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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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Prove that $f^{-1}\mathbb{R}^+f\subsetneqq Aut(\mathbb{I})$!

Let $\mathbb{I}=([0,1],\leq)$ and suppose $Aut(\mathbb{I})=\{f|f:\mathbb{I}\longrightarrow \mathbb{I}, \text{$f$ is 1-1 and onto and $x\leq$ y iff $f(x)\leq f(y)$}\}$. For any $f\in Aut(\mathbb{I})$ ...
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Proof that $x^{n}<y^{n} \Rightarrow x<y $ if $x,y > 0$ [closed]

I need to prove that $x^{n}<y^{n} \Rightarrow x<y $ if $x,y > 0$ $n\in\mathbb{N}$ Been trying to crack this up with no success. would love to get some help
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Examples of non-orderable fields.

I wish to find examples of non-orderable fields. We know that fields with finite characteristics cannot be ordered, especially finite fields. Also $\mathbb{C}$ - the field of complex numbers cannot be ...
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Basic things in Archimedean ordered fields

I'll write theorem first. Theorem: Every Archimedean ordered field can be order-embedded into R. Proof: Let (K, ≤) be Archimedean. Then Q is dense in K. We define φ : K → R as follows. For a ∈ K, ...
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Intuitive Understanding of Order of Hyperreals

I'm trying to understand the ultrapower construction of the hyperreal numbers as described in Wikipedia. The motivation is given as the surprising ability to find a total order of sequences of real ...
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1answer
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Using the order axioms of $\mathbb{R}$ to prove the semi-definite positivity property for the absolute value?

How can I use the order axioms of $\mathbb{R}$ to prove the semi-definite positivity property for the absolute value: For all $x \in \mathbb{R}, |x|\geq0$ and $|x|=0$ if and only if $x=0$?
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A complete ordered field has a unique ordering

A complete ordered field has a unique ordering. Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help! My attempt: Let $\langle A,+,...
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1answer
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The isomorphism between two complete ordered fields is unique

The isomorphism between two complete ordered fields is unique. Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help! My attempt: ...
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The complete ordered field is unique up to isomorphism

The complete ordered field is unique up to isomorphism. I am quite sure that Parts 1,2,3 are correct. Could you please help me verify Parts 4,5,6 at the end of the proof. Do Parts 4,5,6 look fine or ...
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The smallest subfield of a complete ordered field is dense in itself

The smallest subfield of a complete ordered field is dense in itself. This theorem is part of my attempt to prove that A complete ordered field is unique up to isomorphism. My questions: Could you ...
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Let $\mathfrak{A}$ be a ordered field and $\mathfrak{X}$ be the smallest subfield of $\mathfrak{A}$. Is $X$ dense in $A$?

Let $\mathfrak{A}=\langle A,<,+,\cdot,0',1' \rangle$ be an ordered field and $\mathfrak{X}=\langle X,<,+,\cdot,0',1' \rangle$ be the smallest subfield of $\mathfrak{A}$. It is well-known that $\...
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There is a unique isomorphism from $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$ to an ordered field

Let $\mathfrak{Q}=\langle \Bbb Q,<,+,\cdot,0,1 \rangle,\mathfrak{A}=\langle A,<,+,\cdot,0',1' \rangle$ be ordered fields where $\Bbb Q$ is the set of rationals, and $f,g$ be isomorphisms from $\...
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If $\mathfrak{A},\mathfrak{B}$ are ordered fields and $f$ is an isomorphism between them, then $f(0)=0'$ and $f(1)=1'$

If $\mathfrak{A}=\langle A,<,+,\cdot,0,1 \rangle,\mathfrak{B}=\langle B,<,+,\cdot,0',1' \rangle$ are ordered fields and $f:A \to B$ is an isomorphism between $\mathfrak{A}$ and $\mathfrak{B}$, ...
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1answer
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Is it possible for $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$ to be isomorphic to a proper subfield of itself?

It is well-known that every ordered field contains a subfield that is isomorphic to $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$. Let $\langle A,<,+,\cdot,0',1' \rangle$ be an ordered field. I would ...
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1answer
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The smallest subfield of an ordered field is isomorphic to $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$

This is an exercise from Chapter 9. The Sets of Real Numbers from textbook Introduction to Set Theory by Hrbacek and Jech. The textbook does not provide solution and I would like to verify my attempt. ...
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2answers
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Do the axioms of ordered field imply that $a\cdot 0=0$ and $0<1$?

