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I was recently asked to help someone with the following question on their first year analysis course.

Recall that $\mathbb{N}$ is the set of all positive integers. Use the principle of induction to show that $n \ge 1$ for all $ n \in \mathbb{N}.$ [ Hint: Let $ S = \lbrace n \in \mathbb{N} | n \ge 1 \rbrace.$ ]

I am concerned that the first sentence of the question assumes that we already know exactly what the positive integers are, which renders the rest of the question redundant.

I would prefer to see the question written:

Recall that $\mathbb{N}$ is the set of $ x \in \mathbb{R}$ such that $x$ is a member of every inductive set. [ $P$ is an inductive set if (a) $ 1 \in P $ and (b) $ x \in P \Rightarrow x+1 \in P.$ ] Then we can answer as follows:

With $S$ as in the hint, by definition $ S \subseteq \mathbb{N}.$

$1 \in S$ since $1 \in \mathbb{N},$ as it is an inductive set. For $ n \in S, $ $n+1 > n \ge 1,$ therefore $n+1 \in S.$ Hence $S$ is an inductive set, and so $ \mathbb{N} \subseteq S.$ Therefore $ S=\mathbb{N}.$ Therefore $1$ is the least element of $\mathbb{N}.$

Am I being way too picky, or is the question flawed as it stands?

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    "Principle." Whether we know what the natural numbers "are" doesn't necessarily mean we know that they have this property. It all depends on what definitions they started out with and, moreover, it doesn't really matter. (Also, R? What?!)2010-10-29
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    @Qiaochu That's exactly my point; the definition that we start with. Without that, things look kind of circular to me.2010-10-29
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    If the lecturer previously defined $\mathbf{N}$ as the intersection of all inductive subsets of $\mathbf{R}$ then this is perfectly fine. (If I recall correctly, this is what Munkres does in his celebrated textbook on topology.)2010-10-29
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    @Robin Chapman Yes, this is in line with what I was thinking but I was slightly sceptical over how $\mathbb{N}$ was defined (I don't have this information) given that it came up just a few weeks into a first year course. I guess I shouldn't jump to conclusions.2010-10-29

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I agree that if we define $\mathbb{N}$ as a "set of positive integers", the question is redundant. However if we say that $\mathbb{N}$ is a set of natural numbers (and, after all, that is exactly what $\mathbb{N}$ stands for), defined by Peano axioms, then the question is no longer redundant.

As a side note, calling $\mathbb{N}$ a "set of positive integers" is, IMHO, a bad style. Unfortunately, there is no standard convention for "set of positive integers". Many books use $\mathbb{Z}^{+}$ for that, however at least one book I'm familiar with uses $\mathbb{Z}^{+}$ for "group of integers under the regular addition". May be $\mathbb{Z}_{+}$ is better (?)

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I interpret the question as asking for a proof (by induction) that there are no integers strictly between zero and one. I agree with those who point out that how one proves this depends on how one has defined the integers. I remember the question being on a problem set for the Ross program (a summer Math program for high school students, held at Ohio State University) circa 1967. I don't remember the definition of integer used in the program, but I think I remember the solution going something like this: if there are any integers between 0 and 1 then by well-ordering there's a smallest one, call it $p$, then $p^2$ is a smaller positive integer between 0 and 1, contradiction, hence there is no integer between 0 and 1.

OK, so it uses well-ordering instead of induction.

Within the appropriate logical framework, one can go the other way, and use the non-existence of integers between zero and one to establish the correctness of Mathematical Induction (or of well-ordering).

The non-existence of integers between zero and one is the critical observation in many irrationality and transcendence proofs. Indeed, in Burger and Tubbs, Making Transcendence Transparent (page 9), it is referred to as The Fundamental Principle of Number Theory.