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Induction on two variables

  1. Jun 19, 2009 #1
    If I am given a propisition P(m,n) and asked to show that it is true for all integers m and n, how do I go about that?

    My strategy is to fix one of the variables, say m, and then proceed to use induction on n. Once I've shown that P(m,n) holds for all n when m is fixed, I then conclude that P(m,n) holds for all m and n, since m was chosen arbitrarily.

    Is this correct?

    If it helps, the particular problem I'm working on is proving the laws of exponents for a group.

  2. jcsd
  3. Jun 19, 2009 #2
    You show that P(0,0) is true, and that whenever P(m,n) is true, then P(m+1,n) is true and P(m,n+1) is true.

    More generally if you have a set of propositions indexed by a connected metric space, if you show that a given point is true, and that whenever P(x) is true then P(y) is true for all y in a fixed radius of x, then P is true.
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