Prove that gcd(a,b) divides lcm[a,b]

• Feb 15th 2010, 04:33 PM
tarheelborn
Prove that gcd(a,b) divides lcm[a,b]
Prove that gcd(a,b) divides lcm[a,b].
• Feb 15th 2010, 06:18 PM
Bacterius
Hello,

$lcm(a, b) = \frac{|a \times b|}{gcd(a, b)}$.

Can you conclude now ?
• Feb 15th 2010, 07:23 PM
Drexel28
Quote:

Originally Posted by tarheelborn
Prove that gcd(a,b) divides lcm[a,b].

We know that $a\mid [a,b]$ and $(a,b)\mid a$ and so $(a,b)\mid [a,b]$

Quote:

Originally Posted by Bacterius
Hello,

$lcm(a, b) = \frac{|a \times b|}{gcd(a, b)}$.

Can you conclude now ?

That is kind of taken to not be known.
• Feb 15th 2010, 07:39 PM
Bacterius
Quote:

That is kind of taken to not be known.
Sorry, English not being my first language, I don't grasp the meaning of this sentence :(
• Feb 15th 2010, 07:40 PM
Drexel28
Quote:

Originally Posted by Bacterius
Sorry, English not being my first language, I don't grasp the meaning of this sentence :(

It's cool. Clearly the problem is trivial if that fact is known, so it would probably be assumed the OP isn't aware of that fact.
• Feb 15th 2010, 07:44 PM
Bacterius
Oh, right. Well, I think you are right, and I actually think that I ruined everything in the sense that this problem was probably given for him to work out this formula ... that will teach me !