1-1+1-1+1-1+1-1+1-1+1-1 = 1/2

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Anonymous Tue Jun 25 10:17:01 2013 No.5858468 [DELETED]

1-1+1-1+1-1+1-1+1-1+1-1 = 1/2

>>	Anonymous Tue Jun 25 10:19:01 2013 No.5858469 >>5858468 False.

>>	Anonymous Tue Jun 25 10:28:31 2013 No.5858476 >>5858469 1-1+1-1+1-1+1-1+1-1+1-1 ... = 1/2

>>	Anonymous Tue Jun 25 10:33:32 2013 No.5858481 >>5858468 You on let's make a deal. Monty Hall has a woman show you three boxes. One has two goats in it, one has two cars in it, one has a goat and a car it. You pick one of the boxes. What are the odds that the woman has a sister?

>>	Anonymous Tue Jun 25 10:33:47 2013 No.5858480 >>5858468 = zero

Anonymous Tue Jun 25 10:37:16 2013 No.5858487

>>5858476

Depending on your definition for sequence convergence, this can be true.

The problem when people write things as you have is that you are treating "1-1+1-1+1-1+1..." as if without further clarification it is an innately meaningful, physical thing. It is not.

>>	Anonymous Tue Jun 25 10:41:16 2013 No.5858495 By judicious use of parenthesis, you can make this converge to anything you like. Consider that 1=a+1-a and regroup to have convergence to a.

>>	Anonymous Tue Jun 25 10:46:47 2013 No.5858499 >>5858495 This is false; you get this result if you use intuitive, inconsistent, ill-defined ideas of what that sum means.

>>	Anonymous Tue Jun 25 11:36:17 2013 No.5858588 >>5858499 >This is false >Repeat what I said

>>	Anonymous Tue Jun 25 11:39:32 2013 No.5858594 >infinite series >inconsistent we may have a completeness problem here

>>	Anonymous Tue Jun 25 11:40:01 2013 No.5858597 >>5858588 >is obviously not a mathematician Define what you mean by the sum "1-1+1-1+1-1+1..." Then we can talk. Spoiler: "The number you get if you keep adding like that forever" is not a well-posed definition.

>>	Anonymous Tue Jun 25 11:47:16 2013 No.5858613 >>5858597 >is obviously not a mathematician >repeat what I said. Again.

>>	Anonymous Tue Jun 25 11:50:32 2013 No.5858619 >>5858613 >can't even pose a problem properly

>>	Anonymous Tue Jun 25 11:50:48 2013 No.5858620 >>5858597 <span class="math">\sum_{k=0}^{+ \infty} (-1)^k[/spoiler]

>>	Anonymous Tue Jun 25 11:52:48 2013 No.5858623 >>5858619 Christ, read my original comment. I'm obviously giving a counter example. You're just stating over and over that my counter example actually IS a counter example.

>>	Anonymous Tue Jun 25 11:53:33 2013 No.5858626 Can't believe it's taken this long for anyone to say this but terms in a series have to approach zero for the thing to have a possibility to converge ...

Anonymous Tue Jun 25 11:55:02 2013 No.5858629

>>5858623

You haven't provided a counter-example, because your original comment /does not formally mean anything/.

If you don't define what you mean by convergence of an infinite sum beforehand, saying "you can make it converge to anything" is an empty statement. All you have done is fool your own intuition.

Putting parenthesis in an infinite sum, by the way, is not necessarily a well-posed operation.

>>	Anonymous Tue Jun 25 11:56:18 2013 No.5858635 >>5858626 For the standard definition of sequence convergence, this is true. There are expanded notions of convergence for which it is not.

>>	Anonymous Tue Jun 25 11:57:03 2013 No.5858637 >>5858626 learn2 abel/cesaro/lindelöf convergence

>>	Anonymous Tue Jun 25 11:57:31 2013 No.5858639 >>5858629 >You haven't provided a counter-example >Point out what my counter example is a counter example to.

>>	Anonymous Tue Jun 25 11:59:03 2013 No.5858643 >>5858639 Unless you were being facetious in your original comment, I don't see what you're getting at.

>>	Anonymous Tue Jun 25 12:02:46 2013 No.5858652 >>5858468 I knew this would surface here as soon as I saw it. Hoping on seeing some clever ideas.

Anonymous Tue Jun 25 12:05:02 2013 No.5858658

>>5858643
I was manipulting the series formally (i.e. supposing that it converged, say, absolutely) and arrived at a contradiction. The argument I gave with a=1/2 is the one that you typically see when people argue that the alternating series above converges to 1/2.

