Probably true, but provably unprovable












31












$begingroup$


I'm wondering if anyone has found, or can find, a sequence of statements $P(n)$ ($n in mathbb{N}$) such that:




  1. Heuristic arguments using probability theory suggest that all the statements $P(n)$ are true.


  2. One can prove, preferably by making the probabilistic arguments rigorous, that infinitely many of the statements $P(n)$ are true.


  3. One can prove that only finitely many of the statements $P(n)$ are provable (say in Peano arithmetic or ZFC).



The goal here would be to find statements that are true 'just because they are probably true', not because we can put our finger on a reason why any individual one must be true.



An example of such a sequence of statements might be 'if $p_n$ is the $n$th prime, there are infinitely many repunit primes in base $p_n$'. However while this example meets condition 1 we seem far from having enough understanding of mathematics to prove 2, and 3 seems hopeless. I think we need a less charismatic sequence of statements that are more carefully crafted to the task at hand.






nt.number-theory lo.logic







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    It's not a fixed sequence but you can generate arbitrarily many probably true, ZFC independent statements by stating lower bounds on the Kolmogorov complexity of randomly generated binary strings. You can tune the probability that they're true by choosing the length of the string and the lower bound carefully.
    $endgroup$
    – James Hanson
    Nov 16 '18 at 17:19






  • 3




    $begingroup$
    What's the difference between "true" and "provable"?
    $endgroup$
    – YCor
    Nov 16 '18 at 18:13






  • 1




    $begingroup$
    Do statements like the Paris-Hamilton theorem count?
    $endgroup$
    – Ilya Bogdanov
    Nov 16 '18 at 19:37








  • 1




    $begingroup$
    How could you "prove that only finitely many of the statements $P(n)$ are provable" (condition 3) and also "prove that infinitely many of the statements $P(n)$ are true" (condition 2)? Is the latter proof supposed to depend on large cardinals or something?
    $endgroup$
    – Nik Weaver
    Nov 16 '18 at 20:53






  • 3




    $begingroup$
    The Parris-Harrington theorem gives an infinite sequence of statements $P(n)$ that are individually provable in Peano arithmetic, for which $forall n P(n)$ is not provable in PA. It also has nothing to do with probabilistic heuristics. So this fails to meet all three criteria I listed.
    $endgroup$
    – John Baez
    Nov 16 '18 at 22:11
















31












$begingroup$


I'm wondering if anyone has found, or can find, a sequence of statements $P(n)$ ($n in mathbb{N}$) such that:




  1. Heuristic arguments using probability theory suggest that all the statements $P(n)$ are true.


  2. One can prove, preferably by making the probabilistic arguments rigorous, that infinitely many of the statements $P(n)$ are true.


  3. One can prove that only finitely many of the statements $P(n)$ are provable (say in Peano arithmetic or ZFC).



The goal here would be to find statements that are true 'just because they are probably true', not because we can put our finger on a reason why any individual one must be true.



An example of such a sequence of statements might be 'if $p_n$ is the $n$th prime, there are infinitely many repunit primes in base $p_n$'. However while this example meets condition 1 we seem far from having enough understanding of mathematics to prove 2, and 3 seems hopeless. I think we need a less charismatic sequence of statements that are more carefully crafted to the task at hand.






nt.number-theory lo.logic







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    It's not a fixed sequence but you can generate arbitrarily many probably true, ZFC independent statements by stating lower bounds on the Kolmogorov complexity of randomly generated binary strings. You can tune the probability that they're true by choosing the length of the string and the lower bound carefully.
    $endgroup$
    – James Hanson
    Nov 16 '18 at 17:19






  • 3




    $begingroup$
    What's the difference between "true" and "provable"?
    $endgroup$
    – YCor
    Nov 16 '18 at 18:13






  • 1




    $begingroup$
    Do statements like the Paris-Hamilton theorem count?
    $endgroup$
    – Ilya Bogdanov
    Nov 16 '18 at 19:37








