Questions tagged [ordered-fields]

Ordered fields are fields which have an additional structure, a linear order compatible with the field structure. This tag is for questions regarding ordered fields and their properties, as well proofs related to un-orderability of certain fields.

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Does every partially ordered commutative ring admit an order preserving homomorphism to a real closed field?

I hope the answer is yes. It would suffice to construct a homomorphism to a formally real field. I think it would then suffice to extend the partial order to a total ring order, but I'm not sure if ...
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Characterzation of the complex numbers

There is a characterization of the real number system:complete ordered field. And the complete ordered field is unique up to isomorphism. I'm trying to charaterise the complex numbers in a similar way....
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Non-lattice ordered fields that are not totally ordered

For the purposes of this question, totally ordered fields are ordered fields in which every element is comparable under the ordering: $x \geq y$ or $y \geq x$. Thus in this terminology, in an ordered ...
David M's user avatar
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Different intervals in an ordered vector space over $\mathbb{Q}$

Consider an ordered vector space $V$ over $\mathbb{Q}$ in the usual language of ordered vector spaces $(<,0,-,+,\lambda\cdot)_{\lambda \in \mathbb{Q}}$. For the duration of this question, the word ...
Z. A. K.'s user avatar
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Is it possible to replace hyperreal numbers with "good enough" alternatives?

The hyperreal numbers are undoubtedly interesting, generalizable, and have many nice properties, but are they really needed to solve the problems they solve? Would other, smaller fields work too? ...
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What is the order of a function in distribution theory?

Let $\mathscr{D}(\mathcal{O})$ be the space of basic functions, $\phi$, which is equipped with convergence, and where $\phi$ vanish outside of the set $\mathcal{O}$. Let $\mathscr{D}'(\mathcal{O})$ be ...
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Ordering of $\mathbb{R}[X]$ such that $0 < X < a$ for all positive numbers $a$

If $\mathbb{R}[X]$ denotes the field of real valued rational functions, then according to Bochnak, Coste and Roy's Real Algebraic Geometry, the unique ordering for which $X > 0$ and $X < a$ for ...
Anu's user avatar
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Real-closed fields are the existentially closed ordered fields

While reading the proof in Shorter Hodges (Thm 7.4.4) that the theory of real-closed fields has quantifier-elimination, I came across the following claim (paraphased): "Let $A$ and $B$ be real-...
PoorPanda's user avatar
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Compatibility of formally real extensions of an ordered field

My question If $E$ is a formally real extension of an ordered field $F$, does $E$ always admit an ordering compatible with $F$? Less ambitious: What if $E$ is a formally real simple algebraic ...
Ningxin's user avatar
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Positive subset $\mathbb{R}_{>0}$ of $\mathbb{R}$ satisfying the order property

A field $\mathbb{F}$ is an ordered field if $\exists P\subseteq\mathbb{F}$ such that: 1. $\forall a,b\in P$, $a+b,ab\in P$. 2. Only one of the following is true $\forall a\in\mathbb{F}$, 1) $a\in P$ 2)...
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Adjoining greatest and smallest elements to uncountable ordered field

As titled. With $\mathbb{R}$, we can adjoin $+\infty$ and $-\infty$ as the greatest and smallest elements and define $\overline{\mathbb{R}}:=[-\infty, +\infty]$. But if we have an ordered field $\...
user760's user avatar
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Upper bounds for the set $\lbrace x\in {\mathbb Q}: x^2<2\rbrace$

Consider an ordered field $F$ containing the rational numbers $\mathbb Q$, without assuming the completeness axiom. In particular, this means that the integers may well be bounded. (see this nice ...
Math101's user avatar
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Proof of uniqueness of real-closure of an ordered field

I'm reading the proof of uniqueness of real-closure of an ordered field $F$, that is an algebraic extension $R$ of $F$ such that $R$ is real-closed and the unique order on $R$ extends that of $F$. I ...
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Theorems of calculus for real-closed fields [duplicate]

Theorems such as Intermediate value theorem Rolle's theorem Mean value theorem Positive/negative derivative implies strictly increasing/decreasing Hold for polynomials over real-closed fields. If ...
Jakobian's user avatar
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Strong homogeneity of order type of real closed fields

In Keisler's Infinitary Logic Book, he notes at one point that if we take a real closed field of cardinality $\kappa$, the ordered set is countably strongly homogeneous, in the sense we can find an ...
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Does there exist a Cauchy-complete, non-Archimedean ordered field that is not isomorphic to a field extension of $\mathbb{R}$?

Does there exist a metrizable non-Archimedean ordered field $\mathbb{F}$ that is Cauchy-complete under some metric $d$, where $\mathbb{F}$ is not isomorphic to any field extension of $\mathbb{R}$? I ...
MathNeophyte's user avatar
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Does every real-closed field satisfy the first-order least upper bound axiom schema?

