Proving that if $m \in \mathbb{Z}$, then $\left]m,m+1\right[ = \emptyset$ in $\mathbb{Z}$.

Asked Apr 08 '20 at 16:31

Active Apr 08 '20 at 16:35

Viewed 29 times

Using the fact that $\mathbb{Z}$ is an ordered ring.

Is this the same as proving that $n>m \Leftrightarrow n \geq m+1$ (again, in $\mathbb{Z}$)? I take it that it is, but, I also have trouble proving that.

I'd appreciate any sort of guidance or hints.

edited Apr 08 '20 at 16:35

Michael Hardy

asked Apr 08 '20 at 16:31

Rodrigo

1

Do you have available the fact that $\Bbb Z$ is a discrete ordered ring, i.e., that there is nothing strictly between $0$ and $1$? – Brian M. Scott Apr 08 '20 at 16:37
You need more than just an ordered ring structure. $\mathbb{Q}$ is an ordered ring but it does not have this property. – Cameron Williams Apr 08 '20 at 16:42
@BrianM.Scott, yes, I have that available. I know of a proof for that fact. How would one derive this proof from that? – Rodrigo Apr 08 '20 at 16:44
@Rodrigo: Suppose that $m<x<m+1$ and add $-m$ to everything. – Brian M. Scott Apr 08 '20 at 16:45
@Rodrigo You could do it inductively. Show no element exists between $1$ and $2$ (by subtracting $1$..) and so on. – Cameron Williams Apr 08 '20 at 16:45
@Rodrigo the point was more that you need an extra assumption to make it work since the order+ring properties are not enough, but Brian got it covered and you confirmed. – Cameron Williams Apr 08 '20 at 16:45
Yes, it was quite easy. Thanks a bunch, Brian and Cameron! – Rodrigo Apr 08 '20 at 16:46
@Rodrigo: You’re welcome! – Brian M. Scott Apr 08 '20 at 16:47
@Rodrigo Feel free to answer your own question so this post doesn't go unanswered and get bumped by Community for no reason :) – Cameron Williams Apr 08 '20 at 16:57
@Cameron It's part of a characterization of $,\Bbb Z,,$ viz. any ordered ring R whose positives P are well-ordered in R is isomorphic to $\Bbb Z$ as an ordered ring. See here for an easy proof. – Bill Dubuque Apr 08 '20 at 20:46
@BillDubuque I'm aware. It's more that OP couldn't prove it since they were clearly overlooking the final piece of the puzzle. – Cameron Williams Apr 08 '20 at 20:50

Proving that if $m \in \mathbb{Z}$, then $\left]m,m+1\right[ = \emptyset$ in $\mathbb{Z}$.

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