  • 1




    $begingroup$
    How could you "prove that only finitely many of the statements $P(n)$ are provable" (condition 3) and also "prove that infinitely many of the statements $P(n)$ are true" (condition 2)? Is the latter proof supposed to depend on large cardinals or something?
    $endgroup$
    – Nik Weaver
    Nov 16 '18 at 20:53






  • 3




    $begingroup$
    The Parris-Harrington theorem gives an infinite sequence of statements $P(n)$ that are individually provable in Peano arithmetic, for which $forall n P(n)$ is not provable in PA. It also has nothing to do with probabilistic heuristics. So this fails to meet all three criteria I listed.
    $endgroup$
    – John Baez
    Nov 16 '18 at 22:11














31












31








31


9



$begingroup$


I'm wondering if anyone has found, or can find, a sequence of statements $P(n)$ ($n in mathbb{N}$) such that:




  1. Heuristic arguments using probability theory suggest that all the statements $P(n)$ are true.


  2. One can prove, preferably by making the probabilistic arguments rigorous, that infinitely many of the statements $P(n)$ are true.


  3. One can prove that only finitely many of the statements $P(n)$ are provable (say in Peano arithmetic or ZFC).



The goal here would be to find statements that are true 'just because they are probably true', not because we can put our finger on a reason why any individual one must be true.



An example of such a sequence of statements might be 'if $p_n$ is the $n$th prime, there are infinitely many repunit primes in base $p_n$'. However while this example meets condition 1 we seem far from having enough understanding of mathematics to prove 2, and 3 seems hopeless. I think we need a less charismatic sequence of statements that are more carefully crafted to the task at hand.






nt.number-theory lo.logic







share|cite|improve this question











$endgroup$




I'm wondering if anyone has found, or can find, a sequence of statements $P(n)$ ($n in mathbb{N}$) such that:




  1. Heuristic arguments using probability theory suggest that all the statements $P(n)$ are true.


  2. One can prove, preferably by making the probabilistic arguments rigorous, that infinitely many of the statements $P(n)$ are true.


  3. One can prove that only finitely many of the statements $P(n)$ are provable (say in Peano arithmetic or ZFC).



The goal here would be to find statements that are true 'just because they are probably true', not because we can put our finger on a reason why any individual one must be true.



An example of such a sequence of statements might be 'if $p_n$ is the $n$th prime, there are infinitely many repunit primes in base $p_n$'. However while this example meets condition 1 we seem far from having enough understanding of mathematics to prove 2, and 3 seems hopeless. I think we need a less charismatic sequence of statements that are more carefully crafted to the task at hand.






nt.number-theory lo.logic




nt.number-theory lo.logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 16 '18 at 22:52







John Baez

















asked Nov 16 '18 at 17:08









John BaezJohn Baez

8,6174694




8,6174694








  • 5




    $begingroup$
    It's not a fixed sequence but you can generate arbitrarily many probably true, ZFC independent statements by stating lower bounds on the Kolmogorov complexity of randomly generated binary strings. You can tune the probability that they're true by choosing the length of the string and the lower bound carefully.
    $endgroup$
    – James Hanson
    Nov 16 '18 at 17:19






  • 3




    $begingroup$
    What's the difference between "true" and "provable"?
    $endgroup$
    – YCor
    Nov 16 '18 at 18:13






  • 1




    $begingroup$
    Do statements like the Paris-Hamilton theorem count?
    $endgroup$
    – Ilya Bogdanov
    Nov 16 '18 at 19:37








  • 1




    $begingroup$
    How could you "prove that only finitely many of the statements $P(n)$ are provable" (condition 3) and also "prove that infinitely many of the statements $P(n)$ are true" (condition 2)? Is the latter proof supposed to depend on large cardinals or something?
    $endgroup$
    – Nik Weaver
    Nov 16 '18 at 20:53






  • 3




    $begingroup$
    The Parris-Harrington theorem gives an infinite sequence of statements $P(n)$ that are individually provable in Peano arithmetic, for which $forall n P(n)$ is not provable in PA. It also has nothing to do with probabilistic heuristics. So this fails to meet all three criteria I listed.
    $endgroup$
    – John Baez
    Nov 16 '18 at 22:11














