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Math Help - Prove: GCD(GCF) of (a,b) * LCM (a,b) = product a*b

  1. #1
    Newbie panglot's Avatar
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    Question Prove: GCD(GCF) of (a,b) * LCM (a,b) = product a*b

    Hello All,

    I have begun working on this topic's proof but I am not sure where to go after this:

    Given:

    (a*b), a,b elements of the natural numbers
    There exists d such that d is the GCD (a,b) and is a divisor of a and a divisor of b.
    There exists e such that e is the LCM (a,b).

    Prove:

    (e*d) = (a*b)

    ...

    At this point, I want to use the Well-Ordering Principle to prove that the set of multiples contains a smallest possible multiple, but I am not sure how to prove this through contradiction. Would there be a way to do this, and then to prove the initial premise?

    Thank you very much for your time,

    Panglot
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Dec 2009
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    Russia
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    Re: Prove: GCD(GCF) of (a,b) * LCM (a,b) = product a*b

    Quote Originally Posted by panglot View Post
    Hello All,

    I have begun working on this topic's proof but I am not sure where to go after this:

    Given:

    (a*b), a,b elements of the natural numbers
    There exists d such that d is the GCD (a,b) and is a divisor of a and a divisor of b.
    There exists e such that e is the LCM (a,b).

    Prove:

    (e*d) = (a*b)

    ...

    At this point, I want to use the Well-Ordering Principle to prove that the set of multiples contains a smallest possible multiple, but I am not sure how to prove this through contradiction. Would there be a way to do this, and then to prove the initial premise?

    Thank you very much for your time,

    Panglot
    Here:

    Product of GCD and LCM - ProofWiki
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