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As the title says. I encounter this problem in Bernd Schroeder's book of "Mathematical Analysis: A Concise Introduction", p.15. It essentially characterizes natural number from the axioms regarding real number, i.e. the axioms of addition and multiplication of $\mathbb{R}$ as a field. My attempt is the following using Principle of Induction.

Let $S:=\{n \in \mathbb{N}: n \ge 1\}$. Check if $ 1 \in S$. Since $1 \ge 1$ holds, therefore $1 \in S$. Suppose $n \in S$, we want to show $n+1 \in S$. $n+1 \ge 1 \Rightarrow n \ge 0$.However from the induction step, we have $ n \ge 1 > 0$, therefore $n+1 \in S$. By Principle of Induction, $S=\mathbb{N}$ $\Box$

EDIT1: There is a mistake in the proof as pointed out by talmid.

EDIT2: The author of the above mentioned book constructs natural number, integers and rationals from Completeness Axiom. A similar approach can be found in Royden's Real Analysis

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    $0\not\in\Bbb N$ ?2012-06-21
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    @Lierre This is a common convention, I think2012-06-21
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    @Lierre Neither is really more common, from what I've seen. Essentially, $0$ is the first cardinal number, while $1$ is the first ordinal number. (Nobody comes in zero place in a race...) Depends really on the definition used by the book/class.2012-06-21
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    @ThomasAndrews — Good to know, in France zero is always in $\Bbb N$.2012-06-21
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    how do you define $x\geq y$?2012-06-21
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    @Lierre yes $\mathbb{N}$ in the book is defined by ${1,2,3,...}$ unless it is said to union with $0$.2012-06-21
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    @tomasz it is defined as $x>y$ or $x=y$2012-06-21
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    @jsk: I think what tomasz meant was, how do you define $x>y$?2012-06-21
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    @EricStucky: In that case then $x>y$ iff $x-y \in \mathbb{R}^{+}$, the positive reals2012-06-21
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    Given that you don't even have elementary properties of the natural numbers, it seems unreasonable to expect that you have the machinery of $\mathbb{R^+}$. A fairly standard definition is that $x\geq y$ if there is a natural $n$ such that $x=y+n$; do you have any reason to believe you must use something else? (for example, if you have not defined +, if the order is axiomized, etc.)2012-06-21
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    Small correction: since your $\mathbb{N}$ does not contain 0; that is actually a definition of $\gt$, not $\geq$.2012-06-21
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    @EricStucky In the book as I mentioned in the post, natural number is characterized from the reals using all successor sets in $\mathbb{R}$, where the successor set $A \subset \mathbb{R}$ is defined as $1 \in A$, and if $a\in A$, then $a+1 \in A$ which $\mathbb{R}$ is an example of a successor set. I am actually pretty confused too when I read this approach in the book.2012-06-21

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I think the inductive step is a bit fuzzy, because you assume $n + 1\geq 1$ (which is what you want to prove - fallacious), deduce that $n \geq 0$, and since this holds then $n+1 \geq 1$, which would be affirming the consequent (i.e., $(P \rightarrow Q) \wedge Q \vdash P$) - also fallacious. Maybe your $\Rightarrow$ is backwards.

Also, I would add: you seem to have tried to arrive at a contradiction between $n \geq 0$ and $n \geq 1$ ("$n \geq 0$. However [...] $n \geq 1$"), but you can perfectly have an $n$ greater than $0$ and greater than $1$ at the same time. In these cases, try to convince yourself, formally and informally, that you arrived at a contradiction, for example, by trying to come up with counterexamples. I think you'll find it very easy to find an $n$ greater than both $0$ and $1$.

You could say that $n\geq 1$ (inductive hypothesis), and then $n+1 \geq n \geq 1$, so $n + 1\geq 1$ (this is assuming that you know that $n + 1 \geq n$, i.e., $1 \geq 0$).

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    Oh yes thanks for pointing the mistake out2012-06-21
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    I am using the induction step to show that $n+1 \in S$ which I am really not sure if it is a valid step since I am supposing $n$ is true.2012-06-21
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    @jsk You can definitely assume $n \geq 1$, since that's the inductive step, and to show that $n + 1\geq 1$, you could add $1$ to both sides of $n \geq 1$, and get $n + 1 \geq 2 \geq 1$, so $n+1 \geq 1$.2012-06-21