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How to show gcd(ac, bd) = gcd(a,d) * gcd(b,c)
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Given that $gcd(a,b) = 1$ and $gcd(c,d) = 1$, show that $gcd(ac,bd) = gcd(a,d) * gcd(b,c)$
The work that I have done so far goes as follows.
We write $gcd(ac,bd) = acv + bdu$
then $gcd(a,d) = ax + dy$ and $gcd(b,c) = bs + ct$
then $gcd(a,d) * gcd(b,c) = (ax + dy) * (bs + ct) = axbs + axct + dybs + dyct$
I tried to factor out some terms, and get $ax(bs+ct) + dy(bs+ct)$
but then I am stuck here. I don't know what else to use to prove the answer
elementary-number-theory
edited 44 mins ago
greedoid
29.1k93879
asked 50 mins ago
Mario G
242
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Mario G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
4
down vote
favorite
Given that $gcd(a,b) = 1$ and $gcd(c,d) = 1$, show that $gcd(ac,bd) = gcd(a,d) * gcd(b,c)$
The work that I have done so far goes as follows.
We write $gcd(ac,bd) = acv + bdu$
then $gcd(a,d) = ax + dy$ and $gcd(b,c) = bs + ct$
then $gcd(a,d) * gcd(b,c) = (ax + dy) * (bs + ct) = axbs + axct + dybs + dyct$
I tried to factor out some terms, and get $ax(bs+ct) + dy(bs+ct)$
but then I am stuck here. I don't know what else to use to prove the answer
elementary-number-theory
edited 44 mins ago
greedoid
29.1k93879
asked 50 mins ago
Mario G
242
New contributor
Mario G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Please consider using MathJax to format your question - it will make your formulas much more readable
– Zubin Mukerjee
47 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given that $gcd(a,b) = 1$ and $gcd(c,d) = 1$, show that $gcd(ac,bd) = gcd(a,d) * gcd(b,c)$
The work that I have done so far goes as follows.
We write $gcd(ac,bd) = acv + bdu$
then $gcd(a,d) = ax + dy$ and $gcd(b,c) = bs + ct$
then $gcd(a,d) * gcd(b,c) = (ax + dy) * (bs + ct) = axbs + axct + dybs + dyct$
I tried to factor out some terms, and get $ax(bs+ct) + dy(bs+ct)$
but then I am stuck here. I don't know what else to use to prove the answer
elementary-number-theory
edited 44 mins ago
greedoid
29.1k93879
asked 50 mins ago
Mario G
242
New contributor
Mario G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Given that $gcd(a,b) = 1$ and $gcd(c,d) = 1$, show that $gcd(ac,bd) = gcd(a,d) * gcd(b,c)$
The work that I have done so far goes as follows.
We write $gcd(ac,bd) = acv + bdu$
then $gcd(a,d) = ax + dy$ and $gcd(b,c) = bs + ct$
then $gcd(a,d) * gcd(b,c) = (ax + dy) * (bs + ct) = axbs + axct + dybs + dyct$
I tried to factor out some terms, and get $ax(bs+ct) + dy(bs+ct)$
but then I am stuck here. I don't know what else to use to prove the answer
elementary-number-theory
elementary-number-theory
edited 44 mins ago
greedoid
29.1k93879
asked 50 mins ago
Mario G
242
New contributor
Mario G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 44 mins ago
greedoid
29.1k93879
asked 50 mins ago
Mario G
242
New contributor
Mario G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 44 mins ago
greedoid
29.1k93879
edited 44 mins ago
greedoid
29.1k93879
edited 44 mins ago
greedoid
29.1k93879
29.1k93879
asked 50 mins ago
Mario G
242
New contributor
Mario G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 50 mins ago
Mario G
242
asked 50 mins ago
Mario G
242
242
New contributor
Mario G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mario G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mario G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Please consider using MathJax to format your question - it will make your formulas much more readable
– Zubin Mukerjee
47 mins ago
add a comment |Â
1
Please consider using MathJax to format your question - it will make your formulas much more readable
– Zubin Mukerjee
47 mins ago
1
1
Please consider using MathJax to format your question - it will make your formulas much more readable
– Zubin Mukerjee
47 mins ago
Please consider using MathJax to format your question - it will make your formulas much more readable
– Zubin Mukerjee
47 mins ago
add a comment |Â
3 Answers
3
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oldest
votes
up vote
3
down vote
Take all prime factors in turn. We have the following relations between the multiplicities:
$$gcd(a,b) = 1iffmin(alpha,beta)=0,$$
$$gcd(a,b) = 1iffmin(gamma,delta)=0,$$
$$gcd(ac,bd) = gcd(a,d) gcd(b,c)iffmin(alpha+gamma,beta+delta)=min(alpha,delta)+min(beta,gamma).$$
If we consider the case $alpha=gamma=0$,
$$min(0,beta+delta)=min(0,delta)+min(beta,0)$$ does hold.
