Relation between GCD and LCM [duplicate]
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This question already has an answer here:
Identity involving LCM and GCD
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When I was learning about GCD and LCM I found this relation somewhere. But I didn't get proof for this relation. $$frac{gcd(a,b,c)^2}{gcd(a,b)*gcd(b,c)*gcd(c,a)}=frac{operatorname{lcm}(a,b,c)^2}{operatorname{lcm}(a,b)*operatorname{lcm}(b,c)*operatorname{lcm}(c,a)} $$ where GCD represents Greatest Common Divisor and LCM represents lowest common multiple.
greatest-common-divisor least-common-multiple
asked Dec 26 '18 at 16:24
Dev SanDev San
12
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marked as duplicate by Namaste, Dietrich Burde, Bill Dubuque, quid♦ Dec 26 '18 at 18:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Identity involving LCM and GCD
1 answer
When I was learning about GCD and LCM I found this relation somewhere. But I didn't get proof for this relation. $$frac{gcd(a,b,c)^2}{gcd(a,b)*gcd(b,c)*gcd(c,a)}=frac{operatorname{lcm}(a,b,c)^2}{operatorname{lcm}(a,b)*operatorname{lcm}(b,c)*operatorname{lcm}(c,a)} $$ where GCD represents Greatest Common Divisor and LCM represents lowest common multiple.
greatest-common-divisor least-common-multiple
asked Dec 26 '18 at 16:24
Dev SanDev San
12
$endgroup$
marked as duplicate by Namaste, Dietrich Burde, Bill Dubuque, quid♦ Dec 26 '18 at 18:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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Write all the multiples and divisors using the primes that occur in the prime factorizations of $a$, $b$ and $c$. (This will look ugly in complete generality. If you write it out for, say $6, 15, 24$ you should see the argument.)
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– Ethan Bolker
Dec 26 '18 at 16:28
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There are many similar question at this site, where you can see how the argument goes, e.g., here. Have a look yourself at further (more similar) links.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 16:31
add a comment |
$begingroup$
This question already has an answer here:
Identity involving LCM and GCD
1 answer
When I was learning about GCD and LCM I found this relation somewhere. But I didn't get proof for this relation. $$frac{gcd(a,b,c)^2}{gcd(a,b)*gcd(b,c)*gcd(c,a)}=frac{operatorname{lcm}(a,b,c)^2}{operatorname{lcm}(a,b)*operatorname{lcm}(b,c)*operatorname{lcm}(c,a)} $$ where GCD represents Greatest Common Divisor and LCM represents lowest common multiple.
greatest-common-divisor least-common-multiple
asked Dec 26 '18 at 16:24
Dev SanDev San
12
$endgroup$
This question already has an answer here:
Identity involving LCM and GCD
1 answer
When I was learning about GCD and LCM I found this relation somewhere. But I didn't get proof for this relation. $$frac{gcd(a,b,c)^2}{gcd(a,b)*gcd(b,c)*gcd(c,a)}=frac{operatorname{lcm}(a,b,c)^2}{operatorname{lcm}(a,b)*operatorname{lcm}(b,c)*operatorname{lcm}(c,a)} $$ where GCD represents Greatest Common Divisor and LCM represents lowest common multiple.
This question already has an answer here:
Identity involving LCM and GCD
1 answer
greatest-common-divisor least-common-multiple
greatest-common-divisor least-common-multiple
asked Dec 26 '18 at 16:24
Dev SanDev San
12
asked Dec 26 '18 at 16:24
Dev SanDev San
12
asked Dec 26 '18 at 16:24
Dev SanDev San
12
asked Dec 26 '18 at 16:24
Dev SanDev San
12
asked Dec 26 '18 at 16:24
Dev SanDev San
12
12
marked as duplicate by Namaste, Dietrich Burde, Bill Dubuque, quid♦ Dec 26 '18 at 18:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Namaste, Dietrich Burde, Bill Dubuque, quid♦ Dec 26 '18 at 18:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Write all the multiples and divisors using the primes that occur in the prime factorizations of $a$, $b$ and $c$. (This will look ugly in complete generality. If you write it out for, say $6, 15, 24$ you should see the argument.)
$endgroup$
– Ethan Bolker
Dec 26 '18 at 16:28
$begingroup$
There are many similar question at this site, where you can see how the argument goes, e.g., here. Have a look yourself at further (more similar) links.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 16:31
add a comment |
1
$begingroup$
Write all the multiples and divisors using the primes that occur in the prime factorizations of $a$, $b$ and $c$. (This will look ugly in complete generality. If you write it out for, say $6, 15, 24$ you should see the argument.)
