Is there a property in $mathbb{N}$ that we know some number must satisfy but don't know which one?












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I have two questions.




$(1.)$ Is there a property of the natural numbers such that we know at least one number satisfies it but we don't know which one?




Even more,




$(2.)$ Is there a property of the natural numbers such that we know at least one number satisfies but we don't know which one and moreover we cannot bound any such number by another known number?




Now the question might arise of what does it mean to know a number. To be able to write it down in finite time given an infinite amount ink and paper would suffice.










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  • $begingroup$
    Well, I'm not sure if that will help, but I've heard that number theorists know that one of the numbers $3$, $5$ or $7$ (not sure now if these are the numbers, but there are three of them) is a primitive root $operatorname{mod}: p$ for an infinite amount of prime numbers $p$, but they don't know which one of them.
    $endgroup$
    – Shoutre
    Oct 24 '15 at 20:59








  • 1




    $begingroup$
    This question is similar but narrower, and has numerous good answers that fit well here too: math.stackexchange.com/questions/1315615/…
    $endgroup$
    – Travis
    Oct 24 '15 at 21:19
















15












$begingroup$


I have two questions.




$(1.)$ Is there a property of the natural numbers such that we know at least one number satisfies it but we don't know which one?




Even more,




$(2.)$ Is there a property of the natural numbers such that we know at least one number satisfies but we don't know which one and moreover we cannot bound any such number by another known number?




Now the question might arise of what does it mean to know a number. To be able to write it down in finite time given an infinite amount ink and paper would suffice.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, I'm not sure if that will help, but I've heard that number theorists know that one of the numbers $3$, $5$ or $7$ (not sure now if these are the numbers, but there are three of them) is a primitive root $operatorname{mod}: p$ for an infinite amount of prime numbers $p$, but they don't know which one of them.
    $endgroup$
    – Shoutre
    Oct 24 '15 at 20:59








  • 1




    $begingroup$
    This question is similar but narrower, and has numerous good answers that fit well here too: math.stackexchange.com/questions/1315615/…
    $endgroup$
    – Travis
    Oct 24 '15 at 21:19














15












15








15


3



$begingroup$


I have two questions.




$(1.)$ Is there a property of the natural numbers such that we know at least one number satisfies it but we don't know which one?




Even more,




$(2.)$ Is there a property of the natural numbers such that we know at least one number satisfies but we don't know which one and moreover we cannot bound any such number by another known number?




Now the question might arise of what does it mean to know a number. To be able to write it down in finite time given an infinite amount ink and paper would suffice.










share|cite|improve this question











$endgroup$




I have two questions.




$(1.)$ Is there a property of the natural numbers such that we know at least one number satisfies it but we don't know which one?




Even more,




$(2.)$ Is there a property of the natural numbers such that we know at least one number satisfies but we don't know which one and moreover we cannot bound any such number by another known number?




Now the question might arise of what does it mean to know a number. To be able to write it down in finite time given an infinite amount ink and paper would suffice.







elementary-number-theory big-list






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edited Dec 6 '18 at 16:09







Anguepa

















asked Oct 24 '15 at 20:55









AnguepaAnguepa

1,391819




1,391819












  • $begingroup$
    Well, I'm not sure if that will help, but I've heard that number theorists know that one of the numbers $3$, $5$ or $7$ (not sure now if these are the numbers, but there are three of them) is a primitive root $operatorname{mod}: p$ for an infinite amount of prime numbers $p$, but they don't know which one of them.
    $endgroup$
    – Shoutre
    Oct 24 '15 at 20:59








  • 1




    $begingroup$
    This question is similar but narrower, and has numerous good answers that fit well here too: math.stackexchange.com/questions/1315615/…
    $endgroup$
    – Travis
    Oct 24 '15 at 21:19


















  • $begingroup$
    Well, I'm not sure if that will help, but I've heard that number theorists know that one of the numbers $3$, $5$ or $7$ (not sure now if these are the numbers, but there are three of them) is a primitive root $operatorname{mod}: p$ for an infinite amount of prime numbers $p$, but they don't know which one of them.
    $endgroup$
    – Shoutre
    Oct 24 '15 at 20:59








  • 1




    $begingroup$
    This question is similar but narrower, and has numerous good answers that fit well here too: math.stackexchange.com/questions/1315615/…
    $endgroup$
    – Travis
    Oct 24 '15 at 21:19
















$begingroup$
Well, I'm not sure if that will help, but I've heard that number theorists know that one of the numbers $3$, $5$ or $7$ (not sure now if these are the numbers, but there are three of them) is a primitive root $operatorname{mod}: p$ for an infinite amount of prime numbers $p$, but they don't know which one of them.
$endgroup$
– Shoutre
Oct 24 '15 at 20:59






$begingroup$
Well, I'm not sure if that will help, but I've heard that number theorists know that one of the numbers $3$, $5$ or $7$ (not sure now if these are the numbers, but there are three of them) is a primitive root $operatorname{mod}: p$ for an infinite amount of prime numbers $p$, but they don't know which one of them.
$endgroup$
– Shoutre
Oct 24 '15 at 20:59






1




1




$begingroup$
This question is similar but narrower, and has numerous good answers that fit well here too: math.stackexchange.com/questions/1315615/…
$endgroup$
– Travis
Oct 24 '15 at 21:19




$begingroup$
This question is similar but narrower, and has numerous good answers that fit well here too: math.stackexchange.com/questions/1315615/…
$endgroup$
– Travis
Oct 24 '15 at 21:19










6 Answers
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Goldbach's conjecture is the statement that, if $n$ is an even natural number and $n>2$ then $n$ is the sum of two primes. It is not known whether this conjecture is true, nor is any bound known on how big the first counterexample could be if the conjecture is false. In view of this situation, consider the following property of a natural number $n$: Either $n$ is the first counterexample to Goldbach's conjecture or Goldbach's conjecture is true and $n=0$. We know that there is exactly one $n$ with this property (because either Goldbach's conjecture is true or it has exactly one first counterexample), but we don't have any reasonable bound for how big $n$ could be. (The weasely word "reasonable" is there to exclude "bounds" like "$7+$ the first counterexample to Goldbach's conjecture or $13$ if Goldbach's conjecture is true".)






