Problem regarding this proof of Chinese Remainder Theorem The 2019 Stack Overflow Developer Survey Results Are InApplication of the Chinese Remainder TheoremChinese Remainder Theorem clarificationExample involving the Chinese Remainder TheoremProblem in proof of Chinese remainder theorem, and applying it.Chinese remainder theorem applicationGeneral Chinese remainder theorem proofChinese Remainder Theorem ConsequenceSolving Chinese Remainder Theorem AlgebraicallyUsing the Chinese Remainder Theorem to prove that a system of equations has a solution in $mathbbN$ if and only if certain conditions are metChinese remainder theorem method

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Problem regarding this proof of Chinese Remainder Theorem



The 2019 Stack Overflow Developer Survey Results Are InApplication of the Chinese Remainder TheoremChinese Remainder Theorem clarificationExample involving the Chinese Remainder TheoremProblem in proof of Chinese remainder theorem, and applying it.Chinese remainder theorem applicationGeneral Chinese remainder theorem proofChinese Remainder Theorem ConsequenceSolving Chinese Remainder Theorem AlgebraicallyUsing the Chinese Remainder Theorem to prove that a system of equations has a solution in $mathbbN$ if and only if certain conditions are metChinese remainder theorem method










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I am facing problem in understanding the last part of the proof of Theorem 6 (Chinese Remainder Theorem. I cannot understand why $(m_j,n_j)=1$. I do not understand the line "Solving $m_j xequiv 1 (mod n_j)$ using Theorem 5, we have a unique solution $xequiv m_j (mod n_j)$". What is $m_j'$? I also do not understand the rest of the proof. Please help.






number-theory elementary-number-theory modular-arithmetic chinese-remainder-theorem







share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    enter image description here



    enter image description here



    enter image description here



    I am facing problem in understanding the last part of the proof of Theorem 6 (Chinese Remainder Theorem. I cannot understand why $(m_j,n_j)=1$. I do not understand the line "Solving $m_j xequiv 1 (mod n_j)$ using Theorem 5, we have a unique solution $xequiv m_j (mod n_j)$". What is $m_j'$? I also do not understand the rest of the proof. Please help.






    number-theory elementary-number-theory modular-arithmetic chinese-remainder-theorem







    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      enter image description here



      enter image description here



      enter image description here



      I am facing problem in understanding the last part of the proof of Theorem 6 (Chinese Remainder Theorem. I cannot understand why $(m_j,n_j)=1$. I do not understand the line "Solving $m_j xequiv 1 (mod n_j)$ using Theorem 5, we have a unique solution $xequiv m_j (mod n_j)$". What is $m_j'$? I also do not understand the rest of the proof. Please help.






      number-theory elementary-number-theory modular-arithmetic chinese-remainder-theorem







      share|cite|improve this question











      $endgroup$




      enter image description here



      enter image description here



      enter image description here



      I am facing problem in understanding the last part of the proof of Theorem 6 (Chinese Remainder Theorem. I cannot understand why $(m_j,n_j)=1$. I do not understand the line "Solving $m_j xequiv 1 (mod n_j)$ using Theorem 5, we have a unique solution $xequiv m_j (mod n_j)$". What is $m_j'$? I also do not understand the rest of the proof. Please help.






      number-theory elementary-number-theory modular-arithmetic chinese-remainder-theorem




      number-theory elementary-number-theory modular-arithmetic chinese-remainder-theorem






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 24 at 17:14







      MrAP

















      asked Mar 24 at 7:14









      MrAPMrAP

      1,29821432




      1,29821432




















          1 Answer
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          1












          $begingroup$

          The numbers $n_1,ldots,n_k$ are relatively prime.
          Put $M=prod_i=1^k n_i$ and $m_j=M/n_j$ for $1leq jleq k$. Then clearly $n_j$ and $m_j$ are relatively prime, i.e., $(m_j,n_j)=1$. Then by Bezout's theorem, $1 = x_jm_j+y_jn_j$ for some numbers $x_j,y_j$. Then $m_jx_jequiv 1 mod n_j$. These solutions $x_j$ (which are called $m'_j$) are then put together to give a solution $x_0$ for the system in question.






          share|cite|improve this answer









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            $begingroup$

            The numbers $n_1,ldots,n_k$ are relatively prime.
            Put $M=prod_i=1^k n_i$ and $m_j=M/n_j$ for $1leq jleq k$. Then clearly $n_j$ and $m_j$ are relatively prime, i.e., $(m_j,n_j)=1$. Then by Bezout's theorem, $1 = x_jm_j+y_jn_j$ for some numbers $x_j,y_j$. Then $m_jx_jequiv 1 mod n_j$. These solutions $x_j$ (which are called $m'_j$) are then put together to give a solution $x_0$ for the system in question.






            share|cite|improve this answer









            $endgroup$

















              1












              $begingroup$

              The numbers $n_1,ldots,n_k$ are relatively prime.
              Put $M=prod_i=1^k n_i$ and $m_j=M/n_j$ for $1leq jleq k$. Then clearly $n_j$ and $m_j$ are relatively prime, i.e., $(m_j,n_j)=1$. Then by Bezout's theorem, $1 = x_jm_j+y_jn_j$ for some numbers $x_j,y_j$. Then $m_jx_jequiv 1 mod n_j$. These solutions $x_j$ (which are called $m'_j$) are then put together to give a solution $x_0$ for the system in question.






              share|cite|improve this answer









              $endgroup$















                1












                1








                1





                $begingroup$

                The numbers $n_1,ldots,n_k$ are relatively prime.
                Put $M=prod_i=1^k n_i$ and $m_j=M/n_j$ for $1leq jleq k$. Then clearly $n_j$ and $m_j$ are relatively prime, i.e., $(m_j,n_j)=1$. Then by Bezout's theorem, $1 = x_jm_j+y_jn_j$ for some numbers $x_j,y_j$. Then $m_jx_jequiv 1 mod n_j$. These solutions $x_j$ (which are called $m'_j$) are then put together to give a solution $x_0$ for the system in question.






                share|cite|improve this answer









                $endgroup$



                The numbers $n_1,ldots,n_k$ are relatively prime.
                Put $M=prod_i=1^k n_i$ and $m_j=M/n_j$ for $1leq jleq k$. Then clearly $n_j$ and $m_j$ are relatively prime, i.e., $(m_j,n_j)=1$. Then by Bezout's theorem, $1 = x_jm_j+y_jn_j$ for some numbers $x_j,y_j$. Then $m_jx_jequiv 1 mod n_j$. These solutions $x_j$ (which are called $m'_j$) are then put together to give a solution $x_0$ for the system in question.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 24 at 7:38









                WuestenfuxWuestenfux

                5,5131513




                5,5131513



























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