Mixing
Prove $gcd(a,m) mid gcd(ab,m)$ $forall a,b,m in Bbb Z$
$begingroup$
I named $gcd(a,m) = d$ and $gcd(ab,m) = d' $
So I know that $dmid a$, $dmid m $ and $d'mid ab $ , $d' mid m$
But I can't use the transitive property of divisibility here.
How can I prove that $d mid d'$?
elementary-number-theory divisibility greatest-common-divisor
edited Jan 29 at 20:03
J. W. Tanner
3,7451320
asked Jan 27 at 10:25
Computer DreamComputer Dream
84
$endgroup$
add a comment |
$begingroup$
I named $gcd(a,m) = d$ and $gcd(ab,m) = d' $
So I know that $dmid a$, $dmid m $ and $d'mid ab $ , $d' mid m$
But I can't use the transitive property of divisibility here.
How can I prove that $d mid d'$?
elementary-number-theory divisibility greatest-common-divisor
edited Jan 29 at 20:03
J. W. Tanner
3,7451320
asked Jan 27 at 10:25
Computer DreamComputer Dream
84
$endgroup$
1
$begingroup$
If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:37
add a comment |
$begingroup$
I named $gcd(a,m) = d$ and $gcd(ab,m) = d' $
So I know that $dmid a$, $dmid m $ and $d'mid ab $ , $d' mid m$
But I can't use the transitive property of divisibility here.
How can I prove that $d mid d'$?
elementary-number-theory divisibility greatest-common-divisor
edited Jan 29 at 20:03
J. W. Tanner
3,7451320
asked Jan 27 at 10:25
Computer DreamComputer Dream
84
$endgroup$
I named $gcd(a,m) = d$ and $gcd(ab,m) = d' $
So I know that $dmid a$, $dmid m $ and $d'mid ab $ , $d' mid m$
But I can't use the transitive property of divisibility here.
How can I prove that $d mid d'$?
elementary-number-theory divisibility greatest-common-divisor
elementary-number-theory divisibility greatest-common-divisor
edited Jan 29 at 20:03
J. W. Tanner
3,7451320
asked Jan 27 at 10:25
Computer DreamComputer Dream
84
edited Jan 29 at 20:03
J. W. Tanner
3,7451320
asked Jan 27 at 10:25
Computer DreamComputer Dream
84
edited Jan 29 at 20:03
J. W. Tanner
3,7451320
edited Jan 29 at 20:03
J. W. Tanner
3,7451320
edited Jan 29 at 20:03
J. W. Tanner
3,7451320
3,7451320
asked Jan 27 at 10:25
Computer DreamComputer Dream
84
asked Jan 27 at 10:25
Computer DreamComputer Dream
84
asked Jan 27 at 10:25
Computer DreamComputer Dream
84
84
1
$begingroup$
If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:37
add a comment |
1
$begingroup$
If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:37
1
1
$begingroup$
If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:37
$begingroup$
If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:37
add a comment |
1 Answer
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$begingroup$
For each $xmid gcd(a,m)$, $xmid a,m$ so $xmid ab,m$ and $xmidgcd(ab,m)$.
answered Jan 27 at 10:30
YiFanYiFan
4,7861727
$endgroup$
add a comment |
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$begingroup$
For each $xmid gcd(a,m)$, $xmid a,m$ so $xmid ab,m$ and $xmidgcd(ab,m)$.
answered Jan 27 at 10:30
YiFanYiFan
4,7861727
$endgroup$
add a comment |
$begingroup$
For each $xmid gcd(a,m)$, $xmid a,m$ so $xmid ab,m$ and $xmidgcd(ab,m)$.
answered Jan 27 at 10:30
YiFanYiFan
4,7861727
$endgroup$
add a comment |
$begingroup$
For each $xmid gcd(a,m)$, $xmid a,m$ so $xmid ab,m$ and $xmidgcd(ab,m)$.
answered Jan 27 at 10:30
YiFanYiFan
4,7861727
$endgroup$
For each $xmid gcd(a,m)$, $xmid a,m$ so $xmid ab,m$ and $xmidgcd(ab,m)$.
answered Jan 27 at 10:30
YiFanYiFan
4,7861727
edited Jan 27 at 10:44
answered Jan 27 at 10:30
YiFanYiFan
4,7861727
answered Jan 27 at 10:30
YiFanYiFan
4,7861727
answered Jan 27 at 10:30
YiFanYiFan
4,7861727
4,7861727
add a comment |
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$begingroup$
If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:37