現代数学の系譜工学物理雑談古典ガロア理論も読む62

[過去ﾛｸﾞ] 現代数学の系譜工学物理雑談古典ガロア理論も読む62 (1002ﾚｽ)
上下前次1-新
通常表示 512ﾊﾞｲﾄ分割ﾚｽ栞
抽出解除ﾚｽ栞

このｽﾚｯﾄﾞは過去ﾛｸﾞ倉庫に格納されています｡
次ｽﾚ検索歴削→次ｽﾚ栞削→次ｽﾚ過去ﾛｸﾞﾒﾆｭｰ

ﾘﾛｰﾄﾞ規制です｡10分ほどで解除するので､他のﾌﾞﾗｳｻﾞへ避難してください｡

865: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2019/03/25(月) 23:36:53.22 ID:whxE2Y+u

>>856 補足

Sergiu Hart氏PDFより
game2 (without using the Axiom of Choice )
Theorem 2
Proof. ”The proof is the same as for Theorem 1, except that here we do not use the Axiom of Choice."

＜参考引用＞
Sergiu Hart氏のPDF http://www.ma.huji.ac.il/hart/puzzle/choice.pdf
(抜粋)
Consider the following two-person game game1:

Theorem 1 For every ε > 0 Player 2 has a mixed strategy in game1 guaranteeing him a win with probability at least 1 － ε.
Remark. The proof uses the Axiom of Choice.

Proof. Fix an integer K. We will construct K pure strategies of Player 2 such
that against every sequence x of Player 1 at least K － 1 of these strategies yield a win for Player 2.
略

A similar result, but now without using the Axiom of Choice.^2
Consider the following two-person game game2:
^2 Due to Phil Reny.（＝Phil Reny氏より）

Theorem 2 For every ε > 0 Player 2 has a mixed strategy in game2 guaranteeing him a win with probability at least 1 － ε.

Proof. The proof is the same as for Theorem 1, except that here we do not use the Axiom of Choice.
Because there are only countably many sequences x ∈ {0, ・・・, 9}^N that Player 1 may choose
(namely, those x that become eventually periodic),
we can order them － say x^(1), x^(2), ・・・, x^(m), ... － and then choose in
each equivalence class the element with minimal index (thus F(x) = x^(m) iff m
is the minimal natural number such that*4 x ～ x^(m)).

*4 Explicit strategies σj may also be constructed, based on R^j being the index where the
sequence y^j becomes periodic.
(引用終り)

http://rio2016.5ch.net/test/read.cgi/math/1551963737/865

868: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2019/03/26(火) 08:02:31.64 ID:4i9G7Ghx

>>865 補足

私スレ主が、「なにを理解できているか」を証明するには、”このスレの余白は狭すぎる！” （by Fermat）
まあ、理解できていないのかも知れないね（＾＾

でもそれで良いじゃないｗ（＾＾
他の人が理解できるように、コピペは出来ていれば！（＾＾
そして、サイコパスピエロが理解出来ていないことが示せればねｗ

game1 (using the Axiom of Choice )
Theorem 1 For every ε > 0 Player 2 has a mixed strategy in game1 guaranteeing him a win with probability at least 1 － ε.
Remark. The proof uses the Axiom of Choice.

game2 (without using the Axiom of Choice )
Theorem 2 For every ε > 0 Player 2 has a mixed strategy in game2 guaranteeing him a win with probability at least 1 － ε.
Proof. ”The proof is the same as for Theorem 1, except that here we do not use the Axiom of Choice."

上記コピペで、game1 とgame2において、Theorem 1＝Theorem 2であること
及び、game1 とgame2との違いは、(using the Axiom of Choice )と(without using the Axiom of Choice )だということを示せればねｗ
そして、サイコパスピエロが理解出来ていないことが示せればね！！ｗ（＾＾

＜参考再掲＞
Sergiu Hart氏のPDF http://www.ma.huji.ac.il/hart/puzzle/choice.pdf

＜参考＞
https://dic.pixiv.net/a/%E3%83%95%E3%82%A7%E3%83%AB%E3%83%9E%E3%83%BC%E3%81%AE%E6%9C%80%E7%B5%82%E5%AE%9A%E7%90%86
フェルマーの最終定理 とは【ピクシブ百科事典】
(抜粋)
「私は真に驚くべき証明を見つけたが、この余白はそれを書くには狭すぎる。」 - ピエール・ド・フェルマー
(引用終り)

http://rio2016.5ch.net/test/read.cgi/math/1551963737/868

上下前次1-新書関写板覧索設栞歴

ｽﾚ情報赤ﾚｽ抽出画像ﾚｽ抽出歴の未読ｽﾚ AAｻﾑﾈｲﾙ

ぬこの手ぬこTOP 0.032s