  • 5




    $begingroup$
    It's not a fixed sequence but you can generate arbitrarily many probably true, ZFC independent statements by stating lower bounds on the Kolmogorov complexity of randomly generated binary strings. You can tune the probability that they're true by choosing the length of the string and the lower bound carefully.
    $endgroup$
    – James Hanson
    Nov 16 '18 at 17:19






  • 3




    $begingroup$
    What's the difference between "true" and "provable"?
    $endgroup$
    – YCor
    Nov 16 '18 at 18:13






  • 1




    $begingroup$
    Do statements like the Paris-Hamilton theorem count?
    $endgroup$
    – Ilya Bogdanov
    Nov 16 '18 at 19:37








  • 1




    $begingroup$
    How could you "prove that only finitely many of the statements $P(n)$ are provable" (condition 3) and also "prove that infinitely many of the statements $P(n)$ are true" (condition 2)? Is the latter proof supposed to depend on large cardinals or something?
    $endgroup$
    – Nik Weaver
    Nov 16 '18 at 20:53






  • 3




    $begingroup$
    The Parris-Harrington theorem gives an infinite sequence of statements $P(n)$ that are individually provable in Peano arithmetic, for which $forall n P(n)$ is not provable in PA. It also has nothing to do with probabilistic heuristics. So this fails to meet all three criteria I listed.
    $endgroup$
    – John Baez
    Nov 16 '18 at 22:11








5




5




$begingroup$
It's not a fixed sequence but you can generate arbitrarily many probably true, ZFC independent statements by stating lower bounds on the Kolmogorov complexity of randomly generated binary strings. You can tune the probability that they're true by choosing the length of the string and the lower bound carefully.
$endgroup$
– James Hanson
Nov 16 '18 at 17:19




$begingroup$
It's not a fixed sequence but you can generate arbitrarily many probably true, ZFC independent statements by stating lower bounds on the Kolmogorov complexity of randomly generated binary strings. You can tune the probability that they're true by choosing the length of the string and the lower bound carefully.
$endgroup$
– James Hanson
Nov 16 '18 at 17:19




3




3




$begingroup$
What's the difference between "true" and "provable"?
$endgroup$
– YCor
Nov 16 '18 at 18:13




$begingroup$
What's the difference between "true" and "provable"?
$endgroup$
– YCor
Nov 16 '18 at 18:13




1




1




$begingroup$
Do statements like the Paris-Hamilton theorem count?
$endgroup$
– Ilya Bogdanov
Nov 16 '18 at 19:37






$begingroup$
Do statements like the Paris-Hamilton theorem count?
$endgroup$
– Ilya Bogdanov
Nov 16 '18 at 19:37






1




1




$begingroup$
How could you "prove that only finitely many of the statements $P(n)$ are provable" (condition 3) and also "prove that infinitely many of the statements $P(n)$ are true" (condition 2)? Is the latter proof supposed to depend on large cardinals or something?
$endgroup$
– Nik Weaver
Nov 16 '18 at 20:53




$begingroup$
How could you "prove that only finitely many of the statements $P(n)$ are provable" (condition 3) and also "prove that infinitely many of the statements $P(n)$ are true" (condition 2)? Is the latter proof supposed to depend on large cardinals or something?
$endgroup$
– Nik Weaver
Nov 16 '18 at 20:53




3




3




$begingroup$
The Parris-Harrington theorem gives an infinite sequence of statements $P(n)$ that are individually provable in Peano arithmetic, for which $forall n P(n)$ is not provable in PA. It also has nothing to do with probabilistic heuristics. So this fails to meet all three criteria I listed.
$endgroup$
– John Baez
Nov 16 '18 at 22:11




