# Induction on two variables

1. Jun 19, 2009

### samkolb

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.

Sam

2. Jun 19, 2009

### Dragonfall

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.