>>	Anonymous Tue Jun 25 12:07:17 2013 No.5858660 >>5858658 You were crap at communicating what you were trying to do. All you really pointed out was that notating the series as "1 + 1 - 1 +1 - 1 ..." makes it easy to do things which look intuitively correct but are formally meaningless.

>>	Anonymous Tue Jun 25 12:15:46 2013 No.5858690 >>5858660 That was my point exactly. Seems I succeeded.

Anonymous Tue Jun 25 12:17:17 2013 No.5858697

>>5858660
But really, my counter-example was deeper than that. I've met people who thought that judicious use of parenthesis could only produce the result 1/2, thereby somehow singling out a probable result. My example shows that judicious use of parenthesis can produce any result, and therefore doesn't make sense.

>>	Anonymous Tue Jun 25 12:20:02 2013 No.5858705 >>5858703 confusion* Christ, I can't type today.

Anonymous Tue Jun 25 12:20:32 2013 No.5858703

>>5858697

Ah, I understand you now.

Communication through text sucks.

At any rate, confusing over that series pretty much boils down to confusion over what series convergence formally means. The only people who puzzle over it are those who haven't had any real math education.

>>	Anonymous Tue Jun 25 12:25:31 2013 No.5858719 >>5858703 Agreed.

>>	Anonymous Tue Jun 25 12:29:46 2013 No.5858735 >>5858495 could you be more specific ? i don't see how you do that.

Anonymous Tue Jun 25 12:35:17 2013 No.5858756

>>5858703
what does the word "formally" means to you ? i'm asking that because the way i see it a " formal " manipulation is a manipulation of the form, ie a symbolic manipulation ( as in symbolic computation ).

Anyway, those confusions of people without real math education have lead to interesting theory http://en.wikipedia.org/wiki/Divergent_series..

>>	Anonymous Tue Jun 25 12:37:16 2013 No.5858762 >>5858756 I've always taken "formally" to mean "following strictly from a set of definitions."

>>	Anonymous Tue Jun 25 15:12:02 2013 No.5859097 >>5858735 1-1+1-1+1-1... =(a+1-a)-(a+1-a)+(a+1-a)-(a+1-a)+(a+1-a)-(a+1-a)... =a+(1-a-a+1+a+a)+(1-a-a+1+a+a)+(1-a-a+1+a... =a

>>	Anonymous Tue Jun 25 15:13:01 2013 No.5859099 >>5859097 Bah 1-1+1-1+1-1... =(a+1-a)-(a+1-a)+(a+1-a)-(a+1-a)+(a+1-a)-(a+1-a)... =a+(1-a-a-1+a+a)+(1-a-a-1+a+a)+(1-a-a-1+a... =a

>>	Anonymous Tue Jun 25 15:13:01 2013 No.5859100 Cesaro summation.

>>	Anonymous Tue Jun 25 15:31:26 2013 No.5859137 >>5858481 50/50. She eithers' got's one er she don't.

>>	Anonymous Tue Jun 25 15:39:11 2013 No.5859159 A = 1 - 1 + 1 - 1 + 1 - 1 + ... 1 - A = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...) 1 - A = A 2A = 1 A = 0.5

Anonymous Tue Jun 25 15:45:13 2013 No.5859172

>>5859159
You can get any rational number (possibly any real but that is hard to prove).
ie if you want 3:
take the first 3 +1s and group the 1st -1 and the 4th +1, the 2nd -1 and 5th +1, each group equals 0 so you get 3.

If you want a fraction,
A = 1 - 1 + 1 - 1 + 1 - 1 + ...
1 - A = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...)
1 - A - A= 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...) - (1 - 1 + 1 - 1 + 1 - 1 + ...)
3A=1.. .etc

>>	Anonymous Tue Jun 25 18:05:56 2013 No.5859605 1-1+1-1+1-1+1-1+1-1+1-1 = 0 FTFY

>>	Anonymous Tue Jun 25 18:10:12 2013 No.5859622 >>5859159 Holy shit, this is wrong on so many levels I think you just made me feel American.

>>	Anonymous Tue Jun 25 18:28:26 2013 No.5859657 >>5859622 Explain then!

>>	Anonymous Tue Jun 25 18:35:56 2013 No.5859668 https://en.wikipedia.org/wiki/Grandi%27s_series

>>	Anonymous Tue Jun 25 18:54:27 2013 No.5859699 >>5859172 see >>5859099

>>	Anonymous Tue Jun 25 19:24:58 2013 No.5859777 >>5858487 ding ding ding ding ding... winnner.

