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  1. #1
    Senior Member Sampras's Avatar
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    equation

    Show that $\displaystyle (m,n) = 1 \Rightarrow (m) \cap (n) = (mn) $. so $\displaystyle mx+ny = 1 $. Now $\displaystyle (m) = \{cm: c \in R \} $ and $\displaystyle (n) = \{dm: d \in R \} $. So show that there is some common element which implies that they must be equal?
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    Quote Originally Posted by Sampras View Post
    Show that $\displaystyle (m,n) = 1 \Rightarrow (m) \cap (n) = (mn) $. so $\displaystyle mx+ny = 1 $. Now $\displaystyle (m) = \{cm: c \in R \} $ and $\displaystyle (n) = \{dm: d \in R \} $. So show that there is some common element which implies that they must be equal?

    I think you meant $\displaystyle (n) = \{dn: d \in Z \} $

    Let me prove the following for you

    if $\displaystyle m|x$ and $\displaystyle n|x$ then $\displaystyle mn|[\gcd(m,n)x]$

    we have $\displaystyle k_1m+k_2n = \gcd(m,n)$
    = $\displaystyle xk_1m+xk_2n = x\gcd(m,n)$

    Now mn|LHS of the equation above. Hence the result.

    In this case $\displaystyle gcd(m,n)=1$, so $\displaystyle mn|x$ Hence the result
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  3. #3
    Senior Member Sampras's Avatar
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    Quote Originally Posted by aman_cc View Post
    I think you meant $\displaystyle (n) = \{dn: d \in Z \} $

    Let me prove the following for you

    if $\displaystyle m|x$ and $\displaystyle n|x$ then $\displaystyle mn|[\gcd(m,n)x]$

    we have $\displaystyle k_1m+k_2n = \gcd(m,n)$
    = $\displaystyle xk_1m+xk_2n = x\gcd(m,n)$

    Now mn|LHS of the equation above. Hence the result.

    In this case $\displaystyle gcd(m,n)=1$, so $\displaystyle mn|x$ Hence the result
    Can you use fact that either $\displaystyle (m) \cap (n) $ is disjoint or equal? Thus show that if there is some element in $\displaystyle (m) \cap (n) $ then $\displaystyle (m) = (n) $? Thus $\displaystyle (m) \cap (n) = (mn) $?
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    Sorry - but unable to follow you question completely.
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  5. #5
    Senior Member Sampras's Avatar
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    Quote Originally Posted by aman_cc View Post
    I think you meant $\displaystyle (n) = \{dn: d \in Z \} $

    Let me prove the following for you

    if $\displaystyle m|x$ and $\displaystyle n|x$ then $\displaystyle mn|[\gcd(m,n)x]$

    we have $\displaystyle k_1m+k_2n = \gcd(m,n)$
    = $\displaystyle xk_1m+xk_2n = x\gcd(m,n)$

    Now mn|LHS of the equation above. Hence the result.

    In this case $\displaystyle gcd(m,n)=1$, so $\displaystyle mn|x$ Hence the result
    How does this show the result?
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  6. #6
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    Quote Originally Posted by Sampras View Post
    How does this show the result?
    Hi - Maybe I am not getting your question, but here is the logic I have used.

    We have proved if m|x and n|x then => mn|x

    So let
    1. x in (m) and x in (n) => x in (mn)
    2. x in (mn) => x in (m) and x in (n)

    Hence
    $\displaystyle
    (m) \cap (n) = (mn)$

    Essentially all I am saying is $\displaystyle (m) \cap (n) = (LCM(mn))

    $
    Last edited by aman_cc; Oct 14th 2009 at 05:16 AM. Reason: latex tag error
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