伯特蘭·羅素 - 維基百科,自由的百科全書 伯特蘭·亞瑟·威廉·羅素,第三代羅素伯爵(英語:Bertrand Arthur William Russell, 3rd Earl Russell,1872年5月18日-1970年2月2日),OM,FRS,英國哲學家、數學家和邏輯學家,致力於哲學的大眾化、普及化。很多人將羅素視為這個時代的先知,而與此同時羅素的 ...
羅素悖論- 維基百科,自由的百科全書 - Wikipedia 羅素悖論(Russell's paradox),也稱為理髮師悖論,是羅素於1901年提出的悖論, ... 羅素悖論當時的提出,造成第三次數學危機。 ... 理髮師悖論和羅素悖論是等價的:.
理髮師悖論- 維基百科,自由的百科全書 - Wikipedia 理髮師悖論(Barber paradox)是羅素用來比喻羅素悖論的一個通俗說法,是由伯特蘭· ... 用集合論的語言來描述理髮師悖論是這樣的:小城裡的人構成集合 A=\{a | a\ ...
Russell's paradox - Wikipedia, the free encyclopedia In 1908, two ways of avoiding the paradox were proposed, Russell's type theory and the Zermelo set theory, the first constructed axiomatic set theory. Zermelo's axioms went well beyond Frege's axioms of extensionality and unlimited set abstraction, and ev
Russell's Paradox (Stanford Encyclopedia of Philosophy) Russell's paradox is the most famous of the logical or set-theoretical paradoxes. Also known as the Russell-Zermelo paradox, the paradox arises within naïve set theory by considering the set of all sets that are not members of themselves. Such a set appea
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Russell - Wikipedia, the free encyclopedia Russell may refer to:
What is Russell's paradox? - Scientific American John T. Baldwin and Olivier Lessmann of the Department of Mathematics, Statistics and Computer Science at the University of Illinois at Chicago offer the following explanation. Russell's paradox is based on examples like this: Consider a group of barbers
Russell’s Paradox | Internet Encyclopedia of Philosophy Russell's Paradox Russell's paradox represents either of two interrelated logical antinomies. The most commonly discussed form is a contradiction arising in the logic of sets or classes. Some classes (or sets) seem to be members of themselves, while some
Russell's Paradox - Interactive Mathematics Miscellany and Puzzles The set of all subsets of a given set. Self-reference. ... Poincaré disliked Peano's work on a formal language for mathematics, then called "logistic." He wrote of Russell's paradox, with evident satisfaction, "Logistic has finally proved that it is not s