Mixing
Prove $x equiv a pmod{p}$ and $x equiv a pmod{q}$ then $x equiv apmod{pq}$
$begingroup$
$p$ and $q$ are distinct primes.
Where can I start with this proof?
It looks similar to the Chinese Remainder Theorem, but that deals with two different a values.
number-theory
edited Sep 3 '12 at 16:08
Thomas Russell
7,89632552
asked Sep 3 '12 at 15:57
TakkunTakkun
268311
$endgroup$
add a comment |
$begingroup$
$p$ and $q$ are distinct primes.
Where can I start with this proof?
It looks similar to the Chinese Remainder Theorem, but that deals with two different a values.
number-theory
edited Sep 3 '12 at 16:08
Thomas Russell
7,89632552
asked Sep 3 '12 at 15:57
TakkunTakkun
268311
$endgroup$
1
$begingroup$
but that deals with two different $a$ values - What makes you think that?
$endgroup$
– anon
Sep 3 '12 at 15:59
$begingroup$
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
$endgroup$
– Takkun
Sep 3 '12 at 16:03
3
$begingroup$
A couple of variables denoted with two different letters may take on two different values, but not necessarily unless they are explicitly stated to have distinct values.
$endgroup$
– anon
Sep 3 '12 at 16:04
add a comment |
$begingroup$
$p$ and $q$ are distinct primes.
Where can I start with this proof?
It looks similar to the Chinese Remainder Theorem, but that deals with two different a values.
number-theory
edited Sep 3 '12 at 16:08
Thomas Russell
7,89632552
asked Sep 3 '12 at 15:57
TakkunTakkun
268311
$endgroup$
$p$ and $q$ are distinct primes.
Where can I start with this proof?
It looks similar to the Chinese Remainder Theorem, but that deals with two different a values.
number-theory
number-theory
edited Sep 3 '12 at 16:08
Thomas Russell
7,89632552
asked Sep 3 '12 at 15:57
TakkunTakkun
268311
edited Sep 3 '12 at 16:08
Thomas Russell
7,89632552
asked Sep 3 '12 at 15:57
TakkunTakkun
268311
edited Sep 3 '12 at 16:08
Thomas Russell
7,89632552
edited Sep 3 '12 at 16:08
Thomas Russell
7,89632552
edited Sep 3 '12 at 16:08
Thomas Russell
7,89632552
7,89632552
asked Sep 3 '12 at 15:57
TakkunTakkun
268311
asked Sep 3 '12 at 15:57
TakkunTakkun
268311
asked Sep 3 '12 at 15:57
TakkunTakkun
268311
268311
1
$begingroup$
but that deals with two different $a$ values - What makes you think that?
$endgroup$
– anon
Sep 3 '12 at 15:59
$begingroup$
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
$endgroup$
– Takkun
Sep 3 '12 at 16:03
3
$begingroup$
A couple of variables denoted with two different letters may take on two different values, but not necessarily unless they are explicitly stated to have distinct values.
$endgroup$
– anon
Sep 3 '12 at 16:04
add a comment |
1
$begingroup$
but that deals with two different $a$ values - What makes you think that?
$endgroup$
– anon
Sep 3 '12 at 15:59
$begingroup$
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
$endgroup$
– Takkun
Sep 3 '12 at 16:03
3
$begingroup$
A couple of variables denoted with two different letters may take on two different values, but not necessarily unless they are explicitly stated to have distinct values.
$endgroup$
– anon
Sep 3 '12 at 16:04
1
1
$begingroup$
but that deals with two different $a$ values - What makes you think that?
$endgroup$
– anon
Sep 3 '12 at 15:59
$begingroup$
but that deals with two different $a$ values - What makes you think that?
$endgroup$
– anon
Sep 3 '12 at 15:59
$begingroup$
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
$endgroup$
– Takkun
Sep 3 '12 at 16:03
$begingroup$
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
$endgroup$
– Takkun
Sep 3 '12 at 16:03
3
3
$begingroup$
A couple of variables denoted with two different letters may take on two different values, but not necessarily unless they are explicitly stated to have distinct values.
$endgroup$
– anon
Sep 3 '12 at 16:04
$begingroup$
A couple of variables denoted with two different letters may take on two different values, but not necessarily unless they are explicitly stated to have distinct values.
$endgroup$
– anon
Sep 3 '12 at 16:04
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint $ $ Below are few proofs of this constant-case CCRT of Chinese Remainder Theorem (CRT). The first three proofs work for for arbitrary coprime naturals $rm,p,q.$
$rm(1) x equiv apmod {pq}:$ is clearly a solution, and the solution is $color{#C00}{unique}$ mod $rm,pq,$ by CRT.
$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a:$ by the Universal Property of $rm lcm$.
