proof dealing with lcm
Hey all, got a homework problem which I'm not quite sure how to go about:
Basically, it wants me to show that an integer M >= 0 is the lcm of the numbers a1, a2, ... an, if and only if it is 1) a common multiple of them all, and 2) if it divides every other common multiple.
It seems easy to prove that if it divides all common multiples and is a common multiple itself, then it is the lcm (since that's the definition of lcm), but the other way around seems a bit trickier - or maybe I'm just getting hung up on some little thing... any idea how to get through this?
Then there is nothing to prove, you must have a different definition!
Originally Posted by Pigzorkly
clic-clac, I don't understant your point, could you explain more? Are you thinking in terms of prime factorization?
Pigzorkly, yes, if c divides all other multiples of M, then it must be smaller than they and so is the least common multiple. To go the other way, you need to show that, if c is te least common multiple, then every other multiple of M must be a multiple of c- it must contain c as a factor. If b is a multiple of M, then b= Mx for some x. Since c is a multiple of M, c= My. Then M= c/y so b= cx/y. Can you show that x/y must be an integer?
@HallsofIvy: It's just that definitions depend on people, since equivalent statements can be chosen to define the same object. If you see the part in bold that I quoted, what I'm reading means that lcm's definition is precisely what we want it to be equivalent...(I hope you understand what I want to say) So I was asking if Pigzorkly had an other definition for lcm. But I may have misunderstood!