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If lcm(a,b) = ab, why is gcd(a,b) = 1?

  1. Oct 26, 2011 #1
    1. If lcm(a,b)=ab, show that gcd(a,b) = 1



    2. Relevant equations

    We can't use the fact that lcm(a,b) = ab / gcd(a,b)

    3. The attempt at a solution
    I've already shown that gcd(a,b) = 1 → lcm(a,b) = ab but I can't figure out the other direction!

    Any hints?
     
  2. jcsd
  3. Oct 26, 2011 #2
    [itex]lcm[a,b]=\prod p_i^{\max{\alpha_i,\beta_i}}[/itex]

    Where [itex]a=\prod p_i^{\alpha_i}[/itex]

    and [itex]b=\prod p_i^{\beta_i}[/itex]

    And the gcd is the min.

    Use that fact.
     
  4. Oct 26, 2011 #3
    We haven't really used that notation. I imagine it deals with divisibility rules just like the other direction.
     
  5. Oct 26, 2011 #4
    I think one way to approach it is to prove the contrapositive. If c is an integer not equal to 1 that divides a and divides b, then a = mc and b = nc for positive integers m and n. Then see if you can show that lcm (a,b) is not ab.
     
  6. Oct 27, 2011 #5
    Awesome, got it, thanks!
     
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