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You are not logged in. #1 20060730 11:22:21
gcd and lcmAnyone know how to prove this: "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #2 20060730 11:56:58
Re: gcd and lcmAlright, captain. Consider the set of all primes that divide a and/or b: Then for some suitable nonnegative exponents we have Now p^{q} divides p^{r} if and only if q ≤ r, so But we can also tell that for ab ≠ 0, Now consider the product Also, consider the product Because of the fact that we have Thus we can conclude that the two products are equal, so Is that good enough? Last edited by Zhylliolom (20060730 13:05:30) #3 20060730 15:14:43
Re: gcd and lcmIt certainly seems valid, but it also seems too complex. Maybe it's just me, but I hate dealing with primes when I don't have to. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #4 20060824 17:44:31
Re: gcd and lcmLet If we use the identities: , we can have interestiong results: pp: lots of LaTeX! Last edited by krassi_holmz (20060824 18:05:30) IPBLE: Increasing Performance By Lowering Expectations. 