About fractional modulus












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I was reading paper and it had some expression modulo $1$, and modulo $frac12$. Could someone explain me what this $2$ things even mean? I'm confused mainly by the fact of divisibility over rationals. Is it maybe something more related to congruence as a property of groups?










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    $begingroup$


    I was reading paper and it had some expression modulo $1$, and modulo $frac12$. Could someone explain me what this $2$ things even mean? I'm confused mainly by the fact of divisibility over rationals. Is it maybe something more related to congruence as a property of groups?










    share|cite|improve this question









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      0





      $begingroup$


      I was reading paper and it had some expression modulo $1$, and modulo $frac12$. Could someone explain me what this $2$ things even mean? I'm confused mainly by the fact of divisibility over rationals. Is it maybe something more related to congruence as a property of groups?










      share|cite|improve this question









      $endgroup$




      I was reading paper and it had some expression modulo $1$, and modulo $frac12$. Could someone explain me what this $2$ things even mean? I'm confused mainly by the fact of divisibility over rationals. Is it maybe something more related to congruence as a property of groups?







      group-theory elementary-number-theory






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      asked Jan 7 at 9:49









      Bruno AndradesBruno Andrades

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          $begingroup$

          Without any more context I can't be certain, but I don't see any reason it should be different from regular modulus:
          $$xequiv y pmod a iff exists nin Bbb Z: n(x-y) = a$$






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            $begingroup$

            Without any more context I can't be certain, but I don't see any reason it should be different from regular modulus:
            $$xequiv y pmod a iff exists nin Bbb Z: n(x-y) = a$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Without any more context I can't be certain, but I don't see any reason it should be different from regular modulus:
              $$xequiv y pmod a iff exists nin Bbb Z: n(x-y) = a$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Without any more context I can't be certain, but I don't see any reason it should be different from regular modulus:
                $$xequiv y pmod a iff exists nin Bbb Z: n(x-y) = a$$






                share|cite|improve this answer









                $endgroup$



                Without any more context I can't be certain, but I don't see any reason it should be different from regular modulus:
                $$xequiv y pmod a iff exists nin Bbb Z: n(x-y) = a$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 10:02









                ArthurArthur

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