The theory of real-closed fields is usually axiomatized by the axioms of ordered fields, along with an axiom for existence of square roots of positive elements, and also an axiom schema stating that ...
user107952's user avatar
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Can the field of real numbers be ordered in more than one way? [duplicate]

Consider the field $\mathbb R$ of real numbers under the usual addition and multiplication operations. Can this field be equipped with a total order $\prec$ which is not equal to the usual one $<$, ...
Joe's user avatar
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Why is the Positive cone (P) not equal to the ordered field (F)

Given that a positive cone is $$\forall x,y \in P, x+y \in P$$ $$\forall x,y \in P, xy \in P$$ $$\forall x \in F, \space \text{either} \space x \in P \space \text{or} -x \in P \space \text{or} \space ...
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Convexity for a non-Archimedean ordered field

I understand that for $\mathbb{R}$, a set $A\subseteq\mathbb{R}$ is convex iff for any two points $x,y\in A$, the point $kx+(1-k)y\in A$ for all $k\in[0,1]$. Here’s my question. Does the same ...
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First derivative test for polynomials over ordered fields

I am interested in the following generalization of the first derivative test for real functions: Let $K$ be an ordered field and $p\in K[X]$ a polynomial. Consider an interval $I\subseteq K$ on which ...
Tim Seifert's user avatar
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Confusion about subtleties in the proof of "negative times positive is negative" and the like?

I am answering exercise $1.2.4:$ I'm letting the other problems in here just for reference of what was already done so that it's clear what we can use. The answer to $1.2.4$ is: I am a bit confused: ...
Red Banana's user avatar
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Visualize the Completion of (the Ordered Field) of Rational Functions

Every ordered abelian group $G$ can be completed to give a larger ordered abelian group $\bar{G}$. The original abelian group $G$ embeds into $\bar{G}$ as a dense subset, and every non-empty subset of ...
Theone's user avatar
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Prove $\beta$ = inf E iff for every $\epsilon>0$ there exists a $y \in E$ such that $y<\beta + \epsilon$

Here is my question. $S$ is an ordered field and $E \subseteq S$ is bounded below, and $\beta$ is a lower bound of E. Prove $\beta = \inf E\ $ iff for every $\epsilon>0$ there exists a $y \in E$ ...
user13121312's user avatar
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Less then or Equal to Relation on Ordered Fields

If we consider the natural numbers, we have for all $a,b\in\mathbb{N}$ that $a\leq b\Longleftrightarrow (\exists x\in\mathbb{N})[a+x=b]$. (For example if $a=2$ and $b=5$ there exists $3\in\mathbb{N}$ ...
PAT's user avatar
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Is the square root axiom redundant in the standard set of axioms for real closed fields?

Real closed fields, that is, ordered fields $(F;+,-,*,0,1,<)$ which are elementarily equivalent to the ordered field of the real numbers, can be axiomatized by the standard axioms for ordered ...
user107952's user avatar
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"Multiplicative" Archimedean property in ordered fields

There following are two equivalent formulations of the Archimedean property of an ordered field $F$: $\mathbb N$ is unbounded in $F$. (Formulation for the ordered group $(F, +)$). For any $x, \...
Atom's user avatar
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Is there an ordered field whose natural/integer/rational part is definable?

Does there exist an ordered field $(F;+,-,*,0,1,<)$, which is not $\mathbb{Q}$, such that its natural part and/or its integer part and/or its rational part is definable without parameters? By ...
user107952's user avatar
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Reference Request for Axiomatic/Algebraic Big $\mathcal{O}$ and Little $o$

I have seen the formal definitions of big $\mathcal{O}$ and little $o$, and do all right working with them. Still, I have some questions that a good reference might help clear up. In what level of ...
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Prove $a^{n+1} < a$ if and only if $a < 1$ for all $a > 0$ in any ordered field

Prove $a^{n+1} < a$ if and only if $a < 1$ for all $a > 0$ in any ordered field. Proof: We first prove that for $a, b > 0$, then $ab < b$ if and only if $a < 1$. If $a > 1$, then $...
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Prove $b > 0 \implies b^3 + 3b > 3b^2$ in any ordered field

Prove $b > 0 \implies b^3 + 3b > 3b^2$ in any ordered field. Prove this algebraically (i.e. without continuity or IVT). Using analysis, this is easy. Is there a way to prove it without ...
SRobertJames's user avatar
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Let $R$ be a ring. Is there at most one ordering $\leq$ such that $(R,\leq)$ is an ordered ring?

I was wondering whether it is equivalent to consider an ordered ring or a ring that can be ordered. I assume that there are plenty of people who already thought about this issue, hence my question...I ...
Filippo's user avatar
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Does every continuous injection of real numbers preserve or reverse order?

Let $f:(a,b)\to\mathbb R$ be a continuous injection. Must $f$ be order-preserving ($x<y\Rightarrow f(x)<f(y)$) or order-reversing ($x<y\Rightarrow f(x)>f(y)$)?
Steven Clontz's user avatar
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1 answer
127 views

What do you get if you perform the Dedekind Cuts procedure on $\mathbb Q(x)$?

