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You are not logged in. #1 20060422 16:57:54
m does not divide n, m^2 does not divide n^2It seems like such an easy proof, but I just can't seem to get it... "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 20060423 01:29:58
Re: m does not divide n, m^2 does not divide n^2You can also assume that m is strictly less than n, if it helps. "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..." #3 20060423 13:40:47
Re: m does not divide n, m^2 does not divide n^2This is outside my area but maybe this is a starting point for you. #4 20060423 13:48:06
Re: m does not divide n, m^2 does not divide n^2you just need to factor m and n into products of prime numbers. Last edited by George,Y (20060423 13:49:12) X'(yXβ)=0 #5 20060424 00:00:12
Re: m does not divide n, m^2 does not divide n^2Contraversial: so: n^2=n.n => nn^2. and n^2m^2, so we get that : nm^2. But: 2. If a(bc) then (ab)(ac), . then we get that: n^2m^2 => nm^2 =>nm.m=>(nm)or(nm)=>nm. But, by definition, n ! m, so : n^2 can't divide m^2 IPBLE: Increasing Performance By Lowering Expectations. #6 20060424 00:08:42
Re: m does not divide n, m^2 does not divide n^2(ab) means: there exists an integer k, for which ak=b: IPBLE: Increasing Performance By Lowering Expectations. #7 20060424 00:37:24
Re: m does not divide n, m^2 does not divide n^2
6  (3*2) but 6 ! 3 and 6 ! 2 "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..." #8 20060424 10:14:56
Re: m does not divide n, m^2 does not divide n^2thank you Ricky for correcting me.
Why? If we could, the proof is easy: IPBLE: Increasing Performance By Lowering Expectations. #9 20060424 10:40:49
Re: m does not divide n, m^2 does not divide n^2Because I'm using this proof to prove that: Last edited by Ricky (20060424 11:05:13) "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..." #10 20060424 18:14:17
Re: m does not divide n, m^2 does not divide n^2
6  (3*2) but 6 ! 3 and 6 ! 2 but , so for i=1 to infty So for int i: Now let see int k=n^2/m^2: By (1) delta_igamma_i is always >=0, so is always a square. But the product of squares is square, so k is square. But then , int j and: We know that: But now a,b>=0, so . So , where j,m,n are int, but this implies that: . Last edited by krassi_holmz (20060424 18:53:19) IPBLE: Increasing Performance By Lowering Expectations. 