$endgroup$
– Ethan Bolker
Dec 26 '18 at 16:28
$begingroup$
There are many similar question at this site, where you can see how the argument goes, e.g., here. Have a look yourself at further (more similar) links.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 16:31
1
1
$begingroup$
Write all the multiples and divisors using the primes that occur in the prime factorizations of $a$, $b$ and $c$. (This will look ugly in complete generality. If you write it out for, say $6, 15, 24$ you should see the argument.)
$endgroup$
– Ethan Bolker
Dec 26 '18 at 16:28
$begingroup$
Write all the multiples and divisors using the primes that occur in the prime factorizations of $a$, $b$ and $c$. (This will look ugly in complete generality. If you write it out for, say $6, 15, 24$ you should see the argument.)
$endgroup$
– Ethan Bolker
Dec 26 '18 at 16:28
$begingroup$
There are many similar question at this site, where you can see how the argument goes, e.g., here. Have a look yourself at further (more similar) links.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 16:31
$begingroup$
There are many similar question at this site, where you can see how the argument goes, e.g., here. Have a look yourself at further (more similar) links.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 16:31
add a comment |
2 Answers
2
active
oldest
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Let $a=prod{p_i^{alpha_i}}, b=prod{p_i^{beta_i}}$ and $c=prod{p_i^{gamma_i}}$ where $p_i$ denote the prime factors of a, b, c
We know that $gcd(a,b)=prod{p_i^{max(alpha_i,beta_i)}}$, $operatorname{lcm}(a,b)=prod{p_i^{min(alpha_i,beta_i)}}$
Therefore we have to show that $$2max(alpha_i,beta_i,gamma_i)-max(alpha_i,beta_i)-max(beta_i,gamma_i)-max(gamma_i,alpha_i)=2min(alpha_i,beta_i,gamma_i)-min(alpha_i,beta_i)-min(beta_i,gamma_i)-min(gamma_i,alpha_i)$$ for each index i.
Without loss of generality we can assume $alpha_igeqbeta_igeqgamma_i $ for any particular index i. Then above equation can reduce to $$2alpha_i-alpha_i-beta_i-alpha_i=2gamma_i-beta_i-gamma_i-gamma_i $$ which is the identity.
answered Dec 26 '18 at 16:55
Sai Satwik KuppiliSai Satwik Kuppili
658
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Suppose $a=dxza',b=dxyb',c=dyzc'$ and those numbers are realtive prime
so $GCD(a,b,c)=d,GCD(a,b)=dx,GCD(b,c)=dy,GCD(c,a)=dz,LCM(a,b,c)=dxyza'b'c',LCM(a,b)=dxyza'b',LCM(b,c)=dxyzb'c',LCM(c,a)=dxyza'c'$
$$frac{GCD(a,b,c)^2}{GCD(a,b)GCD(b,c)GCD(c,a)}=frac{d^2}{dxcdot dycdot dz}=frac{1}{dxyz}$$
$$frac{LCM(a,b,c)}{LCM(a,b)LCM(b,c)LCM(c,a)}=frac{d^2x^2y^2z^2a'^2b'^2c'^2}{dxyza'b'cdot dxyzb'c'cdot dxyza'c'}=frac{1}{dxyz}$$
answered Dec 26 '18 at 16:37
Reynan HenryReynan Henry
853
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $a=prod{p_i^{alpha_i}}, b=prod{p_i^{beta_i}}$ and $c=prod{p_i^{gamma_i}}$ where $p_i$ denote the prime factors of a, b, c
We know that $gcd(a,b)=prod{p_i^{max(alpha_i,beta_i)}}$, $operatorname{lcm}(a,b)=prod{p_i^{min(alpha_i,beta_i)}}$
Therefore we have to show that $$2max(alpha_i,beta_i,gamma_i)-max(alpha_i,beta_i)-max(beta_i,gamma_i)-max(gamma_i,alpha_i)=2min(alpha_i,beta_i,gamma_i)-min(alpha_i,beta_i)-min(beta_i,gamma_i)-min(gamma_i,alpha_i)$$ for each index i.
