Mixing
Proving a gcd equation: $gcd(ab,c) mid gcd(a,c) cdot gcd(b,c)$
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How would I be able to prove that $$gcd(ab,c) mid gcd(a,c) cdot gcd(b,c)$$
I'm assuming I can start by saying $gcd(ab,c) cdot n = gcd(a,c) cdot gcd(b,c)$ for some $n in Z$ but I'm not sure how I can represent the $gcd$ value as an integer combination of $a$ and $b$ to prove that the left side can divide the right side
elementary-number-theory discrete-mathematics greatest-common-divisor
edited Jan 27 at 11:00
Martin Sleziak
44.9k10122276
asked Apr 19 '17 at 17:49
Spencer ChowSpencer Chow
1417
$endgroup$
add a comment |
$begingroup$
How would I be able to prove that $$gcd(ab,c) mid gcd(a,c) cdot gcd(b,c)$$
I'm assuming I can start by saying $gcd(ab,c) cdot n = gcd(a,c) cdot gcd(b,c)$ for some $n in Z$ but I'm not sure how I can represent the $gcd$ value as an integer combination of $a$ and $b$ to prove that the left side can divide the right side
elementary-number-theory discrete-mathematics greatest-common-divisor
edited Jan 27 at 11:00
Martin Sleziak
44.9k10122276
asked Apr 19 '17 at 17:49
Spencer ChowSpencer Chow
1417
$endgroup$
1
$begingroup$
Easiest, I think, to use unique factorization and the fact that the order of a prime $p$ in $gcd (m,n)$ is the lesser of the orders of $p$ in $m,n$.
$endgroup$
– lulu
Apr 19 '17 at 17:54
$begingroup$
Hint: if $dmid x$ and $dmid y$, then $gcd(x,y) = dcdot gcd(frac xd,frac yd)$. Apply this fact with $d=gcd(a,c)$ and $x=ab$ and $y=c$.
$endgroup$
– Greg Martin
Apr 19 '17 at 18:11
add a comment |
$begingroup$
How would I be able to prove that $$gcd(ab,c) mid gcd(a,c) cdot gcd(b,c)$$
I'm assuming I can start by saying $gcd(ab,c) cdot n = gcd(a,c) cdot gcd(b,c)$ for some $n in Z$ but I'm not sure how I can represent the $gcd$ value as an integer combination of $a$ and $b$ to prove that the left side can divide the right side
elementary-number-theory discrete-mathematics greatest-common-divisor
edited Jan 27 at 11:00
Martin Sleziak
44.9k10122276
asked Apr 19 '17 at 17:49
Spencer ChowSpencer Chow
1417
$endgroup$
How would I be able to prove that $$gcd(ab,c) mid gcd(a,c) cdot gcd(b,c)$$
I'm assuming I can start by saying $gcd(ab,c) cdot n = gcd(a,c) cdot gcd(b,c)$ for some $n in Z$ but I'm not sure how I can represent the $gcd$ value as an integer combination of $a$ and $b$ to prove that the left side can divide the right side
elementary-number-theory discrete-mathematics greatest-common-divisor
elementary-number-theory discrete-mathematics greatest-common-divisor
edited Jan 27 at 11:00
Martin Sleziak
44.9k10122276
asked Apr 19 '17 at 17:49
Spencer ChowSpencer Chow
1417
edited Jan 27 at 11:00
Martin Sleziak
44.9k10122276
asked Apr 19 '17 at 17:49
Spencer ChowSpencer Chow
1417
edited Jan 27 at 11:00
Martin Sleziak
44.9k10122276
edited Jan 27 at 11:00
Martin Sleziak
44.9k10122276
edited Jan 27 at 11:00
Martin Sleziak
44.9k10122276
44.9k10122276
asked Apr 19 '17 at 17:49
Spencer ChowSpencer Chow
1417
asked Apr 19 '17 at 17:49
Spencer ChowSpencer Chow
1417
asked Apr 19 '17 at 17:49
Spencer ChowSpencer Chow
1417
1417
1
$begingroup$
Easiest, I think, to use unique factorization and the fact that the order of a prime $p$ in $gcd (m,n)$ is the lesser of the orders of $p$ in $m,n$.
