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Let a,b be non zero integers and let m be an integer with m greater than or equal to 0.
Prove that gcd(am,bm)=m*gcd(a,b)
Prove that gcd(am,bm)=m*gcd(a,b)
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I have to prove it using the three rules.
I've done the first rule.
It's the second rule I'm stuck on.
I've done the first rule.
It's the second rule I'm stuck on.
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#3
(Original post by Adil2400)
I have to prove it using the three rules.
I've done the first rule.
It's the second rule I'm stuck on.
I have to prove it using the three rules.
I've done the first rule.
It's the second rule I'm stuck on.
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(Original post by RDKGames)
What three rules? If you tell us then we can guide you along the solution you're expected to follow.
What three rules? If you tell us then we can guide you along the solution you're expected to follow.
1) d divides a and d divides b
2)if e divides a and e divides b, then e is less than or equal to d
3)d is greater than or equal to 0
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d=gcd(a,b)=as+bt
So md=m(as+bt)=am(s)+bm(t)
This is what I've done so far. Don't know what to do from here.
So md=m(as+bt)=am(s)+bm(t)
This is what I've done so far. Don't know what to do from here.
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#7
(Original post by Adil2400)
d=gcd(a,b)=as+bt
So md=m(as+bt)=am(s)+bm(t)
This is what I've done so far. Don't know what to do from here.
d=gcd(a,b)=as+bt
So md=m(as+bt)=am(s)+bm(t)
This is what I've done so far. Don't know what to do from here.
You can just apply Bezout's once more. You know that there exist integers s,t such that
... so what does Bezout's say about this? How are md, am, and bm related?
Last edited by RDKGames; 1 year ago
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(Original post by RDKGames)
You can just apply Bezout's once more. You know that there exist integers s,t such that... so what does Bezout's say about this? How are md, am, and bm related?
You can just apply Bezout's once more. You know that there exist integers s,t such that
... so what does Bezout's say about this? How are md, am, and bm related?
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#9
(Original post by Adil2400)
That means that gcd(am,bm)=md but that still doesn't show that rule 2 is satisfied
That means that gcd(am,bm)=md but that still doesn't show that rule 2 is satisfied
and 
then 
.If
then 
.If also
then 
.
.... hence the result follows.
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#11
(Original post by Adil2400)
How do i show that rule 3 is satisfied?
How do i show that rule 3 is satisfied?
and by defining 
you know that 
...
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