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Math Help - Show gcf(a,b) = gcf(a,c) = 1 => gcf(a,bc) = 1

  1. #1
    Senior Member MacstersUndead's Avatar
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    Show gcf(a,b) = gcf(a,c) = 1 => gcf(a,bc) = 1

    (i) Let a, b, and c be integers with gcf(a,b) = gcf(a,c) = 1. Show that gcf(a,bc) = 1.

    There's a more general case to this in (ii) of the same problem, where if [Math]{a_i}[/tex] is relatively prime with b for 1 \le i \le n then the product of [Math]{a_i}[/tex] is relatively prime with b.

    If you could give me a hint on how to prove (i), that would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by MacstersUndead View Post
    (i) Let a, b, and c be integers with gcf(a,b) = gcf(a,c) = 1. Show that gcf(a,bc) = 1.

    There's a more general case to this in (ii) of the same problem, where if [Math]{a_i}[/tex] is relatively prime with b for 1 \le i \le n then the product of [Math]{a_i}[/tex] is relatively prime with b.

    If you could give me a hint on how to prove (i), that would be greatly appreciated.
    Since gcd(a,b)=1 and gcd(a,c)=1, there exist integers s, t, u, and v such that as+bt=1 and au+cv=1. So (as+bt)(au+cv)=a(su+csv+btu)+bc(tv)=1.

    The general case: Suppose that gcd(a_i,b)=1, where 1\le i\le n, then gcd(a_1a_2\cdots a_n,b)=1.
    It's a corrolary of problem (i) and can be proved using induction on n ; the case n=2 is the main result.
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  3. #3
    Senior Member MacstersUndead's Avatar
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    Is it usually the case that for gcf proofs, you recourse back to the integral sums? Thanks again for your help!
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  4. #4
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    can someone show how the induction would go for part (2) in this question?
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