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  • $begingroup$
    I've been rethinking about your answer and noticed I can be somehow picky about it. We do not know whether zero satisfies the property you've described. What about a property $p$ that answers my second question and such that for any $nin mathbb{N}$ we can answer whether $p(n)$ or $neg p(n)$? If you can think of one feel free to share it :)
    $endgroup$
    – Anguepa
    Apr 16 '16 at 11:19










  • $begingroup$
    @Anguepa By "we can answer whether $p(n)$ or $neg p(n)$", do you mean that we know an algorithm which, given any $n$, eventually decides whether $p(n)$ or $neg p(n)$?
    $endgroup$
    – Andreas Blass
    Apr 16 '16 at 19:00










  • $begingroup$
    Yes. Sorry I should've made that clear.
    $endgroup$
    – Anguepa
    Apr 17 '16 at 18:02










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    @Anguepa There's still a loophole, allowing an answer to Question 1 that's almost surely not what you really want. Consider the property "$n$ is the exact number of primes below $10^{10^{10}}$." There's exactly one $n$ with this property, and we know an algorithm to decide whether any proposed $n$ is the right one; indeed, we have an algorithm that computes the right $n$. But the algorithm is awfully slow, and I don't think we actually know the right $n$. (If we do, extend the tower of exponents a few more steps.)
    $endgroup$
    – Andreas Blass
    Apr 17 '16 at 23:02










  • $begingroup$
    This seems to be a more simple version of AJ Stas's answer to question 1. I am however curious about a property that answers question 2, like the one you gave in your answer, but having the additional characteristic we've discussed.
    $endgroup$
    – Anguepa
    Apr 18 '16 at 10:38





















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For the first question, we know there is a smallest natural number $n$ such that $pi(n)>text{li}(n)$ where $pi(x)$ is the prime counting function and li$(x)$ is the logarithmic integral, but we don't know what the number is.



EDIT: For the sake of having a more complete answer, I will include our discussion in the comments.



There has been much progress in determining an upper bound for the first integer at which this switch occurs, but still the number eludes us. Strangely enough, we know that there are infinitely many numbers for which this inequality holds.






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    This is what I had in mind; not only do we know there is a smallest, we know that $pi(x)-mathrm{Li}(x)$ changes sign infinitely many times, and we don't have an example yet of when it is positive.
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    – Clayton
    Oct 24 '15 at 21:02












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    Very interesting @AJStas. Doesn't this answer the second question too?
    $endgroup$
    – Anguepa
    Oct 24 '15 at 21:06






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    @Anguepa: We have bounds for where such a number must live, but the bounds are far too large to expect a modern computer to be able to handle it in practical time.
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    – Clayton
    Oct 24 '15 at 21:08










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    @Anguepa There have been many upper bounds for this number though, so I wasn't really sure.
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    – Rocket Man
    Oct 24 '15 at 21:08






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    @Anguepa: A quick Google search yields a page with some bounds along the lines of $10^{316}$.
    $endgroup$
    – Clayton
    Oct 24 '15 at 21:12



















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Here's an interesting example that is the subject of recent (and celebrated) research:



Consider the set ${p_1, p_2, p_3, ldots}$ of prime numbers, listed in increasing order. The $n$th prime gap is the difference $p_{n + 1} - p_n$. In 2013, Yitang Zhang showed that there are infinitely many prime gaps of size $leq 7 cdot 10^7$. Since there are only finitely many positive integers smaller than this size, at least one of them must occur infinitely often, but we don't know which. Since then, the Polymath project has improved that upper bound to $246$ (as of last December). The famous and longstanding Twin Prime Conjecture says that $2$ occurs infinitely often.






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    6












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    It is known that one of
    ζ(5), ζ(7), ζ(9), and ζ(11) must be irrational,
    but no one knows which one:



    W. Zudilin (2001). "One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational". Russ. Math. Surv. 56 (4): 774–776. doi:10.1070/RM2001v056n04ABEH000427.






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    • 2




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      Probably all of them are irrational.
      $endgroup$
      – azimut
      Oct 30 '15 at 21:20






    • 2




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      Heck, probably all of them are transcendental, normal to all bases, and have all the properties that almost all numbers have.
      $endgroup$
      – marty cohen
      Oct 30 '15 at 21:38



















    1












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    Consider $Sigma(m,n)$ to be the busy beaver function with $m$ states and $n$ colors.



    The following is then very hard to find:




    What is the smallest $m$ such that $Sigma(m,2)>Sigma(10,3)$?




    Since there is a color reduction method for reducing a $k$ state, 3 color machine to a $7k$ state, 2 color machine, it is known that $mleq 70$. On the other hand, it is clear that $m>10$. But it is very hard to say anything between this.






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      Since we do not know , for example $Sigma(5)$ , we do not know which number $n$ satisfies $n=Sigma(5)$, but we know that some number $n$ must satisfy the property. In this case we do not even have an upper bound.