Without loss of generality we can assume $alpha_igeqbeta_igeqgamma_i $ for any particular index i. Then above equation can reduce to $$2alpha_i-alpha_i-beta_i-alpha_i=2gamma_i-beta_i-gamma_i-gamma_i $$ which is the identity.
answered Dec 26 '18 at 16:55
Sai Satwik KuppiliSai Satwik Kuppili
658
$endgroup$
add a comment |
$begingroup$
Let $a=prod{p_i^{alpha_i}}, b=prod{p_i^{beta_i}}$ and $c=prod{p_i^{gamma_i}}$ where $p_i$ denote the prime factors of a, b, c
We know that $gcd(a,b)=prod{p_i^{max(alpha_i,beta_i)}}$, $operatorname{lcm}(a,b)=prod{p_i^{min(alpha_i,beta_i)}}$
Therefore we have to show that $$2max(alpha_i,beta_i,gamma_i)-max(alpha_i,beta_i)-max(beta_i,gamma_i)-max(gamma_i,alpha_i)=2min(alpha_i,beta_i,gamma_i)-min(alpha_i,beta_i)-min(beta_i,gamma_i)-min(gamma_i,alpha_i)$$ for each index i.
Without loss of generality we can assume $alpha_igeqbeta_igeqgamma_i $ for any particular index i. Then above equation can reduce to $$2alpha_i-alpha_i-beta_i-alpha_i=2gamma_i-beta_i-gamma_i-gamma_i $$ which is the identity.
answered Dec 26 '18 at 16:55
Sai Satwik KuppiliSai Satwik Kuppili
658
$endgroup$
add a comment |
$begingroup$
Let $a=prod{p_i^{alpha_i}}, b=prod{p_i^{beta_i}}$ and $c=prod{p_i^{gamma_i}}$ where $p_i$ denote the prime factors of a, b, c
We know that $gcd(a,b)=prod{p_i^{max(alpha_i,beta_i)}}$, $operatorname{lcm}(a,b)=prod{p_i^{min(alpha_i,beta_i)}}$
Therefore we have to show that $$2max(alpha_i,beta_i,gamma_i)-max(alpha_i,beta_i)-max(beta_i,gamma_i)-max(gamma_i,alpha_i)=2min(alpha_i,beta_i,gamma_i)-min(alpha_i,beta_i)-min(beta_i,gamma_i)-min(gamma_i,alpha_i)$$ for each index i.
Without loss of generality we can assume $alpha_igeqbeta_igeqgamma_i $ for any particular index i. Then above equation can reduce to $$2alpha_i-alpha_i-beta_i-alpha_i=2gamma_i-beta_i-gamma_i-gamma_i $$ which is the identity.
answered Dec 26 '18 at 16:55
Sai Satwik KuppiliSai Satwik Kuppili
658
$endgroup$
Let $a=prod{p_i^{alpha_i}}, b=prod{p_i^{beta_i}}$ and $c=prod{p_i^{gamma_i}}$ where $p_i$ denote the prime factors of a, b, c
We know that $gcd(a,b)=prod{p_i^{max(alpha_i,beta_i)}}$, $operatorname{lcm}(a,b)=prod{p_i^{min(alpha_i,beta_i)}}$
Therefore we have to show that $$2max(alpha_i,beta_i,gamma_i)-max(alpha_i,beta_i)-max(beta_i,gamma_i)-max(gamma_i,alpha_i)=2min(alpha_i,beta_i,gamma_i)-min(alpha_i,beta_i)-min(beta_i,gamma_i)-min(gamma_i,alpha_i)$$ for each index i.
Without loss of generality we can assume $alpha_igeqbeta_igeqgamma_i $ for any particular index i. Then above equation can reduce to $$2alpha_i-alpha_i-beta_i-alpha_i=2gamma_i-beta_i-gamma_i-gamma_i $$ which is the identity.
