*NEW DIGIT SEQUENCE* is an OXYMORON!
in [Math]
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From: George Greene on 14 Jun 2010 02:05 On Jun 14, 12:43 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Do you agree that all the digits of pi are in this set? > > 3 > 31 > 314 THIS IS NOT A SET, DUMBASS! THIS IS A LIST! AND PI IS NOT ON this list, since every digit string that IS on this list IS FINITE, while PI IS INFINITELY long!
From: |-|ercules on 14 Jun 2010 02:18 "George Greene" <greeneg(a)email.unc.edu> wrote .. > On Jun 14, 12:43 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> Do you agree that all the digits of pi are in this set? >> >> 3 >> 31 >> 314 > > > THIS IS NOT A SET, DUMBASS! > THIS IS A LIST! > AND PI IS NOT ON this list, since every digit string that IS on this > list IS FINITE, > while PI IS INFINITELY long! Lists are not sets now? Herc
From: |-|ercules on 14 Jun 2010 02:19 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 14, 12:43 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> Do you agree that all the digits of pi are in this set? >> >> 3 >> 31 >> 314 > > > Of course we do, but the > infinite digit string > 111111.... > NEVER OCCURS IN THIS SET, DUMBASS. > NOR DOES PI ITSELF, DUMBASS, because EVERY > string in this set IS FINITE, DUMBASS! > THERE ARE NO INFINITE DIGIT STRINGS IN THIS SET, > DUMBASS! <font size=oo> SO WHAT! </font> Every digit of pi is on that list dumbass! You know what that means dumbass? Herc
From: Colin on 14 Jun 2010 13:15 Wow, such vitriol! Yet everyone seems to be ingoring the fundamental error that Herc is making. What he's saying is this: Let S be an infinite sequence of digits. Herc claims that (1) for every natural number n, there is an algorithm f s.t. f(n)= the nth digit in S implies (2) there is an algorithm f s.t. for every natural number n, f(n)= the nth digit in S. But (1) definitely does not imply (2), and Herc is committing a very basic quantifier shift fallacy.
From: Graham Cooper on 14 Jun 2010 13:52
On Jun 15, 3:15 am, Colin <colinpoa...(a)hotmail.com> wrote: > Wow, such vitriol! Yet everyone seems to be ingoring the fundamental > error that Herc is making. What he's saying is this: > > Let S be an infinite sequence of digits. Herc claims that > > (1) for every natural number n, there is an algorithm f s.t. f(n)= the > nth digit in S > > implies > > (2) there is an algorithm f s.t. for every natural number n, f(n)= the > nth digit in S. > > But (1) definitely does not imply (2), and Herc is committing a very > basic quantifier shift fallacy. There are special cases where 1 does not imply 2 3 31 314 ... This list contains every digit of pi similarly computable sequences contain every digit of every possible sequence therefore this argument does not hold 123 456 789 diag = 159 antidiag = 260 new digit sequence Herc |