      Another less trivial example : The $20$-th Fermat-number $2^{2^{20}}+1$ is known to be composite, but no factor is known.



      Denote $n$ to be the smallest prime factor of the $20$-th Fermat-number.
      This number is well-defined but we do not know it, but in this case, we can bound it.



      If this does not hit your intention, please precise which kind of properties you mean.






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      • $begingroup$
        what means $Sigma$
        $endgroup$
        – miracle173
        Jun 10 '17 at 6:28













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      6 Answers
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      6 Answers
      6






      active

      oldest

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      active

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      2












      $begingroup$

      Goldbach's conjecture is the statement that, if $n$ is an even natural number and $n>2$ then $n$ is the sum of two primes. It is not known whether this conjecture is true, nor is any bound known on how big the first counterexample could be if the conjecture is false. In view of this situation, consider the following property of a natural number $n$: Either $n$ is the first counterexample to Goldbach's conjecture or Goldbach's conjecture is true and $n=0$. We know that there is exactly one $n$ with this property (because either Goldbach's conjecture is true or it has exactly one first counterexample), but we don't have any reasonable bound for how big $n$ could be. (The weasely word "reasonable" is there to exclude "bounds" like "$7+$ the first counterexample to Goldbach's conjecture or $13$ if Goldbach's conjecture is true".)






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I've been rethinking about your answer and noticed I can be somehow picky about it. We do not know whether zero satisfies the property you've described. What about a property $p$ that answers my second question and such that for any $nin mathbb{N}$ we can answer whether $p(n)$ or $neg p(n)$? If you can think of one feel free to share it :)
        $endgroup$
        – Anguepa
        Apr 16 '16 at 11:19










      • $begingroup$
        @Anguepa By "we can answer whether $p(n)$ or $neg p(n)$", do you mean that we know an algorithm which, given any $n$, eventually decides whether $p(n)$ or $neg p(n)$?
        $endgroup$
        – Andreas Blass
        Apr 16 '16 at 19:00










      • $begingroup$
        Yes. Sorry I should've made that clear.
        $endgroup$
        – Anguepa
        Apr 17 '16 at 18:02










      • $begingroup$
        @Anguepa There's still a loophole, allowing an answer to Question 1 that's almost surely not what you really want. Consider the property "$n$ is the exact number of primes below $10^{10^{10}}$." There's exactly one $n$ with this property, and we know an algorithm to decide whether any proposed $n$ is the right one; indeed, we have an algorithm that computes the right $n$. But the algorithm is awfully slow, and I don't think we actually know the right $n$. (If we do, extend the tower of exponents a few more steps.)
        $endgroup$
        – Andreas Blass
        Apr 17 '16 at 23:02










      • $begingroup$
        This seems to be a more simple version of AJ Stas's answer to question 1. I am however curious about a property that answers question 2, like the one you gave in your answer, but having the additional characteristic we've discussed.
        $endgroup$
        – Anguepa
        Apr 18 '16 at 10:38


















      2












      $begingroup$

      Goldbach's conjecture is the statement that, if $n$ is an even natural number and $n>2$ then $n$ is the sum of two primes. It is not known whether this conjecture is true, nor is any bound known on how big the first counterexample could be if the conjecture is false. In view of this situation, consider the following property of a natural number $n$: Either $n$ is the first counterexample to Goldbach's conjecture or Goldbach's conjecture is true and $n=0$. We know that there is exactly one $n$ with this property (because either Goldbach's conjecture is true or it has exactly one first counterexample), but we don't have any reasonable bound for how big $n$ could be. (The weasely word "reasonable" is there to exclude "bounds" like "$7+$ the first counterexample to Goldbach's conjecture or $13$ if Goldbach's conjecture is true".)






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I've been rethinking about your answer and noticed I can be somehow picky about it. We do not know whether zero satisfies the property you've described. What about a property $p$ that answers my second question and such that for any $nin mathbb{N}$ we can answer whether $p(n)$ or $neg p(n)$? If you can think of one feel free to share it :)
        $endgroup$
        – Anguepa
        Apr 16 '16 at 11:19










      • $begingroup$
        @Anguepa By "we can answer whether $p(n)$ or $neg p(n)$", do you mean that we know an algorithm which, given any $n$, eventually decides whether $p(n)$ or $neg p(n)$?
        $endgroup$
        – Andreas Blass
        Apr 16 '16 at 19:00










      • $begingroup$
        Yes. Sorry I should've made that clear.
        $endgroup$
        – Anguepa
        Apr 17 '16 at 18:02










      • $begingroup$
        @Anguepa There's still a loophole, allowing an answer to Question 1 that's almost surely not what you really want. Consider the property "$n$ is the exact number of primes below $10^{10^{10}}$." There's exactly one $n$ with this property, and we know an algorithm to decide whether any proposed $n$ is the right one; indeed, we have an algorithm that computes the right $n$. But the algorithm is awfully slow, and I don't think we actually know the right $n$. (If we do, extend the tower of exponents a few more steps.)
        $endgroup$
        – Andreas Blass
        Apr 17 '16 at 23:02










      • $begingroup$
        This seems to be a more simple version of AJ Stas's answer to question 1. I am however curious about a property that answers question 2, like the one you gave in your answer, but having the additional characteristic we've discussed.
        $endgroup$
        – Anguepa
        Apr 18 '16 at 10:38
