$begingroup$
The Parris-Harrington theorem gives an infinite sequence of statements $P(n)$ that are individually provable in Peano arithmetic, for which $forall n P(n)$ is not provable in PA. It also has nothing to do with probabilistic heuristics. So this fails to meet all three criteria I listed.
$endgroup$
– John Baez
Nov 16 '18 at 22:11










1 Answer
1






active

oldest

votes


















11












$begingroup$

Let $c$ be a constant such that
$$mathrm{PA}notvdash K(x)ge c$$
for all binary strings $x$, where $K$ is Kolmogorov complexity. Such a $c$ exists by Chaitin's Incompleteness Theorem and the linked Q&A discusses how to find it in $mathrm{PA}+mathrm{Con}(mathrm{PA})$.



For any $d$, let $x_d$ be the indicator string for the primes among nonnegative integers $le d$. So for instance, $x_5=001101$ and $x_{9}=0011010100$.



Let $P_{d}(n)$ be the statement
$$K(x_{n+d})ge c.$$



Let $d$ be large and then let $P(n)=P_d(n)$ for all $n$.




  • If $d$ is large enough then heuristically all the $P(n)$ are true.*


  • By the Pigeonhole Principle in $mathrm{PA}+mathrm{Con}(mathrm{PA})$ we prove that all but finitely many $P(n)$ are true.


  • We cannot prove any particular $P(n)$ in $mathrm{PA}$.



(*) Heuristics may suggest something like $dge ccdot mathcal H(1/log c)$ where $mathcal H$ is the entropy function, and we heuristically assume the primes are "random modulo the Prime Number Theorem", but the primes are computable so that heuristic is not quite right (thanks @EmilJerabek).






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    But primes are not random at all, they are computable. Thus the Kolmogorov complexity of $x_d$ is a constant plus the Kolmogorov complexity of $d$. If you defined $x_d$ as a string of $d$ zeros, it would behave in an essentially identical way.
    $endgroup$
    – Emil Jeřábek
    Nov 16 '18 at 22:11










  • $begingroup$
    @EmilJeřábek right, I'll update...
    $endgroup$
    – Bjørn Kjos-Hanssen
    Nov 16 '18 at 22:16











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









11












$begingroup$

Let $c$ be a constant such that
$$mathrm{PA}notvdash K(x)ge c$$
for all binary strings $x$, where $K$ is Kolmogorov complexity. Such a $c$ exists by Chaitin's Incompleteness Theorem and the linked Q&A discusses how to find it in $mathrm{PA}+mathrm{Con}(mathrm{PA})$.



For any $d$, let $x_d$ be the indicator string for the primes among nonnegative integers $le d$. So for instance, $x_5=001101$ and $x_{9}=0011010100$.



Let $P_{d}(n)$ be the statement
$$K(x_{n+d})ge c.$$



Let $d$ be large and then let $P(n)=P_d(n)$ for all $n$.




  • If $d$ is large enough then heuristically all the $P(n)$ are true.*


  • By the Pigeonhole Principle in $mathrm{PA}+mathrm{Con}(mathrm{PA})$ we prove that all but finitely many $P(n)$ are true.


  • We cannot prove any particular $P(n)$ in $mathrm{PA}$.



(*) Heuristics may suggest something like $dge ccdot mathcal H(1/log c)$ where $mathcal H$ is the entropy function, and we heuristically assume the primes are "random modulo the Prime Number Theorem", but the primes are computable so that heuristic is not quite right (thanks @EmilJerabek).






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    But primes are not random at all, they are computable. Thus the Kolmogorov complexity of $x_d$ is a constant plus the Kolmogorov complexity of $d$. If you defined $x_d$ as a string of $d$ zeros, it would behave in an essentially identical way.
    $endgroup$
    – Emil Jeřábek
    Nov 16 '18 at 22:11










  • $begingroup$
    @EmilJeřábek right, I'll update...
    $endgroup$
    – Bjørn Kjos-Hanssen
    Nov 16 '18 at 22:16
















11












$begingroup$

Let $c$ be a constant such that
$$mathrm{PA}notvdash K(x)ge c$$
for all binary strings $x$, where $K$ is Kolmogorov complexity. Such a $c$ exists by Chaitin's Incompleteness Theorem and the linked Q&A discusses how to find it in $mathrm{PA}+mathrm{Con}(mathrm{PA})$.