answered Dec 26 '18 at 16:55
Sai Satwik KuppiliSai Satwik Kuppili
658
answered Dec 26 '18 at 16:55
Sai Satwik KuppiliSai Satwik Kuppili
658
answered Dec 26 '18 at 16:55
Sai Satwik KuppiliSai Satwik Kuppili
658
answered Dec 26 '18 at 16:55
Sai Satwik KuppiliSai Satwik Kuppili
658
658
add a comment |
add a comment |
$begingroup$
Suppose $a=dxza',b=dxyb',c=dyzc'$ and those numbers are realtive prime
so $GCD(a,b,c)=d,GCD(a,b)=dx,GCD(b,c)=dy,GCD(c,a)=dz,LCM(a,b,c)=dxyza'b'c',LCM(a,b)=dxyza'b',LCM(b,c)=dxyzb'c',LCM(c,a)=dxyza'c'$
$$frac{GCD(a,b,c)^2}{GCD(a,b)GCD(b,c)GCD(c,a)}=frac{d^2}{dxcdot dycdot dz}=frac{1}{dxyz}$$
$$frac{LCM(a,b,c)}{LCM(a,b)LCM(b,c)LCM(c,a)}=frac{d^2x^2y^2z^2a'^2b'^2c'^2}{dxyza'b'cdot dxyzb'c'cdot dxyza'c'}=frac{1}{dxyz}$$
answered Dec 26 '18 at 16:37
Reynan HenryReynan Henry
853
$endgroup$
add a comment |
$begingroup$
Suppose $a=dxza',b=dxyb',c=dyzc'$ and those numbers are realtive prime
so $GCD(a,b,c)=d,GCD(a,b)=dx,GCD(b,c)=dy,GCD(c,a)=dz,LCM(a,b,c)=dxyza'b'c',LCM(a,b)=dxyza'b',LCM(b,c)=dxyzb'c',LCM(c,a)=dxyza'c'$
$$frac{GCD(a,b,c)^2}{GCD(a,b)GCD(b,c)GCD(c,a)}=frac{d^2}{dxcdot dycdot dz}=frac{1}{dxyz}$$
$$frac{LCM(a,b,c)}{LCM(a,b)LCM(b,c)LCM(c,a)}=frac{d^2x^2y^2z^2a'^2b'^2c'^2}{dxyza'b'cdot dxyzb'c'cdot dxyza'c'}=frac{1}{dxyz}$$
answered Dec 26 '18 at 16:37
Reynan HenryReynan Henry
853
$endgroup$
add a comment |
$begingroup$
Suppose $a=dxza',b=dxyb',c=dyzc'$ and those numbers are realtive prime
so $GCD(a,b,c)=d,GCD(a,b)=dx,GCD(b,c)=dy,GCD(c,a)=dz,LCM(a,b,c)=dxyza'b'c',LCM(a,b)=dxyza'b',LCM(b,c)=dxyzb'c',LCM(c,a)=dxyza'c'$
$$frac{GCD(a,b,c)^2}{GCD(a,b)GCD(b,c)GCD(c,a)}=frac{d^2}{dxcdot dycdot dz}=frac{1}{dxyz}$$
$$frac{LCM(a,b,c)}{LCM(a,b)LCM(b,c)LCM(c,a)}=frac{d^2x^2y^2z^2a'^2b'^2c'^2}{dxyza'b'cdot dxyzb'c'cdot dxyza'c'}=frac{1}{dxyz}$$
answered Dec 26 '18 at 16:37
Reynan HenryReynan Henry
853
$endgroup$
Suppose $a=dxza',b=dxyb',c=dyzc'$ and those numbers are realtive prime
so $GCD(a,b,c)=d,GCD(a,b)=dx,GCD(b,c)=dy,GCD(c,a)=dz,LCM(a,b,c)=dxyza'b'c',LCM(a,b)=dxyza'b',LCM(b,c)=dxyzb'c',LCM(c,a)=dxyza'c'$
$$frac{GCD(a,b,c)^2}{GCD(a,b)GCD(b,c)GCD(c,a)}=frac{d^2}{dxcdot dycdot dz}=frac{1}{dxyz}$$
$$frac{LCM(a,b,c)}{LCM(a,b)LCM(b,c)LCM(c,a)}=frac{d^2x^2y^2z^2a'^2b'^2c'^2}{dxyza'b'cdot dxyzb'c'cdot dxyza'c'}=frac{1}{dxyz}$$
answered Dec 26 '18 at 16:37
Reynan HenryReynan Henry
853
answered Dec 26 '18 at 16:37
Reynan HenryReynan Henry
853
answered Dec 26 '18 at 16:37
Reynan HenryReynan Henry
853
answered Dec 26 '18 at 16:37
Reynan HenryReynan Henry
853
853
add a comment |
add a comment |
1
$begingroup$
Write all the multiples and divisors using the primes that occur in the prime factorizations of $a$, $b$ and $c$. (This will look ugly in complete generality. If you write it out for, say $6, 15, 24$ you should see the argument.)
$endgroup$
– Ethan Bolker
Dec 26 '18 at 16:28
$begingroup$
There are many similar question at this site, where you can see how the argument goes, e.g., here. Have a look yourself at further (more similar) links.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 16:31