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Math Help - [SOLVED] Simple proof

  1. #1
    Member diddledabble's Avatar
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    Post [SOLVED] Simple proof

    Show that if a\mid{c} and b\mid{c} then ab\mid{c}

    I got started and have Let c=ae and c=bd where d and e are integers.
    c=ae=bd and a and b are not equal then e=b and d=a thus ab\mid{c} where the result would be 1.
    I am pretty sure my logic is really messed up though.
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  2. #2
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    Quote Originally Posted by diddledabble View Post
    Show that if a\mid{c} and b\mid{c} then ab\mid{c}
    3\mid12~\&~6\mid 12\text{ but is it true that }18\mid 12?
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  3. #3
    Member diddledabble's Avatar
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    Post

    I forgot to add that gcd (a,b)=1 Sorry.
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  4. #4
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    In your equations c=ae and c=bd try solving for a and b then look and the product of the two.

    That should get ya there
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  5. #5
    Member diddledabble's Avatar
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    So then a=c/e and b=c/d then ab=e/d Where e/d=c Is that correct?
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  6. #6
    Super Member Gamma's Avatar
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    Proof

    (a,b)=1\Rightarrow \exists s,t\in \mathbb{Z} s.t. as+bt=1

    Multiply through by c to arrive at
    asc+btc=c

    ab|asc because a|a and b|c

    ab|btc because a|c and b|b

    so it divides the sum which is c as desired.
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  7. #7
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    Quote Originally Posted by diddledabble View Post
    So then a=c/e and b=c/d then ab=e/d Where e/d=c Is that correct?

    I take back my original post. Let's try another approach.

    Since gcd(a,b) = 1 then there exists integers x,y such that 1=ax+by

    Multiply this equation by c to yield
    1) c*1 = c = c(ax+by)=axc+byc

    Now c=ak=bl where k,l are integers

    Substitute these values in equation 1)
    c=ax(bl)+by(ak)=abxl+abyk=ab(xl+yk)=abm where m is an integer

    Thus c=abm or equivalently ab|c
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