      2












      2








      2





      $begingroup$

      Goldbach's conjecture is the statement that, if $n$ is an even natural number and $n>2$ then $n$ is the sum of two primes. It is not known whether this conjecture is true, nor is any bound known on how big the first counterexample could be if the conjecture is false. In view of this situation, consider the following property of a natural number $n$: Either $n$ is the first counterexample to Goldbach's conjecture or Goldbach's conjecture is true and $n=0$. We know that there is exactly one $n$ with this property (because either Goldbach's conjecture is true or it has exactly one first counterexample), but we don't have any reasonable bound for how big $n$ could be. (The weasely word "reasonable" is there to exclude "bounds" like "$7+$ the first counterexample to Goldbach's conjecture or $13$ if Goldbach's conjecture is true".)






      share|cite|improve this answer









      $endgroup$



      Goldbach's conjecture is the statement that, if $n$ is an even natural number and $n>2$ then $n$ is the sum of two primes. It is not known whether this conjecture is true, nor is any bound known on how big the first counterexample could be if the conjecture is false. In view of this situation, consider the following property of a natural number $n$: Either $n$ is the first counterexample to Goldbach's conjecture or Goldbach's conjecture is true and $n=0$. We know that there is exactly one $n$ with this property (because either Goldbach's conjecture is true or it has exactly one first counterexample), but we don't have any reasonable bound for how big $n$ could be. (The weasely word "reasonable" is there to exclude "bounds" like "$7+$ the first counterexample to Goldbach's conjecture or $13$ if Goldbach's conjecture is true".)







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Oct 25 '15 at 11:00









      Andreas BlassAndreas Blass

      49.4k351108




      49.4k351108












      • $begingroup$
        I've been rethinking about your answer and noticed I can be somehow picky about it. We do not know whether zero satisfies the property you've described. What about a property $p$ that answers my second question and such that for any $nin mathbb{N}$ we can answer whether $p(n)$ or $neg p(n)$? If you can think of one feel free to share it :)
        $endgroup$
        – Anguepa
        Apr 16 '16 at 11:19










      • $begingroup$
        @Anguepa By "we can answer whether $p(n)$ or $neg p(n)$", do you mean that we know an algorithm which, given any $n$, eventually decides whether $p(n)$ or $neg p(n)$?
        $endgroup$
        – Andreas Blass
        Apr 16 '16 at 19:00










      • $begingroup$
        Yes. Sorry I should've made that clear.
        $endgroup$
        – Anguepa
        Apr 17 '16 at 18:02










      • $begingroup$
        @Anguepa There's still a loophole, allowing an answer to Question 1 that's almost surely not what you really want. Consider the property "$n$ is the exact number of primes below $10^{10^{10}}$." There's exactly one $n$ with this property, and we know an algorithm to decide whether any proposed $n$ is the right one; indeed, we have an algorithm that computes the right $n$. But the algorithm is awfully slow, and I don't think we actually know the right $n$. (If we do, extend the tower of exponents a few more steps.)
        $endgroup$
        – Andreas Blass
        Apr 17 '16 at 23:02










      • $begingroup$
        This seems to be a more simple version of AJ Stas's answer to question 1. I am however curious about a property that answers question 2, like the one you gave in your answer, but having the additional characteristic we've discussed.
        $endgroup$
        – Anguepa
        Apr 18 '16 at 10:38




















      • $begingroup$
        I've been rethinking about your answer and noticed I can be somehow picky about it. We do not know whether zero satisfies the property you've described. What about a property $p$ that answers my second question and such that for any $nin mathbb{N}$ we can answer whether $p(n)$ or $neg p(n)$? If you can think of one feel free to share it :)
        $endgroup$
        – Anguepa
        Apr 16 '16 at 11:19










      • $begingroup$
        @Anguepa By "we can answer whether $p(n)$ or $neg p(n)$", do you mean that we know an algorithm which, given any $n$, eventually decides whether $p(n)$ or $neg p(n)$?
        $endgroup$
        – Andreas Blass
        Apr 16 '16 at 19:00










      • $begingroup$
        Yes. Sorry I should've made that clear.
        $endgroup$
        – Anguepa
        Apr 17 '16 at 18:02










      • $begingroup$
        @Anguepa There's still a loophole, allowing an answer to Question 1 that's almost surely not what you really want. Consider the property "$n$ is the exact number of primes below $10^{10^{10}}$." There's exactly one $n$ with this property, and we know an algorithm to decide whether any proposed $n$ is the right one; indeed, we have an algorithm that computes the right $n$. But the algorithm is awfully slow, and I don't think we actually know the right $n$. (If we do, extend the tower of exponents a few more steps.)
        $endgroup$
        – Andreas Blass
        Apr 17 '16 at 23:02










      • $begingroup$
        This seems to be a more simple version of AJ Stas's answer to question 1. I am however curious about a property that answers question 2, like the one you gave in your answer, but having the additional characteristic we've discussed.
        $endgroup$
        – Anguepa
        Apr 18 '16 at 10:38


















      $begingroup$
      I've been rethinking about your answer and noticed I can be somehow picky about it. We do not know whether zero satisfies the property you've described. What about a property $p$ that answers my second question and such that for any $nin mathbb{N}$ we can answer whether $p(n)$ or $neg p(n)$? If you can think of one feel free to share it :)
      $endgroup$
      – Anguepa
      Apr 16 '16 at 11:19




      $begingroup$
      I've been rethinking about your answer and noticed I can be somehow picky about it. We do not know whether zero satisfies the property you've described. What about a property $p$ that answers my second question and such that for any $nin mathbb{N}$ we can answer whether $p(n)$ or $neg p(n)$? If you can think of one feel free to share it :)
      $endgroup$
      – Anguepa
      Apr 16 '16 at 11:19












