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From: George Greene on 14 Jun 2010 02:04
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!

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.
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