1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    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


    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Discrete math Date
Discrete Math - posets Nov 24, 2015
Help with simplifying boolean expression Jan 29, 2015
Basic Discrete math question Jan 8, 2015
Discrete Math Question Oct 23, 2014
Determine Big O of a function (Discrete math) Oct 9, 2013