      $begingroup$
      @Anguepa By "we can answer whether $p(n)$ or $neg p(n)$", do you mean that we know an algorithm which, given any $n$, eventually decides whether $p(n)$ or $neg p(n)$?
      $endgroup$
      – Andreas Blass
      Apr 16 '16 at 19:00




      $begingroup$
      @Anguepa By "we can answer whether $p(n)$ or $neg p(n)$", do you mean that we know an algorithm which, given any $n$, eventually decides whether $p(n)$ or $neg p(n)$?
      $endgroup$
      – Andreas Blass
      Apr 16 '16 at 19:00












      $begingroup$
      Yes. Sorry I should've made that clear.
      $endgroup$
      – Anguepa
      Apr 17 '16 at 18:02




      $begingroup$
      Yes. Sorry I should've made that clear.
      $endgroup$
      – Anguepa
      Apr 17 '16 at 18:02












      $begingroup$
      @Anguepa There's still a loophole, allowing an answer to Question 1 that's almost surely not what you really want. Consider the property "$n$ is the exact number of primes below $10^{10^{10}}$." There's exactly one $n$ with this property, and we know an algorithm to decide whether any proposed $n$ is the right one; indeed, we have an algorithm that computes the right $n$. But the algorithm is awfully slow, and I don't think we actually know the right $n$. (If we do, extend the tower of exponents a few more steps.)
      $endgroup$
      – Andreas Blass
      Apr 17 '16 at 23:02




      $begingroup$
      @Anguepa There's still a loophole, allowing an answer to Question 1 that's almost surely not what you really want. Consider the property "$n$ is the exact number of primes below $10^{10^{10}}$." There's exactly one $n$ with this property, and we know an algorithm to decide whether any proposed $n$ is the right one; indeed, we have an algorithm that computes the right $n$. But the algorithm is awfully slow, and I don't think we actually know the right $n$. (If we do, extend the tower of exponents a few more steps.)
      $endgroup$
      – Andreas Blass
      Apr 17 '16 at 23:02












      $begingroup$
      This seems to be a more simple version of AJ Stas's answer to question 1. I am however curious about a property that answers question 2, like the one you gave in your answer, but having the additional characteristic we've discussed.
      $endgroup$
      – Anguepa
      Apr 18 '16 at 10:38






      $begingroup$
      This seems to be a more simple version of AJ Stas's answer to question 1. I am however curious about a property that answers question 2, like the one you gave in your answer, but having the additional characteristic we've discussed.
      $endgroup$
      – Anguepa
      Apr 18 '16 at 10:38













      11












      $begingroup$

      For the first question, we know there is a smallest natural number $n$ such that $pi(n)>text{li}(n)$ where $pi(x)$ is the prime counting function and li$(x)$ is the logarithmic integral, but we don't know what the number is.



      EDIT: For the sake of having a more complete answer, I will include our discussion in the comments.



      There has been much progress in determining an upper bound for the first integer at which this switch occurs, but still the number eludes us. Strangely enough, we know that there are infinitely many numbers for which this inequality holds.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        This is what I had in mind; not only do we know there is a smallest, we know that $pi(x)-mathrm{Li}(x)$ changes sign infinitely many times, and we don't have an example yet of when it is positive.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:02












      • $begingroup$
        Very interesting @AJStas. Doesn't this answer the second question too?
        $endgroup$
        – Anguepa
        Oct 24 '15 at 21:06






      • 1




        $begingroup$
        @Anguepa: We have bounds for where such a number must live, but the bounds are far too large to expect a modern computer to be able to handle it in practical time.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:08










      • $begingroup$
        @Anguepa There have been many upper bounds for this number though, so I wasn't really sure.
        $endgroup$
        – Rocket Man
        Oct 24 '15 at 21:08






      • 1




        $begingroup$
        @Anguepa: A quick Google search yields a page with some bounds along the lines of $10^{316}$.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:12
















      11












      $begingroup$

      For the first question, we know there is a smallest natural number $n$ such that $pi(n)>text{li}(n)$ where $pi(x)$ is the prime counting function and li$(x)$ is the logarithmic integral, but we don't know what the number is.



      EDIT: For the sake of having a more complete answer, I will include our discussion in the comments.



      There has been much progress in determining an upper bound for the first integer at which this switch occurs, but still the number eludes us. Strangely enough, we know that there are infinitely many numbers for which this inequality holds.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        This is what I had in mind; not only do we know there is a smallest, we know that $pi(x)-mathrm{Li}(x)$ changes sign infinitely many times, and we don't have an example yet of when it is positive.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:02












      • $begingroup$
        Very interesting @AJStas. Doesn't this answer the second question too?
        $endgroup$
        – Anguepa
        Oct 24 '15 at 21:06






      • 1




        $begingroup$
        @Anguepa: We have bounds for where such a number must live, but the bounds are far too large to expect a modern computer to be able to handle it in practical time.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:08










      • $begingroup$
        @Anguepa There have been many upper bounds for this number though, so I wasn't really sure.
        $endgroup$
        – Rocket Man
        Oct 24 '15 at 21:08






      • 1




        $begingroup$
        @Anguepa: A quick Google search yields a page with some bounds along the lines of $10^{316}$.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:12














      11












      11








      11





      $begingroup$

      For the first question, we know there is a smallest natural number $n$ such that $pi(n)>text{li}(n)$ where $pi(x)$ is the prime counting function and li$(x)$ is the logarithmic integral, but we don't know what the number is.



