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Math Help - proof about gcd

  1. #1
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    proof about gcd

    I have to prove that if a,b,c\in\mathbb{Z} Show that gcd(a,b)=1$ and $gcd(a,c)=1 iff gcd(a,lcm[b,c])=1 I've tried proving the forward direction and im stuck. Here is what I have so far.

    proof: \rightarrow: Assume gcd(a,b)=1$ and $gcd(a,c)=1 . Assume further that  gcd(a,[b,c])\neq 1 . Let d=(a,[b,c]) which implies that d|a$ and $d|[b,c] \implies d|b$ or $d|c
    Case 1: Suppose d|b$ and $d\nmid c . Since d|a$ and $d|b and since gcd(a,b)=1 then d is a common multiple of  gcd(a,b) thus d=1 A contradiction.

    Case 2: Suppose d\nmid b$ and $d|c . Same here as case 1 and d=1. A contradiction.
    Case 3: Suppose  d|b$ and $d|c Since gcd(a,b)=1$ and $gcd(a,c)=1$ and $d|a,  d|b$, and $d|c then d=1 A contradiction.\

    Would this be right?
    Last edited by bonfire09; August 5th 2013 at 12:36 PM.
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  2. #2
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    Re: proof about gcd

    Quote Originally Posted by bonfire09 View Post
    I have to prove that if a,b,c,d \in\mathbb{Z} Show that gcd(a,b)=1$ and $gcd(a,c)=1 iff gcd(a,lcm[b,c])=1 I've tried proving the forward direction and im stuck. Here is what I have so far.
    Please review what you have posted. What does d have to do with anything in the given?

    Please correct it.
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  3. #3
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    Re: proof about gcd

    ok ive fixed it. The part that that I think needs to be explained further is the part where I say since  d|[b,c]\implies d|b$ or $d|c.
    Last edited by bonfire09; August 5th 2013 at 01:02 PM.
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  4. #4
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    Re: proof about gcd

    Actually I take that back. Its wrong. The claim that  d|[a,b]\implies d|a$ pr $d|b is wrong.
    Last edited by pokerface; August 5th 2013 at 03:01 PM.
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