Let $p$ be prime, and $n < p$. Then $ntimes i$ “hits” every residue class for $0 < i < p$. (Searching for clean proof)

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Intuitively, I'm trying to show that the first $p-1$ multiples of n give every possible remainder after division by p. Or equivalently,



$$ T = ; i in mathbbZ, ; 1 leq i < p $$



$$S = a ; $$



are equal.



Showing $T subseteq S$ is straightforward, but my proof for $ S subseteq T$ is two handwritten pages long and (even by my standards) convoluted.



This has the feel of well-known theorem, but I can't find it anywhere. Could someone either provide or link to a short/clean proof for the above?




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  • This is a similar question to the one that I asked here
    – Bill Wallis
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up vote
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favorite












Intuitively, I'm trying to show that the first $p-1$ multiples of n give every possible remainder after division by p. Or equivalently,



$$ T = ; i in mathbbZ, ; 1 leq i < p $$



$$S = a ; $$



are equal.



Showing $T subseteq S$ is straightforward, but my proof for $ S subseteq T$ is two handwritten pages long and (even by my standards) convoluted.



This has the feel of well-known theorem, but I can't find it anywhere. Could someone either provide or link to a short/clean proof for the above?




elementary-number-theory




share|cite|improve this question





















  • This is a similar question to the one that I asked here
    – Bill Wallis
    2 days ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Intuitively, I'm trying to show that the first $p-1$ multiples of n give every possible remainder after division by p. Or equivalently,



$$ T = ; i in mathbbZ, ; 1 leq i < p $$



$$S = a ; $$



are equal.



Showing $T subseteq S$ is straightforward, but my proof for $ S subseteq T$ is two handwritten pages long and (even by my standards) convoluted.



This has the feel of well-known theorem, but I can't find it anywhere. Could someone either provide or link to a short/clean proof for the above?




elementary-number-theory




share|cite|improve this question













Intuitively, I'm trying to show that the first $p-1$ multiples of n give every possible remainder after division by p. Or equivalently,



$$ T = ; i in mathbbZ, ; 1 leq i < p $$



$$S = a ; $$



are equal.



Showing $T subseteq S$ is straightforward, but my proof for $ S subseteq T$ is two handwritten pages long and (even by my standards) convoluted.



This has the feel of well-known theorem, but I can't find it anywhere. Could someone either provide or link to a short/clean proof for the above?





elementary-number-theory





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edited 2 days ago









pointguard0

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asked 2 days ago









Noah Caplinger

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  • This is a similar question to the one that I asked here
    – Bill Wallis
    2 days ago

















  • This is a similar question to the one that I asked here
    – Bill Wallis
    2 days ago
















This is a similar question to the one that I asked here
– Bill Wallis
2 days ago





This is a similar question to the one that I asked here
– Bill Wallis
2 days ago











2 Answers
2






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Suppose that two numbers fell into the same bin: $acdot nequiv bcdot npmod p$. Then $(acdot n-bcdot n)equiv 0bmod p$, so there's some number $c$ with $(a-b)cdot n=ccdot p$. Now, can you finish up from here? (This is where it's essential that $p$ is prime.)



Once you know that no two numbers fall into the same bin, then the rest is just an application of the pigeonhole principle: there are $p$ distinct equivalence classes of $acdot nbmod p$ for $0leq alt p$, and there are $p$ distinct representatives $langle brangle$ of those equivalence classes with $0leq blt p$, so the former must cover all the latter.






share|cite|improve this answer





















  • Just so I'm understanding fully: p divides c*p, so p also divides (a-b)*n. But n does not divide p (p is prime), so p divides (a-b), meaning a = b mod p. Is this correct?
    – Noah Caplinger
    2 days ago










  • Pretty close - $p$ divides $(a-b)cdot n$, so either $p|(a-b)$ or $p|n$, and the latter is impossible since $nlt p$.
    – Steven Stadnicki
    2 days ago











  • So why is it essential that p is prime?
    – Noah Caplinger
    2 days ago










  • Because $a|bcdot c$ doesn't necessarily imply $a|b$ or $a|c$ if $a$ isn't a prime; consider, e.g., $4|6cdot10$.
    – Steven Stadnicki
    2 days ago










  • Thanks for the help!
    – Noah Caplinger
    2 days ago

















up vote
0
down vote













This follows from the fact that $n$ is invertible in the ring $mathbbZ_p$ so you can solve the equation



$$ni = y$$



as



$$i = n^-1y$$






share|cite|improve this answer





















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    2 Answers
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    2 Answers
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    Suppose that two numbers fell into the same bin: $acdot nequiv bcdot npmod p$. Then $(acdot n-bcdot n)equiv 0bmod p$, so there's some number $c$ with $(a-b)cdot n=ccdot p$. Now, can you finish up from here? (This is where it's essential that $p$ is prime.)



