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

Printable View

- October 1st 2006, 04:54 PMJimmyTLeast 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 - October 1st 2006, 05:28 PMQuick
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. - October 1st 2006, 06:02 PMJimmyT
Thank you so much.

Can anyone help me with (b)?