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Thread: Define gcd(a,b,c)

  1. #1
    Mar 2009

    Arrow Define gcd(a,b,c)


    Assuming that the gcd(a,b) of any two integers a and b is defined, show that the gcd(a,b,c) of any three integers is also defined, and that (a,b,c)=((a,b),c).

    My Answer so far: ( i think its totally wrong)

    Let gcd(a,b)=d so d|b, d|a.

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

    We want to prove that


    Since (a,b)=d and e|a and e|b => e|d

    and from ((1)) e|c

    As e|d and e|c we have gcd(d,c)=e

    But i think i still need to show that (a,b,c) is definded and im not sure that i proved the statement correctly, i think i need to show that it is the greatest common divisor but i only showed that it was a divisor, im not sure. If anyone has a completly different solution that actually makes sense that would be super helpful!

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  2. #2
    o_O is offline
    Primero Espada
    o_O's Avatar
    Mar 2008
    Let $\displaystyle d_1 = \gcd(a,b,c) \geq 1$ and $\displaystyle d_2 = \gcd ( \gcd(a,b), c) \geq 1$.

    The idea is to show that $\displaystyle d_1 \mid d_2$ and $\displaystyle d_2 \mid d_1$ which implies $\displaystyle d_1 = d_2$.

    It shouldn't be too troublesome to see why this is the case.
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