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bio website von-eitzen.de/math/tntrep.xml
location Bonn, Germany
age 48
visits member for 2 years, 2 months
seen 2 hours ago

I did study math and had a knack for it, but I am sooo out of that business now ...


Nov
23
comment The order of a cyclic subgroup, generated by a permutation
What is the definition of order of a permutation? What is the definition of order of a group?
Nov
23
comment Axioms for ordered fields
One can define that $(F,0,1,+,\cdot,P)$ is an ordered field if these four axioms hold: OF1 $(F,0,1,+,\cdot)$ is a field. OF2 $P+P\subseteq P$. OF3 $P\cdot P\subseteq P$. OF4 For $a\in F$ exactly one of $a\in P$, $a=0$, $\exists b\in P\colon a+b=0$ is true. - Then for $F=\mathbb R_{\ge0}$ and $0,1,+,\cdot$ as usual at least OF1 is violated. The fate of the other axioms depends on $P$ (which you didn't specify).
Nov
23
comment Axioms for ordered fields
No. The axiom of additive inverse is violated.
Nov
23
comment Why those division by zero are formalized?
As the question is tagged "nonstandard-analysis" and "infinitesimals", you can save the argument by replacing your $0$s with an infinitesimal $\epsilon$, expand suitably, cancel what can be canceled, then strike out all that is still a multiple of $\epsilon$ (i.e. round to the nearest standard real)
Nov
23
comment Axioms for ordered fields
The answer may depend on the precise formulation o fthe axioms, but essentially the lack of additive inverses is the main thing missing (and of course one axiom violated is enough)
Nov
23
comment A subset that is not open
Why look for an open set contained in $(U\setminus\{x\})\cup\{y\}$ when the assumption is already that $(U\setminus\{x\})\cup\{y\}$ itself is open?
Nov
23
answered “The limit of a sequence is insensitive to finite changes in the sequence” - help me understand this sentence!
Nov
23
revised “The limit of a sequence is insensitive to finite changes in the sequence” - help me understand this sentence!
deleted 9 characters in body
Nov
23
reviewed Edit suggested edit on Show: Every continuous function on the set E is bounded this implies E is compact
Nov
23
revised Show: Every continuous function on the set E is bounded this implies E is compact
Improved the title to contain the proposition
Nov
23
comment Show: Every continuous function on the set E is bounded this implies E is compact
Is $E$ at least supposed to be Hausdorff maybe?
Nov
23
comment Show that $\mathbb{Z} [\sqrt p]$ is an ordered Integral Domain.
You are also supposed to show that $P$ is closed under addition and multiplication. Both properties also follow from g1cs observation.
Nov
23
comment Show that $\mathbb{Z} [\sqrt p]$ is an ordered Integral Domain.
Stylistic side-note: "$p$ is prime" should be outside the curly braces in the definition of $\mathbb Z[\sqrt p]$
Nov
23
comment Is there really anything wrong with Bourbaki's Set Theory?
I suppose he doesn't mean "wrong" in the sense of "contains false statements as proclaimed 'theorems'" or "contains invalid proofs", but rather "is not the right book to learn se ttheory from, just as Russel-Whitehead's Principia Mathematica are not the right place to learn $2+2=4$ from"
Nov
23
answered Elementary combinatorics problem: which answer is the right one?
Nov
23
awarded  Nice Answer
Nov
22
answered Show that for any integer not divisible by 2 or 5, there is a multiple of it which is a string of 1s.
Nov
22
answered Two quotient morphisms and universal property
Nov
22
answered Formal power series question
Nov
22
answered How to determine the smallest interpolation degree required?