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Offline brian0918

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« on: March 01, 2004, 05:07:04 pm »
« Last Edit: March 27, 2004, 10:23:23 pm by brian0918 »

(No subject)
« Reply #1 on: March 02, 2004, 03:17:23 am »
Sonic Center Alert:  Record has been broken! OMFG WTF!!!111

This alert is to inform you that user ybbun has just beaten your old record on Sonic 1 Green Hill 1 with a kick ass new time of 0.01 seconds using only save-states and a bot program!  You've been dropped to 2nd place.  Suicide is the only option.


OMFG You sound like half of LUE. The wiser half.

Good idea/10

Offline SadisticMystic

(No subject)
« Reply #2 on: March 02, 2004, 04:11:46 am »
meow
« Last Edit: March 29, 2004, 01:27:29 am by SadisticMystic »

Offline ybbun

(No subject)
« Reply #3 on: March 05, 2004, 11:33:39 am »
Yeah my use of save states on my genesis really is something to admire. I am so good at using them that I can even add in milliseconds to my times. Using my knowledge I can traverse the level in almost instantaneous fashion resulting in sonic blowing up because of the massive air resistance.

(No subject)
« Reply #4 on: March 06, 2004, 05:58:26 am »
Quote
meow
<3

Pure <3


Here's another:

.999~ = sigma(.9*[.1]^[n-1]).

.999~ = .9 + .09 + .009 + .0009 + .00009 + .000009 .... (ad infinitum)
Let a, 1st term = .9
Let r, common ratio = 10 ^ -1
Infinite geometric progression: a + ar + ar²... (ad infinitum)
Sum to infinity = a / (1 - r) = .9 / (1 - .1) = .9 / .9 = 1
.999~ = 1
http://www.geocities.com/admiralmussina/SigmaNinev2.rtf

sigma(n = 1, n -> inf.) 9/(10^n) = Definition of geometric series.
9 * sigma(n = 1, n -> inf.)(1/10)^n = Property of a series.
9 * 1/9 = (r/[1 - r], r = .1)
3/9 = 1/3
.999~ = 1

Let Sequence A = (.9, .99, .999... ad infinitum)
Let Sequence B = (1, 1, 1...)
If sequences A and B were equivalent, then they would have the same limit. In sequence A, the infinith degree holds .999~. For sequence A to be equal to sequence B, then sequence B must have the infinith degree be equal to .999~. Since sequence B holds only 1, then for sequence A and B to be equal, .999~ must equal 1.
Let S = .999~
S = .9 + .09 + .009 + .0009 + .00009 + .000009 .... (ad infinitum)
S = 0.9 + (1/10)S
(9/10)S = .9
1 = S
Sequence A = Sequence B
.999~ = 1

For two numbers to be different, there must be a number between them.
.999~ < N < 1
Find N. Because .999~ has infinite nines, there are no numbers between them, thus N cannot exist. .999~ = 1.

.999~ = .9 + .09 + .009 + .0009 + .00009 + .000009 .... (ad infinitum)
sigma[-1:0 -> inf.](.9 * .10^-1)
omega -> sigma = .9 / (1 - .1) = .9 / .9 = 1
.999~ = 1

.(42)~ = 42/99
.(402)~ = 402/999
.000~ = 0/9
.111~ = 1/9
.222~ = 2/9
.333~ = 3/9
.444~ = 4/9
.555~ = 5/9
.666~ = 6/9
.777~ = 7/9
.888~ = 8/9
.999~ = 9/9
9/9 = 1
.999~ = 1

.333~ = sigma(n = 1, n -> inf.) 3/(10^n) = Definition of geometric series.
3 * sigma(n = 1, n -> inf.) (1/10)^n = Property of a series.
3 * 1/9 = (common ratio, r/[1 - r], r = .1)
3/9 = 1/3
1/3 = .333~
1/3 * 2 = 2/3
.333~ * 2 = .666~
2/3 = .666~
.333~ + .666~ = .999~
1/3 + 2/3 = 3/3
3/3 = .999~
3/3 = 1
.999~ = 1

--- Corollary Proof ---
.999~/3 = .333~
1/3 = .333~
.333~ = .333~
.999~/3 = 1/3
.999~ = 1

If Rolken asks why you didn't get the record, tell him it's because you were getting stoned. He'll understand.
« Last Edit: March 29, 2004, 01:27:50 am by SadisticMystic »

Offline ybbun

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« Reply #5 on: March 06, 2004, 10:46:55 am »
I used to visit LUE, back in my darker days, but I missed the sign-up list. I've seen that proof. Also the - 4 squared = -16 was hilarious too.
Oh, and gotta love that smiley face in your picture.

Offline Cybrax

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« Reply #6 on: March 06, 2004, 11:18:26 am »
:ph34r:  oh god, my head, ow... :wacko:

Offline Rolken

(No subject)
« Reply #7 on: March 06, 2004, 01:10:59 pm »
Quote
.999~ = 9/9
9/9 = 1
.999~ = 1
why don't they just say this and be done with it?
StH JJ1 WkS+ Sal++^i Rbk++i Knu- McS+++ P++ D[af]opw $++++ E03 A24 GM CoUT
What fun is it being cool if you can't wear a sombrero?

