Are Different Proofs of a Theorem Really the Same?Date: 07/05/2006 at 20:42:28 From: Eric Subject: Are all mathematical proofs basically the same When I was in Junior High School I remember being surprised to find out that there are many different proofs of the Pythagorean Theorem. I remember wondering if all of those proofs are in some way really the same? A few nights ago it occurred to me that there is a better way to ask this question. If you have a mathematical system with several axioms (call them A, B, C, D, E and F), is it possible to have two proofs of a theorem in this system where one proof uses only axioms A, B and C and the other proof uses only axioms D, E and F? In other words is it possible for two proofs to use no common axioms? Date: 07/05/2006 at 22:43:58 From: Doctor Tom Subject: Re: Are all mathematical proofs basically the same Hi Eric - Very nice question! The study of this sort of thing is called "metamathematics", or "foundations of mathematics", and is incredibly deep and interesting. Let's take a simple example. Suppose we have a system that consists of objects and an operator * on those objects, and contains the following two axioms (and perhaps others): A) There exists an identity element e in the system such that if a is any object in the system, e*a = a*e = a. B) The system contains at least two objects. Here's my "theorem": There is at least one object in the system. Clearly this can be proven from A alone or from B alone, and clearly axioms A and B are independent. Now this seems trivial, but it satisfies your conditions, and it's possible to construct more interesting axiom systems and theorems that also satisfy your requirements. So the answer is no--not all proofs are essentially equivalent. A much more common situation is that a theorem can be proved starting from different subsets (possibly overlapping) of the axioms, but if the axioms are independent, meaning that none of them is a consequence of the fact that some subset of others forces it to be true, then the two proofs are obviously not equivalent. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ |
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