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Some textbooks say that proof by mathematical induction is applicable when $n$ is a positive integer, while others say $n\in\mathbb{N}$. Which one of ‘em is correct tho? (since set of positive integers doesn't include 0)

HERO
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  • an argument of induction is applicable for any subset of the integers, as long as it has a smallest element. – Donlans Donlans Sep 18 '19 at 16:03
  • Proof by mathematical induction is applicable in any countable context, including all integers, even the rational numbers, and so on. You just need a clear way to go from the "$n$'th term in the sequence" to the "$n+1$'st term" and have the validity of the statement for the $n$'th term imply the validity of the statement for the $n+1$'st term. – JMoravitz Sep 18 '19 at 16:04
  • @JMoravitz That's not quite right. You do need to be able to get to any particular element from the basis case by finitely many operations, or else you need transfinite induction. So I suppose the precise terminology should be "order-isomorphic to the natural numbers" or similar. – Chappers Sep 18 '19 at 16:09
  • @Chappers Umm... would you kindly simplify what you just said? – HERO Sep 18 '19 at 16:49
  • @AmandaMacaurenni The properties of the natural numbers that are used in induction are that 1.) there is a smallest one (Let's say it's 0 in the naturals), 2.) each element has a successor (n has successor s(n)=n+1 for the naturals), 3.) each element that is not 0 is the successor of another and 4.) each natural is $s(s(\dots s(0)))$ for finitely many applications of the successor function. The first three are used to construct the induction (true for 0, true for n implies true for n+1), the fourth is required because our proof of the nth case should only be finitely many lines long. [...] – Chappers Sep 22 '19 at 12:57
  • But, there are structures that start off looking like the natural numbers, in that they have a smallest element, and each element is reached as the successor of the successor of ... the successor of the basis case (with finitely many successor operations), but they have "infinite" elements that you can't get to by applying the successor operation finitely many times ("ordinal numbers"). Ordinary induction doesn't work on these, so one formulates a new type of induction ("transfinite induction") to avoid the problem of not being able to get to the infinite elements with finitely many operations – Chappers Sep 22 '19 at 13:00
  • As for "order-isomorphic", an order on a set is a generalisation of "<". An isomorphism is a one-to-one function between two objects that maintains the structure on one in the new structure on the other: here, it means that $x<y \implies f(x) < f(y)$ for every $x,y$. Isomorphisms allow us to categorise objects that "look the same". So an object is "order-isomorphic to the naturals" means that there is a one-to-one mapping from the naturals to this object where the order on the new object "looks the same" as the ordering on the naturals.[...] – Chappers Sep 22 '19 at 13:06
  • For example, the map $n \mapsto n-k$ is an order isomorphism on the integers, so ${ -k,-k+1,\dotsc }$ is order-isomorphic to the natural numbers. If you want to know more about this, ask a new question. – Chappers Sep 22 '19 at 13:06

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As far as the general structure of an arbitrary induction proof goes, the only thing you're doing with the different $n$ is comparing them (you're not adding or multiplying them or something like that). And in that respect, the two sets (natural numbers with $0$ or without $0$) are essentially indistinguishable. You can use whichever suits your purposes best.

Also note that it is not universally assumed that $0\in \Bbb N$. I'm not even consistent myself.

Arthur
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