mcwang719

04-04-2007, 12:18 AM

Hi I'm working on a proof I have the solution but don't understand it fully. The problem is:

Let a and b be natural numbers and LCM(a,b)=m. Prove that m=b if and only if a divides b.

Proof: Suppose LCM(a,b)=b. Then by definition of LCM, a divides b. Suppose a divides b. Then b is a common multiple of a and b; so b>or equal to m. On the other had b divdes m, so b < or equal to m. Therefore b=m.

How can a divide b if b is the LCM, that means b<a?

Let a and b be natural numbers and LCM(a,b)=m. Prove that m=b if and only if a divides b.

Proof: Suppose LCM(a,b)=b. Then by definition of LCM, a divides b. Suppose a divides b. Then b is a common multiple of a and b; so b>or equal to m. On the other had b divdes m, so b < or equal to m. Therefore b=m.

How can a divide b if b is the LCM, that means b<a?