Warren, Goodstein's theorem can be stated in Peano arithmetic using two quantifiers (for all n there exists a solution t to a Diophantine equation modelling a Turing machine that checks whether the nth Goodstein process halts within t steps), but cannot be proved in Peano arithmetic. The same applies to Kruskal's tree theorem (which is strictly harder) and the Robertson-Seymour theorem (which is even harder still). Fermat's Last Theorem can be stated using one quantifier (there does not exist a solution t to a Diophantine equation modelling a Turing machine that systematically checks all quadruples {a,b,c,n} of naturals and halts whenever a^(n+2) + b^(n+2) = c^(n+2)), so I suppose it's potentially 'easier' than the aforementioned three theorems. Anyway, you can't define the reals in Peano arithmetic, since it's a first-order theory and therefore incapable of quantifying over sets. Second-order logic with the Peano axioms *can* define and prove many facts about the reals (maybe that's what you meant), but this is not the same as Peano arithmetic. Sincerely, Adam P. Goucher http://cp4space.wordpress.com
----- Original Message ----- From: Warren Smith Sent: 02/22/13 04:37 AM To: math-fun Subject: [math-fun] Fermat's Last Theorem question
Peano's axioms suffice to define the integers, and hence (?) real & complex numbers, eh? And you are claiming you'd be "surprised" if Fermat's Last Thm was proven using those alone?!?! I'm freaking surprised if more is needed. In fact, I'm shaken.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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