Curse

Kattar · Jun 10, 2008, 07:23 PM // 19:23

See, that's where it all falls apart.

Logic =/= Maths.

FlamingMetroid · Jun 10, 2008, 07:26 PM // 19:26

Quote:

Originally Posted by Katsumi

See, that's where it all falls apart.

Logic =/= Maths.

yes
these threads depress me, because this really doesn't matter >__>

around · Jun 10, 2008, 08:25 PM // 20:25

Quote:

Attempted proofs are then saying that 10^-∞ = 0 WTF! It's obviously not consistent with the rest of mathematics because 10^-∞ is the infinitessimal and in mathematics this is greater than zero or calculus wouldn't work.

LERN2LIMITS.

In fact, infinitesimals don't exist.

Quote:

Originally Posted by spawnofebil

How so? Infinitesimals do not exist (in the standard formulation for real numbers), because there is an axiom which states that if x < z, there exists a y such that x < y and y < z.

poobert · Jun 10, 2008, 09:45 PM // 21:45

Quote:

Originally Posted by spawnofebil

LERN2LIMITS.

In fact, infinitesimals don't exist.

Correct.

Quote:

Only if you're rounding, which was established at the beginning to not be legitimate.

You are not rounding. It is a limit imposed by the infinite series. Think of it as a Gaussian, which is continuous but has a finite integral.

Rushin Roulette · Jun 11, 2008, 04:32 AM // 04:32

Quote:

Originally Posted by spawnofebil

LERN2LIMITS.

LERN2LIMITS =/= ∞

wu is me · Jun 11, 2008, 07:40 AM // 07:40

Quote:

Originally Posted by fenix

Let x = 0.999∞

1x = 0.999∞
10x = 9.999∞
10x-1x = 9x = 9.999∞ - 0.999∞
9x = 9
x = 1

0.999∞ = 1

PROVE US WRONG

I haz induction proof that it isn't true =P :

0.8 + 0.1 =0.9 = a
let r=0.1
a*r^1 +a = 0.99
@ a*r^n + a*r(n-1) ...... +a = 0. 999999999.....(n)
!= 1
and for n+1
a*r^(n+1)....a = 0.9....(n+1)...9 != 1
so.... since for n=1 f(1) != 1 [ f(x) = a( 1 - r )^x/(1-r)] and f(n+1) !=1 if f(n) !=1 ...yaddayadddayadda, f(n) != 1 for all cases of n.
in particular f(∞) = 0.999....∞ != 1

amirite?

edit: taking the idea of a geometric sum further....
for n = ∞
f(x) = a/(1-r) = 0.9 / ( 1- 0.1) = 0.9/0.9 = 1 = wtf???( formula derived from the use of limits?)

**Marty Silverblade** · Jun 11, 2008, 10:05 AM // 10:05

Quote:

Originally Posted by Snow Bunny

Only if you're rounding, which was established at the beginning to not be legitimate.
I'm pretty sure Marty's right here. I haven't done maths for 3 years, but I'm pretty sure his logic there wins.

I just realised that he's not talking about rounding. He's saying that 0.499=0.5 just like 0.999 supposedly =1. Just checked, it's mathematically correct.

Let x=0.499
Thus 10x=4.999
10x-x=4.999-0.499
9x=4.5
x=0.5

Back to square one I guess. Then again, I could use the 0.5/2=0.25 and 0.499=/=0.25 logic again, but meh. It'll never end.

Taurus · Jun 11, 2008, 10:10 AM // 10:10

you all really dont like GW anymore...

Monkey Slayer · Jun 11, 2008, 07:26 PM // 19:26

May I ask what the point of this is?

MrSlayer · Jun 11, 2008, 07:49 PM // 19:49

It's old news to me..

Kanyatta · Jun 11, 2008, 09:31 PM // 21:31

Quote:

Originally Posted by Monkey Slayer

May I ask what the point of this is?

So people can display their incredible ability to read Wikipedia.

Lord Sojar · Jun 15, 2008, 04:50 AM // 04:50

Quote:

Originally Posted by Akuma

do you go to yale too

Nah, just have my doctrine in nano electrical physics with a specialization in microprocessor architecture and design, employed by nvidia as a senior physical design engineer. But beyond that, I have no knowledge of simple mathematics and physics.

Oh, and no, I didn't attend Yale. I am more a Purdue University and Cornell guy.

Oh, and to all of those people "proving" .999 repeating = 1. You are using defined mathematics and set numerals to prove it. Thus, your proof is, in and of itself, invalid and baseless. You have to use infinite numbers to prove infinite possibilities.