Results 1 to 3 of 3

Math Help - (a,b,c) = ((a,b),c)

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    6

    Unhappy (a,b,c) = ((a,b),c)

    Hey! the question is:
    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).


    This is what i have done:

    let d1 = ((a,b),c)

    where (a,b) = gcd(a,b) and is defined

    hence let e = (a,b)


    thus, d1 = (e,c)
    since this is the gcd of two numbers, then the gcd(e,c)is also defined.

    hence, d1 = ((a,b), c) is defined


    now let d = (a,b,c)

    d = ax0 +by0 +cz0
    let a = d1f, b = d1g, c = d1h
    then,

    d = d1fx0 + d1gy0 + d1hz0
    = d1(fx0 + gy0 + hz0)
    therefore, d | d1



    i know i need to show that d=(a,b,c) = ((a,b),c)=d1, and dat for this to occur d|d1 and d1|d, but im not sure if what i have done is right especially this part:
    d = ax0 +by0 +cz0
    let a = d1f, b = d1g, c = d1h
    then,

    d = d1fx0 + d1gy0 + d1hz0
    = d1(fx0 + gy0 + hz0)
    therefore, d | d1

    Plus i also dont know how to continue -_-'.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    You got the right idea and not much work is really needed. All you need is the fact that any common divisor of two numbers divides the GCD of those two numbers.

    Let d_1 = \gcd (a,b,c) \geq 1 and d_2 = \gcd (\gcd(a,b) , c) \geq 1.

    You want to show d_1 \mid d_2 and d_2 \mid d_1 which will imply d_1 = d_2.

    By definition, we have: d_1 \mid a, \ d_1 \mid b, \ d_1 \mid c. But this means d_1 \mid \gcd (a,b). But this also means d_1 is a common divisor of both \gcd(a,b) and c which means d_1 \mid d_2.

    Can you prove the other case?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    6
    Quote Originally Posted by o_O View Post
    You got the right idea and not much work is really needed. All you need is the fact that any common divisor of two numbers divides the GCD of those two numbers.

    Let d_1 = \gcd (a,b,c) \geq 1 and d_2 = \gcd (\gcd(a,b) , c) \geq 1.

    You want to show d_1 \mid d_2 and d_2 \mid d_1 which will imply d_1 = d_2.

    By definition, we have: d_1 \mid a, \ d_1 \mid b, \ d_1 \mid c. But this means d_1 \mid \gcd (a,b). But this also means d_1 is a common divisor of both \gcd(a,b) and c which means d_1 \mid d_2.

    Can you prove the other case?

    ... hang on don't you need to prove/show that gcd(a,b,c) exists/is defined? or can u just assume?

    really sorry im starting to get confused
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum