Here is what we have got to work with:
A subset $E$ of the set $\mathbb{R}$ of real numbers is said to be inductive if $1 \in E$ and if for every $x \in E$ the number $x+1 \in E$ also.
The intersection of all the inductive sets is the set $\mathbb{N}$ of natural numbers, which is also inductive.
Moreover, since the sets $\{ x \in \mathbb{R} | x > 0 \}$ and $\{ x \in \mathbb{R} | x \geq 1 \}$ are inductive, the set $\mathbb{N}$ consists of only positive real numbers $\geq 1$.
Last but not least, we of course have the field, the order, and the completeness axioms of the real number system at our disposal.
Using the above machinery, how can we prove (using induction) the following facts?
If $n > 1$ is a natural number, then $n-1$ is also a natural number.
If $n$ and $m$ are natural numbers with $n > m$, then $n-m$ is also a natural number.
For each $n\in\Bbb N$, let $P(n)$ be the assertion “$n=1$ or $n-1\in\Bbb N$”. Clearly, $P(1)$ holds. Now let $n\in\Bbb N$ and suppose that $n=1$ or that $n-1\in\Bbb N$. Then $(n+1)-1=n\in\Bbb N$, and therefore $P(n+1)$ holds.