We introduce some definitions: A structure $\mathfrak{A}=\langle A,<,+,\cdot,0,1 \rangle$ where $<$ is a linear ordering, $+$ and $\cdot$ are binary operations, and $0,1$ are constants such ...
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0answers
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Countable ordered subfield of any Ordered Field

When we talk about any ordered field($\mathbb{k}$), we can always generate the rational elements of the field($\mathbb{Q_{k}}$) by first generating the natural elements($\mathbb{N_{k}}$) and the ...
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2answers
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Seeking references on ordered fields

I have a result that uses ordered fields. However, I am ignorant of the literature surrounding ordered fields. I have read the basic facts about them, such as those contained in the wikipedia page. ...
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1answer
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Bolzano's theorem for real closed field.

Take any real closed field $K$ and denote its unique ordering by $\leq$. Prove if $f \in K[X]$ and $a,b \in K$ such that $a<b$ and $f(a) < 0 < f(b)$, then there is $c$ that $f(c)=0$. I want ...
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Ordered subfields of $\mathbb{Q}_p$

I recently read about real ordered fields. Using real closures, I figured out that for each algebraically closed field $C$ of characteristic $0$, there exists a real closed subfield $R\subseteq C$ ...
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every ordered field $K$ has a natural valuation $v$, whose residue field is an archimedean ordered field.

Natural valuation has a convex valuation ring, in fact the valuation ring is the convex hull of $\mathbb{Z}$. How to prove that every ordered field $K$ has a natural valuation $v$, whose residue field ...
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1answer
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Definition of rational numbers

We can uniquely define the set of real numbers as a complete ordered field. But can we do something similar with the set of rational numbers ? I think we need to change the completness axiom with some ...
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1answer
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The ladder of the real closure of an ordered field

$\DeclareMathOperator{\lad}{Lad}$Let $(K,<)$ be an ordered field, and $(R,<)$ be the real closure of $(K,<)$. Let $\lad(K)$ denote the ladder of $(K,<)$ (for the definition of the ladder, ...
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1answer
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Suppose $x$ is a non-negative real number such that for all $\epsilon >0$ we have $x < \epsilon$. Then $x=0$

Is the following statement true? Suppose $x$ is a nonnegative real number such that for all $\epsilon >0$ we have $x < \epsilon$. Then $x=0$. Intuitively speaking yes since $0$ is an infimum ...
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1answer
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$K \neq \sum K^2 \implies K$ admits an ordering

$K$ is a field. $K^2 = \{a^2 | a \in K \}$, $\sum K^2$ is set of all sum of squares. I have to prove an implication: $K \neq \sum K^2 \implies K$ admits an ordering $K \neq \sum K^2$, so $-1 \...
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1answer
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I don't understand why this axiom makes the difference between a completely ordered field and an ordered field

I've been reading about ordered fields and completely ordered fields, and I'm stuck on the difference between the rational and real numbers that makes the rational numbers an ordered field and the ...
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1answer
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Could Someone Explain an Ordered Field in Layman's Terms?

High school student here... I am currently working on Spivak and I'm not very far (just reached questions in chapter 1). I was discussing my progress with someone recently and they asked me to ...
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Definition of < in Construction of Reals

In the section of Spivak's Calculus on the construction of the reals, < is defined: if alpha and beta are real numbers, then a < b means that a is contained in b (that is, every element of a is ...
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1answer
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Archimedian Property of a somewhat weird field

This exercise is part 3 of some connected exercises. Important for my question will be several things: $P[x]$ is the set of all polynomials $\sum_{i=0}^{N}a_ix^{i}, N \in \mathbb{N} \cup \{0\}, a_i \...
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1answer
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Problem about ordered fields

Let $K_0$, $K_1$, $K_2$ be ordered fields and $i_k$:$K_0$ $\rightarrow K_k$ are ordered inclusions for $k=1,2$. Show that there exists ordered field $K$ and ordered insclusions $j_k$ : $K_k \...
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2answers
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Transform a totally ordered set to a structure that is isomorphic to (R,+,.,≤)

So let $(M,\le_M)$ be a totally ordered set. Can we define $+$ and $.$ to make $M$ isomorphic to $(\mathbb{R},+,.,\le)$? I mean the well known axioms. To let this possible: $M$ is not bounded above ...
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1answer
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Can we define an order such that all complex numbers are equal?

There is an exercise "Prove that no order can be defined in the complex field that turns it into a field." in baby Rudin's book. And I can get a contradiction if $i>0$ or $i<0$. However, I ...
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In ordered field every sum of squares is non-negative, but $-1$ is always negative

Show that in an ordered field every sum of squares is non-negative, but $-1$ is always negative, so $-1$ is never a sum of squares. $K$ is an ordered field, $P=\{a \in K | 0 \leq a \}$, $-P=\{ -a | ...
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Does the above non-Archimedean but ordered field satisfy Nested interval property?

