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Euclidean algorithm

  1. Sep 25, 2003 #1
    I am little confused on this issue, actually the thing that is confusing me is my book.
    We all know the formula m=qn+r and 0<r<n
    In the formula we go until r=0, then you will know relativley prime.
    When given a problem, for example, (862,347) we have m=862 while n=347
    so we have the equation 862=347(q)+r. Then we solve till r=0
    For some reason in my book it gives the example (1251,1976)
    the way they set it up is 1976=1251(1)+725 according to this
    m=1976 and n=1251, isn't that backwards? I thought it went (m,n), with that in mind shouldn't it be 1251=1976(1)+r?
    Or do you place the numbers accordingly so 0<r?
  2. jcsd
  3. Sep 26, 2003 #2


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    In the question of whether two numbers are relatively prime, the order of the two numbers is not important.

    In determining whether or not 1251 and 1976 are relatively prime, you text is looking for 1976= 1251(q)+ r because 1251< 1976.

    Yes, r has to be positive: that's part of the definition (in writing b= aq+ r you want 0<= r< a). You don't want to look at
    1251= 1976q+ r because that has the obvious solution q= 0, r= 1251 which doesn't help.
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