$qquad! $ Further $rm:(p,q)=1iff:lcm(p,q) = pq,,$ by this answer.
$(3) , $ By Euclid's Lemma: $rm:(p,q)=1, p:|:qn =:x!-!a:Rightarrow:p:|:n:Rightarrow:pq:|:nq = x!-!a.$
$rm(4) , $ Since the prime factorization of $rm,x!-!a,$ is $color{#C00}{unique}$, and the prime $rm,p,$ occurs in one factorization, and the distinct prime $rm,q,$ occurs in another, both factorizations are the same up to order, hence contain both $rm,p,$ and $rm,q,:$ hence have $rm,pq,$ as a divisor.
Remark $ $ This constant-case optimization of CRT arises frequently in practice so is well-worth memorizing, e.g. see some prior posts for many examples.
Quite frequently, $color{#C00}{uniqueness} theorems$ provide powerful tools for proving equalities.
answered Sep 3 '12 at 16:18
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
$begingroup$
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
$endgroup$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
$begingroup$
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
$endgroup$
– Bill Dubuque
Sep 23 '12 at 23:37
add a comment |
$begingroup$
Let $[A,B]=lcm(A,B)$ and $(A,B)=gcd(A,B)$
If $p,q$ are different integers, $pmid(x-a)$ and $qmid(x-a)implies [p,q]mid(x-a)$
We know $[p,q]cdot (p,q)=pcdot q$
If $(p,q)=1, [p,q]=pcdot q$
If $p,q$ are distinct primes, $(p,q)=1$
answered Sep 3 '12 at 16:01
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
add a comment |
$begingroup$
Let $y=x-a$. We want to show that if $p$ divides $y$ and $q$ divides $y$ then $pq$ divides $y$.
Since $p$ divides $y$, we have $y=pz$ for some $z$. Thus $q$ divides $pz$. Since $q$ is prime, this implies $q$ divides $p$ or $q$ divides $z$. But $q$ cannot divide $p$, so $q$ divides $z$. Suppose that $z=qw$. Then $y=pqw$.
answered Sep 3 '12 at 16:08
André NicolasAndré Nicolas
454k36432819
$endgroup$
add a comment |
Your Answer
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3 Answers
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3 Answers
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$begingroup$
Hint $ $ Below are few proofs of this constant-case CCRT of Chinese Remainder Theorem (CRT). The first three proofs work for for arbitrary coprime naturals $rm,p,q.$
$rm(1) x equiv apmod {pq}:$ is clearly a solution, and the solution is $color{#C00}{unique}$ mod $rm,pq,$ by CRT.
$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a:$ by the Universal Property of $rm lcm$.
$qquad! $ Further $rm:(p,q)=1iff:lcm(p,q) = pq,,$ by this answer.
$(3) , $ By Euclid's Lemma: $rm:(p,q)=1, p:|:qn =:x!-!a:Rightarrow:p:|:n:Rightarrow:pq:|:nq = x!-!a.$
$rm(4) , $ Since the prime factorization of $rm,x!-!a,$ is $color{#C00}{unique}$, and the prime $rm,p,$ occurs in one factorization, and the distinct prime $rm,q,$ occurs in another, both factorizations are the same up to order, hence contain both $rm,p,$ and $rm,q,:$ hence have $rm,pq,$ as a divisor.
Remark $ $ This constant-case optimization of CRT arises frequently in practice so is well-worth memorizing, e.g. see some prior posts for many examples.
Quite frequently, $color{#C00}{uniqueness} theorems$ provide powerful tools for proving equalities.
answered Sep 3 '12 at 16:18
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
$begingroup$
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
$endgroup$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
$begingroup$
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
$endgroup$
– Bill Dubuque
Sep 23 '12 at 23:37
add a comment |
$begingroup$
Hint $ $ Below are few proofs of this constant-case CCRT of Chinese Remainder Theorem (CRT). The first three proofs work for for arbitrary coprime naturals $rm,p,q.$
$rm(1) x equiv apmod {pq}:$ is clearly a solution, and the solution is $color{#C00}{unique}$ mod $rm,pq,$ by CRT.
$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a:$ by the Universal Property of $rm lcm$.
$qquad! $ Further $rm:(p,q)=1iff:lcm(p,q) = pq,,$ by this answer.
$(3) , $ By Euclid's Lemma: $rm:(p,q)=1, p:|:qn =:x!-!a:Rightarrow:p:|:n:Rightarrow:pq:|:nq = x!-!a.$
$rm(4) , $ Since the prime factorization of $rm,x!-!a,$ is $color{#C00}{unique}$, and the prime $rm,p,$ occurs in one factorization, and the distinct prime $rm,q,$ occurs in another, both factorizations are the same up to order, hence contain both $rm,p,$ and $rm,q,:$ hence have $rm,pq,$ as a divisor.
