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I have just begun using Judson's 2018 Abstract Algebra: theory and applications. In the text, there is a Lemma with the following statement and proof:

The Principle of Mathematical Induction implies that 1 is the least positive natural number.

Proof. Let $S=\{n \in \mathbb{N} | n \ge 1\}$. Then $1 \in S$. Assume that $n \in S$. Since $0<1$, it must be the case that $n=n+0<n+1$. Therefore, $1 \le n < n+1$. Consequently, if $n \in S$, then $n+1$ must also be in $S$, and by the Principle of Mathematical Induction, and $S=\mathbb{N}$. QED.

I am having some issues with this proof.

  1. The last sentence doesn't seem coherent.
  2. S is defined as the set of natural numbers $\ge1$, but I don't understand how the proof shows that $S = \mathbb{N}$ to arrive at the conclusion that 1 is the least positive natural number.

Prior to this Lemma we are given the definition of natural numbers $\mathbb{N}=\{1,2,3,...\}$, and as propositions the First and Second Principle of Mathematical Induction, as well as the Principle of Well-Ordering.

Can anyone either help me understand why this proof is correct or otherwise help me fill in what may be missing?

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    $\begingroup$ The last sentence should be something like “... and by the Principle of Mathematical Induction, we have $S=\Bbb N$. Otherwise, the proof is completely fine. $\endgroup$ – Maximilian Janisch May 20 at 19:51
  • $\begingroup$ Not sure why (1) is incoherent. In induction starting at $0$ (and hence when $0$ is considered a natural number) this same proof shows that for all natural $n,$ you either have $n=0$ or $n\geq 1.$ $\endgroup$ – Thomas Andrews May 20 at 19:53
  • $\begingroup$ It is true that you don't actually need to property $n\in S$ to prove $n+1\in S$ in this case. A simpler lemma is (now assuming the natural numbers/induction start at $1$): For all $n,$ either $n=1$ or $n=m+1$ for some natural number $m.$ $\endgroup$ – Thomas Andrews May 20 at 19:55
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The proof showed that

  • $1\in S$ and
  • for every $n\in S$, we have $n+1\in S$.

By using Peano axiomatic (aka the principle of induction), the proof correctly concludes that $S=\Bbb N$. Thus every natural number is $\geq 1$. Hence, by definition, $1$ is a smallest number of $\Bbb N$ (it is in fact also the only smallest number of $\Bbb N$.)

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You might think that since $\mathbb{N}$ is defined to be $\{1,2,\dots\}$ this proof is pointless. Although Judson doesn't make this clear, the point of this proof is that it gives us an alternative more formal way of describing the properties of $\mathbb{N}$ without using the informal ellipsis; if we take an appropriate version of induction as an axiom then we don't need to define $\mathbb{N}$ to be $\{1,2,\dots\}$.

Unfortunately, the form of induction stated by Judson does not give a correct proof of this theorem. Judson's version only proves that $n\ge1$ for all $n\ge1$. The version of induction that's needed for this theorem is something like the following:

Let $S(n$) be a statement about integers for $n \in \mathbb{N}$ and suppose $S(1)$ is true. If for all integers $k$ with $k \ge1$, $S(k)$ implies that $S(k + 1)$ is true, then $S(n)$ is true for all integers $n\in\mathbb{N}$.

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