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Divisibility property

  1. Jul 31, 2010 #1
    1. The problem statement, all variables and given/known data

    proof the theorem

    if a l b and b l a then a=+-b

    2. Relevant equations

    3. The attempt at a solution

    there exist integer p,q such that ap=b and bq=a, then i've no idea how i can relate it to a=+-b.. clue please T_T
  2. jcsd
  3. Jul 31, 2010 #2
    [tex] ap = b = (\frac{a}{q}) [/tex]

    Multiply by q and divide be a (since b|a <--> a is not 0), giving us:

    [tex] pq = 1 [/tex]

    Do you follow?
  4. Jul 31, 2010 #3
    yea yah, i thought that too, but don't know to continue from there too, owhoho, more clue please, ;P
  5. Jul 31, 2010 #4
    wait, let me think first
  6. Jul 31, 2010 #5


    Staff: Mentor

    So b = pa and a = qb, for some integers p and q.
    Then b = pa = p(qb) = (pq)b.

    What can you say about pq?
  7. Jul 31, 2010 #6
    Remember, p and q are both nonzero integers.
  8. Jul 31, 2010 #7
    hmm, so pq=1 , hence, p=1 and q=1, hence a=b and b=a, still cannot get +-b, T_T
  9. Jul 31, 2010 #8
    WAIIITTTT (-1)(-1) also equal 1, wait wait let me think again
  10. Jul 31, 2010 #9
    so i get

    a=b or (a=-b and b=-a)

    => (a=b or a=-b) and (a=b or b=-a)

    (a=b or a=-b) is enough to verify it right?
  11. Jul 31, 2010 #10
    Yep, though technically you're proving, not verifying. "Proving" is a stronger word...makes you sound "smarter/cooler" :P!
  12. Jul 31, 2010 #11
    owho, thankyou very much
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