$endgroup$
– lulu
Apr 19 '17 at 17:54
$begingroup$
Hint: if $dmid x$ and $dmid y$, then $gcd(x,y) = dcdot gcd(frac xd,frac yd)$. Apply this fact with $d=gcd(a,c)$ and $x=ab$ and $y=c$.
$endgroup$
– Greg Martin
Apr 19 '17 at 18:11
add a comment |
1
$begingroup$
Easiest, I think, to use unique factorization and the fact that the order of a prime $p$ in $gcd (m,n)$ is the lesser of the orders of $p$ in $m,n$.
$endgroup$
– lulu
Apr 19 '17 at 17:54
$begingroup$
Hint: if $dmid x$ and $dmid y$, then $gcd(x,y) = dcdot gcd(frac xd,frac yd)$. Apply this fact with $d=gcd(a,c)$ and $x=ab$ and $y=c$.
$endgroup$
– Greg Martin
Apr 19 '17 at 18:11
1
1
$begingroup$
Easiest, I think, to use unique factorization and the fact that the order of a prime $p$ in $gcd (m,n)$ is the lesser of the orders of $p$ in $m,n$.
$endgroup$
– lulu
Apr 19 '17 at 17:54
$begingroup$
Easiest, I think, to use unique factorization and the fact that the order of a prime $p$ in $gcd (m,n)$ is the lesser of the orders of $p$ in $m,n$.
$endgroup$
– lulu
Apr 19 '17 at 17:54
$begingroup$
Hint: if $dmid x$ and $dmid y$, then $gcd(x,y) = dcdot gcd(frac xd,frac yd)$. Apply this fact with $d=gcd(a,c)$ and $x=ab$ and $y=c$.
$endgroup$
– Greg Martin
Apr 19 '17 at 18:11
$begingroup$
Hint: if $dmid x$ and $dmid y$, then $gcd(x,y) = dcdot gcd(frac xd,frac yd)$. Apply this fact with $d=gcd(a,c)$ and $x=ab$ and $y=c$.
$endgroup$
– Greg Martin
Apr 19 '17 at 18:11
add a comment |
1 Answer
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$begingroup$
From $gcd(ab,c)$, we conclude that $gcd(ab,c) mid ab$ and $gcd(ab,c) mid c$. Now we will have two cases:
Case 1: $gcd(ab,c) mid a$ and $gcd(ab,c) mid c$:
In this case, we see that $gcd(ab,c)$ is a common divisor of $a$ and $c$, so by definition of gcd, we have $gcd(ab,c) mid gcd(a,c)$.
Case 2: $gcd(ab,c) mid b$ and $gcd(ab,c) mid c$:
Similarly, in this case we have $gcd(ab,c)$ to be a common divisor of $b$ and $c$, so $gcd(ab,c) mid gcd(b,c)$.
In either case, we will have $gcd(ab,c) mid gcd(a,c) cdot gcd(b,c)$, and we are done.
answered Apr 19 '17 at 18:09
Chenyang YuChenyang Yu
31728
$endgroup$
add a comment |
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$begingroup$
From $gcd(ab,c)$, we conclude that $gcd(ab,c) mid ab$ and $gcd(ab,c) mid c$. Now we will have two cases:
Case 1: $gcd(ab,c) mid a$ and $gcd(ab,c) mid c$:
In this case, we see that $gcd(ab,c)$ is a common divisor of $a$ and $c$, so by definition of gcd, we have $gcd(ab,c) mid gcd(a,c)$.
Case 2: $gcd(ab,c) mid b$ and $gcd(ab,c) mid c$:
Similarly, in this case we have $gcd(ab,c)$ to be a common divisor of $b$ and $c$, so $gcd(ab,c) mid gcd(b,c)$.
