Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Proof of a lemma of BÉZOUT’S THEOREM

  1. Sep 11, 2016 #1
    One silly thing is bothering me. As per one lemma, If a, b, and c are positive integers such that gcd(a, b) = 1 and a | bc, then a | c. This is intuitively obvious. i.e.
    Since GCD is 1 'a' does not divide 'b'. Now, 'a' divides 'bc' so, 'a' divides 'c'. Proved.

    What is bothering me is : suppose bc/a = s. Then as = bc. Thus a = (b/s) c ..... (1)
    Now, if c/a is an integer so is s/b. Which means b/s is not an integer. Putting this in (1) - how 'a' divides 'c'?
    Thanks in advance
  2. jcsd
  3. Sep 11, 2016 #2


    User Avatar
    2016 Award

    Staff: Mentor

    So what? Where does b/s appear?

    Maybe it is easier to understand with a numerical example:
    a=5, b=3, c=10
    5 | 30 is true, 5 | 10 is true as well.
    bc/a = s gives us s = 30/5 = 6.
    s/b=2, and b/s=1/2 is not an integer. So what?
  4. Sep 12, 2016 #3
    Thanks for your reply.
    Sorry...don't know why i asked this question..... a (s/b) = c ....i dont know why I was thinking the other way.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Proof of a lemma of BÉZOUT’S THEOREM
  1. Proofs of theorems (Replies: 8)

  2. Banach lemma proof (Replies: 4)