2013年10月13日星期日 文章後繼

又一個人被我列入拒絕討論黑名單裡面了
我過去曾經讚賞過的人
在2013年10月13日星期日文章中提到過

信念歸信念
信念不能被證明
所以你跟我說你的信念是什麼就是什麼
這沒什麼

當你企圖把信念放在理性層次討論時
你就必然的要接受批判檢視
這是理性層面的優點，同時也是缺點

當你把所有不利於你論點的可能性都刪除時
你不會得到知識或真理
你只會的到自己永遠是對的結論

當討論雙方無法達成共識，有可能是因為前提或推論過程沒有交集
那討論下去沒有意義，要試著回頭看沒有交集之處
或者直接回到信念層次

當被扣上某種帽子，因為你的智商，你的意識形態時
完全沒有繼續討論下去的意義
因為他們只是要給自己台階下
"因為你是白癡，所以你講的都是錯的，我贏了"
"因為你是某黨的人，所以你說的話都不可信"

你們想要什麼?
想證明自己很聰明永遠不會錯?
想證明自己很有智慧?

你們想要就給你們吧
我不在乎那些東西
我只想知道那些是對的那些是假的

給科學家的一封信

什麼才是成為優秀科學家的條件?
數學能力?紀律?智商?
在著名的科學家中
有些人數學及強，有些人極差
有些人一絲不苟，有些人生活糜爛
但是都有一個特點
如果他們都覺得某個領域太過無聊
那麼他們根本就不會投注時間在那領域，更何況有重大發現
所以，答案很明顯了
就是因為我們笨，我們好奇，所以我們是科學家

今天是希爾伯特的忌日

特此紀念

Hilbert's twenty-three problems are:

