If $2$ numbers are co-prime, does it imply that their difference is also prime to those numbers?

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Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.



If $p$ and $q$ are coprime, the fraction cannot be simplified. Therefore, we can rewrite $p/q$ as $(q+d)/q$, and we obtain $1+d/q$. As the fraction for $p/q$ cannot be simplified, $d/q$ can also not be simplified, therefore $d$ is also prime to $q$. [We can do same arguement for $p$ too]. Is this correct?




elementary-number-theory coprime




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    Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
    – DanielWainfleet
    Aug 6 at 20:27















up vote
6
down vote

favorite
1












Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.



If $p$ and $q$ are coprime, the fraction cannot be simplified. Therefore, we can rewrite $p/q$ as $(q+d)/q$, and we obtain $1+d/q$. As the fraction for $p/q$ cannot be simplified, $d/q$ can also not be simplified, therefore $d$ is also prime to $q$. [We can do same arguement for $p$ too]. Is this correct?




elementary-number-theory coprime




share|cite|improve this question

















  • 1




    Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
    – DanielWainfleet
    Aug 6 at 20:27













up vote
6
down vote

favorite
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up vote
6
down vote

favorite
1






1





Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.



If $p$ and $q$ are coprime, the fraction cannot be simplified. Therefore, we can rewrite $p/q$ as $(q+d)/q$, and we obtain $1+d/q$. As the fraction for $p/q$ cannot be simplified, $d/q$ can also not be simplified, therefore $d$ is also prime to $q$. [We can do same arguement for $p$ too]. Is this correct?




elementary-number-theory coprime




share|cite|improve this question













Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.



If $p$ and $q$ are coprime, the fraction cannot be simplified. Therefore, we can rewrite $p/q$ as $(q+d)/q$, and we obtain $1+d/q$. As the fraction for $p/q$ cannot be simplified, $d/q$ can also not be simplified, therefore $d$ is also prime to $q$. [We can do same arguement for $p$ too]. Is this correct?





elementary-number-theory coprime





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edited Aug 7 at 0:21









Daniel Buck

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asked Aug 6 at 18:15









Dean Yang

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  • 1




    Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
    – DanielWainfleet
    Aug 6 at 20:27













  • 1




    Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
    – DanielWainfleet
    Aug 6 at 20:27








1




1




Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
– DanielWainfleet
Aug 6 at 20:27





Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
– DanielWainfleet
Aug 6 at 20:27











2 Answers
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Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.






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  • Alright, thanks :)
    – Dean Yang
    Aug 6 at 18:20

















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4
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Of course it is. If $q$ and $d$ have a common factor $f$:



$q = Qf$

$d = Df$



then



$p = q + d = (Q +D)f$



and more generally



$kqpm md = (kQ pm mD)f$



will have that factor, too.






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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    up vote
    16
    down vote













    Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.






    share|cite|improve this answer





















    • Alright, thanks :)
      – Dean Yang
      Aug 6 at 18:20














    up vote
    16
    down vote













    Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.






    share|cite|improve this answer





















    • Alright, thanks :)
      – Dean Yang
      Aug 6 at 18:20












    up vote
    16
    down vote










    up vote
    16
    down vote









    Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.






    share|cite|improve this answer













    Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Aug 6 at 18:19









    Ross Millikan

    276k21187352




    276k21187352











    • Alright, thanks :)
      – Dean Yang
      Aug 6 at 18:20
















    • Alright, thanks :)
      – Dean Yang
      Aug 6 at 18:20















    Alright, thanks :)
    – Dean Yang
    Aug 6 at 18:20




    Alright, thanks :)
    – Dean Yang
    Aug 6 at 18:20










    up vote
    4
    down vote













    Of course it is. If $q$ and $d$ have a common factor $f$:



    $q = Qf$

    $d = Df$



    then



    $p = q + d = (Q +D)f$



    and more generally



    $kqpm md = (kQ pm mD)f$



    will have that factor, too.






    share|cite|improve this answer



























      up vote
      4
      down vote













      Of course it is. If $q$ and $d$ have a common factor $f$:



      $q = Qf$

      $d = Df$



      then



      $p = q + d = (Q +D)f$



      and more generally



      $kqpm md = (kQ pm mD)f$



      will have that factor, too.






      share|cite|improve this answer

























        up vote
        4
        down vote










        up vote
        4
        down vote









        Of course it is. If $q$ and $d$ have a common factor $f$:



        $q = Qf$

        $d = Df$



        then



        $p = q + d = (Q +D)f$



        and more generally



        $kqpm md = (kQ pm mD)f$



        will have that factor, too.






        share|cite|improve this answer















        Of course it is. If $q$ and $d$ have a common factor $f$:



        $q = Qf$

        $d = Df$



        then



        $p = q + d = (Q +D)f$



        and more generally



        $kqpm md = (kQ pm mD)f$



        will have that factor, too.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 8 at 10:25


























        answered Aug 6 at 18:21









        CiaPan

        9,93311044




        9,93311044






















             

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