I am currently studying Von Neumann integers. But how can we prove the induction theorem without reductio ad absurdum ?
Let $P_n$ be a property on $n$, an integer then if $P_0$ holds and $(P_n \implies P_{n+1})$ Then $P_n$ is true for any integer.
(when I say integer I'm talking of $ n \in \mathbb{N}$) Thank you.
On
First of all, a remark: when studying ordinals and related stuff (well-orderings) there is often a lot of reudctio ad absurdum going on. So as in the comments: why do you want to avoid it ?
Here's what I can do, I hope there will be better answers (I'll take your "without reduction ad absurdum" in the strongest possible sense, that is "can we prove that constructively ?", or rather "intuitionistically ?"). I'll use $s$ to denote the successor.
Let $P$ be such a property and consider $A=\{n\in \mathbb{N}\mid \neg P(n)\}$.
$0\notin A$. Let $n\in A$. Then $n\neq 0$, hence $n=s(m)$ for some $m$.
But $P(m) \implies P(s(m))$ so (constructively, there is no reductio ad absurdum here !) $\neg P(s(m))\implies \neg P(m)$, hence $\neg P(m)$.
Thus $m\in A$ and $m<n$.
We have proved $\forall n \in A, \exists m\in A, m<n$.
This implies (constructively) $\neg (\exists n\in A, \forall m\in A, n\leq m)$.
But we know (proved, or assumed) the well-ordering of $\mathbb{N}$ (it is an ordinal, hence it's well-ordered). Therefore $(\exists n, n\in A )\implies (\exists n\in A, \forall m\in A, n\leq m) (*)$, hence again by contraposition (the constructive part of it) $\neg(\exists n\in A, \forall m\in A, n\leq m)\implies \neg(\exists n, n\in A)$, hence there does not exist $n$ such that $\neg P(n)$.
This implies that for all $n$, $\neg\neg P(n)$. To get from there to $P(n)$ is exactly reductio ad absurdum.
A second thing is the $(*)$ bit: usually we say a set is well-ordered when any nonempty subset has a minimum element: this reads $\neg\neg(\exists n, n\in A) \implies (\exists n\in A, \forall m\in A, n\leq m)$. Here I've worded it as to say "an inhabited subset has a minimum element"; but that's a weaker statement so we're fine (note that the conclusion wouldn't have been changed since by contraposing we lose some constructive strength, indeed $\neg\neg\neg = \neg$)
Now, can we do better ? I.e. can we prove constructively that $\forall n, P(n)$ ? Well for starters, if $P$ is $\neg\neg$-stable (essentially of the form $\neg Q$ for some $Q$ - it means that $\neg\neg P \implies P$), then what we've done suffices.
Essentially what we're asking is whether the sequent $\forall A\subset \mathbb{N}, \neg\neg(\exists n, n\in A) \implies (\exists n\in A, \forall m\in A, n\leq m), \forall n, \neg (n=0) \implies \exists m, m=s(n), \forall n, n<s(n)\vdash Rec(P) \implies \forall n, P(n)$ is intuitionistically provable (where $Rec(P)\equiv P(0)\land \forall n, P(n) \implies P(s(n))$. I would think it's not; but at the moment I can't answer that.
I believe you are asking how we prove this in set theory. The answer is that we have a particular fact about the set $\omega$ of finite von Neumann ordinals: if we have any set $W$ so that $0 \in W$ and the successor $S(x)$ is in $W$ whenever $x$ is an ordinal in $W$, then every element of $\omega$ is in $W$. This is part of the definition of $\omega$.
So how do we prove induction for a specific property $P$? We simply form the set $W$ of ordinals that satisfy $P$. Then, the hypotheses on $P$ used for induction, together with the fact about $\omega$ from above, show us that every element of $\omega$ satisfies $P$.
There is no reductio ad absurdum here and no mention of the well ordering of $\omega$. The key point is just the definition of $\omega$ in set theory as the intersection of every set of ordinals $W$ that contains $0$ and is closed under successor.
Of course, we can only prove this for properties for which we can form a set - properties expressed in the language of set theory. This is part and parcel of formalizing induction. Whenever we talk about induction formally, we will have a formal language at hand, and we can only talk about induction for properties expressible in that formal language.