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From: George Greene on 20 Jun 2010 02:40
On Jun 17, 9:53 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> Consider this LOGIC LOOP as an infinite sequence of propositions.

NO, DUMBASS, we DO NOT consider infinite sequences of "propositions"
here. Our rules of inference operate FROM FINITE SETS of premises.


> AD = 0
> D = 1
> WHILE TRUE
>    AD = AD + a different digit than the Dth digit of the Dth real) / 10^D
>    YOU HAVEN'T PRODUCED A NEW DIGIT SEQUENCE YET!
>    D++
> WEND

You CANNOT COUNT TO INFINITY, DIPSHIT.
This loop NEVER ENDS. So of course IT doesn't produce
an infinite sequence.

Try THIS one instead:
D=0
WHILE TRUE
PRINT D
D++
YOU HAVEN'T PRINTED OUT ALL THE NATURAL NUMBERS YET
WEND

This is just bullshit.
The fact that something is operating on an infinite set does NOT allow
you to invoke time
to claim that it can never finish. If it did, you could never have
constructed L in the first place,
since L is both infinitely long AND infinitely wide. If L can be
finished then obviously its anti-
diagonal can be finished THE SAME way, and in 1/infinitieth of the
time, since L has infinitely
MANY infinitely long rows, while the anti-diagonal is only ONE
infinitely long string.


> How does this make a *new digit sequence* that is not present on the
> list of computable reals?
>
> Herc
> --
> If you ever rob someone, even to get your own stuff back, don't use the phrase
> "Nobody leave the room!" ~ OJ Simpson

From: |-|ercules on 20 Jun 2010 03:07
"George Greene" <greeneg(a)email.unc.edu> wrote
> On Jun 19, 4:21 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote:
>
>> Yes I know
>>
>> An AD(n) = (L(n,n) + 1) mod 9
>
> No, you don't.
> ACTUALLY,
> AD(n) = 9 - L(n,n)
>

Here is proof that George will argue against any valid point
and replace it with an equivalent rewording for no reason.

Herc
From: |-|ercules on 20 Jun 2010 03:08
"George Greene" <greeneg(a)email.unc.edu> wrote
On Jun 19, 4:21 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> The digit-string IS INFINITE! The increasing finite prefixes ARE
> FINITE!
>

But the length of the set of finite prefixes is infinite.

Herc
From: |-|ercules on 20 Jun 2010 06:19
"|-|ercules" <radgray123(a)yahoo.com> wrote
> "George Greene" <greeneg(a)email.unc.edu> wrote
>> On Jun 19, 4:21 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote:
>>
>>> Yes I know
>>>
>>> An AD(n) = (L(n,n) + 1) mod 9
>>
>> No, you don't.
>> ACTUALLY,
>> AD(n) = 9 - L(n,n)
>>
>
> Here is proof that George will argue against any valid point
> and replace it with an equivalent rewording for no reason.


Actually George's formula is not equivalent to mine, his doesn't work!

Diag = 0.500000...
George's_Anti_Diag = 0.499999...

Well done George, way to construct a new number!

Herc
From: George Greene on 20 Jun 2010 13:42
On Jun 20, 3:07 am, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> Here is proof that George will argue against any valid point
> and replace it with an equivalent rewording for no reason.

The fact that somebody does this one time does not imply that it
generally happens.
1 instance IS NOT a proof of the general case, DUMBASS.
1 instance is only good for DISproving generalizations.


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