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Let $a, b, c in Z$ such that $gcd(a,c) = d$ for some integer $d$. Prove if $amid bc$ then $amid bd$.
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up vote
2
down vote
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Here is what I have tried.
If $gcd (a,c) = d$ then you can pick $x, y$ such that $d = ax + cy$
So to show $bd = la$, multiply $b$ into above to get $bd = bax + bcy$
And since $bc = ma$, $bd = bax + may$
Is this sufficient proof? I think I need to get rid of the $b$ in $bax$
elementary-number-theory
edited Jul 18 at 21:32
Bernard
110k635103
asked Jul 18 at 21:29
SolidSnackDrive
1107
add a comment |Â
up vote
2
down vote
favorite
1
Here is what I have tried.
If $gcd (a,c) = d$ then you can pick $x, y$ such that $d = ax + cy$
So to show $bd = la$, multiply $b$ into above to get $bd = bax + bcy$
And since $bc = ma$, $bd = bax + may$
Is this sufficient proof? I think I need to get rid of the $b$ in $bax$
elementary-number-theory
edited Jul 18 at 21:32
Bernard
110k635103
asked Jul 18 at 21:29
SolidSnackDrive
1107
Yes, it is correct. You don't need to get rid of b in $bax$.
– Hector Blandin
Jul 18 at 21:36
1
It's fine for me. Why should you get rid of the $b $ in $bax$? You can factor out $a$ in the r.h.s., that's the main point.
– Bernard
Jul 18 at 21:36
What is $l$ and $m$? Where did those come from? Oh. $l = frac bda$ and $m = frac bcm$.... Yes, other than the style point you need to introduce $m,l$ the proof is valid. And no, then $b$ is $bax$ is not a problem.
– fleablood
Jul 18 at 21:38
2
alternative proof, the way I tend to do things which is neither easier nor more illuminating than yours, just different: Let $a=a'd;c=c'd$ Then $gcd(a',c')=1$ then $a|bcimplies a'|bc'implies a'|b$. So $a'd=a|bd$. (Mine uses a modified Euclid's Lemma; yours Bezout's-- kind of interesting as your proof is nearly exactly how we would prove Euclid's lemma)
– fleablood
Jul 18 at 21:49
add a comment |Â
up vote
2
down vote
favorite
1
up vote
2
down vote
favorite
1
1
Here is what I have tried.
If $gcd (a,c) = d$ then you can pick $x, y$ such that $d = ax + cy$
So to show $bd = la$, multiply $b$ into above to get $bd = bax + bcy$
And since $bc = ma$, $bd = bax + may$
Is this sufficient proof? I think I need to get rid of the $b$ in $bax$
elementary-number-theory
edited Jul 18 at 21:32
Bernard
110k635103
asked Jul 18 at 21:29
SolidSnackDrive
1107
Here is what I have tried.
If $gcd (a,c) = d$ then you can pick $x, y$ such that $d = ax + cy$
So to show $bd = la$, multiply $b$ into above to get $bd = bax + bcy$
And since $bc = ma$, $bd = bax + may$
Is this sufficient proof? I think I need to get rid of the $b$ in $bax$
elementary-number-theory
edited Jul 18 at 21:32
Bernard
110k635103
asked Jul 18 at 21:29
SolidSnackDrive
1107
edited Jul 18 at 21:32
Bernard
110k635103
edited Jul 18 at 21:32
Bernard
110k635103
edited Jul 18 at 21:32
Bernard
110k635103
110k635103
asked Jul 18 at 21:29
SolidSnackDrive
1107
asked Jul 18 at 21:29
SolidSnackDrive
1107
asked Jul 18 at 21:29
SolidSnackDrive
1107
1107
Yes, it is correct. You don't need to get rid of b in $bax$.
– Hector Blandin
Jul 18 at 21:36
1
It's fine for me. Why should you get rid of the $b $ in $bax$? You can factor out $a$ in the r.h.s., that's the main point.
– Bernard
Jul 18 at 21:36
What is $l$ and $m$? Where did those come from? Oh. $l = frac bda$ and $m = frac bcm$.... Yes, other than the style point you need to introduce $m,l$ the proof is valid. And no, then $b$ is $bax$ is not a problem.
– fleablood
Jul 18 at 21:38
2
alternative proof, the way I tend to do things which is neither easier nor more illuminating than yours, just different: Let $a=a'd;c=c'd$ Then $gcd(a',c')=1$ then $a|bcimplies a'|bc'implies a'|b$. So $a'd=a|bd$. (Mine uses a modified Euclid's Lemma; yours Bezout's-- kind of interesting as your proof is nearly exactly how we would prove Euclid's lemma)
– fleablood
Jul 18 at 21:49
add a comment |Â
Yes, it is correct. You don't need to get rid of b in $bax$.
– Hector Blandin
Jul 18 at 21:36
1
It's fine for me. Why should you get rid of the $b $ in $bax$? You can factor out $a$ in the r.h.s., that's the main point.
