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omg new proof

Name: .999...9 =/= 1 2007-05-12 6:26 ID:/qcJW6wi

Assumptions
If A = B, then
a)  A-B = 0
b) A = (A+B)/2 = B

If A > B, then
a)  A-B > 0
b) A > (A+B)/2 > B


Let A = 1, B = .999...9
a) A-B = 1-.999...9 = .000...1 =/= 0
b) (A+B)/2 = (1+.999...9)/2 = 1.999...9/2 = 1.999...95 < A

Therefore, .999...9 =/= 1

Name: Anonymous 2007-05-16 11:51 ID:+60ZhJaQ

>>79
Pi

Name: Anonymous 2007-05-16 12:14 ID:4hYn09lA

>>81
Pi is not a real number, but rather a process with no corresponding real number. You can only use its symbolic form in equations involving real numbers. The same is true for the summation representing .999...

{Anyone taking a Numerical Analysis class would know the above.}

Name: Anonymous 2007-05-16 12:21 ID:srxeNIT8

>>82
LOL
'Numerical Analysis' has nothing to do with what you are talking about - unless you are confusing machine numbers with real numbers!

Name: Anonymous 2007-05-16 12:34 ID:4hYn09lA

>>83
Wrong again.

Name: Anonymous 2007-05-16 12:58 ID:srxeNIT8

>>84
Well I guess I've got to burn my NA books then.  I'll have to tell the authors to include some chapters they apparently overlooked!

Name: Anonymous 2007-05-16 14:35 ID:Heaven

>>82
Jesus. Go back to your computer science class, you've just failed so epically at math that it gave me a headache. REAL NUMBERS AND RATIONAL NUMBERS ARE DIFFERENT SETS.

Name: Anonymous 2007-05-16 15:07 ID:3R8tcskS

>>82
Oh god it hurts. IT HURTS MAKE HIM STOP

Name: Anonymous 2007-05-16 15:25 ID:qVPJ6ae6

This troll is brilliant.

Name: Anonymous 2007-05-16 16:19 ID:Heaven

This troll is getting boring. Make a better one.

Name: Anonymous 2007-05-17 0:12 ID:EPK2xjzE

>>82
uhmm real numbers are different from floating point numbers. The second set is finite.

Name: 4tran 2007-05-17 4:42 ID:BbONVS68

>>86, 87
I feel your pain.  >__<

Name: Anonymous 2007-05-17 8:09 ID:l61vv0Xm

Oh wow! Constructivism!

Name: Anonymous 2007-05-17 8:23 ID:MjZ2RHX2

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Name: Anonymous 2007-05-17 12:49 ID:Heaven

>>92
It isn't constructivism. Constructivists are at least smart enough to realize that redefining the real numbers to be the rational numbers is pointless, and mainly focus on arguing that we shouldn't use the real numbers.

Name: Anonymous 2007-05-17 17:00 ID:sStkk/IP

>>94
Well, then they are wrong also. Can't use numbers that don't exist.

Name: Anonymous 2007-05-17 17:54 ID:Heaven

>>95
Stop talking.

Name: Anonymous 2007-05-17 18:38 ID:sStkk/IP

>>96
Reality never stops happening; although you can tune it out, if you really want to.

Name: Anonymous 2007-05-17 18:39 ID:rqrvgDVQ

>>94
>>95
>>96

I lol'd

Name: Anonymous 2007-05-17 19:08 ID:Heaven

>>97
Yes, you're living proof of that last part.

Name: Anonymous 2007-05-18 5:39 ID:RzpTVOxN

>>80

I'd like to point out that using dictionary definitions for mathematical terms is a really fucking dumb idea.

Name: 4tran 2007-05-18 9:51 ID:qENmSn+F

>>100
Care to pull out a math textbook with a better definition of finite vs infinite?

Name: Anonymous 2007-05-18 11:09 ID:ijhPKrg1

>>101
Copypasta:

2.3 Definition If there exists a 1-1 mapping of A onto B, we say that A and B can be put in 1-1 correspondence, or that A and B have the same cardinal number, or, briefly, that A and B are equivalent, and we write A ~ B. This relation clearly has the following properties:
It is reflexive: A ~ A.
It is symmetric: If A ~ B, then B ~ A.
It is transitive: If A ~ B and B ~ C, then A ~ C.
Any relation with these three properties is called an quivalence relation.

2.4 Definition For any positive integer n, let J_n be the set whose elements are the integers 1, 2,..., n; let J be the set consisting of all positive integers. For any set A, we say:
(a) A is finite if A ~ J_n for some n (the empty set is also considered to be finite).
(b) A is infinite if A is not finite.
(c) A is countable if A ~ J.
(d) A is uncountable if A is neither finite nor countable.
(e) A is at most countable if A is finite or countable.

