Mixing
Prove $x equiv a pmod{p}$ and $x equiv a pmod{q}$ then $x equiv apmod{pq}$
up vote
3
down vote
favorite
$p$ 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
7,80632550
asked Sep 3 '12 at 15:57
Takkun
263211
add a comment |
up vote
3
down vote
favorite
$p$ 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
7,80632550
asked Sep 3 '12 at 15:57
Takkun
263211
but that deals with two different $a$ values - What makes you think that?
– anon
Sep 3 '12 at 15:59
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
– Takkun
Sep 3 '12 at 16:03
2
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.
– anon
Sep 3 '12 at 16:04
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$p$ 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
7,80632550
asked Sep 3 '12 at 15:57
Takkun
263211
$p$ 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
7,80632550
asked Sep 3 '12 at 15:57
Takkun
263211
edited Sep 3 '12 at 16:08
Thomas Russell
7,80632550
asked Sep 3 '12 at 15:57
Takkun
263211
edited Sep 3 '12 at 16:08
Thomas Russell
7,80632550
edited Sep 3 '12 at 16:08
Thomas Russell
7,80632550
edited Sep 3 '12 at 16:08
Thomas Russell
7,80632550
7,80632550
asked Sep 3 '12 at 15:57
Takkun
263211
asked Sep 3 '12 at 15:57
Takkun
263211
asked Sep 3 '12 at 15:57
Takkun
263211
263211
but that deals with two different $a$ values - What makes you think that?
– anon
Sep 3 '12 at 15:59
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
– Takkun
Sep 3 '12 at 16:03
2
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.
– anon
Sep 3 '12 at 16:04
add a comment |
but that deals with two different $a$ values - What makes you think that?
– anon
Sep 3 '12 at 15:59
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
– Takkun
Sep 3 '12 at 16:03
2
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.
– anon
Sep 3 '12 at 16:04
but that deals with two different $a$ values - What makes you think that?
– anon
Sep 3 '12 at 15:59
but that deals with two different $a$ values - What makes you think that?
– anon
Sep 3 '12 at 15:59
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
– Takkun
Sep 3 '12 at 16:03
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
– Takkun
Sep 3 '12 at 16:03
2
2
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.
– anon
Sep 3 '12 at 16:04
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.
– anon
Sep 3 '12 at 16:04
add a comment |
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
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.
$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a.:$ Further $rm:(p,q)=1iff:lcm(p,q) = pq.$
$(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 Dubuque
206k29189621
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
– Bill Dubuque
Sep 23 '12 at 23:37
add a comment |
up vote
4
down vote
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 bhattacharjee
220k15154271
add a comment |
up vote
3
down vote
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 Nicolas
450k36419801
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
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.
$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a.:$ Further $rm:(p,q)=1iff:lcm(p,q) = pq.$
$(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 Dubuque
206k29189621
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
– Bill Dubuque
Sep 23 '12 at 23:37
add a comment |
up vote
6
down vote
accepted
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.
$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a.:$ Further $rm:(p,q)=1iff:lcm(p,q) = pq.$
$(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 Dubuque
206k29189621
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
– Bill Dubuque
Sep 23 '12 at 23:37
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
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.
$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a.:$ Further $rm:(p,q)=1iff:lcm(p,q) = pq.$
$(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 Dubuque
206k29189621
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.
$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a.:$ Further $rm:(p,q)=1iff:lcm(p,q) = pq.$
$(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 Dubuque
206k29189621
edited 2 days ago
answered Sep 3 '12 at 16:18
Bill Dubuque
206k29189621
answered Sep 3 '12 at 16:18
Bill Dubuque
206k29189621
answered Sep 3 '12 at 16:18
Bill Dubuque
206k29189621
206k29189621
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
– Bill Dubuque
Sep 23 '12 at 23:37
add a comment |
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
– Bill Dubuque
Sep 23 '12 at 23:37
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
in (2) do you mean to say $gcd(p, q) = 1 iff lcm(p, q) = pq$
– CodeKingPlusPlus
Sep 23 '12 at 23:17
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
– Bill Dubuque
Sep 23 '12 at 23:37
@Code Yes, $rm:(x,y):$ means $rm:gcd(x,y):$ in number theory (common notation).
– Bill Dubuque
Sep 23 '12 at 23:37
add a comment |
up vote
4
down vote
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 bhattacharjee
220k15154271
add a comment |
up vote
4
down vote
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 bhattacharjee
220k15154271
add a comment |
up vote
4
down vote
up vote
4
down vote
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 bhattacharjee
220k15154271
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 bhattacharjee
220k15154271
edited Sep 3 '12 at 16:06
answered Sep 3 '12 at 16:01
lab bhattacharjee
220k15154271
answered Sep 3 '12 at 16:01
lab bhattacharjee
220k15154271
answered Sep 3 '12 at 16:01
lab bhattacharjee
220k15154271
220k15154271
add a comment |
add a comment |
up vote
3
down vote
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 Nicolas
450k36419801
add a comment |
up vote
3
down vote
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 Nicolas
450k36419801
add a comment |
up vote
3
down vote
up vote
3
down vote
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 Nicolas
450k36419801
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 Nicolas
450k36419801
edited Sep 3 '12 at 16:30
answered Sep 3 '12 at 16:08
André Nicolas
450k36419801
answered Sep 3 '12 at 16:08
André Nicolas
450k36419801
answered Sep 3 '12 at 16:08
André Nicolas
450k36419801
450k36419801
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f190514%2fprove-x-equiv-a-pmodp-and-x-equiv-a-pmodq-then-x-equiv-a-pmodpq%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
but that deals with two different $a$ values - What makes you think that?
– anon
Sep 3 '12 at 15:59
Because for the explanations of Chinese Remainder Theorem I've read, they use something like a = x (mod p) and a = y (mod p)
– Takkun
Sep 3 '12 at 16:03
2
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.
– anon
Sep 3 '12 at 16:04