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

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

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

2. Hello,

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

Can you conclude now ?

3. Originally Posted by tarheelborn
Prove that gcd(a,b) divides lcm[a,b].
We know that $\displaystyle a\mid [a,b]$ and $\displaystyle (a,b)\mid a$ and so $\displaystyle (a,b)\mid [a,b]$

Originally Posted by Bacterius
Hello,

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

Can you conclude now ?
That is kind of taken to not be known.

4. 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

5. 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.

6. 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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# prove that gcd(a b) divides lcm(a b)

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