Show that $forall_{n,m,kin mathbb{N}}(k|nm$ and gcd$(n,k)=1 Rightarrow k|m)$











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Definitons



gcd := greatest common divisor



Let $k,nin mathbb{Z}$ One says that $k$ divides $n$, in symbols $k|n$ if there exists a $l in mathbb{Z}$ s.t. $n=kl$



For two natural numbers $m,n in mathbb{N}$ the greatest common divisior is defined (symblolic gcd$(k,n)$), the biggest natural number $k$ with $k|m$ and $k|n$






algebra-precalculus prime-numbers







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    up vote
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    down vote

    favorite












    Please help me



    Definitons



    gcd := greatest common divisor



    Let $k,nin mathbb{Z}$ One says that $k$ divides $n$, in symbols $k|n$ if there exists a $l in mathbb{Z}$ s.t. $n=kl$



    For two natural numbers $m,n in mathbb{N}$ the greatest common divisior is defined (symblolic gcd$(k,n)$), the biggest natural number $k$ with $k|m$ and $k|n$






    algebra-precalculus prime-numbers







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Please help me



      Definitons



      gcd := greatest common divisor



      Let $k,nin mathbb{Z}$ One says that $k$ divides $n$, in symbols $k|n$ if there exists a $l in mathbb{Z}$ s.t. $n=kl$



      For two natural numbers $m,n in mathbb{N}$ the greatest common divisior is defined (symblolic gcd$(k,n)$), the biggest natural number $k$ with $k|m$ and $k|n$






      algebra-precalculus prime-numbers







      share|cite|improve this question













      Please help me



      Definitons



      gcd := greatest common divisor



      Let $k,nin mathbb{Z}$ One says that $k$ divides $n$, in symbols $k|n$ if there exists a $l in mathbb{Z}$ s.t. $n=kl$



      For two natural numbers $m,n in mathbb{N}$ the greatest common divisior is defined (symblolic gcd$(k,n)$), the biggest natural number $k$ with $k|m$ and $k|n$






      algebra-precalculus prime-numbers




      algebra-precalculus prime-numbers






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      asked Nov 21 at 11:15









      RM777

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      1528






















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          If $m = 0$ the implication is obvious


          If $m ne 0$

          Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$

          We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$

          And $gcd(mn,mk) = m$
          $$implies k|m$$






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            1 Answer
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            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            If $m = 0$ the implication is obvious


            If $m ne 0$

            Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$

            We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$

            And $gcd(mn,mk) = m$
            $$implies k|m$$






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              If $m = 0$ the implication is obvious


              If $m ne 0$

              Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$

              We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$

              And $gcd(mn,mk) = m$
              $$implies k|m$$






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                If $m = 0$ the implication is obvious


                If $m ne 0$

                Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$

                We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$

                And $gcd(mn,mk) = m$
                $$implies k|m$$






                share|cite|improve this answer












                If $m = 0$ the implication is obvious


                If $m ne 0$

                Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$

                We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$

                And $gcd(mn,mk) = m$
                $$implies k|m$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 21 at 12:39









                TheD0ubleT

                37718




                37718






























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