# Circular reasoning in the hint of proving that if $n$ is a natural number greater than $1$, then $n-1$ is natural number

Problem Show that if $n$ is a natural number greater than 1, then $n-1$ is a natural number.

In Advanced Calculus 2nd edition by Fitzpatrick the author defines the set of natural numbers as the intersection of all inductive sets.

For this problem, he provides the following hint: Show that the set $\{n \, : \, n = 1 \text{ or } n \text{ is a natural number and } n-1 \text{ is a natural number}\}$ is inductive.

Question: Using the hint, it was easy to construct a proof. However, I'm wondering if the hint is circular. If we want to show that $n-1$ is a natural number, why is it okay to define a set with the property that $n-1$ is a natural number? It seems that we are assuming our conclusion to be true in order to demonstrate that the conclusion is true.

Thanks.

Here is the definition of an inductive set: $S$ is an inductive set if

1. $1 \in S$ and
2. If $x \in S$, then $x+1 \in S$.
• "Show that the set {} ..." ... is what? Inductive? – DonAntonio Jun 2 '18 at 15:23
• @DonAntonio Yes. Will update the question now. – Benedict Voltaire Jun 2 '18 at 15:25
• this is strange... in ZFC set theory any set is a subset, so your definition of set $\{x:...\}$ is not correct, it must be $\{x\in A:...\}$ for some set $A$ to make sense. – Masacroso Jun 2 '18 at 15:30
• The general $n$ in the hint is not necessarily the same as the specific $n$ in the problem. Put $m$ instead of $n$ throughout the hint and see this as simply the definition of a set which will be used later to address the problem. – Mark Bennet Jun 2 '18 at 15:30
• @MarkBennet Thanks, Mark. That thought crossed my mind, so I'm glad someone made that explicit. – Benedict Voltaire Jun 2 '18 at 15:34

## 1 Answer

Well, then it is a rather clear and pretty strong hint: if you prove that set that is given there is inductive, then as the natural numbers has been defined to you as the intersection of all inductive sets, then the set of natural numbers is contained in that set. But...(end the argument).

Nothing circular here.