Then with $beta=gamma=0$,
$$min(alpha,delta)=min(alpha,delta).$$
By symmetry other cases hold.
answered 26 mins ago
Yves Daoust
115k667210
add a comment |Â
up vote
2
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Let $x = gcd(a,d)$ and $y= gcd(b,c)$, then $xymid ac$ and $xymid bd$ so $boxedxymid z$ where $z=gcd(ac,bd)$.
Vice versa:
We can write $a=xa'$ and $d=xd'$ where $gcd(a',d')=1$ and $b=yb'$ and $c=yc'$ where $gcd(b',c')=1$
Now since $zmid ac = xya'c'$ and $zmid bd = xyb'd'$
Now if there is prime $p$ such that $pmid z$ and $gcd(xy,p)=1$ then $pmid a'c'$ and $pmid b'd'$. If $pmid a'$ then $pmid b'$ since $gcd(a',d')=1$. But then $pmid a$ and $pmid b$ so $pmid gcd(a,b)=1$ a contradiction. So there is no such $p$ and thus $boxedzmid xy$
answered 32 mins ago
greedoid
29.1k93879
What is the property you are using in the first line where $xy|ac$ and $xy|bd$ so $xy|z$ ?
– Mario G
30 mins ago
$xy$ divide $ac$ and $bd$ so it must divide the greates common divisor of them.
– greedoid
28 mins ago
The greatest common divisor is a multiple of every other common divisor.
– PossiblyDakota
26 mins ago
You did not use the fact that $gcd(a,b) = gcd(c,d) = 1$. As your "counterexample" did not satisfy the assumption.
– Hw Chu
25 mins ago
add a comment |Â
up vote
1
down vote
Let $gcd(a,d) = g$ so that $a = a'g$ and $d = d'g$ and $gcd(a',d') =1$. (why?)
And let $gcd(b,c) = h$ so that $b =b'h$ and $c= c'h$ and $gcd(b',c') =1$ (why?).
Then $gcd(ac,bd) = gcd(a'c'gh, b'd'gh) = gh*gcd(a'c',b'd')$. [$gcd(m*k, n*k) = kgcd(m,n)$. (why?)]
And if you take any prime factor of $a'c'$ then it is a prime factor of $a'$ or $c'$ so it is not a prime factor of either $b'$ or $d'$ (as those both coprime to both $a'$ and $c'$; $a$ and $b$ are coprime, $c$ and $d$ are coprime, and so are $a'$ and $d'$ and so are $b'$ and $c'$). So it is not a prime factor of $b'd'$. Likewise no prime factor of $b'd'$ is a prime factor of $a'c'$. So $gcd(a'c', b'd') = 1$.
So $gcd(ac, bd) = gh = gcd(a,d)gcd(b,c)$.
answered 16 mins ago
fleablood
62.1k22678
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Take all prime factors in turn. We have the following relations between the multiplicities:
$$gcd(a,b) = 1iffmin(alpha,beta)=0,$$
$$gcd(a,b) = 1iffmin(gamma,delta)=0,$$
$$gcd(ac,bd) = gcd(a,d) gcd(b,c)iffmin(alpha+gamma,beta+delta)=min(alpha,delta)+min(beta,gamma).$$
If we consider the case $alpha=gamma=0$,
$$min(0,beta+delta)=min(0,delta)+min(beta,0)$$ does hold.
Then with $beta=gamma=0$,
$$min(alpha,delta)=min(alpha,delta).$$
By symmetry other cases hold.
answered 26 mins ago
Yves Daoust
115k667210
add a comment |Â
up vote
3
down vote
Take all prime factors in turn. We have the following relations between the multiplicities:
$$gcd(a,b) = 1iffmin(alpha,beta)=0,$$
$$gcd(a,b) = 1iffmin(gamma,delta)=0,$$
$$gcd(ac,bd) = gcd(a,d) gcd(b,c)iffmin(alpha+gamma,beta+delta)=min(alpha,delta)+min(beta,gamma).$$
If we consider the case $alpha=gamma=0$,
$$min(0,beta+delta)=min(0,delta)+min(beta,0)$$ does hold.
