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Discrete math (gcd)

  1. Oct 13, 2013 #1
    Corollary from book:

    if d= gcd(a,b), then there exists integers x and y such that ax + by = d.

    This is not an obvious statement to me. Are there any direct proofs to prove this statement? The book proves this by induction.

    My proof:
    Suppose d = gcd(a,b) and a and b are positive integers. a does not necessarily divide b and b does not necessarily divide a so

    let qa and ra be integers such that a = qab + ra. If d|a then d|(qab + ra) therefore d|ra, that is there exists an integer Ra such that ra = Rad.

    Let qb and rb be integers such that b = qba + rb. We can see d|rb so Let Rb be the integer such that rb = Rbd.

    now
    a + b = qab + qba + (Ra + Rb)d and
    a(1-qb)/(Ra + Rb) + b(1-qa)/(Ra + Rb) = d.

    Now I just need to prove that the coef. of a and b are integers...
     
    Last edited: Oct 13, 2013
  2. jcsd
  3. Oct 14, 2013 #2

    verty

    User Avatar
    Homework Helper

    The extended Euclidean algorithm calculates those coefficients, so one avenue is an invariant-based proof based on that algorithm. Otherwise, a proof by contradiction should be doable.
     
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