[math-fun] Floor function question
21 Sep
2005
21 Sep
'05
3:14 a.m.
Starting with a field-based axiomatization of the real numbers, how do we show that floor(x) exists for every x? -------------------------------- - David Wilson
21 Sep
21 Sep
3:50 a.m.
New subject: Antwort: [math-fun] Floor function question
The real numbers are characterised as being the unique totally ordered field, s.t. for every number x there exists a natural number n (= 1+...+1) with n>x. The set of all n with n>x has a unique minimum m =: ceiling(x). floor(x) is then defined in the usual way. A total ordering is necessary (complex numbers), that every number is eventually surpassed by the naturals is necessary as well (surreal numbers). Hope this helps. Cheers, Hartmut
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hartmut.holzwart@allianz.de