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Here's my proof:
Last edited by Stanley_Marsh (2007-04-12 14:07:11)
Numbers are the essence of the Universe
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You're not using what your given. Let |ab| = k. Then (ab)^k = a^k * b^k = 1. Thus, it must be that |a| | k and |b| | k, and k is the least of such numbers. Thus, k = lcm(n, m). But (m, n) = 1, so lcm(n, m) = mn.
I can guarantee your proof is invalid as you didn't use the fact that ab = ba, and this statement is not true in general.
"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..."
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Awww, I dont really understand the importance of commutative in this case?
Numbers are the essence of the Universe
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(ab)^k = a^k * b^k if ab = ba.
"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..."
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To explain Ricky's statement a little more, note that
We cannot rearrange these terms into a[sup]k[/sup]b[sup]k[/sup] if a and b do not commute. Do you see this?
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You are just assuming that the a[sup]i[/sup] and b[sup]j[/sup] are distinct if 0 < i < m, 0 < j < n. Shouldnt you actually show that they are?
Last edited by JaneFairfax (2007-04-12 16:49:48)
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I see how commutative works~
Jane: Can I do that through proving that (a,b)=1 ?
Numbers are the essence of the Universe
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What do you mean by (a,b)?
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I mean , a ,b are relatively prime
Numbers are the essence of the Universe
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a and b are elements in the group.
You mean make use of gcd(|a|,|b|)=1? Of course. In fact, you have to use that condition.
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Yeah , that's what I meant.
Numbers are the essence of the Universe
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