What is a minimal set of rules that determine the usual order on $Bbb{N}$ given that $1 lt p_1 lt p_2 lt...












3












$begingroup$


Let $p_i$ always mean the $i$th prime number.



Given two numbers $a, b in Bbb{N}$ in the form $a = {(p_i, e_i) : p_i^{e_i} mid a, e_i text{ maximal}}$ ie. essentially their unique factorization, what is a natural way to compute whether $a gt b$?



For instance a rule might be $p_{j}^2 gt p_{j+1}$ always. So that given $a = 2^2$, $b = 3$, we immediately have $a gt b$.



Is there a simple set of rules such that given the usual ordering on the primes and $1$: $1 lt p_1 lt p_2 lt dots$ we can always derive $a gt b$ when it is true?



By no means can we include converting the number to a base representation and comparing the last digit.



It's best to work by example sometimes. Take $2 cdot 7 lt 3cdot 5$. Well, $2 lt 3 lt 5 lt 7$ so... ?



I'm stuck on this one, what is the simplest rule you can think of to overcome this example?






elementary-number-theory prime-numbers order-theory natural-numbers







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It's certainly true that, given only the information $p_1<p_2<p_3<p_4$, there's no way to determine which of $p_1p_4$ and $p_2p_3$ is larger.
    $endgroup$
    – Greg Martin
    Dec 20 '18 at 5:31






  • 1




    $begingroup$
    Or how about comparing $2^{11}3^25^97$ with $p_{10 000 000 000}?$ I don't expect there to be a simple rule.
    $endgroup$
    – Ross Millikan
    Dec 20 '18 at 5:39


















3












$begingroup$


Let $p_i$ always mean the $i$th prime number.



Given two numbers $a, b in Bbb{N}$ in the form $a = {(p_i, e_i) : p_i^{e_i} mid a, e_i text{ maximal}}$ ie. essentially their unique factorization, what is a natural way to compute whether $a gt b$?



For instance a rule might be $p_{j}^2 gt p_{j+1}$ always. So that given $a = 2^2$, $b = 3$, we immediately have $a gt b$.



Is there a simple set of rules such that given the usual ordering on the primes and $1$: $1 lt p_1 lt p_2 lt dots$ we can always derive $a gt b$ when it is true?



By no means can we include converting the number to a base representation and comparing the last digit.



It's best to work by example sometimes. Take $2 cdot 7 lt 3cdot 5$. Well, $2 lt 3 lt 5 lt 7$ so... ?



I'm stuck on this one, what is the simplest rule you can think of to overcome this example?






elementary-number-theory prime-numbers order-theory natural-numbers







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It's certainly true that, given only the information $p_1<p_2<p_3<p_4$, there's no way to determine which of $p_1p_4$ and $p_2p_3$ is larger.
    $endgroup$
    – Greg Martin
    Dec 20 '18 at 5:31






  • 1




    $begingroup$
    Or how about comparing $2^{11}3^25^97$ with $p_{10 000 000 000}?$ I don't expect there to be a simple rule.
    $endgroup$
    – Ross Millikan
    Dec 20 '18 at 5:39
















3












3








3


1



$begingroup$


Let $p_i$ always mean the $i$th prime number.



Given two numbers $a, b in Bbb{N}$ in the form $a = {(p_i, e_i) : p_i^{e_i} mid a, e_i text{ maximal}}$ ie. essentially their unique factorization, what is a natural way to compute whether $a gt b$?



For instance a rule might be $p_{j}^2 gt p_{j+1}$ always. So that given $a = 2^2$, $b = 3$, we immediately have $a gt b$.



Is there a simple set of rules such that given the usual ordering on the primes and $1$: $1 lt p_1 lt p_2 lt dots$ we can always derive $a gt b$ when it is true?



By no means can we include converting the number to a base representation and comparing the last digit.



It's best to work by example sometimes. Take $2 cdot 7 lt 3cdot 5$. Well, $2 lt 3 lt 5 lt 7$ so... ?



I'm stuck on this one, what is the simplest rule you can think of to overcome this example?






elementary-number-theory prime-numbers order-theory natural-numbers







share|cite|improve this question











$endgroup$




Let $p_i$ always mean the $i$th prime number.



Given two numbers $a, b in Bbb{N}$ in the form $a = {(p_i, e_i) : p_i^{e_i} mid a, e_i text{ maximal}}$ ie. essentially their unique factorization, what is a natural way to compute whether $a gt b$?



For instance a rule might be $p_{j}^2 gt p_{j+1}$ always. So that given $a = 2^2$, $b = 3$, we immediately have $a gt b$.



Is there a simple set of rules such that given the usual ordering on the primes and $1$: $1 lt p_1 lt p_2 lt dots$ we can always derive $a gt b$ when it is true?



By no means can we include converting the number to a base representation and comparing the last digit.



It's best to work by example sometimes. Take $2 cdot 7 lt 3cdot 5$. Well, $2 lt 3 lt 5 lt 7$ so... ?