      EDIT: For the sake of having a more complete answer, I will include our discussion in the comments.



      There has been much progress in determining an upper bound for the first integer at which this switch occurs, but still the number eludes us. Strangely enough, we know that there are infinitely many numbers for which this inequality holds.






      share|cite|improve this answer











      $endgroup$



      For the first question, we know there is a smallest natural number $n$ such that $pi(n)>text{li}(n)$ where $pi(x)$ is the prime counting function and li$(x)$ is the logarithmic integral, but we don't know what the number is.



      EDIT: For the sake of having a more complete answer, I will include our discussion in the comments.



      There has been much progress in determining an upper bound for the first integer at which this switch occurs, but still the number eludes us. Strangely enough, we know that there are infinitely many numbers for which this inequality holds.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Oct 25 '15 at 16:01

























      answered Oct 24 '15 at 21:01









      Rocket ManRocket Man

      2,123818




      2,123818








      • 1




        $begingroup$
        This is what I had in mind; not only do we know there is a smallest, we know that $pi(x)-mathrm{Li}(x)$ changes sign infinitely many times, and we don't have an example yet of when it is positive.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:02












      • $begingroup$
        Very interesting @AJStas. Doesn't this answer the second question too?
        $endgroup$
        – Anguepa
        Oct 24 '15 at 21:06






      • 1




        $begingroup$
        @Anguepa: We have bounds for where such a number must live, but the bounds are far too large to expect a modern computer to be able to handle it in practical time.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:08










      • $begingroup$
        @Anguepa There have been many upper bounds for this number though, so I wasn't really sure.
        $endgroup$
        – Rocket Man
        Oct 24 '15 at 21:08






      • 1




        $begingroup$
        @Anguepa: A quick Google search yields a page with some bounds along the lines of $10^{316}$.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:12














      • 1




        $begingroup$
        This is what I had in mind; not only do we know there is a smallest, we know that $pi(x)-mathrm{Li}(x)$ changes sign infinitely many times, and we don't have an example yet of when it is positive.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:02












      • $begingroup$
        Very interesting @AJStas. Doesn't this answer the second question too?
        $endgroup$
        – Anguepa
        Oct 24 '15 at 21:06






      • 1




        $begingroup$
        @Anguepa: We have bounds for where such a number must live, but the bounds are far too large to expect a modern computer to be able to handle it in practical time.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:08










      • $begingroup$
        @Anguepa There have been many upper bounds for this number though, so I wasn't really sure.
        $endgroup$
        – Rocket Man
        Oct 24 '15 at 21:08






      • 1




        $begingroup$
        @Anguepa: A quick Google search yields a page with some bounds along the lines of $10^{316}$.
        $endgroup$
        – Clayton
        Oct 24 '15 at 21:12








      1




      1




      $begingroup$
      This is what I had in mind; not only do we know there is a smallest, we know that $pi(x)-mathrm{Li}(x)$ changes sign infinitely many times, and we don't have an example yet of when it is positive.
      $endgroup$
      – Clayton
      Oct 24 '15 at 21:02






      $begingroup$
      This is what I had in mind; not only do we know there is a smallest, we know that $pi(x)-mathrm{Li}(x)$ changes sign infinitely many times, and we don't have an example yet of when it is positive.
      $endgroup$
      – Clayton
      Oct 24 '15 at 21:02














      $begingroup$
      Very interesting @AJStas. Doesn't this answer the second question too?
      $endgroup$
      – Anguepa
      Oct 24 '15 at 21:06




      $begingroup$
      Very interesting @AJStas. Doesn't this answer the second question too?
      $endgroup$
      – Anguepa
      Oct 24 '15 at 21:06




      1




      1




      $begingroup$
      @Anguepa: We have bounds for where such a number must live, but the bounds are far too large to expect a modern computer to be able to handle it in practical time.
      $endgroup$
      – Clayton
      Oct 24 '15 at 21:08




      $begingroup$
      @Anguepa: We have bounds for where such a number must live, but the bounds are far too large to expect a modern computer to be able to handle it in practical time.
      $endgroup$
      – Clayton
      Oct 24 '15 at 21:08












      $begingroup$
      @Anguepa There have been many upper bounds for this number though, so I wasn't really sure.
      $endgroup$
      – Rocket Man
      Oct 24 '15 at 21:08




      $begingroup$
      @Anguepa There have been many upper bounds for this number though, so I wasn't really sure.
      $endgroup$
      – Rocket Man
      Oct 24 '15 at 21:08




      1




      1




      $begingroup$
      @Anguepa: A quick Google search yields a page with some bounds along the lines of $10^{316}$.
      $endgroup$
      – Clayton
      Oct 24 '15 at 21:12




      $begingroup$
      @Anguepa: A quick Google search yields a page with some bounds along the lines of $10^{316}$.
      $endgroup$
      – Clayton
      Oct 24 '15 at 21:12











      8












      $begingroup$

      Here's an interesting example that is the subject of recent (and celebrated) research:



      Consider the set ${p_1, p_2, p_3, ldots}$ of prime numbers, listed in increasing order. The $n$th prime gap is the difference $p_{n + 1} - p_n$. In 2013, Yitang Zhang showed that there are infinitely many prime gaps of size $leq 7 cdot 10^7$. Since there are only finitely many positive integers smaller than this size, at least one of them must occur infinitely often, but we don't know which. Since then, the Polymath project has improved that upper bound to $246$ (as of last December). The famous and longstanding Twin Prime Conjecture says that $2$ occurs infinitely often.






