Suppose that $R$ is a partially ordered commutative ring. The order behaves nicely: We have $0 \leq 1$. If $a \leq b$ and $c > 0$, then $a + c \leq b + c$ and $a \cdot c \leq b \cdot c$. For $n \in \mathbb{Z}$ define $\hat{n} = 1 + \cdots + 1$(n terms), where $1$ is the ring's multiplicative identity. The property of the ring is:

For all $r \in R$ such that $r > 0$ is an $n \in \mathbb{Z}$ such that either $1 < \hat{n} \cdot r$ or $\hat{n} \cdot r \nleq 1$.

The first case is like the Archimedean property. The second case says there are none of the usual infinitesimals.

If this property has a commonly used name, I would like to know it. Any references would be appreciated.

  • $\begingroup$ I am not sure what Exactly you want here. When we have either "$(1 \lt \hat{n} \cdot r)$" or "$(\hat{n} \cdot r \not \leq 1)\equiv(1 \lt \hat{n} \cdot r)$" , we have the Same Equivalent Statement. $\endgroup$
    – Prem
    May 10, 2023 at 18:02
  • $\begingroup$ The order is a partial order not a total order. $\endgroup$
    – Jay
    May 10, 2023 at 18:36
  • $\begingroup$ One Statement says , we can multiply a number such that we get a number larger than 1. Other Statement says , we can multiply a number such that we get a number which is not less than 1 , which means it is larger than 1. Seems Equivalent , though I might be wrong. Thinking ! $\endgroup$
    – Prem
    May 10, 2023 at 18:50
  • $\begingroup$ @Prem In a partial order items can be incomparable. $\endgroup$
    – Jay
    May 10, 2023 at 18:53
  • $\begingroup$ I realize that , but I see "$\le$" , "$\gt$" , "$\not \le$" , where it is not clear which is the Partial Order. In Case that is a valid objection (I am not sure) , maybe you could edit it such that you use only that Partial Order through-out. In Case that is not a valid objection , you can leave it unchanged & maybe somebody will Post some Answer ! $\endgroup$
    – Prem
    May 10, 2023 at 19:16


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