I'm stuck on this one, what is the simplest rule you can think of to overcome this example?






elementary-number-theory prime-numbers order-theory natural-numbers




elementary-number-theory prime-numbers order-theory natural-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 5:01







BananaCats

















asked Dec 20 '18 at 4:55









BananaCatsBananaCats

9,26252558




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  • 2




    $begingroup$
    It's certainly true that, given only the information $p_1<p_2<p_3<p_4$, there's no way to determine which of $p_1p_4$ and $p_2p_3$ is larger.
    $endgroup$
    – Greg Martin
    Dec 20 '18 at 5:31






  • 1




    $begingroup$
    Or how about comparing $2^{11}3^25^97$ with $p_{10 000 000 000}?$ I don't expect there to be a simple rule.
    $endgroup$
    – Ross Millikan
    Dec 20 '18 at 5:39
















  • 2




    $begingroup$
    It's certainly true that, given only the information $p_1<p_2<p_3<p_4$, there's no way to determine which of $p_1p_4$ and $p_2p_3$ is larger.
    $endgroup$
    – Greg Martin
    Dec 20 '18 at 5:31






  • 1




    $begingroup$
    Or how about comparing $2^{11}3^25^97$ with $p_{10 000 000 000}?$ I don't expect there to be a simple rule.
    $endgroup$
    – Ross Millikan
    Dec 20 '18 at 5:39










2




2




$begingroup$
It's certainly true that, given only the information $p_1<p_2<p_3<p_4$, there's no way to determine which of $p_1p_4$ and $p_2p_3$ is larger.
$endgroup$
– Greg Martin
Dec 20 '18 at 5:31




$begingroup$
It's certainly true that, given only the information $p_1<p_2<p_3<p_4$, there's no way to determine which of $p_1p_4$ and $p_2p_3$ is larger.
$endgroup$
– Greg Martin
Dec 20 '18 at 5:31




1




1




$begingroup$
Or how about comparing $2^{11}3^25^97$ with $p_{10 000 000 000}?$ I don't expect there to be a simple rule.
$endgroup$
– Ross Millikan
Dec 20 '18 at 5:39






$begingroup$
Or how about comparing $2^{11}3^25^97$ with $p_{10 000 000 000}?$ I don't expect there to be a simple rule.
$endgroup$
– Ross Millikan
Dec 20 '18 at 5:39












1 Answer
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$begingroup$

Here's some evidence that this may be hard.



In mathematical logic class we learn that the theory of $(mathbb N,times)$ is decidable. Basically, because the fundamental theorem of arithmetic states that the monoid of positive integers under multiplication $times$ is a free commutative monoid on an infinite set of generators, the prime numbers.



But if we add in the relation $x< y$, you get the same definable sets as in all of first-order arithmetic $(mathbb N,+,times)$, which is not decidable. This is described by Alexis Bès.






share|cite|improve this answer









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    $begingroup$

    Here's some evidence that this may be hard.



    In mathematical logic class we learn that the theory of $(mathbb N,times)$ is decidable. Basically, because the fundamental theorem of arithmetic states that the monoid of positive integers under multiplication $times$ is a free commutative monoid on an infinite set of generators, the prime numbers.



    But if we add in the relation $x< y$, you get the same definable sets as in all of first-order arithmetic $(mathbb N,+,times)$, which is not decidable. This is described by Alexis Bès.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Here's some evidence that this may be hard.



      In mathematical logic class we learn that the theory of $(mathbb N,times)$ is decidable. Basically, because the fundamental theorem of arithmetic states that the monoid of positive integers under multiplication $times$ is a free commutative monoid on an infinite set of generators, the prime numbers.



      But if we add in the relation $x< y$, you get the same definable sets as in all of first-order arithmetic $(mathbb N,+,times)$, which is not decidable. This is described by Alexis Bès.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Here's some evidence that this may be hard.



        In mathematical logic class we learn that the theory of $(mathbb N,times)$ is decidable. Basically, because the fundamental theorem of arithmetic states that the monoid of positive integers under multiplication $times$ is a free commutative monoid on an infinite set of generators, the prime numbers.



        But if we add in the relation $x< y$, you get the same definable sets as in all of first-order arithmetic $(mathbb N,+,times)$, which is not decidable. This is described by Alexis Bès.






        share|cite|improve this answer









        $endgroup$



        Here's some evidence that this may be hard.



        In mathematical logic class we learn that the theory of $(mathbb N,times)$ is decidable. Basically, because the fundamental theorem of arithmetic states that the monoid of positive integers under multiplication $times$ is a free commutative monoid on an infinite set of generators, the prime numbers.



        But if we add in the relation $x< y$, you get the same definable sets as in all of first-order arithmetic $(mathbb N,+,times)$, which is not decidable. This is described by Alexis Bès.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 5:40









        Bjørn Kjos-HanssenBjørn Kjos-Hanssen

        2,097918




        2,097918






























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