Partitioning the positive integers into finitely many arithmetic progressions












2












$begingroup$


From Bóna's A Walk through Combinatorics:




Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.




A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.






arithmetic-progression elementary-proofs







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  • 2




    $begingroup$
    What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
    $endgroup$
    – Fedor Petrov
    12 hours ago






  • 1




    $begingroup$
    Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
    $endgroup$
    – John
    11 hours ago
















2












$begingroup$


From Bóna's A Walk through Combinatorics:




Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.




A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.






arithmetic-progression elementary-proofs







share|cite|improve this question









New contributor




John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
    $endgroup$
    – Fedor Petrov
    12 hours ago






  • 1




    $begingroup$
    Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
    $endgroup$
    – John
    11 hours ago














2












2








2





$begingroup$


From Bóna's A Walk through Combinatorics:




Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.




A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.






arithmetic-progression elementary-proofs







share|cite|improve this question









New contributor




John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




From Bóna's A Walk through Combinatorics:




Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.




A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.






arithmetic-progression elementary-proofs




arithmetic-progression elementary-proofs






share|cite|improve this question









New contributor




John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









Rodrigo de Azevedo

1,8422719




1,8422719






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asked 13 hours ago









JohnJohn

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194




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  • 2




    $begingroup$
    What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
    $endgroup$
    – Fedor Petrov
    12 hours ago






  • 1




    $begingroup$
    Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
    $endgroup$
    – John
    11 hours ago














  • 2




    $begingroup$
    What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
    $endgroup$
    – Fedor Petrov
    12 hours ago






  • 1




    $begingroup$
    Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
    $endgroup$
    – John
    11 hours ago








2




2




$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
12 hours ago




$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
12 hours ago




1




1




$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
11 hours ago




$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
11 hours ago










2 Answers
2






active

oldest

votes


















7












$begingroup$

Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term.



    for k from 1 to inf {
    S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf }



    S(1) is odd integers;
    S(2) is 2, 6, 10 - twice an odd integer;
    and continues for powers of 2.



    In all cases the initial term is the difference / 2.






    share|cite|improve this answer








    New contributor




    Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$













    • $begingroup$
      The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
      $endgroup$
      – user44191
      4 hours ago











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    2 Answers
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    active

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    2 Answers
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    active

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    $begingroup$

    Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).






    share|cite|improve this answer











    $endgroup$


















      7












      $begingroup$

      Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).






      share|cite|improve this answer











      $endgroup$
















        7












        7








        7





        $begingroup$

        Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).






        share|cite|improve this answer











        $endgroup$



        Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 7 hours ago

























        answered 10 hours ago









        Fedor PetrovFedor Petrov

        49.6k5114230




        49.6k5114230























            0












            $begingroup$

            Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term.



            for k from 1 to inf {
            S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf }



            S(1) is odd integers;
            S(2) is 2, 6, 10 - twice an odd integer;
            and continues for powers of 2.



            In all cases the initial term is the difference / 2.






            share|cite|improve this answer








            New contributor




            Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













            • $begingroup$
              The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
              $endgroup$
              – user44191
              4 hours ago
















            0












            $begingroup$

            Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term.



            for k from 1 to inf {
            S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf }



            S(1) is odd integers;
            S(2) is 2, 6, 10 - twice an odd integer;
            and continues for powers of 2.



            In all cases the initial term is the difference / 2.






            share|cite|improve this answer








            New contributor




            Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













            • $begingroup$
              The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
              $endgroup$
              – user44191
              4 hours ago














            0












            0








            0





            $begingroup$

            Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term.



            for k from 1 to inf {
            S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf }



            S(1) is odd integers;
            S(2) is 2, 6, 10 - twice an odd integer;
            and continues for powers of 2.



            In all cases the initial term is the difference / 2.






            share|cite|improve this answer








            New contributor




            Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$



            Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term.



            for k from 1 to inf {
            S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf }



            S(1) is odd integers;
            S(2) is 2, 6, 10 - twice an odd integer;
            and continues for powers of 2.



            In all cases the initial term is the difference / 2.







            share|cite|improve this answer








            New contributor




            Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|cite|improve this answer



            share|cite|improve this answer






            New contributor




            Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            answered 4 hours ago









            Judah GreenblattJudah Greenblatt

            1




            1




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            New contributor





            Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.












            • $begingroup$
              The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
              $endgroup$
              – user44191
              4 hours ago


















            • $begingroup$
              The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
              $endgroup$
              – user44191
              4 hours ago
















            $begingroup$
            The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
            $endgroup$
            – user44191
            4 hours ago




            $begingroup$
            The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
            $endgroup$
            – user44191
            4 hours ago










            John is a new contributor. Be nice, and check out our Code of Conduct.










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