>>	Anonymous Tue Jun 25 22:54:23 2013 No.5860241 >>5858468 [spoiler]1-1 ok that's 0. +1 ok 1.-1 ok 0 +1 ok 1 -1 ok 0.... I'm a noob at math, and i want to come to this board in order to be better at it. Why would it ever get to 1/2?

>>	Anonymous Tue Jun 25 23:06:40 2013 No.5860272 >>5860241 I think someone already prooved it, but i guess you didn't see it, so here it is: S=1-1+1-1+1-1+1... 1-S=1-(1-1+1-1+1-1+1...) 1-S=1-1+1-1+1-1+1-1... so: 1-S=S 1=2S then S=1/2

Anonymous Tue Jun 25 23:09:56 2013 No.5860280

>>5860272

That is not a valid proof for why the series of (-1)^n converges to 1/2 under certain definitions of sequence convergence.

The reasons for the invalidity of that "proof" have been discussed at length in the thread. Namely, the manipulation assumes that the sum converges.

If you are confused about that sum, it is because you do not actually understand what is meant by an infinite sum. Take an analysis course, and it will become clear.

Under the most common definition of sequence convergence, that sum is undefined.

Anonymous Tue Jun 25 23:14:40 2013 No.5860296

>>5860280
Yeah I know that it doesn't converge if you look at it as a regular infinite sum (if we can say it like that?), I've seen that in calculus while we were studying series and infinite sums for integrating certain functions. But if we assume that the serie has a sum, then it is the answer to the sum. It just depend on how you see the problem.

Anonymous Tue Jun 25 23:21:26 2013 No.5860330

>>5860296

It is true that if the sum converges, then it must converge to 1/2. Your proof was almost correct there, but your use of parenthesis is abuse of notation and the proper way to notate it is to compare the sequence with a sequence offset by one (which must, by definition, converge to the same value if the series converges).

At any rate, if you want it to converge you first must define what you mean by series convergence in a way in which it does, indeed, converge (Cesaro summation or otherwise).

>>	Anonymous Wed Jun 26 02:28:41 2013 No.5860772 >>5858469 Why?

>>	Anonymous Wed Jun 26 02:54:11 2013 No.5860794 Is there any meaning to considering the limsup and liminf of the partial sums and averaging them? This yields 1/2 in this case as well.

>>	Anonymous Wed Jun 26 16:49:07 2013 No.5862012 Alternating series test. /thread

>>	Anonymous Thu Jun 27 03:13:37 2013 No.5862917 That's the infinite series, not the finite sum.

>>	Anonymous Thu Jun 27 04:10:37 2013 No.5862999 Divergent Partial sums are: 1,0,1,0,1,0,1,0,... Limit of partial sums does not exist.

>>	Anonymous Thu Jun 27 09:42:37 2013 No.5863283 >>5862999 ay, cptn.

>>	Anonymous Thu Jun 27 10:14:37 2013 No.5863315 >>5858468 Integers are closed under multiplication?

>>	Anonymous Thu Jun 27 10:15:37 2013 No.5863316 >>5863315 umm... addition, as well. But, addition seems more relevant.

>>	Anonymous Thu Jun 27 10:18:37 2013 No.5863318 >>5859657 A = 1 - 1 + 1 - 1 + 1 - 1 + ... 1 - A = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...) >1 - A = A dafuq it's 1-A=1-A

>>	Anonymous Thu Jun 27 10:20:37 2013 No.5863320 >>5858637 >>5859100

Anonymous Thu Jun 27 10:20:37 2013 No.5863321

S = 1-1+1-1+1-1+1-1+1-1+1-1 .....
S = 1 -(1-1+1-1+1-1+1-1+1-1+1-1.....)
S = 1 - S
2S = 1
S = 1/2

Sum of values in the sequence = 1/2

S = 1-1+1-1+1-1+1-1+1-1+1-1 .....
S = (1-1) + (1-1) + (1-1) ....
S = 0 + 0 + 0 ....
S= 0

Sum of values in the sequence = 0

S = 1-1+1-1+1-1+1-1+1-1+1-1 .....
S = 1 + (-1+1) + (-1+1) + (-1+1) .....
S = 1 + 0 + 0 +0....
S = 1

Sum of values in the sequence = 1

J = 1-1+1-1+1-1+1-1+1-1+1-1
J = 9 (1/9 - 1/9 + 1/9 ....)
9 + 99 (1/9) = 9 + 11
9 + 11 = 911
J = 911

Jews did 9/11

>>	Anonymous Thu Jun 27 10:24:37 2013 No.5863324 >>5859172 >take the first 3 +1s and group the 1st -1 and the 4th +1, the 2nd -1 and 5th +1, each group equals 0 so you get 3. Just fyi: you cant do this.

