All Questions
22
questions
3
votes
0
answers
112
views
Proof that for all nonzero real numbers $a$, $\frac{1}{a}$ is nonzero
I was wondering if someone could check my proof that "For all $a\in\mathbb{R}$, if $a\neq 0$ then $\frac{1}{a}\neq 0$". The definitions/assumptions I am basing the proof off of come from &...
1
vote
1
answer
44
views
What does Artin mean by "real numbers are the *only* ones needed for the usual for the usual algebraic operations?"
In page 81 of the 2nd edition Michael Artin's Algebra, he introduces fields and presents $\mathbb{R}$ as a familiar example, but goes on to say that "the fact that they are the only ones needed ...
0
votes
1
answer
81
views
Characterizations of the reals
I know that one characterization of the reals is that it is the only Dedekind-complete ordered field. Are there any other characterizations of the reals as a field?
1
vote
0
answers
67
views
What's the proof that the only Dedekind-complete field is the reals? [duplicate]
I know that the field of the rational numbers is ordered but not Dedekind-complete. What's the proof that the only Dedekind-complete field is the reals?
0
votes
0
answers
95
views
Is every formally real field isomorphic to a subfield of the reals?
A formally real field is a field $K$ such that $-1$ is not a sum of squares in $K$. Clearly subfields of $\mathbb{R}$ are formally real. I also know finite fields and algebraically closed fields are ...
1
vote
2
answers
72
views
Deriving the addition and product on $\mathbb C$
I am studying Stillwell's Elements of Algebra. In Chapter 3, Section 3.6, he writes
There is a unique extension of $+$, $-$, $\times$, $\div$ to $\mathbb C$ satisfying the field properties since if ...
1
vote
0
answers
90
views
Proof that $\mathbb Q$ and $\mathbb R$ are Archimedean ordered fields
I searched for "archimedean ordered field" on this website and Google but didn't find much.
Exercises:
(pages 90 and 101 of Analysis I by Amann and Escher)
My attempt:
These exercises seem ...
1
vote
1
answer
91
views
Alternative to the proof on Wolfram Mathworld that $\mathbb Q$ is the smallest subfield of $\mathbb R$
Exercise 2 (b) on page 100 of Analysis I by Amann and Escher asks me to show that $\mathbb Q$ is the smallest subfield of $\mathbb R$.
Wolfram MathWorld gives the following reasoning:
I don't know ...
3
votes
1
answer
136
views
Is there an ordered field with distinct subfields isomorphic to the reals?
Is there an ordered field with distinct subfields isomorphic to the field $\mathbb R$ of real numbers?
1
vote
2
answers
104
views
Proof that $\frac{1}{2} + \frac{1}{2} = 1$ using just the algebraic properties of $\mathbb R$
Like the title says, can you prove rigorously that $\frac{1}{2} + \frac{1}{2} = 1$ using only the nine field properties of $\mathbb R$? I don't know if addition and multiplication are supposed to be ...
2
votes
1
answer
635
views
Is an automorphism of the field of real numbers without ordering the identity map? [duplicate]
We know an automorphism on $\mathbb{R}$ must fix $\mathbb{Q}$. If we assume the usual order structure and topology on $\mathbb{R}$, then we can use the density of $\mathbb{Q}$ to show an automorphism ...
1
vote
1
answer
140
views
On the existence of an algebraically closed field containing other fields
This question arose while I was reading a paper I found in the web.
It might be very simple, but I don't know the answer.
Let $\mathbb{R}$ be the set of real numbers and $\mathbb{Q}_p$ the set of all $...
0
votes
1
answer
165
views
The properties of real numbers field [closed]
I know, that the multiplicative group of $\mathbb{R}$ is create on the set $\mathbb{R}\setminus \{0\}$. But how we can multiply real numbers on the $0$ after this?
This point was unswered, I think. ...
10
votes
2
answers
1k
views
Are the real numbers the unique Dedekind-complete ordered set?
A totally ordered set is Dedekind-complete if any subset which has an upper bound also has a least upper bound. Now any two ordered fields which are Dedekind-complete are order-isomorphic as well as ...
6
votes
4
answers
207
views
Is $\mathbb Q$ a quotient of $\mathbb R[X]$?
Is there some ideal $I \subseteq \mathbb R[X]$ such that $\mathbb R[X]/I \cong \mathbb Q$?
$I$ is clearly not a principal ideal.