    Once you know that no two numbers fall into the same bin, then the rest is just an application of the pigeonhole principle: there are $p$ distinct equivalence classes of $acdot nbmod p$ for $0leq alt p$, and there are $p$ distinct representatives $langle brangle$ of those equivalence classes with $0leq blt p$, so the former must cover all the latter.






    share|cite|improve this answer





















    • Just so I'm understanding fully: p divides c*p, so p also divides (a-b)*n. But n does not divide p (p is prime), so p divides (a-b), meaning a = b mod p. Is this correct?
      – Noah Caplinger
      2 days ago










    • Pretty close - $p$ divides $(a-b)cdot n$, so either $p|(a-b)$ or $p|n$, and the latter is impossible since $nlt p$.
      – Steven Stadnicki
      2 days ago











    • So why is it essential that p is prime?
      – Noah Caplinger
      2 days ago










    • Because $a|bcdot c$ doesn't necessarily imply $a|b$ or $a|c$ if $a$ isn't a prime; consider, e.g., $4|6cdot10$.
      – Steven Stadnicki
      2 days ago










    • Thanks for the help!
      – Noah Caplinger
      2 days ago














    up vote
    1
    down vote



    accepted










    Suppose that two numbers fell into the same bin: $acdot nequiv bcdot npmod p$. Then $(acdot n-bcdot n)equiv 0bmod p$, so there's some number $c$ with $(a-b)cdot n=ccdot p$. Now, can you finish up from here? (This is where it's essential that $p$ is prime.)



    Once you know that no two numbers fall into the same bin, then the rest is just an application of the pigeonhole principle: there are $p$ distinct equivalence classes of $acdot nbmod p$ for $0leq alt p$, and there are $p$ distinct representatives $langle brangle$ of those equivalence classes with $0leq blt p$, so the former must cover all the latter.






    share|cite|improve this answer





















    • Just so I'm understanding fully: p divides c*p, so p also divides (a-b)*n. But n does not divide p (p is prime), so p divides (a-b), meaning a = b mod p. Is this correct?
      – Noah Caplinger
      2 days ago










    • Pretty close - $p$ divides $(a-b)cdot n$, so either $p|(a-b)$ or $p|n$, and the latter is impossible since $nlt p$.
      – Steven Stadnicki
      2 days ago











    • So why is it essential that p is prime?
      – Noah Caplinger
      2 days ago










    • Because $a|bcdot c$ doesn't necessarily imply $a|b$ or $a|c$ if $a$ isn't a prime; consider, e.g., $4|6cdot10$.
      – Steven Stadnicki
      2 days ago










    • Thanks for the help!
      – Noah Caplinger
      2 days ago












    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    Suppose that two numbers fell into the same bin: $acdot nequiv bcdot npmod p$. Then $(acdot n-bcdot n)equiv 0bmod p$, so there's some number $c$ with $(a-b)cdot n=ccdot p$. Now, can you finish up from here? (This is where it's essential that $p$ is prime.)



    Once you know that no two numbers fall into the same bin, then the rest is just an application of the pigeonhole principle: there are $p$ distinct equivalence classes of $acdot nbmod p$ for $0leq alt p$, and there are $p$ distinct representatives $langle brangle$ of those equivalence classes with $0leq blt p$, so the former must cover all the latter.






    share|cite|improve this answer













    Suppose that two numbers fell into the same bin: $acdot nequiv bcdot npmod p$. Then $(acdot n-bcdot n)equiv 0bmod p$, so there's some number $c$ with $(a-b)cdot n=ccdot p$. Now, can you finish up from here? (This is where it's essential that $p$ is prime.)



    Once you know that no two numbers fall into the same bin, then the rest is just an application of the pigeonhole principle: there are $p$ distinct equivalence classes of $acdot nbmod p$ for $0leq alt p$, and there are $p$ distinct representatives $langle brangle$ of those equivalence classes with $0leq blt p$, so the former must cover all the latter.