----------
copied straight from the first book I found, Rudin, Principles of Mathematical Analysis.  Honestly, finite vs infinite ain't hard.  Just wait till you get to toposes without NNOs.  Then it gets hard.

Name: 4tran 2007-05-18 11:26 ID:qENmSn+F

>>102
Yes, that's a very good definition of finite vs infinite (thanks), but it doesn't apply.  You provided a detailed and thorough definition of finite vs infinite for sets, but we're not dealing with sets.  We're arguing over a single element (that may or may not belong to some set).  The original argument was whether .999... was finite or not.  I claim it is.  Random person denies that.

Name: Anonymous 2007-05-18 12:15 ID:FE/YS3+W

>>48
Of course 0.333... is not a real number

I WANT TO KICK YOUR ASS FOR BEING SUCH A MORON

Name: Anonymous 2007-05-18 12:30 ID:ijhPKrg1

>>103
lulz, >>101 asked for a better definition of finite vs infinite.  So I gave it.

Anyone who talks about infinite real numbers does not understand the real number system.  All(*) reals are finite!

(*): boolean 2-valued logic, ZFC, cauchy/dedekind reals assumed ofc.

Name: Anonymous 2007-05-18 12:50 ID:Heaven

>>103
The person saying 0.999... was infinite was claiming that, because it has an infinite number of digits after the decimal, it is not a real number. He was, of course, wrong.

Name: 4tran 2007-05-18 13:07 ID:qENmSn+F

>>106
Yes, that's true, but it somehow seems that he believes that infinite digits = infinite number it represents (I probably should have been more explicit).  Glad you recognize his error.

Name: Anonymous 2007-05-18 15:19 ID:YDTd2OD7

fapfapfapfap

Name: Anonymous 2007-05-18 15:33 ID:Heaven

0.999... doesn't represent shit

Name: Anonymous 2007-05-18 16:26 ID:7SWGt+F6

>>109

And shit =/= 1.

Name: Anonymous 2007-05-18 22:28 ID:WuvEeAHT

>>110
So 0.999... must equal 1
Sounds logic to me

Name: Anonymous 2007-05-19 0:10 ID:Heaven

apparently you haven't learned the idiom concerning shit

Name: Anonymous 2007-05-19 2:30 ID:jqaiHHIB

Infinity is not a number at all it is merely the absence of an end. Saying that k is infinity is not valid. The glimmer of hope that you see is merely the fact that 1-1/k calculated in an infinite number of ways would yield an infinite number of results, however infinity does not do the impossible. The only other argument you could make is that the universe does not actually divide like math does. For example when a decaying radioactive isotope gets to less than one atom the mathematical model collapses and the atom just decays. However the truth of that matter is very much the same. Lets say that if you were counting in atoms, but due to the strings that make up the subatomic particles you are limited to an absolute calculation in (for the sake of argument) lets say the .000000000001 place. In this case the farthest possible you could go while describing atoms in a 1/k function is .999999999999. If you cant go any farther due to it not existing then the only logical step is to reach 1.

Name: Anonymous 2007-05-19 3:16 ID:Heaven

>>113
Listen: Math isn't physics.

Name: 4tran 2007-05-19 9:30 ID:Heaven

>>113
fail.  Particle decays are expressed as a statistical average.  When the mathematical model says that there is less than 1 particle remaining, it does not always mean the particle will instantly decay.

Name: Anonymous 2007-05-19 12:29 ID:jqaiHHIB

No it doesnt, but thats quite irrelevant. The point is that the mathematical model, even if it is only probability, cannot go farther than one atom and still express something other than the probability of that atom decaying. If there was an indivisible level of reality, it logically follows that it would be the end of all irrational numbers, and the farthest a .9999... can go.

Name: Anonymous 2007-05-19 12:31 ID:jqaiHHIB

The real question is: do subatomic particles have components that are smaller and smaller until infinity? Is there any other force that can make this irrelevant? Finally, you must ask your self if infinity exists in this universe. I don't care either way but I prefer to think that it doesn't.

Name: Anonymous 2007-05-19 12:46 ID:Heaven

>>116
>>117
LISTEN: MATH ISN'T PHYSICS.

Name: Anonymous 2007-05-19 17:52 ID:l7RPRTJw

What is the limit of sum[i=1,+inf](phail^i)?

Name: Anonymous 2007-05-19 17:54 ID:7Q3Xpb9q

FAPFAPFAPFAPFAPFAP

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