# Least Common Multiple

• Oct 1st 2006, 05:54 PM
JimmyT
Least Common Multiple
Could someone help me prove the following:

I'm denoting the least common multiple as [a,b].

(a) whenever a divides k and b divides k, then [a,b] divides k.
(b) [a,b]=(ab)/(gcd(a,b)) if a,b are greater than zero.

Thanks for the help.

Jimmy
• Oct 1st 2006, 06:28 PM
Quick
Quote:

Originally Posted by JimmyT
Could someone help me prove the following:

I'm denoting the least common multiple as [a,b].

(a) whenever a divides k and b divides k, then [a,b] divides k.

You know that there exists two integers a and x such that: ax=k

You know that there exists two intergers b and y such that: by=k

Let us say that the least common multiple of a and b is n

Thus there exists two integers n and c such that: nc=a

Thus there exists two integers n and d such that: nd=b

Go back to: ax=k

thus: x=k/a

then: x=k/(nc)

multiply both sides by c: xc=k/n

Now, two integers multiplied together equal an integer, so k/n is an integer, thus k is divisible by n.
• Oct 1st 2006, 07:02 PM
JimmyT
Thank you so much.

Can anyone help me with (b)?