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Is it possible to define Peano arithmetic using ZFC solely?

According to my knowledge of ZFC, it seems impossible to define Peano arithmetic using solely ZFC as it only includes few relations. Is a model required to do this?

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    Not sure what you mean... the Peano Axioms are theorems of ZFC as they apply to $\omega$ and the successor function on the ordinals. Addition, multiplication, and order of the natural numbers are just the restriction of the addition, multiplication, and order of the cardinals.2012-05-09
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    @ArturoMagidin But I do not see cardinal addition defined explicitly in the ZFC axiom... It only seems to talk about equivalence relation... what am I getting wrong?2012-05-09
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    Not everything needs to be an axiom; and you have the language of functions and so on derived from First Order Logic. One can define addition of cardinals within ZFC and prove theorems about it, after all!2012-05-09
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    @ArturoMagidin I get it. Thanks for your help. :)2012-05-09

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Yes, it can be done, routinely, for example in the usual way. It can even be done in a variant of ZFC in which the Axiom of Infinity is negated.

If we develop Number Theory within ZFC, we obtain a theory which is strictly stronger than (first-order) Peano Arithmetic. For one thing, within ZFC one can prove the formal consistency of PA (but naturally not of ZFC itself, unless ZFC is inconsistent, one does not escape the Second Incompleteness Theorem). More interestingly, there are "natural" sentences of PA, such as the strengthened Ramsey's Theorem which can be stated but not proved in first-order Peano arithmetic, but which are theorems of Number Theory as developed within ZFC.