      share|cite|improve this answer











      $endgroup$


















        8












        $begingroup$

        Here's an interesting example that is the subject of recent (and celebrated) research:



        Consider the set ${p_1, p_2, p_3, ldots}$ of prime numbers, listed in increasing order. The $n$th prime gap is the difference $p_{n + 1} - p_n$. In 2013, Yitang Zhang showed that there are infinitely many prime gaps of size $leq 7 cdot 10^7$. Since there are only finitely many positive integers smaller than this size, at least one of them must occur infinitely often, but we don't know which. Since then, the Polymath project has improved that upper bound to $246$ (as of last December). The famous and longstanding Twin Prime Conjecture says that $2$ occurs infinitely often.






        share|cite|improve this answer











        $endgroup$
















          8












          8








          8





          $begingroup$

          Here's an interesting example that is the subject of recent (and celebrated) research:



          Consider the set ${p_1, p_2, p_3, ldots}$ of prime numbers, listed in increasing order. The $n$th prime gap is the difference $p_{n + 1} - p_n$. In 2013, Yitang Zhang showed that there are infinitely many prime gaps of size $leq 7 cdot 10^7$. Since there are only finitely many positive integers smaller than this size, at least one of them must occur infinitely often, but we don't know which. Since then, the Polymath project has improved that upper bound to $246$ (as of last December). The famous and longstanding Twin Prime Conjecture says that $2$ occurs infinitely often.






          share|cite|improve this answer











          $endgroup$



          Here's an interesting example that is the subject of recent (and celebrated) research:



          Consider the set ${p_1, p_2, p_3, ldots}$ of prime numbers, listed in increasing order. The $n$th prime gap is the difference $p_{n + 1} - p_n$. In 2013, Yitang Zhang showed that there are infinitely many prime gaps of size $leq 7 cdot 10^7$. Since there are only finitely many positive integers smaller than this size, at least one of them must occur infinitely often, but we don't know which. Since then, the Polymath project has improved that upper bound to $246$ (as of last December). The famous and longstanding Twin Prime Conjecture says that $2$ occurs infinitely often.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Oct 24 '15 at 21:17

























          answered Oct 24 '15 at 21:07









          TravisTravis

          59.9k767146




          59.9k767146























              6












              $begingroup$

              It is known that one of
              ζ(5), ζ(7), ζ(9), and ζ(11) must be irrational,
              but no one knows which one:



              W. Zudilin (2001). "One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational". Russ. Math. Surv. 56 (4): 774–776. doi:10.1070/RM2001v056n04ABEH000427.






              share|cite|improve this answer









              $endgroup$









              • 2




                $begingroup$
                Probably all of them are irrational.
                $endgroup$
                – azimut
                Oct 30 '15 at 21:20






              • 2




                $begingroup$
                Heck, probably all of them are transcendental, normal to all bases, and have all the properties that almost all numbers have.
                $endgroup$
                – marty cohen
                Oct 30 '15 at 21:38
















              6












              $begingroup$

              It is known that one of
              ζ(5), ζ(7), ζ(9), and ζ(11) must be irrational,
              but no one knows which one:



              W. Zudilin (2001). "One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational". Russ. Math. Surv. 56 (4): 774–776. doi:10.1070/RM2001v056n04ABEH000427.






              share|cite|improve this answer









              $endgroup$









              • 2




                $begingroup$
                Probably all of them are irrational.
                $endgroup$
                – azimut
                Oct 30 '15 at 21:20






              • 2




                $begingroup$
                Heck, probably all of them are transcendental, normal to all bases, and have all the properties that almost all numbers have.
                $endgroup$
                – marty cohen
                Oct 30 '15 at 21:38














              6












              6








              6





              $begingroup$

              It is known that one of
              ζ(5), ζ(7), ζ(9), and ζ(11) must be irrational,
              but no one knows which one:



              W. Zudilin (2001). "One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational". Russ. Math. Surv. 56 (4): 774–776. doi:10.1070/RM2001v056n04ABEH000427.






              share|cite|improve this answer









              $endgroup$



              It is known that one of
              ζ(5), ζ(7), ζ(9), and ζ(11) must be irrational,
              but no one knows which one:



              W. Zudilin (2001). "One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational". Russ. Math. Surv. 56 (4): 774–776. doi:10.1070/RM2001v056n04ABEH000427.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Oct 24 '15 at 22:37









              marty cohenmarty cohen

              72.9k549128




              72.9k549128








              • 2




                $begingroup$
                Probably all of them are irrational.
                $endgroup$
                – azimut
                Oct 30 '15 at 21:20






              • 2




                $begingroup$
                Heck, probably all of them are transcendental, normal to all bases, and have all the properties that almost all numbers have.
                $endgroup$
                – marty cohen
                Oct 30 '15 at 21:38














              • 2




                $begingroup$
                Probably all of them are irrational.
                $endgroup$
                – azimut
                Oct 30 '15 at 21:20






              • 2




                $begingroup$
                Heck, probably all of them are transcendental, normal to all bases, and have all the properties that almost all numbers have.
                $endgroup$
                – marty cohen
                Oct 30 '15 at 21:38








              2




              2




              $begingroup$
              Probably all of them are irrational.
              $endgroup$
              – azimut
              Oct 30 '15 at 21:20




              $begingroup$
              Probably all of them are irrational.
              $endgroup$
              – azimut
              Oct 30 '15 at 21:20




              2




              2




              $begingroup$
              Heck, probably all of them are transcendental, normal to all bases, and have all the properties that almost all numbers have.
              $endgroup$
              – marty cohen
              Oct 30 '15 at 21:38




              $begingroup$
              Heck, probably all of them are transcendental, normal to all bases, and have all the properties that almost all numbers have.
              $endgroup$
              – marty cohen
              Oct 30 '15 at 21:38











              1












              $begingroup$

              Consider $Sigma(m,n)$ to be the busy beaver function with $m$ states and $n$ colors.



