Well-Ordering

Quote:

Originally Posted by GoldendoodleMom

Prove: If y>0 then there exists a natural number, n, such that n-1 is less than or equal to y < n.
I'm supposed to prove using the well-ordering property on the set {m as an element of the natural numbers : m > y}

To use that, you must know that $\mathbb{N}$ is not bounded above.
Because that is true, $T = \left\{ {m \in \mathbb{N}:y < m} \right\} \ne \emptyset$ .
Every nonempty subset of $\mathbb{N}$ has a first term.
Let $n = \min \left( T \right) \Rightarrow n \in T \Rightarrow y < n.$
Moreover, because $n - 1 < n\quad \Rightarrow \quad n - 1 \notin T\quad \Rightarrow \quad n - 1 \leqslant y < n$ .