Question

$K \neq \sum K^2 \implies K$ admits an ordering

score 79 · Answer 1 · 2018-11-14 20:53:13

79

Views

$K \neq \sum K^2 \implies K$ admits an ordering

Published on 14 Nov 2018 - 20:53

score 49 · Answer 2 · 2018-11-14 21:46:20

49

Views

Suppose $x$ is a non-negative real number such that for all $\epsilon >0$ we have $x < \epsilon$. Then $x=0$

Published on 14 Nov 2018 - 21:46

score 91 · Answer 3 · 2018-11-22 07:25:25

91

Views

The ladder of the real closure of an ordered field

Published on 22 Nov 2018 - 7:25

score 117 · Answer 4 · 2018-12-02 12:15:55

117

Views

Definition of rational numbers

Published on 02 Dec 2018 - 12:15

score 100 · Answer 5 · 2018-12-02 13:02:47

100

Views

every ordered ﬁeld $K$ has a natural valuation $v$, whose residue ﬁeld is an archimedean ordered ﬁeld.

Published on 02 Dec 2018 - 13:02

score 67 · Answer 6 · 2018-12-09 21:35:33

67

Views

Ordered subfields of $\mathbb{Q}_p$

Published on 09 Dec 2018 - 21:35

score 94 · Answer 7 · 2018-12-12 19:29:24

94

Views

Bolzano's theorem for real closed field.

Published on 12 Dec 2018 - 19:29

score 240 · Answer 8 · 2018-12-16 17:32:20

240

Views

Seeking references on ordered fields

Published on 16 Dec 2018 - 17:32

score 141 · Answer 9 · 2018-12-22 07:04:06

141

Views

Countable ordered subfield of any Ordered Field

Published on 22 Dec 2018 - 7:04

score 70 · Answer 10 · 2019-01-08 00:49:42

70

Views

Do the axioms of ordered field imply that $a\cdot 0=0$ and $0<1$?

Published on 08 Jan 2019 - 0:49

score 677 · Answer 11 · 2019-01-08 14:23:30

677

Views

The smallest subfield of an ordered field is isomorphic to $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$

Published on 08 Jan 2019 - 14:23

score 33 · Answer 12 · 2019-01-09 00:33:51

33

Views

Is it possible for $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$ to be isomorphic to a proper subfield of itself?

Published on 09 Jan 2019 - 0:33

score 26 · Answer 13 · 2019-01-09 02:26:03

26

Views

If $\mathfrak{A},\mathfrak{B}$ are ordered fields and $f$ is an isomorphism between them, then $f(0)=0'$ and $f(1)=1'$

Published on 09 Jan 2019 - 2:26

score 46 · Answer 14 · 2019-01-09 02:57:38

46

Views

There is a unique isomorphism from $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$ to an ordered field

Published on 09 Jan 2019 - 2:57

score 43 · Answer 15 · 2019-01-09 04:20:57

Let $\mathfrak{A}$ be a ordered field and $\mathfrak{X}$ be the smallest subfield of $\mathfrak{A}$. Is $X$ dense in $A$?

List Question

$K \neq \sum K^2 \implies K$ admits an ordering

Suppose $x$ is a non-negative real number such that for all $\epsilon >0$ we have $x < \epsilon$. Then $x=0$

The ladder of the real closure of an ordered field

Definition of rational numbers

every ordered ﬁeld $K$ has a natural valuation $v$, whose residue ﬁeld is an archimedean ordered ﬁeld.

Ordered subfields of $\mathbb{Q}_p$

Bolzano's theorem for real closed field.

Seeking references on ordered fields

Countable ordered subfield of any Ordered Field

Do the axioms of ordered field imply that $a\cdot 0=0$ and $0<1$?

The smallest subfield of an ordered field is isomorphic to $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$

Is it possible for $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$ to be isomorphic to a proper subfield of itself?

If $\mathfrak{A},\mathfrak{B}$ are ordered fields and $f$ is an isomorphism between them, then $f(0)=0'$ and $f(1)=1'$

There is a unique isomorphism from $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$ to an ordered field

Let $\mathfrak{A}$ be a ordered field and $\mathfrak{X}$ be the smallest subfield of $\mathfrak{A}$. Is $X$ dense in $A$?

Trending Questions

Popular # Hahtags

Popular Questions