Then with $beta=gamma=0$,
$$min(alpha,delta)=min(alpha,delta).$$
By symmetry other cases hold.
answered 26 mins ago
Yves Daoust
115k667210
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Take all prime factors in turn. We have the following relations between the multiplicities:
$$gcd(a,b) = 1iffmin(alpha,beta)=0,$$
$$gcd(a,b) = 1iffmin(gamma,delta)=0,$$
$$gcd(ac,bd) = gcd(a,d) gcd(b,c)iffmin(alpha+gamma,beta+delta)=min(alpha,delta)+min(beta,gamma).$$
If we consider the case $alpha=gamma=0$,
$$min(0,beta+delta)=min(0,delta)+min(beta,0)$$ does hold.
Then with $beta=gamma=0$,
$$min(alpha,delta)=min(alpha,delta).$$
By symmetry other cases hold.
answered 26 mins ago
Yves Daoust
115k667210
Take all prime factors in turn. We have the following relations between the multiplicities:
$$gcd(a,b) = 1iffmin(alpha,beta)=0,$$
$$gcd(a,b) = 1iffmin(gamma,delta)=0,$$
$$gcd(ac,bd) = gcd(a,d) gcd(b,c)iffmin(alpha+gamma,beta+delta)=min(alpha,delta)+min(beta,gamma).$$
If we consider the case $alpha=gamma=0$,
$$min(0,beta+delta)=min(0,delta)+min(beta,0)$$ does hold.
Then with $beta=gamma=0$,
$$min(alpha,delta)=min(alpha,delta).$$
By symmetry other cases hold.
answered 26 mins ago
Yves Daoust
115k667210
edited 8 mins ago
answered 26 mins ago
Yves Daoust
115k667210
answered 26 mins ago
Yves Daoust
115k667210
answered 26 mins ago
Yves Daoust
115k667210
115k667210
add a comment |Â
add a comment |Â
up vote
2
down vote
Let $x = gcd(a,d)$ and $y= gcd(b,c)$, then $xymid ac$ and $xymid bd$ so $boxedxymid z$ where $z=gcd(ac,bd)$.
Vice versa:
We can write $a=xa'$ and $d=xd'$ where $gcd(a',d')=1$ and $b=yb'$ and $c=yc'$ where $gcd(b',c')=1$
Now since $zmid ac = xya'c'$ and $zmid bd = xyb'd'$
Now if there is prime $p$ such that $pmid z$ and $gcd(xy,p)=1$ then $pmid a'c'$ and $pmid b'd'$. If $pmid a'$ then $pmid b'$ since $gcd(a',d')=1$. But then $pmid a$ and $pmid b$ so $pmid gcd(a,b)=1$ a contradiction. So there is no such $p$ and thus $boxedzmid xy$
answered 32 mins ago
greedoid
29.1k93879
What is the property you are using in the first line where $xy|ac$ and $xy|bd$ so $xy|z$ ?
– Mario G
30 mins ago
$xy$ divide $ac$ and $bd$ so it must divide the greates common divisor of them.
– greedoid
28 mins ago
The greatest common divisor is a multiple of every other common divisor.
– PossiblyDakota
26 mins ago
You did not use the fact that $gcd(a,b) = gcd(c,d) = 1$. As your "counterexample" did not satisfy the assumption.
– Hw Chu
25 mins ago
add a comment |Â
up vote
2
down vote
Let $x = gcd(a,d)$ and $y= gcd(b,c)$, then $xymid ac$ and $xymid bd$ so $boxedxymid z$ where $z=gcd(ac,bd)$.
Vice versa:
We can write $a=xa'$ and $d=xd'$ where $gcd(a',d')=1$ and $b=yb'$ and $c=yc'$ where $gcd(b',c')=1$
Now since $zmid ac = xya'c'$ and $zmid bd = xyb'd'$
Now if there is prime $p$ such that $pmid z$ and $gcd(xy,p)=1$ then $pmid a'c'$ and $pmid b'd'$. If $pmid a'$ then $pmid b'$ since $gcd(a',d')=1$. But then $pmid a$ and $pmid b$ so $pmid gcd(a,b)=1$ a contradiction. So there is no such $p$ and thus $boxedzmid xy$
answered 32 mins ago
greedoid
29.1k93879
What is the property you are using in the first line where $xy|ac$ and $xy|bd$ so $xy|z$ ?
– Mario G
30 mins ago
$xy$ divide $ac$ and $bd$ so it must divide the greates common divisor of them.
– greedoid
28 mins ago
The greatest common divisor is a multiple of every other common divisor.
– PossiblyDakota
26 mins ago
You did not use the fact that $gcd(a,b) = gcd(c,d) = 1$. As your "counterexample" did not satisfy the assumption.
