# Strange Proof: The Principle of Mathematical Induction implies that 1 is the least positive natural number.

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 $$n=n+0. Therefore, $$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?

• 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. – Maximilian Janisch May 20 at 19:51
• 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 $n\geq 1.$ – Thomas Andrews May 20 at 19:53
• 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 $m.$ – Thomas Andrews May 20 at 19:55

• $$1\in S$$ 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 $$\Bbb N$$.)
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 $$n\in\mathbb{N}$$.