Your problem is with the quote:
In this case the induction hypothesis is not used at all, a fact which may make the proof somewhat harder to follow.
This sentence isn't saying, "There is a proof of this statement that doesn't use the induction axiom of Peano." Indeed, there isn't such a thing. What this footnote was trying (and failing) to do was to avoid the confusion caused by the fact that the proof that $1$ is less than every other number is an induction proof whose induction step doesn't use the induction hypothesis. i.e. In order to show that $1<a'$, you don't need to use the assumption that $a=1$ or $1<a$.
(But it is still an induction proof. We are only allowed to write this proof because of the fifth Peano axiom. And it does take the standard form of an induction proof:
- Base Case: $1=1$, so either $1=1$ or $1<1$
- Induction Step: Assume that $a=1$ or $1<a$ (But we actually don't need this). Then $1<a'$. Therefore $a'=1$ or $1<a'$.
If you need to take some time to get your head around an induction proof whose induction step doesn't use the induction hypothesis, don't worry. It is a rather counter-intuitive idea that we'd need such proofs.
Here's a similar induction proof that might help demonstrate this idea. You might have at one point been taught that Weak Induction and Strong Induction are equivalent. This isn't quite true: ordinals larger than the natural numbers are sets which have strong induction but not weak induction. What is true is that Weak Induction is equivalent to (Strong Induction and Every Number is either $1$ or a Successor).
Therefore Every Number is either $1$ or a Successor is a property that's really key to the natural numbers. We should be able to prove it quickly using the Peano Axioms you've been given. Let's prove it by induction.
- Base Case: $1$ is $1$, so it is either $1$ or a successor.
- Induction Step: Assume that $a$ is either $1$ or a successor. (But again, we don't need to) Then $a'$ is a successor, because it's the successor of $a$. Therefore $a'$ is either $1$ or a successor.
By induction, every number is either $1$ or a successor. QED
Again, we didn't need the induction hypothesis in the induction step. But without the induction axiom, we wouldn't be able to prove this. If you play around with some of the other 'obvious' properties of the naturals, trying to prove them just from the five Peano Axioms, you might find more cases of this.
The general concept of an axiom system does not tell you what the objects are that it refers to, it tells you only properties of those objects and of the interactions between them. The axioms do not prescribe what things actually are, they merely describe things.
So, for example, the Peano Axioms do not tell you what $2$ is. Also, they do not tell you what $+$ is. Nor do the tell you how to go about actually performing $x+1$.
In fact that's the whole point of the Peano Axioms. They are not authoritative. Their sole purpose is to give one a tool for building the natural numbers on a simpler foundation.
Our experience of the natural numbers includes many named objects like $1, 2, 3$ and $485096$, and complicated binary operations such as $+$ and $\times$, and complicated binary relations such as $<$, and long lists of useful identities such as $a \times (b+c) = a \times b + a \times c$. The point of the Peano Axioms is that they give us a much less complicated system of objects and properties to accept, from which we, using our powers of logic, can construct the much more complicated system of objects and properties which comprise the theory of natural numbers. Once you accept this simpler foundation (based on your intuition, or on whatever basis you choose to accept), you can then then apply the Peano Axioms to define $2$ and $3$ and so on.
Have you accepted the successor function? Great, now define $2=1'$. Next, define $3=2'$ and $4=3'$ and $5=4'$ and $6=5'$ and $7=8'$ and $9=8'$. That was boring, onwards.
Have you accepted the axioms of the successor function? Great, now define addition, and prove it is commutative and associative, then define multiplication and prove is it commutative and associative and that multiplication distributes over addition, and so on.
What if you don't want to accept the Peano Axioms? What if you really, really want someone to tell you what $2$ is and $3$ is and $+$ is and so on?
Well, there are alternatives.
For example, many modern textbooks of analysis will start instead with axioms for the real numbers, and from those will construct the natural numbers satisfying Peano's Axioms; Fitzpatrick's Advanced Calculus textbook is an example, which I used in my Advanced Calculus class.
For another example, in modern set theory one starts with the ZFC axioms, and then using Von Neumann ordinals one constructs the natural numbers satisfying Peano's Axioms.
However, in both cases it's just a matter of pushing the ball backwards: you have to accept some axioms as your starting point: Peano's axioms; the axioms of the real numbers; the ZFC set theory axioms; or something.
Best Answer
NO.
We state that the successor relation $s(n)$ is defined by a function, in order to guarantee that there are no multiple values for the same argument.
But a function can map two arguments to the same value.
This is why we rquire that :
I.e., by contraposition, if the two arguments $n$ and $m$ are distinct, also their successors must be.
This is why Peano (1889) original formulation stated :
In the modern formulation in the language of first-order logic, the successor relation is expressed with a function symbol $s(n)$. Thus, the "functionality" is built-in into the rules of the language and thus the corresponding axiom amounts to :
This axioms is necessary to ensure the infinity of the number sequence; withou it, we may have some sort of "circularity", like e.g. $s(10)=2$. In this case, we have that both $1$ and $10$ have the same successor.