– Hw Chu
25 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $x = gcd(a,d)$ and $y= gcd(b,c)$, then $xymid ac$ and $xymid bd$ so $boxedxymid z$ where $z=gcd(ac,bd)$.
Vice versa:
We can write $a=xa'$ and $d=xd'$ where $gcd(a',d')=1$ and $b=yb'$ and $c=yc'$ where $gcd(b',c')=1$
Now since $zmid ac = xya'c'$ and $zmid bd = xyb'd'$
Now if there is prime $p$ such that $pmid z$ and $gcd(xy,p)=1$ then $pmid a'c'$ and $pmid b'd'$. If $pmid a'$ then $pmid b'$ since $gcd(a',d')=1$. But then $pmid a$ and $pmid b$ so $pmid gcd(a,b)=1$ a contradiction. So there is no such $p$ and thus $boxedzmid xy$
answered 32 mins ago
greedoid
29.1k93879
Let $x = gcd(a,d)$ and $y= gcd(b,c)$, then $xymid ac$ and $xymid bd$ so $boxedxymid z$ where $z=gcd(ac,bd)$.
Vice versa:
We can write $a=xa'$ and $d=xd'$ where $gcd(a',d')=1$ and $b=yb'$ and $c=yc'$ where $gcd(b',c')=1$
Now since $zmid ac = xya'c'$ and $zmid bd = xyb'd'$
Now if there is prime $p$ such that $pmid z$ and $gcd(xy,p)=1$ then $pmid a'c'$ and $pmid b'd'$. If $pmid a'$ then $pmid b'$ since $gcd(a',d')=1$. But then $pmid a$ and $pmid b$ so $pmid gcd(a,b)=1$ a contradiction. So there is no such $p$ and thus $boxedzmid xy$
answered 32 mins ago
greedoid
29.1k93879
edited 22 mins ago
answered 32 mins ago
greedoid
29.1k93879
answered 32 mins ago
greedoid
29.1k93879
answered 32 mins ago
greedoid
29.1k93879
29.1k93879
What is the property you are using in the first line where $xy|ac$ and $xy|bd$ so $xy|z$ ?
– Mario G
30 mins ago
$xy$ divide $ac$ and $bd$ so it must divide the greates common divisor of them.
– greedoid
28 mins ago
The greatest common divisor is a multiple of every other common divisor.
– PossiblyDakota
26 mins ago
You did not use the fact that $gcd(a,b) = gcd(c,d) = 1$. As your "counterexample" did not satisfy the assumption.
– Hw Chu
25 mins ago
add a comment |Â
What is the property you are using in the first line where $xy|ac$ and $xy|bd$ so $xy|z$ ?
– Mario G
30 mins ago
$xy$ divide $ac$ and $bd$ so it must divide the greates common divisor of them.
– greedoid
28 mins ago
The greatest common divisor is a multiple of every other common divisor.
– PossiblyDakota
26 mins ago
You did not use the fact that $gcd(a,b) = gcd(c,d) = 1$. As your "counterexample" did not satisfy the assumption.
– Hw Chu
25 mins ago
What is the property you are using in the first line where $xy|ac$ and $xy|bd$ so $xy|z$ ?
– Mario G
30 mins ago
What is the property you are using in the first line where $xy|ac$ and $xy|bd$ so $xy|z$ ?
– Mario G
30 mins ago
$xy$ divide $ac$ and $bd$ so it must divide the greates common divisor of them.
– greedoid
28 mins ago
$xy$ divide $ac$ and $bd$ so it must divide the greates common divisor of them.
– greedoid
28 mins ago
The greatest common divisor is a multiple of every other common divisor.
– PossiblyDakota
26 mins ago
The greatest common divisor is a multiple of every other common divisor.
– PossiblyDakota
26 mins ago
You did not use the fact that $gcd(a,b) = gcd(c,d) = 1$. As your "counterexample" did not satisfy the assumption.
– Hw Chu
25 mins ago
You did not use the fact that $gcd(a,b) = gcd(c,d) = 1$. As your "counterexample" did not satisfy the assumption.
– Hw Chu
25 mins ago
add a comment |Â
up vote
1
down vote
Let $gcd(a,d) = g$ so that $a = a'g$ and $d = d'g$ and $gcd(a',d') =1$. (why?)
And let $gcd(b,c) = h$ so that $b =b'h$ and $c= c'h$ and $gcd(b',c') =1$ (why?).
Then $gcd(ac,bd) = gcd(a'c'gh, b'd'gh) = gh*gcd(a'c',b'd')$. [$gcd(m*k, n*k) = kgcd(m,n)$. (why?)]