For any $d$, let $x_d$ be the indicator string for the primes among nonnegative integers $le d$. So for instance, $x_5=001101$ and $x_{9}=0011010100$.



Let $P_{d}(n)$ be the statement
$$K(x_{n+d})ge c.$$



Let $d$ be large and then let $P(n)=P_d(n)$ for all $n$.




  • If $d$ is large enough then heuristically all the $P(n)$ are true.*


  • By the Pigeonhole Principle in $mathrm{PA}+mathrm{Con}(mathrm{PA})$ we prove that all but finitely many $P(n)$ are true.


  • We cannot prove any particular $P(n)$ in $mathrm{PA}$.



(*) Heuristics may suggest something like $dge ccdot mathcal H(1/log c)$ where $mathcal H$ is the entropy function, and we heuristically assume the primes are "random modulo the Prime Number Theorem", but the primes are computable so that heuristic is not quite right (thanks @EmilJerabek).






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    But primes are not random at all, they are computable. Thus the Kolmogorov complexity of $x_d$ is a constant plus the Kolmogorov complexity of $d$. If you defined $x_d$ as a string of $d$ zeros, it would behave in an essentially identical way.
    $endgroup$
    – Emil Jeřábek
    Nov 16 '18 at 22:11










  • $begingroup$
    @EmilJeřábek right, I'll update...
    $endgroup$
    – Bjørn Kjos-Hanssen
    Nov 16 '18 at 22:16














11












11








11





$begingroup$

Let $c$ be a constant such that
$$mathrm{PA}notvdash K(x)ge c$$
for all binary strings $x$, where $K$ is Kolmogorov complexity. Such a $c$ exists by Chaitin's Incompleteness Theorem and the linked Q&A discusses how to find it in $mathrm{PA}+mathrm{Con}(mathrm{PA})$.



For any $d$, let $x_d$ be the indicator string for the primes among nonnegative integers $le d$. So for instance, $x_5=001101$ and $x_{9}=0011010100$.



Let $P_{d}(n)$ be the statement
$$K(x_{n+d})ge c.$$



Let $d$ be large and then let $P(n)=P_d(n)$ for all $n$.




  • If $d$ is large enough then heuristically all the $P(n)$ are true.*


  • By the Pigeonhole Principle in $mathrm{PA}+mathrm{Con}(mathrm{PA})$ we prove that all but finitely many $P(n)$ are true.


  • We cannot prove any particular $P(n)$ in $mathrm{PA}$.



(*) Heuristics may suggest something like $dge ccdot mathcal H(1/log c)$ where $mathcal H$ is the entropy function, and we heuristically assume the primes are "random modulo the Prime Number Theorem", but the primes are computable so that heuristic is not quite right (thanks @EmilJerabek).






share|cite|improve this answer











$endgroup$



Let $c$ be a constant such that
$$mathrm{PA}notvdash K(x)ge c$$
for all binary strings $x$, where $K$ is Kolmogorov complexity. Such a $c$ exists by Chaitin's Incompleteness Theorem and the linked Q&A discusses how to find it in $mathrm{PA}+mathrm{Con}(mathrm{PA})$.



For any $d$, let $x_d$ be the indicator string for the primes among nonnegative integers $le d$. So for instance, $x_5=001101$ and $x_{9}=0011010100$.



Let $P_{d}(n)$ be the statement
$$K(x_{n+d})ge c.$$



Let $d$ be large and then let $P(n)=P_d(n)$ for all $n$.




  • If $d$ is large enough then heuristically all the $P(n)$ are true.*


  • By the Pigeonhole Principle in $mathrm{PA}+mathrm{Con}(mathrm{PA})$ we prove that all but finitely many $P(n)$ are true.


  • We cannot prove any particular $P(n)$ in $mathrm{PA}$.



