I'm new here (and to math in general), so I apologize for any mistake/rule-breaking.
I've reached the Mathematical Induction chapter in Velleman's book How to Prove It, and I asked myself "Is there a way to make induction work for real numbers?".
Around the same time, I started Elementary Calculus: An Infinitesimal Approach by Keisler.
So I had the idea of using infinitesimals in induction.
Let I be the set of nonzero infinitesimals and R* is the set of hyperreals, and P is any statement, the idea is
$$[\forall\epsilon\in I(P(0) \land \forall r \in R^*(P(r) \implies P(r + \epsilon)))] \implies\forall x \in R(P(x))$$
because if we suppose
$$\forall\epsilon\in I(P(0) \land \forall r \in R^*(P(r) \implies P(r + \epsilon)))$$
Then for any real number x and nonzero infinitesimal ε
$$P(0) \implies P(\epsilon )$$
$$P(\epsilon ) \implies P(2\epsilon)$$
$$…$$
$$ P((\frac{x}{\epsilon} – 1) \cdot\epsilon ) \implies P(\frac{x}{\epsilon}\cdot\epsilon )$$
and so P(x)
I think the base case can be any number, not just 0.
Am I missing something, or can this approach be used to prove statements about all real numbers?
Also, can it be used to prove statements about hyperreal numbers?
Thank you in advance.
Edit: it seems the induction hypothesis need not be true for all infinitesimals, only one positive and one negative infinitesimal are sufficient for proving statements about real numbers.
Best Answer
Take $P(x)$ to be the predicate "$x$ is infinitesimally close to $0$", i.e. there is no real (or no rational) number strictly between $x$ and $0$. Then certainly $P(0)$ and for each $x$ if $P(x)$, then $P(x+\varepsilon)$, but it is not the case that $P(1)$.
Similarly, you cannot deduce that $P(\frac{1}{2} \varepsilon)$ from $P(0)$ and the implication $P(x) \implies P(x+\varepsilon)$.