hopkins
    11 hours ago






  • 3




    $begingroup$
    @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.
    $endgroup$
    – Cut the wood
    11 hours ago






  • 1




    $begingroup$
    meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
    $endgroup$
    – Todd Trimble
    6 hours ago






  • 1




    $begingroup$
    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.
    $endgroup$
    – Christian Remling
    2 hours ago






  • 1




    $begingroup$
    @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
    $endgroup$
    – Pietro Majer
    2 hours ago














10












10








10


3



$begingroup$


I am pretty distant from anything analytic, including analytic number theory but I decided to read the Wikipedia page on the Riemann hypothesis 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?










share|cite|improve this question











$endgroup$




I am pretty distant from anything analytic, including analytic number theory but I decided to read the Wikipedia page on the Riemann hypothesis 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?







big-list conjectures






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








asked 11 hours ago


























community wiki





Cut the wood













  • $begingroup$
    Related (and closed) question: mathoverflow.net/questions/312439/…
    $endgroup$
    – Sam Hopkins
    11 hours ago






  • 3




    $begingroup$
    @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.
    $endgroup$
    – Cut the wood
    11 hours ago






  • 1




    $begingroup$
    meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
    $endgroup$
    – Todd Trimble
    6 hours ago






  • 1




    $begingroup$
    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.
    $endgroup$
    – Christian Remling
    2 hours ago






  • 1




    $begingroup$
    @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
    $endgroup$
    – Pietro Majer
    2 hours ago


















  • $begingroup$
    Related (and closed) question: mathoverflow.net/questions/312439/…
    $endgroup$
    – Sam Hopkins
    11 hours ago






  • 3




    $begingroup$
    @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.
    $endgroup$
    – Cut the wood
    11 hours ago






  • 1




    $begingroup$
    meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
    $endgroup$
    – Todd Trimble
    6 hours ago






  • 1




    $begingroup$
    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.
    $endgroup$
    – Christian Remling
    2 hours ago






  • 1




    $begingroup$
    @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
    $endgroup$
    – Pietro Majer
    2 hours ago
















$begingroup$
Related (and closed) question: mathoverflow.net/questions/312439/…
$endgroup$
– Sam Hopkins
11 hours ago




$begingroup$
Related (and closed) question: mathoverflow.net/questions/312439/…
$endgroup$
– Sam Hopkins
11 hours ago




3




3




$begingroup$
@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.
$endgroup$
– Cut the wood
11 hours ago




$begingroup$
@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.
$endgroup$
– Cut the wood
11 hours ago




1




1




$begingroup$
meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
$endgroup$
– Todd Trimble
6 hours ago




$begingroup$
meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
$endgroup$
– Todd Trimble
6 hours ago




1




1




$begingroup$
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.
$endgroup$
– Christian Remling
2 hours ago




$begingroup$
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.
$endgroup$
– Christian Remling
2 hours ago




1




1




$begingroup$
@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
$endgroup$
– Pietro Majer
2 hours ago




$begingroup$
@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
$endgroup$
– Pietro Majer
2 hours ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

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 $neg S$.






share|cite|improve this answer











$endgroup$









  • 5




    $begingroup$
    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.
    $endgroup$
    – Cut the wood
    9 hours ago








  • 1




    $begingroup$
    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.
    $endgroup$
    – Joel David Hamkins
    9 hours ago












  • $begingroup$
    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!).
    $endgroup$
    – Cut the wood
    8 hours ago



















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – Cut the wood
    8 hours ago





















2












$begingroup$

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 polynomial in $n$.



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 pseduo-random generator with the desired properties.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    is the existence of a pseudo-random generator hard?
    $endgroup$
    – Cut the wood
    4 hours ago










  • $begingroup$
    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 $P neq NP$.
    $endgroup$
    – Izaak Meckler
    4 hours ago














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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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 $neg S$.






share|cite|improve this answer











$endgroup$









  • 5




    $begingroup$
    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.
    $endgroup$
    – Cut the wood
    9 hours ago








  • 1




    $begingroup$
    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.
    $endgroup$
    – Joel David Hamkins
    9 hours ago












  • $begingroup$
    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!).
    $endgroup$
    – Cut the wood
    8 hours ago
