Remark $ $ This constant-case optimization of CRT arises frequently in practice so is well-worth memorizing, e.g. see some prior posts for many examples.
Quite frequently, $color{#C00}{uniqueness} theorems$ provide powerful tools for proving equalities.
answered Sep 3 '12 at 16:18
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
$begingroup$
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
$endgroup$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
$begingroup$
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
$endgroup$
– Bill Dubuque
Sep 23 '12 at 23:37
add a comment |
$begingroup$
Hint $ $ Below are few proofs of this constant-case CCRT of Chinese Remainder Theorem (CRT). The first three proofs work for for arbitrary coprime naturals $rm,p,q.$
$rm(1) x equiv apmod {pq}:$ is clearly a solution, and the solution is $color{#C00}{unique}$ mod $rm,pq,$ by CRT.
$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a:$ by the Universal Property of $rm lcm$.
$qquad! $ Further $rm:(p,q)=1iff:lcm(p,q) = pq,,$ by this answer.
$(3) , $ By Euclid's Lemma: $rm:(p,q)=1, p:|:qn =:x!-!a:Rightarrow:p:|:n:Rightarrow:pq:|:nq = x!-!a.$
$rm(4) , $ Since the prime factorization of $rm,x!-!a,$ is $color{#C00}{unique}$, and the prime $rm,p,$ occurs in one factorization, and the distinct prime $rm,q,$ occurs in another, both factorizations are the same up to order, hence contain both $rm,p,$ and $rm,q,:$ hence have $rm,pq,$ as a divisor.
Remark $ $ This constant-case optimization of CRT arises frequently in practice so is well-worth memorizing, e.g. see some prior posts for many examples.
Quite frequently, $color{#C00}{uniqueness} theorems$ provide powerful tools for proving equalities.
answered Sep 3 '12 at 16:18
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
Hint $ $ Below are few proofs of this constant-case CCRT of Chinese Remainder Theorem (CRT). The first three proofs work for for arbitrary coprime naturals $rm,p,q.$
$rm(1) x equiv apmod {pq}:$ is clearly a solution, and the solution is $color{#C00}{unique}$ mod $rm,pq,$ by CRT.
$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a:$ by the Universal Property of $rm lcm$.
$qquad! $ Further $rm:(p,q)=1iff:lcm(p,q) = pq,,$ by this answer.
$(3) , $ By Euclid's Lemma: $rm:(p,q)=1, p:|:qn =:x!-!a:Rightarrow:p:|:n:Rightarrow:pq:|:nq = x!-!a.$
$rm(4) , $ Since the prime factorization of $rm,x!-!a,$ is $color{#C00}{unique}$, and the prime $rm,p,$ occurs in one factorization, and the distinct prime $rm,q,$ occurs in another, both factorizations are the same up to order, hence contain both $rm,p,$ and $rm,q,:$ hence have $rm,pq,$ as a divisor.
Remark $ $ This constant-case optimization of CRT arises frequently in practice so is well-worth memorizing, e.g. see some prior posts for many examples.
Quite frequently, $color{#C00}{uniqueness} theorems$ provide powerful tools for proving equalities.
answered Sep 3 '12 at 16:18
Bill DubuqueBill Dubuque
213k29195654
edited Feb 7 at 16:34
answered Sep 3 '12 at 16:18
Bill DubuqueBill Dubuque
213k29195654
answered Sep 3 '12 at 16:18
Bill DubuqueBill Dubuque
213k29195654
answered Sep 3 '12 at 16:18
Bill DubuqueBill Dubuque
213k29195654
213k29195654
$begingroup$
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
$endgroup$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
$begingroup$
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
$endgroup$
– Bill Dubuque
Sep 23 '12 at 23:37
add a comment |
$begingroup$
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
$endgroup$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
$begingroup$
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
$endgroup$
– Bill Dubuque
Sep 23 '12 at 23:37
$begingroup$
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
$endgroup$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
$begingroup$
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
$endgroup$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
$begingroup$
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
$endgroup$
– Bill Dubuque
Sep 23 '12 at 23:37
$begingroup$
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