Let $\mathbb Q(x)$ denote, as usual, the field of rational functions with rational coefficients. Any element of $\mathbb Q(x)$ can be written in the form $$f=\frac{a_n x^n + \cdots +a_0}{b_m x^m + \...
mweiss's user avatar
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Classification of "complete dense ordered near-semirings"

Let's define a complete dense ordered near-semiring, or a CDON, as a set $A$ equipped with two binary operations $+,\times$ and a binary relation $\leq$ such that: $+$ is a monoid, whose identity is ...
zxcv's user avatar
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In an ordered field, $a< ar + (1-r)b< b$ whenever $a<b$ and $0<r<1$

I am stuck on problem 9 from Berberian- A First Course in Real Analysis which deals with inequalites in an ordered field. Problem: Show that in an ordered field, $a< ar + (1-r)b< b$ whenever $a&...
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Is there a maximal ordered field? What about $\mathbb R$?

In my real analysis class today, we revisited the fact that there is only one complete ordered field (up to isomorphism): $\mathbb R$. But this induced several thoughts I began wondering about. I ...
Atom's user avatar
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5 votes
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How can a field be Cauchy complete and non Archimedean

The Wikipedia page for the completeness of the Real numbers, says that “ there are non Archimedean fields that are ordered and Cauchy complete.” However, in many other places, I’ve read that non ...
Lave Cave's user avatar
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2 votes
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Cardinality of integer parts of real closed fields

Every real closed field $R$ has in integer part $I$. That is, $I$ is a discrete ordered subring of $R$ such that for each $x \in R$ there is $z \in I$ such that $z \leq x < z + 1$. If $R$ is ...
Alex's user avatar
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3 votes
1 answer
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Product of infinitesimal by inverse of non infinitesimal is infinitesimal

This is probably very simple, though I am getting confused because of my inexperience with infinitesimals. Let $K$ be an ordered real-closed field. We say that $a \in K$ is finite if it's absolute ...
user480840's user avatar
5 votes
1 answer
116 views

Convex subring is local

Let $K$ be an ordered field, $R\subseteq K$ a convex subring - that is, for all $a \in K$, $a \in R$ if $x \leq a \leq y$ for some $x,y \in R$. Define $I = \{ x \in R: x^{-1} \notin R\}$. It is clear ...
user480840's user avatar
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Can the hyperreal numbers have a property akin to completeness by considering hypernatural sequences?

I’ve seen that $\mathbb{R}^*$ isn’t complete as many Cauchy sequences won’t converge, and that includes power series. In other stack exchange posts, I’ve seen that even the exponential function won’t ...
Lave Cave's user avatar
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Whether some positive sequence approaches 0 in any ordered field

Consider an ordered field $(\mathbb{F}, +, \times, <)$ with additive identity $0$ and multiplicative identity $1$. It is not hard to show that $1\ne 0$ and then $1=1\times 1>0$. I am aware that ...
Tony Ma's user avatar
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Can the surreal numbers be completed to form an ordered field?

The surreal number line isn’t Cauchy complete as it’s filled with “gaps”. When constructing the real numbers from the rationals, one could take the ring of all Cauchy sequences and take the quotient ...
Lave Cave's user avatar
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2 votes
2 answers
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Can the hyper hyper real numbers be constructed?

The hyperreal numbers can be constructed as $\mathbb{R}^{\mathbb{N}}/U$ given some ultra filter and this allows first-order statements to be transferred over to $\mathbb{R}^*$. Can this be done again ...
Lave Cave's user avatar
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2 votes
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Let $(F,+,\cdot,0_F,1_F,\le)$ be an ordered field. If the set $\{x\in F: 0_F\le x\le 1_F\}$ is complete, is the whole field complete?

Let $(F,+,\cdot,0_F,1_F,\le)$ be an ordered field. If the set $\{x\in F: 0_F\le x\le 1_F\}$ is complete in the sense that the least upper bound property is satisfied, is the whole field complete, in ...
phst's user avatar
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Question about axioms of an ordered field.

I’m currently studying Introduction to Analysis by Ross. I wanted to ask if the either - or is an inclusive or exclusive disjunction in property O1 below. I believe this should just expressing the ...
punk4me's user avatar
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2 answers
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$F(\sqrt d)$ is an ordered field

I'm trying to solve this exercise: (hartshorne Euclid and beyond ex 15.3) Let $F$ be an ordered field, let $d>0$, and suppose that $d$ does not have a square root in $F$. Let $F(\sqrt d)$ denote ...
izzy's user avatar
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Is this way of writing the Field and Order Axioms correct?

I am currently self-studying math and am trying to write a Mathematics Cheat Sheet document that among other things includes the Field and Order Axioms. I want to write them as succinct as possible (...
Farrel Ahmed's user avatar
22 votes
1 answer
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Prove all 4 axioms of "less than" are necessary (for real numbers)

One way to define an ordered field is as a field $F$ with a relation $<$ that satisfies: For all $x,y \in F$, exactly one of $x<y$, $x=y$, $y<x$ holds. For all $x,y,z \in F$, if $x<y$ and ...
Misha Lavrov's user avatar

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