4












$begingroup$

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 $neg S$.






share|cite|improve this answer











$endgroup$









  • 5




    $begingroup$
    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.
    $endgroup$
    – Cut the wood
    9 hours ago








  • 1




    $begingroup$
    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.
    $endgroup$
    – Joel David Hamkins
    9 hours ago












  • $begingroup$
    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!).
    $endgroup$
    – Cut the wood
    8 hours ago














4












4








4





$begingroup$

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 $neg S$.






share|cite|improve this answer











$endgroup$



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 $neg S$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








answered 9 hours ago


























community wiki





Joel David Hamkins









  • 5




    $begingroup$
    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.
    $endgroup$
    – Cut the wood
    9 hours ago








  • 1




    $begingroup$
    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.
    $endgroup$
    – Joel David Hamkins
    9 hours ago












  • $begingroup$
    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!).
    $endgroup$
    – Cut the wood
    8 hours ago














  • 5




    $begingroup$
    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.
    $endgroup$
    – Cut the wood
    9 hours ago








  • 1




    $begingroup$
    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.
    $endgroup$
    – Joel David Hamkins
    9 hours ago












  • $begingroup$
    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!).
    $endgroup$
    – Cut the wood
    8 hours ago








5




5




$begingroup$
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.
$endgroup$
– Cut the wood
9 hours ago






$begingroup$
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.
$endgroup$
– Cut the wood
9 hours ago






1




1




$begingroup$
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.
$endgroup$
– Joel David Hamkins
9 hours ago






$begingroup$
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.
$endgroup$
– Joel David Hamkins
9 hours ago














$begingroup$
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!).
$endgroup$
– Cut the wood
8 hours ago




$begingroup$
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!).
$endgroup$
– Cut the wood
8 hours ago











3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – Cut the wood
    8 hours ago


















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – Cut the wood
    8 hours ago
















3












3








3





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








answered 8 hours ago


























community wiki





user36212













  • $begingroup$
    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.
    $endgroup$
    – Cut the wood
    8 hours ago




















  • $begingroup$
    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.
    $endgroup$
    – Cut the wood
    8 hours ago


















$begingroup$
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.
$endgroup$
– Cut the wood
8 hours ago






$begingroup$
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.
$endgroup$
– Cut the wood
8 hours ago













2












$begingroup$

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 polynomial in $n$.



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 pseduo-random generator with the desired properties.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    is the existence of a pseudo-random generator hard?
    $endgroup$
    – Cut the wood
    4 hours ago










  • $begingroup$
    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 $P neq NP$.
    $endgroup$
    – Izaak Meckler
    4 hours ago


















2












$begingroup$

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 polynomial in $n$.



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 pseduo-random generator with the desired properties.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    is the existence of a pseudo-random generator hard?
    $endgroup$
    – Cut the wood
    4 hours ago










  • $begingroup$
    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 $P neq NP$.
    $endgroup$
    – Izaak Meckler
    4 hours ago
















2












2








2





$begingroup$

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 polynomial in $n$.



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 pseduo-random generator with the desired properties.






share|cite|improve this answer











$endgroup$



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 polynomial in $n$.



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 pseduo-random generator with the desired properties.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








answered 4 hours ago


























community wiki





Izaak Meckler













  • $begingroup$
    is the existence of a pseudo-random generator hard?
    $endgroup$
    – Cut the wood
    4 hours ago










  • $begingroup$
    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 $P neq NP$.
    $endgroup$
    – Izaak Meckler
    4 hours ago




















  • $begingroup$
    is the existence of a pseudo-random generator hard?
    $endgroup$
    – Cut the wood
    4 hours ago










  • $begingroup$
    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 $P neq NP$.
    $endgroup$
    – Izaak Meckler
    4 hours ago


















$begingroup$
is the existence of a pseudo-random generator hard?
$endgroup$
– Cut the wood
4 hours ago




$begingroup$
is the existence of a pseudo-random generator hard?
$endgroup$
– Cut the wood
4 hours ago












$begingroup$
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 $P neq NP$.
$endgroup$
– Izaak Meckler
4 hours ago






$begingroup$
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 $P neq NP$.
$endgroup$
– Izaak Meckler
4 hours ago




















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