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#1 2007-04-12 14:01:44

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

Check

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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#2 2007-04-12 14:54:48

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Check

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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#3 2007-04-12 14:59:29

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Check

Awww, I dont really understand the importance of commutative in this case?


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#4 2007-04-12 15:01:54

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Check

(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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#5 2007-04-12 16:01:31

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

Re: Check

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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#6 2007-04-12 16:48:54

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Check

Stanley_Marsh wrote:

You are just assuming that the a[sup]i[/sup] and b[sup]j[/sup] are distinct if 0 < i < m, 0 < j < n. Shouldn’t you actually show that they are? smile

Last edited by JaneFairfax (2007-04-12 16:49:48)

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#7 2007-04-12 18:10:45

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Check

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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#8 2007-04-12 18:58:54

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Check

What do you mean by (a,b)?

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#9 2007-04-13 02:21:07

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Check

I mean , a ,b are relatively prime


Numbers are the essence of the Universe

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#10 2007-04-13 04:02:16

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Check

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. smile

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#11 2007-04-13 09:05:52

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Check

Yeah , that's what I meant.


Numbers are the essence of the Universe

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