(*) Heuristics may suggest something like $dge ccdot mathcal H(1/log c)$ where $mathcal H$ is the entropy function, and we heuristically assume the primes are "random modulo the Prime Number Theorem", but the primes are computable so that heuristic is not quite right (thanks @EmilJerabek).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 16 '18 at 22:15

























answered Nov 16 '18 at 21:43









Bjørn Kjos-HanssenBjørn Kjos-Hanssen

17.9k33886




17.9k33886








  • 2




    $begingroup$
    But primes are not random at all, they are computable. Thus the Kolmogorov complexity of $x_d$ is a constant plus the Kolmogorov complexity of $d$. If you defined $x_d$ as a string of $d$ zeros, it would behave in an essentially identical way.
    $endgroup$
    – Emil Jeřábek
    Nov 16 '18 at 22:11










  • $begingroup$
    @EmilJeřábek right, I'll update...
    $endgroup$
    – Bjørn Kjos-Hanssen
    Nov 16 '18 at 22:16














  • 2




    $begingroup$
    But primes are not random at all, they are computable. Thus the Kolmogorov complexity of $x_d$ is a constant plus the Kolmogorov complexity of $d$. If you defined $x_d$ as a string of $d$ zeros, it would behave in an essentially identical way.
    $endgroup$
    – Emil Jeřábek
    Nov 16 '18 at 22:11










  • $begingroup$
    @EmilJeřábek right, I'll update...
    $endgroup$
    – Bjørn Kjos-Hanssen
    Nov 16 '18 at 22:16








2




2




$begingroup$
But primes are not random at all, they are computable. Thus the Kolmogorov complexity of $x_d$ is a constant plus the Kolmogorov complexity of $d$. If you defined $x_d$ as a string of $d$ zeros, it would behave in an essentially identical way.
$endgroup$
– Emil Jeřábek
Nov 16 '18 at 22:11




$begingroup$
But primes are not random at all, they are computable. Thus the Kolmogorov complexity of $x_d$ is a constant plus the Kolmogorov complexity of $d$. If you defined $x_d$ as a string of $d$ zeros, it would behave in an essentially identical way.
$endgroup$
– Emil Jeřábek
Nov 16 '18 at 22:11












$begingroup$
@EmilJeřábek right, I'll update...
$endgroup$
– Bjørn Kjos-Hanssen
Nov 16 '18 at 22:16




$begingroup$
@EmilJeřábek right, I'll update...
$endgroup$
– Bjørn Kjos-Hanssen
Nov 16 '18 at 22:16


















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National Museum of Racing and Hall of Fame

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Coordinates: 43°04′35″N 73°46′24″W  /  43.0763°N 73.7734°W  / 43.0763; -73.7734 Professional sports hall of fame in Saratoga Springs, New York National Museum of Racing and Hall of Fame National Museum of Racing and Hall of Fame Established 1951 Location Saratoga Springs, New York Type Professional sports hall of fame Director Cathy Marino Curator Victoria Tokarowski Website racingmuseum.org The National Museum of Racing and Hall of Fame was founded in 1951 in Saratoga Springs, New York, to honor the achievements of American Thoroughbred race horses, jockeys, and trainers. In 1955, the museum moved to its current location on Union Avenue near Saratoga race course, at which time inductions into the hall of fame began. Each spring, following the tabulation of the final votes, the announcement of new inductees is made, usually during Kentucky Derby Week in early May. The actual inductions are held in mid-August during the Saratoga race meeting...
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Guess what letter conforming each word

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6 $begingroup$ This is an entry in the Fortnightly Topic Challenge #41: Short and Sweet. 1 letter only could be mean anything . With an additional letter in front , could be used as medicine component and something you definitely learn or at least heard once. Add another letter in front , you would possibly hear that many times . Add an additional letter in front , you would found an independent country . With an additional letter, in the end , you would found something related to apple . Add another letter in the end , you would found something bad . Hint: Each letter is unique . word knowledge letters chemistry share | improve this question ...
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