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2020-12-07 18:54 - LineareAlgebruh im Themenstart schreibt:
Hallo.
Jede komplexe Zahl lässt sich darstellen als \(a+bi\) mit \(a,b \in \mathbb{R}\). Ich habe mich jetzt gefragt, ob sich dann auch jede reelle Zahl darstellen lässt als \( a+b \sqrt 2\), mit \(a,b \in \mathbb{Q}\)
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"R.P. Steiner. "A theorem on the Syracuse problem". In: ed. by D. McCarthy and H. C. Williams. Congressus numerantium; 20. Proceedings of the 7th Manitoba Conference on Numerical Mathematics and Computation, September 29-October 1, 1977. Winnipeg: Utilitas Mathematica Pub., 1978, pp. 553-559."
2020-06-27 17:16 - Slash in Beitrag No. 1 schreibt:
Ich weiß nicht, ob dieses Resultat passend ist für das was du vorhast.
Berg, L., and Meinardus, G., Functional equations connected with the Collatz problem, Results in Math. 25 (1994)
hier
hier
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\newcommand{\proofofthmn}[1]{\gudl{\sc{P}\!roof\tx{}of\tx{}\sc{T}\!heorem\tx{}#1\colon}}
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\(\endgroup\)
Analysis Eins:
\(\begingroup\)\(\DeclareMathOperator{\mer}{mer}
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2020-05-13 20:13 - helmetzer in Beitrag No. 4 schreibt:
2020-05-13 19:22 - xiao_shi_tou_ in Beitrag No. 3 schreibt:
Beachte auch, dass eine asymmetrische Relation nicht symmetrisch ist. Kann also eine symmetrische Relation asymmetrisch sein?
Auch wenn es an Haarspalterei grenzt: Bei einer leeren Relation sollte man das noch einmal überdenken.
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2020-05-09 13:52 - MePep in Beitrag No. 6 schreibt:
Moment, ist also \(\sim\) die Schnittmenge von \(\sim_{1} und \sim_{2}\) ?
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2020-04-04 16:25 - Nuramon in Beitrag No. 4 schreibt:
2020-04-04 16:09 - ziad38 in Beitrag No. 2 schreibt:
wooow Du hast Recht, warum habe ich nicht auch so gedacht obwohl ich 3 mal gelesen habe.
Ich habe mich auch schon mal über ein [1] am Ende einer Formel in einem Artikel gewundert, bis mir der Betreuer meiner Bachelorarbeit sagte, dass das eine Quellenangabe ist.
Welche Fragen hast du zu dem Beweis?
[Die Antwort wurde nach Beitrag No.2 begonnen.] [Die Antwort wurde nach Beitrag No.11 begonnen.]
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2020-03-25 16:33 - Nuramon in Beitrag No. 14 schreibt:
2020-03-25 16:25 - xiao_shi_tou_ in Beitrag No. 13 schreibt:
Auf deinem Profil gibt es ein Latex Profil.
Da kannst du Abkuerzungen wie
copy \newcommand{\Q}{\mathbb{Q}}
hinschreiben. Aber dort keine Dollarzeichen verwenden.
Dann musst du nur noch
schreiben, wenn du $\Q$ haben willst.
Bei default gibt es auf dem MP schon \IQ, \IR, \IN, \IZ, \IC für $\IQ, \IR, \IN, \IZ, \IC$. Wenn man sich den einen extra Buchstaben leisten kann, dann braucht man muss man sich also dafr gar nicht extra Abkürzungen einrichten.
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Einheiten .TeX
copy $\mathbb{R}$
$\mathrm{ggT}$
[Die Antwort wurde nach Beitrag No.10 begonnen.] copy \newcommand{\Q}{\mathbb{Q}}
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2020-03-25 15:36 - OliverFuchs in Beitrag No. 5 schreibt:
Ja danke nochmal. Zum einen für die Willommensgrüße im Forum zum anderen
für die schnelle Hilfe. Das mit der Linearkombination von a und b,
hätte ich auch sehen können. Sorry
Aber das was der Kollege mit p/ggT(a^2,b^2) gesagt hat, da hat er recht.
Genau so habe ich es auch in meiner Version gemacht nur habe ich es
aus dem Grund nicht verwendet, weil ich mir nicht sicher war ob man
bei der Verwendung der Primzahl nicht Argumentte mit einbaut,
mit denen man auch die Primfaktorzerlegung zeigen kann. Da habe ich
zu wenig parat wie man diese zeigt und welch Argumente da eingehen.
Wenn ich mich aber recht aus der Vorlesung erinnere, so kann man die Primfaktorzerlegung in jedem Ring zeigen, in welchem auch die Divison mit
Rest gilt. Nur muss sie dann nicht eindeutig sein. Dazu, glaube ich wenigstens, benötigt man den Hauptsatz der Arithmetik (p/a1...an =>
p/ai 1<=i<=n). Wenn also bei der alternativen Beweisführung nicht beide
Dinge verwendet werden dann wäre der Beweis zulässig. Das muss
ich mir noch ansehen.
Danke nochmals für die Hilfe und den Willkommensgruß
lg Oliver
Ich muss mich jetzt doch noch korrigieren [Die Antwort wurde nach Beitrag No.6 begonnen.] \(\endgroup\)
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2021-01-18 16:54 U ?
Wie unterstütze ich den Matheplaneten mit Geld?