And if you take any prime factor of $a'c'$ then it is a prime factor of $a'$ or $c'$ so it is not a prime factor of either $b'$ or $d'$ (as those both coprime to both $a'$ and $c'$; $a$ and $b$ are coprime, $c$ and $d$ are coprime, and so are $a'$ and $d'$ and so are $b'$ and $c'$). So it is not a prime factor of $b'd'$. Likewise no prime factor of $b'd'$ is a prime factor of $a'c'$. So $gcd(a'c', b'd') = 1$.
So $gcd(ac, bd) = gh = gcd(a,d)gcd(b,c)$.
answered 16 mins ago
fleablood
62.1k22678
add a comment |Â
up vote
1
down vote
Let $gcd(a,d) = g$ so that $a = a'g$ and $d = d'g$ and $gcd(a',d') =1$. (why?)
And let $gcd(b,c) = h$ so that $b =b'h$ and $c= c'h$ and $gcd(b',c') =1$ (why?).
Then $gcd(ac,bd) = gcd(a'c'gh, b'd'gh) = gh*gcd(a'c',b'd')$. [$gcd(m*k, n*k) = kgcd(m,n)$. (why?)]
And if you take any prime factor of $a'c'$ then it is a prime factor of $a'$ or $c'$ so it is not a prime factor of either $b'$ or $d'$ (as those both coprime to both $a'$ and $c'$; $a$ and $b$ are coprime, $c$ and $d$ are coprime, and so are $a'$ and $d'$ and so are $b'$ and $c'$). So it is not a prime factor of $b'd'$. Likewise no prime factor of $b'd'$ is a prime factor of $a'c'$. So $gcd(a'c', b'd') = 1$.
So $gcd(ac, bd) = gh = gcd(a,d)gcd(b,c)$.
answered 16 mins ago
fleablood
62.1k22678
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $gcd(a,d) = g$ so that $a = a'g$ and $d = d'g$ and $gcd(a',d') =1$. (why?)
And let $gcd(b,c) = h$ so that $b =b'h$ and $c= c'h$ and $gcd(b',c') =1$ (why?).
Then $gcd(ac,bd) = gcd(a'c'gh, b'd'gh) = gh*gcd(a'c',b'd')$. [$gcd(m*k, n*k) = kgcd(m,n)$. (why?)]
And if you take any prime factor of $a'c'$ then it is a prime factor of $a'$ or $c'$ so it is not a prime factor of either $b'$ or $d'$ (as those both coprime to both $a'$ and $c'$; $a$ and $b$ are coprime, $c$ and $d$ are coprime, and so are $a'$ and $d'$ and so are $b'$ and $c'$). So it is not a prime factor of $b'd'$. Likewise no prime factor of $b'd'$ is a prime factor of $a'c'$. So $gcd(a'c', b'd') = 1$.
So $gcd(ac, bd) = gh = gcd(a,d)gcd(b,c)$.
answered 16 mins ago
fleablood
62.1k22678
Let $gcd(a,d) = g$ so that $a = a'g$ and $d = d'g$ and $gcd(a',d') =1$. (why?)
And let $gcd(b,c) = h$ so that $b =b'h$ and $c= c'h$ and $gcd(b',c') =1$ (why?).
Then $gcd(ac,bd) = gcd(a'c'gh, b'd'gh) = gh*gcd(a'c',b'd')$. [$gcd(m*k, n*k) = kgcd(m,n)$. (why?)]
And if you take any prime factor of $a'c'$ then it is a prime factor of $a'$ or $c'$ so it is not a prime factor of either $b'$ or $d'$ (as those both coprime to both $a'$ and $c'$; $a$ and $b$ are coprime, $c$ and $d$ are coprime, and so are $a'$ and $d'$ and so are $b'$ and $c'$). So it is not a prime factor of $b'd'$. Likewise no prime factor of $b'd'$ is a prime factor of $a'c'$. So $gcd(a'c', b'd') = 1$.
So $gcd(ac, bd) = gh = gcd(a,d)gcd(b,c)$.
answered 16 mins ago
fleablood
62.1k22678
answered 16 mins ago
fleablood
62.1k22678
answered 16 mins ago
fleablood
62.1k22678
answered 16 mins ago
fleablood
62.1k22678
62.1k22678
add a comment |Â
add a comment |Â
Mario G is a new contributor. Be nice, and check out our Code of Conduct.
Mario G is a new contributor. Be nice, and check out our Code of Conduct.
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Please consider using MathJax to format your question - it will make your formulas much more readable
– Zubin Mukerjee
47 mins ago