Thread: Well-Ordering

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}

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