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Mathematical Induction on rationals

  1. Mar 22, 2010 #1
    I'm a high school student. I gave a proof for the following theorem, but I was told by some professors that this is trivial and using natural induction twice for the rationals will do the same thing. What do you think? Is it just redundant?


    Let P(r) be a statement about r, then if :
    1) P(1) is true and,
    2) [tex]\forall[/tex] m,n [tex]\in[/tex] [tex]N[/tex] , m[tex]\geq[/tex]n ; P([tex]\frac{m}{n}[/tex])[tex]\rightarrow[/tex] P([tex]\frac{m+1}{n}[/tex])

    Then [tex]\forall[/tex] r[tex]\in[/tex] Q, r[tex]\geq[/tex]1 ; P(r).

    (PS: I apologize for my (probable) mistakes, because I'm neither an English speaker nor familiar with Latex.)
  2. jcsd
  3. Mar 22, 2010 #2


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    It's not quite true as stated; take P(r) = true if r > 0 and false otherwise. But the basic idea is right.

    If you were a college student, I would agree that this is a trivial result by double induction. But as a high-school student I actually think it's pretty good. Most wouldn't think to extend induction to the rationals at all.
  4. Mar 22, 2010 #3
    I can't get why it's not true. Did you notice r[tex]\geq[/tex]1 in the last sentence?

    And thank you for the reply.
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