$endgroup$
– Bill Dubuque
Sep 23 '12 at 23:37
add a comment |
$begingroup$
Let $[A,B]=lcm(A,B)$ and $(A,B)=gcd(A,B)$
If $p,q$ are different integers, $pmid(x-a)$ and $qmid(x-a)implies [p,q]mid(x-a)$
We know $[p,q]cdot (p,q)=pcdot q$
If $(p,q)=1, [p,q]=pcdot q$
If $p,q$ are distinct primes, $(p,q)=1$
answered Sep 3 '12 at 16:01
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
add a comment |
$begingroup$
Let $[A,B]=lcm(A,B)$ and $(A,B)=gcd(A,B)$
If $p,q$ are different integers, $pmid(x-a)$ and $qmid(x-a)implies [p,q]mid(x-a)$
We know $[p,q]cdot (p,q)=pcdot q$
If $(p,q)=1, [p,q]=pcdot q$
If $p,q$ are distinct primes, $(p,q)=1$
answered Sep 3 '12 at 16:01
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
add a comment |
$begingroup$
Let $[A,B]=lcm(A,B)$ and $(A,B)=gcd(A,B)$
If $p,q$ are different integers, $pmid(x-a)$ and $qmid(x-a)implies [p,q]mid(x-a)$
We know $[p,q]cdot (p,q)=pcdot q$
If $(p,q)=1, [p,q]=pcdot q$
If $p,q$ are distinct primes, $(p,q)=1$
answered Sep 3 '12 at 16:01
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
Let $[A,B]=lcm(A,B)$ and $(A,B)=gcd(A,B)$
If $p,q$ are different integers, $pmid(x-a)$ and $qmid(x-a)implies [p,q]mid(x-a)$
We know $[p,q]cdot (p,q)=pcdot q$
If $(p,q)=1, [p,q]=pcdot q$
If $p,q$ are distinct primes, $(p,q)=1$
answered Sep 3 '12 at 16:01
lab bhattacharjeelab bhattacharjee
227k15158278
edited Sep 3 '12 at 16:06
answered Sep 3 '12 at 16:01
lab bhattacharjeelab bhattacharjee
227k15158278
answered Sep 3 '12 at 16:01
lab bhattacharjeelab bhattacharjee
227k15158278
answered Sep 3 '12 at 16:01
lab bhattacharjeelab bhattacharjee
227k15158278
227k15158278
add a comment |
add a comment |
$begingroup$
Let $y=x-a$. We want to show that if $p$ divides $y$ and $q$ divides $y$ then $pq$ divides $y$.
Since $p$ divides $y$, we have $y=pz$ for some $z$. Thus $q$ divides $pz$. Since $q$ is prime, this implies $q$ divides $p$ or $q$ divides $z$. But $q$ cannot divide $p$, so $q$ divides $z$. Suppose that $z=qw$. Then $y=pqw$.
answered Sep 3 '12 at 16:08
André NicolasAndré Nicolas
454k36432819
$endgroup$
add a comment |
$begingroup$
Let $y=x-a$. We want to show that if $p$ divides $y$ and $q$ divides $y$ then $pq$ divides $y$.
Since $p$ divides $y$, we have $y=pz$ for some $z$. Thus $q$ divides $pz$. Since $q$ is prime, this implies $q$ divides $p$ or $q$ divides $z$. But $q$ cannot divide $p$, so $q$ divides $z$. Suppose that $z=qw$. Then $y=pqw$.
answered Sep 3 '12 at 16:08
André NicolasAndré Nicolas
454k36432819
$endgroup$
add a comment |
$begingroup$
Let $y=x-a$. We want to show that if $p$ divides $y$ and $q$ divides $y$ then $pq$ divides $y$.
Since $p$ divides $y$, we have $y=pz$ for some $z$. Thus $q$ divides $pz$. Since $q$ is prime, this implies $q$ divides $p$ or $q$ divides $z$. But $q$ cannot divide $p$, so $q$ divides $z$. Suppose that $z=qw$. Then $y=pqw$.
answered Sep 3 '12 at 16:08
André NicolasAndré Nicolas
454k36432819
$endgroup$
Let $y=x-a$. We want to show that if $p$ divides $y$ and $q$ divides $y$ then $pq$ divides $y$.
Since $p$ divides $y$, we have $y=pz$ for some $z$. Thus $q$ divides $pz$. Since $q$ is prime, this implies $q$ divides $p$ or $q$ divides $z$. But $q$ cannot divide $p$, so $q$ divides $z$. Suppose that $z=qw$. Then $y=pqw$.
answered Sep 3 '12 at 16:08
André NicolasAndré Nicolas
454k36432819
edited Sep 3 '12 at 16:30
answered Sep 3 '12 at 16:08
André NicolasAndré Nicolas
454k36432819
answered Sep 3 '12 at 16:08
André NicolasAndré Nicolas
454k36432819
answered Sep 3 '12 at 16:08
André NicolasAndré Nicolas
454k36432819
454k36432819
add a comment |
add a comment |
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1
$begingroup$
but that deals with two different $a$ values - What makes you think that?
$endgroup$
– anon
Sep 3 '12 at 15:59
$begingroup$
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
$endgroup$
– Takkun
Sep 3 '12 at 16:03
3
$begingroup$
A couple of variables denoted with two different letters may take on two different values, but not necessarily unless they are explicitly stated to have distinct values.
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– anon
Sep 3 '12 at 16:04