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Homework Help: Proof by induction

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

    [tex]\sqrt{ab}[/tex] = [tex]\sqrt{a}[/tex][tex]\sqrt{b}[/tex] for all ab>0

    2. Relevant equations



    3. The attempt at a solution

    let a=0, [tex]\sqrt{0}[/tex] = [tex]\sqrt{0}[/tex]

    assume that it's true for some a, consider a+1

    [tex]\sqrt{(a+1)b}[/tex] = and i'm lost
     
  2. jcsd
  3. Jun 1, 2010 #2

    Mark44

    Staff: Mentor

    Why do you think that this should be done by induction?
     
  4. Jun 1, 2010 #3
    ooops, yea, on second thought, induction is for integer, not real numbers

    is there any ways to proof this?
     
  5. Jun 1, 2010 #4
    but wait,

    [tex]\sqrt{ab}[/tex] = [tex]\sqrt{a}[/tex][tex]\sqrt{b}[/tex] for all integer a and b such that ab>0

    how about that? can i solve this using induction?
     
  6. Jun 1, 2010 #5

    Mark44

    Staff: Mentor

    How is the problem stated? Does it explicitly specify that a and b are integers? Also, does it say ab > 0 or does it say a > 0 and b > 0? There's a difference.
     
  7. Jun 1, 2010 #6
    no, i just made one up.

    but what about

    [tex]\sqrt{ab}[/tex] = [tex]\sqrt{a}[/tex][tex]\sqrt{b}[/tex] for all integer a and b such that ab>0

    can't i prove it using induction?
     
  8. Jun 1, 2010 #7

    Mark44

    Staff: Mentor

    Induction really isn't a good fit here. With an induction proof you have a sequence of statements P(1), P(2), P(3), ..., P(n), ... With the problem you made up, you have two variables a and b, both of which should be real and nonnegative.
     
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