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Math Help - If gcd(a,b)=1, a|c and b|c, then ab|c

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    4

    If gcd(a,b)=1, a|c and b|c, then ab|c

    Suppose that gcd(a,b)=1, a|c and b|c. Prove that ab|c.

    I've already got this part:
    If gcd(a,b)=1 then there are integers r,s such that ra + sb = 1.
    If a|n and b|n then there are integers k,l such that n = ka = lb.
    rka + skb = k,
    b(rl + sk) = k,
    ab(rl + sk) = ka = c,
    so ab | c

    How can I show that the initial assumption of gcd(a,b)=1 is necessary for this to work?
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  2. #2
    Junior Member
    Joined
    Mar 2009
    Posts
    31
    Hello,

    Let's say that gcd(a,b)=d then we have:

    a=a_1*d \Rightarrow (a_1d|c)

    and

     b= b_1*d \Rightarrow (b_1d|c)

    From this 2 realtions results that:

     ab|c \Rightarrow d^2a_1b_1|c \Rightarrow

    d^2a_1|c

    and

     d^2b_1|c

    following the same procedure we will find that

    d^na_1|c

    and

     d^nb_1|c

    this case is absurd if d is not equal to 1.

    so gcd(a,b)=1;

    Have a nice day,
    Hush_Hush.
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