Consider the ordered non-Archimedean field $ \mathbb{R}(t)$, the field of rational function. My question is: $ \text{Does the above non-Archimedean but ordered field satify Nested interval property?} ...
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1answer
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Abstracting Magnitude Measurement Systems (i.e. subsets of ${\mathbb R}^{\ge 0}$) via Archimedean Semirings.

I did some googling but could not find any easily accessible theory so I am going to lay out my ideas and ask if they hold water. Definition: A PM-Semiring $M$ satisfies the following six axioms: (1)...
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1answer
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Proving there is no smallest positive element of any ordered field F

I missed lecture so these notes may be incomplete and I'm trying to fill in the blanks. Really not following this proof though (feel like maybe some stuff was said out loud but not written down) ...
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1answer
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Rudin's Definition of Ordered Fields

In his Principles of Mathematical Analysis, Rudin defines ordered fields as follows: An ordered field is a field $F$ which is also an order set, such that (i) $x+y<x+z$ if $x, y, z\in F$ and $y&...
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Why it is necessary to consider positive $ \ x,y \ $ in $ F\ $?

If $ F $ is an ordered field , then for any positive elements $x $ and $y$ in $F$ there is a natural number $ n$ such that $ nx>y \ $ i.e., \begin{eqnarray*} \underbrace{x+ \cdots+ x}_{n \ \text{...
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1answer
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Proving multiplication is compatible with ordering considering an ordered field.

Definition: Ordered field is a field $\mathcal F=(\Bbb{F},+,\cdot)$ together with relation $<$ which satisfy: $$\forall x,y\in\Bbb{F}: \text{exactly one of these 3 hold: }x<y\vee y<x\vee x=...
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Juggling three non-Archimedean fields

I'm comparing the field of hyperreals, the Levi-Civita field and the Dehn's field for the first time. I'm not very familiar with their properties, so I'm looking for ways to understand and distinguish ...
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1answer
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Doubts in proof of - complex numbers not an ordered field. [closed]

Sorry if have incomplete knowledge, as am new to this topic. I.e., even though have chosen 'ordered-fields' as tag; am not clear about it. It seems (as shown below) that the simple property of ...
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1answer
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Describe an infinite process of denumerable steps, that defines supremum property for the following set X (supremum of any bounded subset of X exists)

Let $X\subset 2^{\mathbb Z}$ be the set that consists of subsets of $\mathbb Z$ that are bounded above. That is to say, $A\in X$ iff $A\subset\mathbb Z$ and $\exists\max A$. Show that the set $X$ has ...
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2answers
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Derivatives for non Real/Complex functions

I recently started on my journey into Abstract Algebra and am interested in learning about derivatives in a generalised sense. By that I mean all of my current understanding (or the overwhelming ...
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1answer
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Check if the following defines a (total) order ($x\leq_p y$ iff the largest prime factor of $x$ is $\leq$ the largest prime factor of $y$)

Let $(\mathbb{N}\setminus\{1\}, \leq_p)$ with $x\leq_p y$ iff the largest prime factor of $x$ is $\leq$ the largest prime factor of $y$. Check if this defines a (total) order by checking the four ...
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Does the complex field satisfy Archimedean Property?

I have two questions regarding the Archimedean Property: I understand that for all $x \in \mathbb{C}$, there exists a positive $n \in \mathbb{N}$ such that $\vert nx \vert >1$ (Please correct me ...
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When $xyz=1$ why is $x+y+z\geq3$? ($x,y,z>0$)

How can I prove the statement below without using the Inequality of arithmetic and geometric means? $\forall x,y,z \in F $ (F is an ordered field) $(x,y,z>0)$ $xyz=1 \implies x+y+z\geqslant3$ ...
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1answer
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Trouble constructing an ordered set that is not directed

I want to try to construct an ordered set that is not directed to give myself a better understanding of what it means to be directed. Any good examples out there?
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1answer
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If $S<G:=\text{Gal}(E/R)$ is a Sylow 2-subgroup, then $[\text{Fix}_E(S):R] = [G:S]$

My question pertains to page 235 in Grillet's 'Abstract Algebra', Second Edition. The setup is as follows: Let $R$ be a formally real field, such that (i) every positive element of $R$ is a square ...
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well ordering principle and ordered field

I am trying to prove that there is no integer between $0$ and $1$. The standard proof uses well ordering principle. But if one takes integers as subset of rationals and rational numbers as an ordered ...
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1answer
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Number of orbits of the natural action of order preserving bijections of $\mathbb Q$ on $\mathbb Q^n$

By an order preserving map $f : \mathbb Q \to \mathbb Q$, let us mean $ x > y \implies f(x) > f(y)$ . Let $Aut (\mathbb Q) =\{ f : \mathbb Q \to \mathbb Q : f $ is bijective and order preserving ...