Can Induction Prove the Laws of Exponents for a Group?

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To prove a proposition P(m,n) for all integers m and n, one effective strategy is to fix one variable, such as m, and use induction on the other variable, n. After demonstrating that P(m,n) holds for all n with m fixed, it can be concluded that P(m,n) is true for all integers m and n, as m was chosen arbitrarily. Additionally, the discussion highlights the importance of establishing a base case, such as P(0,0), and verifying that if P(m,n) is true, then P(m+1,n) and P(m,n+1) are also true. A more general approach involves showing that a proposition holds at a point in a connected metric space and that it holds in a neighborhood around that point. This method can be applied to prove the laws of exponents for a group.
samkolb
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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
 
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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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