(No subject)
« Reply #8 on: March 07, 2004, 02:28:43 am »
Quote
I used to visit LUE, back in my darker days, but I missed the sign-up list. I've seen that proof. Also the - 4 squared = -16 was hilarious too.
Oh, and gotta love that smiley face in your picture.
Heh, it's a lot of fun there. I found the -4^2 thing more annoying than the .999~ thing, but yeah, they were both funny.

And shh! The smiley was intentional!

Brian: The .9~ is identical to .9°, or .9..., etc. It's just used to represent "Goes on forever" by LUE. I just copied and pasted all those proofs from another LUEser.

For some reason it's really hard for people to believe that .999... = 1.

Yeah, I'm still trying to understand why they don't understand too =\

Offline Rolken

(No subject)
« Reply #9 on: March 07, 2004, 03:15:15 am »
x = y
x^2 = xy
x^2 - y^2 = xy - y^2
(x+y) (x-y) = y (x-y)
x + y = y
2y = y
2 = 1
StH JJ1 WkS+ Sal++^i Rbk++i Knu- McS+++ P++ D[af]opw $++++ E03 A24 GM CoUT
What fun is it being cool if you can't wear a sombrero?

(No subject)
« Reply #10 on: March 07, 2004, 04:16:07 am »
YOUR HEAD A SPLODE

Offline SadisticMystic

(No subject)
« Reply #11 on: January 01, 1970, 07:00:00 am »

Offline CosmicFalcon

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« Reply #12 on: March 07, 2004, 04:54:52 am »
Quote
x = y
x^2 = xy
x^2 - y^2 = xy - y^2
(x+y) (x-y) = y (x-y)
x + y = y
2y = y
2 = 1
My old Maths teacher showed us that, and I've been trying for ages to remember it. Thing is, technically that proof doesn't work because here:

(x+y) (x-y) = y (x-y)
x + y = y

You are dividing by 0 (x = y, ergo x - y = 0) so it's WRONG! Denied  :ph34r:

Although, that 0.999... = 1 proof doesn't need to be that complicated.

Let r = 0.999...
10r = 9.999...
9r = 10r - r = 9.999... - 0.999... = 9
r = 9r / 9 = 1

Booshed!

And... what is this -4^2 = -16 thing? HOW DOES THAT WORK?
"A graph of cf's coolness as age increases would be exponential." - Stefan [14:26, 2008/08/23]
"I now realise that CF is complete and utter win." - Cruizer [13:46, 2009/10/23]

Offline CosmicFalcon

(No subject)
« Reply #13 on: March 07, 2004, 05:17:12 pm »
Quote
Here's a nice one:

1 = 2:


-1/1 = 1/-1

Taking the square root of both sides: sqrt(-1/1) = sqrt(1/-1)

Simplifying:  sqrt(-1)/sqrt(1) = sqrt(1)/sqrt(-1)

In other words:  i/1 = 1/i

Therefore, i / 2 = 1 / (2i),

i/2 + 3/(2i) = 1/(2i) + 3/(2i),

i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),

(i²)/2 + (3i)/2i = i/(2i) + (3i)/(2i)

(-1)/2 + 3/2 = 1/2 + 3/2

1 = 2


This one's kinda obvious.  I'm trying to find a more difficult one that I remember seeing.
Is the reason for that the fact that you took the square root of a minus number? I mean... I don't know anything about complex numbers so I can't check up on that.
"A graph of cf's coolness as age increases would be exponential." - Stefan [14:26, 2008/08/23]
"I now realise that CF is complete and utter win." - Cruizer [13:46, 2009/10/23]

Offline SadisticMystic

(No subject)
« Reply #14 on: March 07, 2004, 06:02:23 pm »
It has to do with the limited domain over which square roots (and more generally, exponents) are distributive.

(No subject)
« Reply #15 on: March 10, 2004, 05:55:26 am »
Quote
1) You are dividing by 0 (x = y, ergo x - y = 0) so it's WRONG! Denied  :ph34r:

2) Although, that 0.999... = 1 proof doesn't need to be that complicated.

3) And... what is this -4^2 = -16 thing? HOW DOES THAT WORK?
1) Exactly. But still cool to think about.

2) Try telling that to half of GameFAQs  <_<

3) It works depending on where you put your imaginary brackets. Mine always go around the -4 first, even though logically they shouldn't. Exponents first, they say.

Offline SadisticMystic

(No subject)
« Reply #16 on: March 10, 2004, 06:23:40 am »
In that case, the - sign is a unary operator acting on the 4, and those take precedence over everything.

Offline Cybrax

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« Reply #17 on: March 10, 2004, 07:52:46 am »
notice people like me, and anyone else under 18, are staying out of this one...

Offline yse

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(No subject)
« Reply #18 on: March 11, 2004, 01:45:38 am »
Lol, well pointed out Cybrax.

However, I intended to plunge myself into this one.

Yes, -4^2 = 16, as long as there aren't brackets around the -4. As Popo Nana said, always do indices first.

Quote
Let r = 0.999...
10r = 9.999...
9r = 10r - r = 9.999... - 0.999... = 9
r = 9r / 9 = 1
That was how I did that one. I tried to prove to a lot of people at my school that working, nobody believed me... Just for fun, I added that because of the infinitescimally(sp?) small difference between 0.999... and 1 (which obviously isn't there, but some people are gullible) could be multiplied by any number to prove that any two numbers are the same, and therefore that maths has no value. :P

And no, I've never been to LUE. I'm not a LUEser like you lot :P

<3 Thorn.

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