Use the WOP of N to prove that there is no $\displaystyle n \in N $ such that 0<n<1

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- Aug 18th 2009, 11:50 AMflower3well ordering principle
Use the WOP of N to prove that there is no $\displaystyle n \in N $ such that 0<n<1

- Aug 18th 2009, 12:59 PMputnam120
Well if there were then since $\displaystyle n\in\mathbb{N}$ and $\displaystyle n>0$, it follows that $\displaystyle (n-1)\in\mathbb{N}$. Now you have that $\displaystyle (n-1)\neq{0}$, so continue in this way to get a contradiction.

- Aug 18th 2009, 01:37 PMPlato
I am a bit puzzled by the wording of this question.

I my experience the following is a standard theorem: $\displaystyle \left( {\forall x \in \mathbb{R}} \right)\left( {\exists !n \in \mathbb{Z}} \right)\left[ {n \leqslant x < n + 1} \right]$.

The unique integer is of course the floor of $\displaystyle x$.

I see what putnam120 is trying to do. But I an not at all sure where W.O. be used.

The theorem I quoted above with $\displaystyle x=0$ give the result at once.

PS. Of course, this is all mute if your course includes a rigorous construction of the natural numbers. - Aug 18th 2009, 02:18 PMputnam120
W.O. states that every set of non-negative integers has a least element. Thus since $\displaystyle \mathbb{N}$ only contains non-negative integers it to must have a minimal element. So using the construction I started you show that the set doesn't have a minimal element.

I usually see $\displaystyle \mathbb{N}$ defined as follows:

1) $\displaystyle 0\in\mathbb{N}$

2) $\displaystyle (n\neq{0})\in\mathbb{N}\Longrightarrow (n\pm{1})\in\mathbb{N}$

3) $\displaystyle (n=0)\in\mathbb{N}\Longrightarrow (n+1)\in\mathbb{N}$