In either case, we will have $gcd(ab,c) mid gcd(a,c) cdot gcd(b,c)$, and we are done.
answered Apr 19 '17 at 18:09
Chenyang YuChenyang Yu
31728
$endgroup$
add a comment |
$begingroup$
From $gcd(ab,c)$, we conclude that $gcd(ab,c) mid ab$ and $gcd(ab,c) mid c$. Now we will have two cases:
Case 1: $gcd(ab,c) mid a$ and $gcd(ab,c) mid c$:
In this case, we see that $gcd(ab,c)$ is a common divisor of $a$ and $c$, so by definition of gcd, we have $gcd(ab,c) mid gcd(a,c)$.
Case 2: $gcd(ab,c) mid b$ and $gcd(ab,c) mid c$:
Similarly, in this case we have $gcd(ab,c)$ to be a common divisor of $b$ and $c$, so $gcd(ab,c) mid gcd(b,c)$.
In either case, we will have $gcd(ab,c) mid gcd(a,c) cdot gcd(b,c)$, and we are done.
answered Apr 19 '17 at 18:09
Chenyang YuChenyang Yu
31728
$endgroup$
add a comment |
$begingroup$
From $gcd(ab,c)$, we conclude that $gcd(ab,c) mid ab$ and $gcd(ab,c) mid c$. Now we will have two cases:
Case 1: $gcd(ab,c) mid a$ and $gcd(ab,c) mid c$:
In this case, we see that $gcd(ab,c)$ is a common divisor of $a$ and $c$, so by definition of gcd, we have $gcd(ab,c) mid gcd(a,c)$.
Case 2: $gcd(ab,c) mid b$ and $gcd(ab,c) mid c$:
Similarly, in this case we have $gcd(ab,c)$ to be a common divisor of $b$ and $c$, so $gcd(ab,c) mid gcd(b,c)$.
In either case, we will have $gcd(ab,c) mid gcd(a,c) cdot gcd(b,c)$, and we are done.
answered Apr 19 '17 at 18:09
Chenyang YuChenyang Yu
31728
$endgroup$
From $gcd(ab,c)$, we conclude that $gcd(ab,c) mid ab$ and $gcd(ab,c) mid c$. Now we will have two cases:
Case 1: $gcd(ab,c) mid a$ and $gcd(ab,c) mid c$:
In this case, we see that $gcd(ab,c)$ is a common divisor of $a$ and $c$, so by definition of gcd, we have $gcd(ab,c) mid gcd(a,c)$.
Case 2: $gcd(ab,c) mid b$ and $gcd(ab,c) mid c$:
Similarly, in this case we have $gcd(ab,c)$ to be a common divisor of $b$ and $c$, so $gcd(ab,c) mid gcd(b,c)$.
In either case, we will have $gcd(ab,c) mid gcd(a,c) cdot gcd(b,c)$, and we are done.
answered Apr 19 '17 at 18:09
Chenyang YuChenyang Yu
31728
answered Apr 19 '17 at 18:09
Chenyang YuChenyang Yu
31728
answered Apr 19 '17 at 18:09
Chenyang YuChenyang Yu
31728
answered Apr 19 '17 at 18:09
Chenyang YuChenyang Yu
31728
31728
add a comment |
add a comment |
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$begingroup$
Easiest, I think, to use unique factorization and the fact that the order of a prime $p$ in $gcd (m,n)$ is the lesser of the orders of $p$ in $m,n$.
$endgroup$
– lulu
Apr 19 '17 at 17:54
$begingroup$
Hint: if $dmid x$ and $dmid y$, then $gcd(x,y) = dcdot gcd(frac xd,frac yd)$. Apply this fact with $d=gcd(a,c)$ and $x=ab$ and $y=c$.
$endgroup$
– Greg Martin
Apr 19 '17 at 18:11