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered 2 days ago









    Steven Stadnicki

    40.1k765119




    40.1k765119











    • Just so I'm understanding fully: p divides c*p, so p also divides (a-b)*n. But n does not divide p (p is prime), so p divides (a-b), meaning a = b mod p. Is this correct?
      – Noah Caplinger
      2 days ago










    • Pretty close - $p$ divides $(a-b)cdot n$, so either $p|(a-b)$ or $p|n$, and the latter is impossible since $nlt p$.
      – Steven Stadnicki
      2 days ago











    • So why is it essential that p is prime?
      – Noah Caplinger
      2 days ago










    • Because $a|bcdot c$ doesn't necessarily imply $a|b$ or $a|c$ if $a$ isn't a prime; consider, e.g., $4|6cdot10$.
      – Steven Stadnicki
      2 days ago










    • Thanks for the help!
      – Noah Caplinger
      2 days ago
















    • Just so I'm understanding fully: p divides c*p, so p also divides (a-b)*n. But n does not divide p (p is prime), so p divides (a-b), meaning a = b mod p. Is this correct?
      – Noah Caplinger
      2 days ago










    • Pretty close - $p$ divides $(a-b)cdot n$, so either $p|(a-b)$ or $p|n$, and the latter is impossible since $nlt p$.
      – Steven Stadnicki
      2 days ago











    • So why is it essential that p is prime?
      – Noah Caplinger
      2 days ago










    • Because $a|bcdot c$ doesn't necessarily imply $a|b$ or $a|c$ if $a$ isn't a prime; consider, e.g., $4|6cdot10$.
      – Steven Stadnicki
      2 days ago










    • Thanks for the help!
      – Noah Caplinger
      2 days ago















    Just so I'm understanding fully: p divides c*p, so p also divides (a-b)*n. But n does not divide p (p is prime), so p divides (a-b), meaning a = b mod p. Is this correct?
    – Noah Caplinger
    2 days ago




    Just so I'm understanding fully: p divides c*p, so p also divides (a-b)*n. But n does not divide p (p is prime), so p divides (a-b), meaning a = b mod p. Is this correct?
    – Noah Caplinger
    2 days ago












    Pretty close - $p$ divides $(a-b)cdot n$, so either $p|(a-b)$ or $p|n$, and the latter is impossible since $nlt p$.
    – Steven Stadnicki
    2 days ago





    Pretty close - $p$ divides $(a-b)cdot n$, so either $p|(a-b)$ or $p|n$, and the latter is impossible since $nlt p$.
    – Steven Stadnicki
    2 days ago













    So why is it essential that p is prime?
    – Noah Caplinger
    2 days ago




    So why is it essential that p is prime?
    – Noah Caplinger
    2 days ago












    Because $a|bcdot c$ doesn't necessarily imply $a|b$ or $a|c$ if $a$ isn't a prime; consider, e.g., $4|6cdot10$.
    – Steven Stadnicki
    2 days ago




    Because $a|bcdot c$ doesn't necessarily imply $a|b$ or $a|c$ if $a$ isn't a prime; consider, e.g., $4|6cdot10$.
    – Steven Stadnicki
    2 days ago












    Thanks for the help!
    – Noah Caplinger
    2 days ago




    Thanks for the help!
    – Noah Caplinger
    2 days ago










    up vote
    0
    down vote













    This follows from the fact that $n$ is invertible in the ring $mathbbZ_p$ so you can solve the equation



    $$ni = y$$



    as



    $$i = n^-1y$$






    share|cite|improve this answer

























      up vote
      0
      down vote













      This follows from the fact that $n$ is invertible in the ring $mathbbZ_p$ so you can solve the equation



      $$ni = y$$



      as



      $$i = n^-1y$$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        This follows from the fact that $n$ is invertible in the ring $mathbbZ_p$ so you can solve the equation



        $$ni = y$$



        as



        $$i = n^-1y$$






        share|cite|improve this answer













        This follows from the fact that $n$ is invertible in the ring $mathbbZ_p$ so you can solve the equation



        $$ni = y$$



        as



        $$i = n^-1y$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 2 days ago









        ploosu2

        3,982922




        3,982922






















             

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