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}

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- Oct 15th 2008, 07:33 AMGoldendoodleMomWell-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} - Oct 15th 2008, 08:13 AMPlato
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 $\displaystyle n - 1 < n\quad \Rightarrow \quad n - 1 \notin T\quad \Rightarrow \quad n - 1 \leqslant y < n$.