– Bernard
Jul 18 at 21:36
What is $l$ and $m$? Where did those come from? Oh. $l = frac bda$ and $m = frac bcm$.... Yes, other than the style point you need to introduce $m,l$ the proof is valid. And no, then $b$ is $bax$ is not a problem.
– fleablood
Jul 18 at 21:38
2
alternative proof, the way I tend to do things which is neither easier nor more illuminating than yours, just different: Let $a=a'd;c=c'd$ Then $gcd(a',c')=1$ then $a|bcimplies a'|bc'implies a'|b$. So $a'd=a|bd$. (Mine uses a modified Euclid's Lemma; yours Bezout's-- kind of interesting as your proof is nearly exactly how we would prove Euclid's lemma)
– fleablood
Jul 18 at 21:49
Yes, it is correct. You don't need to get rid of b in $bax$.
– Hector Blandin
Jul 18 at 21:36
Yes, it is correct. You don't need to get rid of b in $bax$.
– Hector Blandin
Jul 18 at 21:36
1
1
It's fine for me. Why should you get rid of the $b $ in $bax$? You can factor out $a$ in the r.h.s., that's the main point.
– Bernard
Jul 18 at 21:36
It's fine for me. Why should you get rid of the $b $ in $bax$? You can factor out $a$ in the r.h.s., that's the main point.
– Bernard
Jul 18 at 21:36
What is $l$ and $m$? Where did those come from? Oh. $l = frac bda$ and $m = frac bcm$.... Yes, other than the style point you need to introduce $m,l$ the proof is valid. And no, then $b$ is $bax$ is not a problem.
– fleablood
Jul 18 at 21:38
What is $l$ and $m$? Where did those come from? Oh. $l = frac bda$ and $m = frac bcm$.... Yes, other than the style point you need to introduce $m,l$ the proof is valid. And no, then $b$ is $bax$ is not a problem.
– fleablood
Jul 18 at 21:38
2
2
alternative proof, the way I tend to do things which is neither easier nor more illuminating than yours, just different: Let $a=a'd;c=c'd$ Then $gcd(a',c')=1$ then $a|bcimplies a'|bc'implies a'|b$. So $a'd=a|bd$. (Mine uses a modified Euclid's Lemma; yours Bezout's-- kind of interesting as your proof is nearly exactly how we would prove Euclid's lemma)
– fleablood
Jul 18 at 21:49
alternative proof, the way I tend to do things which is neither easier nor more illuminating than yours, just different: Let $a=a'd;c=c'd$ Then $gcd(a',c')=1$ then $a|bcimplies a'|bc'implies a'|b$. So $a'd=a|bd$. (Mine uses a modified Euclid's Lemma; yours Bezout's-- kind of interesting as your proof is nearly exactly how we would prove Euclid's lemma)
– fleablood
Jul 18 at 21:49
add a comment |Â
1 Answer
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active
oldest
votes
up vote
2
down vote
accepted
This is a sufficient proof. You have shown $bd = a(bx+my)=al$, which is what you wanted.
answered Jul 18 at 21:36
Sambo
1,2561427
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is a sufficient proof. You have shown $bd = a(bx+my)=al$, which is what you wanted.
answered Jul 18 at 21:36
Sambo
1,2561427
add a comment |Â
up vote
2
down vote
accepted
This is a sufficient proof. You have shown $bd = a(bx+my)=al$, which is what you wanted.
answered Jul 18 at 21:36
Sambo
1,2561427
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is a sufficient proof. You have shown $bd = a(bx+my)=al$, which is what you wanted.
answered Jul 18 at 21:36
Sambo
1,2561427
This is a sufficient proof. You have shown $bd = a(bx+my)=al$, which is what you wanted.
answered Jul 18 at 21:36
Sambo
1,2561427
answered Jul 18 at 21:36
Sambo
1,2561427
answered Jul 18 at 21:36
Sambo
1,2561427
answered Jul 18 at 21:36
Sambo
1,2561427
1,2561427
add a comment |Â
add a comment |Â
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Yes, it is correct. You don't need to get rid of b in $bax$.
– Hector Blandin
Jul 18 at 21:36
1
It's fine for me. Why should you get rid of the $b $ in $bax$? You can factor out $a$ in the r.h.s., that's the main point.
– Bernard
Jul 18 at 21:36
What is $l$ and $m$? Where did those come from? Oh. $l = frac bda$ and $m = frac bcm$.... Yes, other than the style point you need to introduce $m,l$ the proof is valid. And no, then $b$ is $bax$ is not a problem.
– fleablood
Jul 18 at 21:38
2
alternative proof, the way I tend to do things which is neither easier nor more illuminating than yours, just different: Let $a=a'd;c=c'd$ Then $gcd(a',c')=1$ then $a|bcimplies a'|bc'implies a'|b$. So $a'd=a|bd$. (Mine uses a modified Euclid's Lemma; yours Bezout's-- kind of interesting as your proof is nearly exactly how we would prove Euclid's lemma)
– fleablood
Jul 18 at 21:49