1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

Homework Help: Question on the Euclidean Algorithm

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [itex]a,b\in\mathbb{Z}[/itex]. Suppose [itex]r_{0}=a[/itex] and [itex]r_{1}=b[/itex]. By the algorithm, [itex]r_{i}=0[/itex] for some [itex]i\geq 2[/itex] is the first remainder that terminates. Show that [itex]r_{i-1}=\gcd(a,b)[/itex].


    2. Relevant equations



    3. The attempt at a solution
    I've shown that [itex]c|r_{i-1}[/itex], and I know that I should show that [itex]r_{i-1}|a[/itex] and [itex]r_{i-1}|b[/itex]. I just don't know how to show both the latter. I don't know where to continue.

    I don't want full solutions just given to me (obviously), just some insight :)

    Thanks!
     
  2. jcsd
  3. Sep 27, 2011 #2
    Maybe prove first that if a certain c divides [itex]r_k[/itex] and [itex]r_{k+1}[/itex], then it divides [itex]r_{k-1}[/itex].
    Then apply this result for c=gcd(a,b).
     
  4. Sep 27, 2011 #3
    Nevermind, I got it without having to break it down into cases!

    Note, [itex]\gcd(a,b)=\gcd(a+mb,b)[/itex] for any [itex]m\in\mathbb{Z}[/itex].

    Define [itex]P(n): \gcd(a_{n-1},a_{n})=\gcd(a_{n},a_{n+1})[/itex]. Let [itex]n_{0}=1[/itex] be our base case. Since [itex]a_{0}=ma_{1}+a_{2}[/itex] by the division algorithm, we have [itex]\gcd(a_{0},a_{1})=\gcd(ma_{1}+a_{2})=\gcd(ma_{1}-ma_{1}+a_{2},a_{1})=\gcd(a_{2},a_{1})=\gcd(a_{1},a_{2})[/itex], thus [itex]P(n_{0})[/itex] is true.

    Now let [itex]n\geq 1[/itex] such that [itex]P(n)[/itex] is true. Then we have [itex]\gcd(a_{n-1},a_{n})=\gcd(a_{n},a_{n+1})=\gcd(ma_{n+1}+a_{n+2},a_{n+1})=\gcd(ma_{n+1}-ma_{n+1}+a_{n+2},a_{n+1})=\gcd(a_{n+2},a_{n+1})=\gcd(a_{n+1},a_{n+2})[/itex]. Hence, [itex]\gcd(a_{n},a_{n+1})=\gcd(a_{n+1},a_{n+2})[/itex], thus for all [itex]n\geq 1[/itex], [itex]P(n)\implies P(n+1)[/itex]. Hence, by the principle of Mathematical Induction: [itex]\forall n\geq 1,P(n)[/itex].

    Since we know that for some [itex]i[/itex], [itex]a_{i}=0[/itex], it follows that: [tex]\gcd(a,b)=\gcd(a_{0},a_{1})=\cdots =\gcd(a_{i-1},a_{i})=\gcd(a_{i-1},0)=a_{i-1}.[/tex] Therefore, [itex]\gcd(a,b)=a_{i-1}[/itex], as required.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook