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

**tarheelborn** · February 15th 2010, 04:33 PM

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

**Bacterius** · February 15th 2010, 06:18 PM

Hello,

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

Can you conclude now ?

**Drexel28** · February 15th 2010, 07:23 PM

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 $(a,b)\mid [a,b]$

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.

**Bacterius** · February 15th 2010, 07:39 PM

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

**Drexel28** · February 15th 2010, 07:40 PM

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.

**Bacterius** · February 15th 2010, 07:44 PM

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 !

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