              The following is then very hard to find:




              What is the smallest $m$ such that $Sigma(m,2)>Sigma(10,3)$?




              Since there is a color reduction method for reducing a $k$ state, 3 color machine to a $7k$ state, 2 color machine, it is known that $mleq 70$. On the other hand, it is clear that $m>10$. But it is very hard to say anything between this.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Consider $Sigma(m,n)$ to be the busy beaver function with $m$ states and $n$ colors.



                The following is then very hard to find:




                What is the smallest $m$ such that $Sigma(m,2)>Sigma(10,3)$?




                Since there is a color reduction method for reducing a $k$ state, 3 color machine to a $7k$ state, 2 color machine, it is known that $mleq 70$. On the other hand, it is clear that $m>10$. But it is very hard to say anything between this.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Consider $Sigma(m,n)$ to be the busy beaver function with $m$ states and $n$ colors.



                  The following is then very hard to find:




                  What is the smallest $m$ such that $Sigma(m,2)>Sigma(10,3)$?




                  Since there is a color reduction method for reducing a $k$ state, 3 color machine to a $7k$ state, 2 color machine, it is known that $mleq 70$. On the other hand, it is clear that $m>10$. But it is very hard to say anything between this.






                  share|cite|improve this answer









                  $endgroup$



                  Consider $Sigma(m,n)$ to be the busy beaver function with $m$ states and $n$ colors.



                  The following is then very hard to find:




                  What is the smallest $m$ such that $Sigma(m,2)>Sigma(10,3)$?




                  Since there is a color reduction method for reducing a $k$ state, 3 color machine to a $7k$ state, 2 color machine, it is known that $mleq 70$. On the other hand, it is clear that $m>10$. But it is very hard to say anything between this.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 16 '16 at 11:03









                  wythagoraswythagoras

                  21.6k444104




                  21.6k444104























                      1












                      $begingroup$

                      Since we do not know , for example $Sigma(5)$ , we do not know which number $n$ satisfies $n=Sigma(5)$, but we know that some number $n$ must satisfy the property. In this case we do not even have an upper bound.



                      Another less trivial example : The $20$-th Fermat-number $2^{2^{20}}+1$ is known to be composite, but no factor is known.



                      Denote $n$ to be the smallest prime factor of the $20$-th Fermat-number.
                      This number is well-defined but we do not know it, but in this case, we can bound it.



                      If this does not hit your intention, please precise which kind of properties you mean.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        what means $Sigma$
                        $endgroup$
                        – miracle173
                        Jun 10 '17 at 6:28


















                      1












                      $begingroup$

                      Since we do not know , for example $Sigma(5)$ , we do not know which number $n$ satisfies $n=Sigma(5)$, but we know that some number $n$ must satisfy the property. In this case we do not even have an upper bound.



                      Another less trivial example : The $20$-th Fermat-number $2^{2^{20}}+1$ is known to be composite, but no factor is known.



                      Denote $n$ to be the smallest prime factor of the $20$-th Fermat-number.
                      This number is well-defined but we do not know it, but in this case, we can bound it.



                      If this does not hit your intention, please precise which kind of properties you mean.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        what means $Sigma$
                        $endgroup$
                        – miracle173
                        Jun 10 '17 at 6:28
















                      1












                      1








                      1





                      $begingroup$

                      Since we do not know , for example $Sigma(5)$ , we do not know which number $n$ satisfies $n=Sigma(5)$, but we know that some number $n$ must satisfy the property. In this case we do not even have an upper bound.



                      Another less trivial example : The $20$-th Fermat-number $2^{2^{20}}+1$ is known to be composite, but no factor is known.



                      Denote $n$ to be the smallest prime factor of the $20$-th Fermat-number.
                      This number is well-defined but we do not know it, but in this case, we can bound it.



                      If this does not hit your intention, please precise which kind of properties you mean.






                      share|cite|improve this answer









                      $endgroup$



                      Since we do not know , for example $Sigma(5)$ , we do not know which number $n$ satisfies $n=Sigma(5)$, but we know that some number $n$ must satisfy the property. In this case we do not even have an upper bound.



                      Another less trivial example : The $20$-th Fermat-number $2^{2^{20}}+1$ is known to be composite, but no factor is known.



                      Denote $n$ to be the smallest prime factor of the $20$-th Fermat-number.
                      This number is well-defined but we do not know it, but in this case, we can bound it.



                      If this does not hit your intention, please precise which kind of properties you mean.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered May 31 '16 at 13:17









                      PeterPeter

                      46.9k1039125




                      46.9k1039125












                      • $begingroup$
                        what means $Sigma$
                        $endgroup$
                        – miracle173
                        Jun 10 '17 at 6:28




















                      • $begingroup$
                        what means $Sigma$
                        $endgroup$
                        – miracle173
                        Jun 10 '17 at 6:28


















                      $begingroup$
                      what means $Sigma$
                      $endgroup$
                      – miracle173
                      Jun 10 '17 at 6:28






                      $begingroup$
                      what means $Sigma$
                      $endgroup$
                      – miracle173
                      Jun 10 '17 at 6:28




















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