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
$begingroup$
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
asked Mar 24 at 7:14
MrAPMrAP
1,29821432
$endgroup$
add a comment |
$begingroup$
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
asked Mar 24 at 7:14
MrAPMrAP
1,29821432
$endgroup$
add a comment |
$begingroup$
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
asked Mar 24 at 7:14
MrAPMrAP
1,29821432
$endgroup$
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
asked Mar 24 at 7:14
MrAPMrAP
1,29821432
asked Mar 24 at 7:14
MrAPMrAP
1,29821432
edited Mar 24 at 17:14
MrAP
asked Mar 24 at 7:14
MrAPMrAP
1,29821432
asked Mar 24 at 7:14
MrAPMrAP
1,29821432
asked Mar 24 at 7:14
MrAPMrAP
1,29821432
1,29821432
add a comment |
add a comment |
1 Answer
1
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oldest
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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.
answered Mar 24 at 7:38
WuestenfuxWuestenfux
5,5131513
$endgroup$
add a comment |
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1 Answer
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1 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.
answered Mar 24 at 7:38
WuestenfuxWuestenfux
5,5131513
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 24 at 7:38
WuestenfuxWuestenfux
5,5131513
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 24 at 7:38
WuestenfuxWuestenfux
5,5131513
$endgroup$
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.
answered Mar 24 at 7:38
WuestenfuxWuestenfux
5,5131513
answered Mar 24 at 7:38
WuestenfuxWuestenfux
5,5131513
answered Mar 24 at 7:38
WuestenfuxWuestenfux
5,5131513
answered Mar 24 at 7:38
WuestenfuxWuestenfux
5,5131513
5,5131513
add a comment |
add a comment |
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