Problem	Brief explanation	Status	Year Solved
1st	The continuum hypothesis (that is, there is no set whose cardinality is strictly between that of the integers and that of the real numbers)	Resolved. Proven to be impossible to prove or disprove within the Zermelo–Fraenkel set theory with or without the Axiom of Choice (provided the Zermelo–Fraenkel set theory with or without the Axiom of Choice is consistent, i.e., contains no two theorems such that one is a negation of the other). There is general consensus that this solves the problem, although there have been proposals which would give a definitive truth value (see Ω-logic).	1963
2nd	Prove that the axioms of arithmetic are consistent.	There is no consensus on whether results of Gödel and Gentzen give a solution to the problem as stated by Hilbert. Gödel'ssecond incompleteness theorem, proved in 1931, shows that no proof of its consistency can be carried out within arithmetic itself. Gentzen proved in 1936 that the consistency of arithmetic follows from the well-foundedness of the ordinal ε₀.	1936?
3rd	Given any two polyhedra of equal volume, is it always possible to cut the first into finitely many polyhedral pieces which can be reassembled to yield the second?	Resolved. Result: no, proved using Dehn invariants.	1900
4th	Construct all metrics where lines are geodesics.	Too vague to be stated resolved or not.[n 1]	–
5th	Are continuous groups automatically differential groups?	Resolved by Andrew Gleason, depending on how the original statement is interpreted. If, however, it is understood as an equivalent of the Hilbert–Smith conjecture, it is still unsolved.	1953?
6th	Mathematical treatment of the axioms of physics	Partially resolved depending on how the original statement is interpreted.[13] In particular, in a further explanation Hilbert proposed two specific problems: (i) axiomatic treatment of probability with limit theorems for foundation of statistical physics and (ii) the rigorous theory of limiting processes "which lead from the atomistic view to the laws of motion of continua". Kolmogorov’s axiomatics (1933) is now accepted as standard. There is some success on the way from the "atomistic view to the laws of motion of continua".[14]	1933-2002?
7th	Is a b transcendental, for algebraic a ≠ 0,1 and irrational algebraic b ?	Resolved. Result: yes, illustrated by Gelfond's theorem or the Gelfond–Schneider theorem.	1935
8th	The Riemann hypothesis ("the real part of any non-trivial zero of the Riemann zeta function is ½") and other prime number problems, among them Goldbach's conjecture and the twin prime conjecture	Unresolved.	–
9th	Find the most general law of the reciprocity theorem in any algebraic number field.	Partially resolved.[n 2]	–
10th	Find an algorithm to determine whether a given polynomial Diophantine equation with integer coefficients has an integer solution.	Resolved. Result: impossible, Matiyasevich's theorem implies that there is no such algorithm.	1970
11th	Solving quadratic forms with algebraic numerical coefficients.	Partially resolved.[citation needed]	–
12th	Extend the Kronecker–Weber theorem on abelian extensions of the rational numbers to any base number field.	Unresolved.	–
13th	Solve 7-th degree equation using continuous functions of two parameters.	The problem was partially solved by Vladimir Arnold based on work by Andrei Kolmogorov. [n 4]	1957
14th	Is the ring of invariants of an algebraic group acting on a polynomial ring always finitely generated?	Resolved. Result: no, counterexample was constructed by Masayoshi Nagata.	1959
15th	Rigorous foundation of Schubert's enumerative calculus.	Partially resolved.[citation needed]	–
16th	Describe relative positions of ovals originating from a real algebraic curve and as limit cycles of a polynomial vector field on the plane.	Unresolved.	–
17th	Express a nonnegative rational function as quotient of sums of squares.	Resolved. Result: yes, due to Emil Artin. Moreover, an upper limit was established for the number of square terms necessary.[citation needed]	1927
18th	(a) Is there a polyhedron which admits only an anisohedral tiling in three dimensions? (b) What is the densest sphere packing?	(a) Resolved. Result: yes (by Karl Reinhardt). (b) Widely believed to be resolved, by computer-assisted proof (by Thomas Callister Hales). Result: Highest density achieved by close packings, each with density approximately 74%, such as cubic close packing and hexagonal close packing.[n 5][citation needed]	(a) 1928 (b) 1998
19th	Are the solutions of regular problems in the calculus of variations always necessarily analytic?	Resolved. Result: yes, proven by Ennio de Giorgi and, independently and using different methods, by John Forbes Nash.	1957
20th	Do all variational problems with certain boundary conditions have solutions?	Resolved. A significant topic of research throughout the 20th century, culminating in solutions[citation needed] for the non-linear case.	?
21st	Proof of the existence of linear differential equations having a prescribed monodromic group	Resolved. Result: Yes or no, depending on more exact formulations of the problem.[citation needed]	?
22nd	Uniformization of analytic relations by means of automorphic functions	Resolved.[citation needed]	?
23rd	Further development of the calculus of variations	Unresolved.	–

叔本華說，一個人可以做他想要做的，但不能意志他想要意志的。這句話伴隨我度過生命所有的際遇，且使我容易順從他人的行為，即使那些行為讓我很煩惱。

艾伯特‧愛因斯坦，對德國人權聯盟演講，1932年秋，柏林—

Hilbert's tomb:
Wir müssen wissen
Wir werden wissen

Paul Mccartney 

Maybe I'm Amazed

Baby I'm amazed at the way you love me all the time
Maybe I'm afraid of the way I love you
Baby I'm amazed at the the way you pulled me out of time
Hung me on a line
Maybe I'm amazed at the way I really need you

Baby I'm a man and maybe I'm a lonely man
Who's in the middle of something
That he dosen't really understand
Babe I'm a man and maybe you're the only woman
Who could ever help me
Baby won't you help to me understand

Baby I'm a man and maybe I'm a lonely man
Who's in the middle of something
That he dosen't really understand

Babe I'm a man and maybe you're the only woman
Who could ever help me
Baby won't you help me understand

Baby I'm amazed at the way you're with me all the time
Maybe I'm afraid of the way I leave you
Baby I'm amazed at the way you help me sing my song
You right me when I'm wrong
Maybe I'm amazed at the way I really need you

大家都相信自己先天是完全自由的，甚至涵蓋個人行動，而且認為在任何時間他都可以開始另一種生活方式...。但後天，從經驗上，他會驚訝地發現自己並不自由，而是受制於必需品，而且不顧他的所有決心，他無法改變自己的行為，而這就形成從他生命開始到結束的生活，他必須扮演自己譴責的角色...。