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Math Help - Proof involving GCD of numbers

  1. #1
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    Proof involving GCD of numbers

    I would like to know of the line of approach one needs to take in the following (and similar problems involving elementary concepts of GCD of numbers)problem.

    The problem is-

    For any 2 integers a,b, prove that the (a+b,a-b)>=(a,b)
    where (x,y) represents the GCD of the numbers x,y.
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  2. #2
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    Quote Originally Posted by p vs np View Post
    I would like to know of the line of approach one needs to take in the following (and similar problems involving elementary concepts of GCD of numbers)problem.

    The problem is-

    For any 2 integers a,b, prove that the (a+b,a-b)>=(a,b)
    where (x,y) represents the GCD of the numbers x,y.
    Any common divisor of a and b is a divisor of a+b and a-b, hence the greatest
    common divisor of a and b is a divisor of both a+b and a-b, and so less than
    or equal to the gcd of a+b and a-b.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    Any common divisor of a and b is a divisor of a+b and a-b, hence the greatest
    common divisor of a and b is a divisor of both a+b and a-b, and so less than
    or equal to the gcd of a+b and a-b.

    RonL
    My apologies for the late response.
    Considering the reply, i am now forced to ask this question-
    Can you prove the statement "Any common divisor of a and b is a divisor of a+b and a-b", because at face value, it seems to be quite logical.
    But is there a mathematical treatment this statement can be given?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by p vs np View Post
    My apologies for the late response.
    Considering the reply, i am now forced to ask this question-
    Can you prove the statement "Any common divisor of a and b is a divisor of a+b and a-b", because at face value, it seems to be quite logical.
    But is there a mathematical treatment this statement can be given?
    c \in \mathbb{Z} is a common divisor of a \in \mathbb{Z} and b \in \mathbb{Z} if (and only if) there exists u and v in \mathbb{Z} such that:

    a=uc,\ b=vc

    Now:

    a+b=uc+vc=(u+v)c

    but (u+v) \in \mathbb{Z}, so c is a divisor of (a+b).

    Same type of argument for (a-b) applies.


    RonL
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  5. #5
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    Oh...after the solution has been provided, it looks rather elementary(as in every case).
    Thanks for the solution.
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