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Mathematical induction in natural number is proved with axiom of natural number. I'm curious about is it possible to extent mathematical induction into real number range. Can the mathematical induction be established in axiom of actual number? Or is it impossible?

ABC
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    This is unclear. What do you mean by "actual number range"? – Simply Beautiful Art Jun 12 '17 at 12:25
  • What is the 'actual number range'? – Stefan Mesken Jun 12 '17 at 12:25
  • I guess you mean real numbers. If you manage to define a well-ordering on them, you can, sort of. –  Jun 12 '17 at 12:26
  • @Lehs Without any concrete feedback, there is no reason to assume "actual number" means "real number" to the degree of editing the question. This is not merely a typo or lack of MathJax; it might genuinely be that they mean something else, and that something else was lost in translation (although I believe the probability to be so low that I wrote an answer anyways, with an assumption clause at the beginning). Thus I believe editing to be wrong, and I have rolled it back. – Arthur Jun 12 '17 at 12:30
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    Related (but probably not an answer for any likely interpretation of "actual number"): https://en.wikipedia.org/wiki/Transfinite_induction – Travis Willse Jun 12 '17 at 12:40

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When you say "actual numbers", do you mean real numbers? In that case, there are assumptions that would make it work, in theory (usually the Axiom of Choice in the form of the Well-ordering Principle, although the Continuum Hypothesis is a popular second chioce). However, even if you make such an assumption, there is no real arithmetical or algebraical connection between one number and the "next one" the way it is for natural numbers, so it's basically useless in practice.

Even induction on the rational numbers, where we can construct concrete inductable orderings (so-called "well-orderings") without any such assumptions, is very difficult to make work, since no well-ordering of the rationals plays nicely with addition or multiplication in any way.

Arthur
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I think you mean that whether we can use numbers instead of assuming true for n=k in the Inductive Hypothesis and proving n = k+1 for the Inductive Step.

The answer is yes we can BUT the purpose of mathematical induction is to prove for ALL numbers. So while we could show infinite numbers (which is neither sustainable nor a good math practice), it is better that we show it holds true for ALL numbers.

Doing so allows our proof to be more rigorous and account for any side cases. This is useful later in discrete maths where you can use the inductive hypothesis with more versatility.

mathy-girl
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There is a natural extension of induction to well ordered sets called transfinite induction(well ordered means any non-empty set has a least element). As mentioned before in order to ensure there even exists a well ordering on the index set one must assume an axiom such as the Axiom of Choice. In transfinite induction there are three cases, similar to the two cases of induction.

Zero Case: Prove the property holds for the least element $0$.

Successor Case: Prove if the property holds for an element $\alpha$ then it holds for the successor $\alpha +1$.

Limit Case: Prove if the property holds for all $\beta < \gamma$ then the property holds for $\gamma$.

Birch Bryant
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