I have the book the art of proof by Beck and Geohgegan. Proposition 2.21 is that there is no integer x such that $$0<x<1$$ Is there any proof using only the axioms of order of Z and algebraic properties of Z, without using the well ordering principle?
Asked
Active
Viewed 103 times
-1
-
An ordered ring $R$ whose positives $P$ are well-ordered is isomorphic to $\Bbb Z$ as an ordered ring, cf. easy proof. $\ \ $ – Bill Dubuque Aug 05 '24 at 20:55
-
This question is similar to: Axiomatization of $\mathbb{Z}$ via well-ordering of positives.. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Rob Arthan Aug 05 '24 at 21:35
-
How to prove something that is obvious depends a lot on how you define things. So what exactly are your definitions/axioms for "integer", "0", "1", and "<"? – Karl Aug 05 '24 at 21:51
1 Answers
1
There cannot be such a proof because other rings like $\mathbb{Q}$ satisfy the same algebraic and order axioms, but do have an element $x$ with $0 < x < 1$. The well ordering principle is what distinguishes $\mathbb{Z}$ from these other rings.
Ted
- 35,732
-
I'm a highschool student so sadly I don't understand these rings :") Can we say that $\log(t)≥0 \forall t\in\mathbb{Z^+}$? Because then if $0<x<1$, then $\log(x)<0$ and so $x$ can't be an integer. This is a naive proof, and I'm sure it's wrong or uses cyclic reasoning somehow. But can you help me find out my mistake. Thanks in advance – Gwen Aug 05 '24 at 20:47
-
That its positives are well-ordered characterizes $\Bbb Z$ among ordered rings, cf. my comment on question. – Bill Dubuque Aug 05 '24 at 21:10