I'm really sorry if this question has been asked before, I looked but couldn't find anything.
The least number principle states that every non-empty set of positive integers contains a least element.
I need to find some examples of theorems, involving natural numbers, that are consequences of the least number principle.
One example is given by the Dickson lemma: we have an infinite sequence $l_0,l_1,l_2, \ldots$ of natural numbers and we want to show that exist two indexes $i$ and $j$ such that $i<j$ and $l_i\leq l_j$.
The proof is: let $i^*$ be the minimum of the sequence that exists by the least number principle. Then define $i=i^*$ and $j=i^*+1$.
Thanks!
The principle of induction is a consequence of the least number principle and hence everything, that you can proof via induction. Here is why:
Suppose $N\subseteq \mathbb{N}, 1\in N$ and $\forall n\in N : n+1\in N$. Let $M = \mathbb{N}\setminus N$. Suppose $M$ is non-empty. Then it has a least element $m$. Then $m-1\notin M$, so $m-1\in N$. But then $m\in N$. (Contradiction). Hence $M$ is empty, therefore $N = \mathbb{N}$.
Similarly, you can prove statements normally proved by induction. Let's show, that every positive integer greater than $1$ has a prime divisor: Let $n\in \mathbb{N}$ be greater then $1$. Then $n$ has a divisor greater than $1$ (namely $n$), hence it has a least divisor $p$ greater then $1$. If $p$ is not prime, then it has a divisor $t$ with $1<t<p$. But in that case $t$ is divisor of $n$, greater then $1$ with $t<p$ (Contradiction, since $p$ was minimal with that property). Therefore $p$ is prime and $n$ has a prime divisor.