A "successor function" could also be written as s(x)= x+ 1butsince, in Peano's axioms, addition is defined in terms of the "successor fuction" this would be "circular". (For any x, x+ 1 is defined as s(x). If b is not 1, then there exist c such that s(c)= b. In that case a+ b is defined as s(a+ c).)

It must also have the property that there exist a unique member of N (typically called "1") such that the successor function is "onto" N-{1}.2) A successor function is certainly a one-to-one function but is that sufficient and if not is there another property that a successor function must possess?

I don't know what "true" means in regard to a3) Given

“a” is a real number,

“n” is a natural number and the recursive definition:

a^{1}= a, and a^{n+1}= a^{n}a, it can be proved that this recursive definition is true for all n.

Does that mean that the here stated recursive definition is itself proved to be a “successor function”? If yes, does this mean all recursive functions proved true for all “n” are successor functions?definition. You can determine whether or not is "well defined"- that is, whether it defines a unique object, for every real number a and every positive integer, n. You seem to have a very vague idea of what a "successor function" is. No, it does not mean that any recursive function is a successor function- a successor function is exactly what I said and what was probably given in your text: It is a one-to-one function from N onto N-{1}.

To prove that the given definition is "well defined" you must show that you can apply it to any positive integer n. (a is a real number and we are only concerned with integers here so we accept it.) As usual with the natural numbers, which are, with Peano's axioms, defined "inductively", we use "proof by induction". Let "S" be the set of all natural numbers such that is defined.

(i) is defined to be 1 so 1 is in S.

(ii) Suppose n is in S. Then . Since n is in S, is defined and multiplication of two members of N is defined so is defined. That is, is defined so a is in S.

By induction then, S= N so is defined for all N.

Thanks.