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Math Help - LCM[|x|,|y|]

  1. #1
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    LCM[|x|,|y|]

    Assume |x| = n and |y| = m. Suppose that xy = yx. Prove that |xy||\text{LCM}[n,m].

    \text{LCM}[n,m]=\frac{nm}{\text{gcd}(n,m)}

    x^ny^m=e

    What is the commutativity used for?
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  2. #2
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    Re: LCM[|x|,|y|]

    to establish that (xy)^k = x^ky^k

    by the way, what you are trying to prove is false, without further conditions. consider m = n, and y = x^{-1}.
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    Re: LCM[|x|,|y|]

    Quote Originally Posted by Deveno View Post
    to establish that (xy)^k = x^ky^k

    by the way, what you are trying to prove is false, without further conditions. consider m = n, and y = x^{-1}.
    That would work. The order of |xy| = 1. LCM[m,n] = n.

    Then 1|n true.
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    Re: LCM[|x|,|y|]

    my bad...didn't see the extra vertical line. in any case, what you do is show that (xy)^lcm(m,n) = e, and you're done!
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  5. #5
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    Re: LCM[|x|,|y|]

    Ok so here is what I have.

    x^ky^k=e

    k=nq_1+r_1 \ \text{and} \ k=mq_2+r_2

    q_1,q_2\in\mathbb{Z}, \ 0\leq r_1 <n, \ 0\leq r_2<m

    x^k=\cdots x^{r_1} \ \text{and} \ y^k=\cdots y^{r_2}

    But m and n are least so r_1,r_2=0.

    So we have n|k \ \text{and} \ m|k

    Can I just say k is a common multiple and by definition of LCM, k|LCM?
    Last edited by dwsmith; October 9th 2011 at 09:10 AM.
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    Re: LCM[|x|,|y|]

    Quote Originally Posted by dwsmith View Post
    Ok so here is what I have.

    x^ky^k=e

    k=nq_1+r_1 \ \text{and} \ k=mq_2+r_2

    q_1,q_2\in\mathbb{Z}, \ 0\leq r_1 <n, \ 0\leq r_2<m

    x^k=\cdots x^{r_1} \ \text{and} \ y^k=\cdots y^{r_2}

    But m and n are least so r_1,r_2=0.

    So we have n|k \ \text{and} \ m|k

    Can I just say k is a common multiple and by definition of LCM, k|LCM?
    By definition of LCM we have that LCM divides every commom multiple, so you can say that LCM/k - not k/LCM.
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  7. #7
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    Re: LCM[|x|,|y|]

    Quote Originally Posted by dwsmith View Post
    Assume |x| = n and |y| = m. Suppose that xy = yx. Prove that |xy||\text{LCM}[n,m].
    Let k = LCM[m,n].

    Then m/k and n/k. So, there exist \lambda_1, \lambda _2 such that: k = \lambda_1 m and k = \lambda_2 n.

    Now, (xy)^k \overset{xy=yx}{==} x^k y^k = (x^n)^{\lambda_2} (y^m)^{\lambda_1}=e^{\lambda_2} e^{\lambda_1} = e \implies |xy|/k.
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