You wrote:
i understand how to do ordinary induction proofs and i understand that strong induction proofs are the same as ordinary with the exception that you have to show that the theorem holds for all numbers up to and including some n (starting at the base case) then we try and show: theorem holds for $n+1$
No, not at all: in strong induction you assume as your induction hypothesis that the theorem holds for all numbers from the base case up through some $n$ and try to show that it holds for $n+1$; you don’t try to prove the induction hypothesis.
In your example the simple induction hypothesis that the result is true for $n$ is already enough to let you prove that it’s true for $n+1$, so there’s neither need nor reason to use a stronger induction hypothesis. The proof by ordinary induction can be seen as a proof by strong induction in which you simply didn’t use most of the induction hypothesis.
I suggest that you read this question and my answer to it and see whether that clears up some of your confusion; at worst it may help you to pinpoint exactly where you’re having trouble.
Added: Here’s an example of an argument that really does want strong induction. Consider the following solitaire ‘game’. You start with a stack of $n$ pennies. At each move you pick a stack that has at least two pennies in it and split it into two non-empty stacks; your score for that move is the product of the numbers of pennies in the two stacks. Thus, if you split a stack of $10$ pennies into a stack of $3$ and a stack of $7$, you get $3\cdot7=21$ points. The game is over when you have $n$ stacks of one penny each.
Claim: No matter how you play, your total score at the end of the game will be $\frac12n(n-1)$.
If $n=1$, you can’t make any move at all, so your final score is $0=\frac12\cdot1\cdot0$, so the theorem is certainly true for $n=1$. Now suppose that $n>1$ and the theorem is true for all positive integers $m<n$. (This is the strong induction hypothesis.) You make your first move; say that you divide the pile into a pile of $m$ pennies and another pile of $n-m$ pennies, scoring $m(n-m)$ points. You can now think of the rest of the game as splitting into a pair of subgames, one starting with $m$ pennies, the other with $n-m$.
Since $m<n$, by the induction hypothesis you’ll get $\frac12m(m-1)$ points from the first subgame. Similarly, $n-m<n$, so by the induction hypothesis you’ll get $\frac12(n-m)(n-m-1)$ points from the second subgame. (Note that the two subgames really do proceed independently: the piles that you create in one have no influence on what you can do in the other.)
Your total score is therefore going to be
$$m(n-m)+\frac12m(m-1)+\frac12(n-m)(n-m-1)\;,$$
which (after a bit of algebra) simplifies to $\frac12n(n-1)$, as desired, and the result follows by (strong) induction.
It is possible indeed, the principle of induction is much more general than you may think. There are many induction theorems you can prove for the most intuitive of them if you understand the principal idea which is well-foundedness, or well-orderedness. (but maybe it is too soon for you to benefit from the effort you would have to make to understand these notions)
The usual induction theorem for $\mathbb{Z}$ can be proven using regular induction. It can be stated as:
If a property $P(x)$ satisfies $P(n)$ for some integer $n$ and $\forall k \in \mathbb{Z}(P(k) \longrightarrow P(k-1) $ and $P(k+1))$, then it is true for every integer.
If you don't know how to prove PMI using the fact that every non-empty subset of $\mathbb{N}$ has a least element, I suggest you look for a proof. I remember I was very happy to discover this at the time I thought PMI was just some kind of common belief among mathematicians; and it is still a very interesting theorem for me.
Best Answer
Okay, I can't resist: here is a quick answer.
I am construing the question in the following way: "Is there some criterion for a subset of $[0,\infty)$ to be all of $[0,\infty)$ which is (a) analogous to the principle of mathematical induction on $\mathbb{N}$ and (b) useful for something?" The answer is yes, at least to (a).
Let me work a little more generally: let $(X,\leq)$ be a totally ordered set which has
$\bullet $a least element, called $0$, and no greatest element.
$\bullet$ The greatest lower bound property: any nonempty subset $Y$ of $X$ has a greatest lower bound.
Principle of Induction on $(X,\leq)$: Let $S \subset X$ satisfy the following properties:
(i) $0 \in S$.
(ii) For all $x$ such that $x \in S$, there exists $y > x$ such that $[x,y] \subset S$.
(iii) If for any $y \in X$, the interval $[0,y) \subset S$, then also $y \in S$.
Then $S = X$.
Indeed, if $S \neq X$, then the complement $S' = X \setminus S$ is nonempty, so has a greatest lower bound, say $y$. By (i), we cannot have $y = 0$, since $y \in S$. By (ii), we cannot have $y \in S$, and by (iii) we cannot have $y \in S'$. Done!
Note that in case $(X,\leq)$ is a well-ordered set, this is equivalent to the usual statement of transfinite induction.
It also applies to an interval in $\mathbb{R}$ of the form $[a,\infty)$. It is not hard to adapt it to versions applicable to any interval in $\mathbb{R}$.
Note that I believe that some sort of converse should be true: i.e., an ordered set with a principle of induction should have the GLB / LUB property. [Added: yes, this is true. A totally ordered set with minimum element $0$ satisfies the principle of ordered induction as stated above iff every nonempty subset has an infimum.]
Added: as for usefulness, one can use "real induction" to prove the three basic Interval Theorems of honors calculus / basic real analysis. These three theorems assert three fundamental properties of any continuous function $f: [a,b] \rightarrow \mathbb{R}$.
First Interval Theorem: $f$ is bounded.
Inductive Proof: Let $S = \{x \in [a,b] \ | \ f|_{[a,x]} \text{ is bounded} \}$. It suffices to show that $S = [a,b]$, and we prove this by induction.
(i) Of course $f$ is bounded on $[a,a]$. (ii) Suppose that $f$ is bounded on $[a,x]$. Then, since $f$ is continuous at $x$, $f$ is bounded near $x$, i.e., there exists some $\epsilon > 0$ such that $f$ is bounded on $(x-\epsilon,x+\epsilon)$, so overall $f$ is bounded on $[0,x+\epsilon)$.
(iii) If $f$ is bounded on $[0,y)$, of course it is bounded on $[0,y]$. Done!
Corollary: $f$ assumes its maximum and minimum values.
Proof: Let $M$ be the least upper bound of $f$ on $[a,b]$. If $M$ is not a value of $f$, then $f(x)-M$ is never zero but takes values arbitrarily close to $0$, so $g(x) = \frac{1}{f(x)-M}$ is continuous and unbounded on $[a,b]$, contradiction.
(Unfortunately in the proof I said "least upper bound", and I suppose the point of proofs by induction is to remove explicit appeals to LUBs. Perhaps someone can help me out here.)
Second Interval Theorem (Intemediate Value Theorem): Suppose that $f(a) < 0$ and $f(b) > 0$. Then there exists $c \in (a,b)$ such that $f(c) = 0$.
Proof: Define $S = \{x \in [a,b] \ | \ f(x) \leq 0\}$. In this case we are given that $S \neq [a,b]$, so at least one of the hypotheses of real induction must fail. But which?
(i) Certainly $a \in S$.
(iii) If $f(x) \leq 0$ on $[a,y)$ and $f$ is continuous at $y$, then we must have $f(y) \leq 0$ as well: otherwise, there is a small interval about $y$ on which $f$ is positive.
So it must be that (ii) fails: there exists some $x \in (a,b)$ such that $f \leq 0$ on $[a,x]$ but there is no $y > x$ such that $f \leq 0$ on $[a,y]$. As above, since $f$ is continuous at $x$, we must have $f(x) = 0$!
Third Interval Theorem: $f$ is uniformly continuous.
(Proof left to the interested reader.)
Moreover, one can give even easier inductive proofs of the following basic theorems of analysis: that the interval $[a,b]$ is connected, that the interval $[a,b]$ is compact (Heine-Borel), that every infinite subset of $[a,b]$ has an accumulation point (Bolzano-Weierstrass).
Acknowledgement: My route to thinking about real induction was the paper by I. Kalantari "Induction over the Continuum".
His setup is slightly different from mine -- instead of (ii) and (iii), he has the single axiom that $[0,x) \subset S$ implies there exists $y > x$ such that $[0,y) \subset S$ -- and I am sorry to say that I was initially confused by it and didn't think it was correct. But I corresponded with Prof. Kalantari this morning and he kindly set me straight on the matter.
For that matter, several other papers exist in the literature doing real induction (mostly in the same setup as Kalantari's, but also with some other variants such as assuming (ii) and that the subset $S$ be closed). The result goes back at least to a 1922 paper of Khinchin, who in fact later used real induction as a basis for an axiomatic treatment of real analysis. It is remarkable to me that this appealing concept is not more widely known -- there seems to be a serious PR problem here.
Added Later: I wrote an expository article on real induction soon after giving this answer: it's very similar to what I wrote above, but longer and more polished. It is available here if you want to see it.