>>	Anonymous Thu Jun 27 10:25:37 2013 No.5863326 File: 431 KB, 474x599, utilitarism.png [View same] [iqdb] [saucenao] [google] >>5863321 my sides

>>	Anonymous Thu Jun 27 10:25:37 2013 No.5863327 >>5859137 Actually, it was a trick question. It is a goat.

>>	Anonymous Thu Jun 27 10:28:37 2013 No.5863329 >>5863318 1 - A = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...) = 1 - 1 + 1 - 1 + 1 - 1 + ... = A

>>	Anonymous Thu Jun 27 10:33:37 2013 No.5863334 >>5863329 >1 - A = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...) = = 1-1+1-1+1-1+1 - ... and if i repeat this sequence i get: 1 - 1 - 1 + 1 ....

>>	Anonymous Thu Jun 27 10:39:37 2013 No.5863341 >>5863334 The brackets arent there for nothing you know. Expanding - (1 - 1 + 1 - 1 + 1 - 1 + ...) Gets you - 1 + 1 -1 + 1 - 1 + 1 - .... So 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...) Gets you 1 - 1 + 1 -1 + 1 - 1 + 1 - .... = A

>>	Anonymous Thu Jun 27 17:55:37 2013 No.5864141 >>5863341 Now all you need to do is solve for A.

>>	Anonymous Fri Jun 28 01:55:37 2013 No.5864808 >>5864141 But I only know how to solve for x.

>>	Anonymous Fri Jun 28 17:20:34 2013 No.5865935 >>5864808 Take a class on advanced calculus.

>>	Anonymous Sat Jun 29 02:30:34 2013 No.5866834 >>5865935 What is advanced calculus and why do we need it?

>>	Anonymous Sat Jun 29 03:04:19 2013 No.5866896 File: 49 KB, 624x420, glorious winged faggot.jpg [View same] [iqdb] [saucenao] [google] >>5858468 No, it's 0.

>>	Anonymous Sat Jun 29 15:59:10 2013 No.5867893 >>5866834 Advanced calculus is the first higher math class. It'll teach you equivariant cohomologies and K-theory.

>>	Anonymous Sat Jun 29 20:00:55 2013 No.5868517 >>5866896 No, it's indeterminate.

Anonymous Sun Jun 30 00:33:10 2013 No.5869054

>>5858468
The answer's 1. Because infinity-infinity=0, so 1-0=1. Just group the numbers up into infinites. Monty's fatal mistake was that he added a false rule- implying there is a finite limit. If that were the case then the stated limit would determine the result. The only way he's right is if it were a finite pattern that was unknown.

BTW- 0 divided by 0=infinity, for a division is but a ratio comparison and 0 has no value to require rationing with, deeming the division infinite irregardless the value of the divisor. That means any number divides infinitely into 0 and 0 divides 0 into any value that isn't 0 itself.

You fucking dumbasses.

>>	Anonymous Sun Jun 30 00:59:50 2013 No.5869125 >>5867893 Thank you.

>>	Anonymous Sun Jun 30 01:16:20 2013 No.5869189 >>5869054 >infinity-infinity=0 >0 divided by 0=infinity How to spot a american.

>>	Anonymous Sun Jun 30 01:21:50 2013 No.5869201 >>5858468 well... on the fact that % is just - - - ... you could just as easily say that it = 1x2

>>	Anonymous Sun Jun 30 01:44:51 2013 No.5869229 >>5858499 >>5858735 http://en.wikipedia.org/wiki/Conditional_convergence

>>	Anonymous Sun Jun 30 18:17:31 2013 No.5870654 >>5869229 Doesn't apply here.

>>	Anonymous Mon Jul 1 02:09:33 2013 No.5871385 >>5869189 She is right. Are you implying Americans are knowledgeable in mathematics?

>>	Anonymous Mon Jul 1 19:01:38 2013 No.5873168 >>5871385 Didn't an american win the maths nobel prize?

>>	Anonymous Mon Jul 1 22:14:38 2013 No.5873494 >>5871385 what? how do you know anon is a gril? wth is up with people on here? uh nvm it's not important just strange

>>	Anonymous Mon Jul 1 22:45:06 2013 No.5873527 File: 88 KB, 250x250, 1369732810018.png [View same] [iqdb] [saucenao] [google] >>5873168 >maths nobel prize

>>	Anonymous Mon Jul 1 23:32:22 2013 No.5873566 >>5873168 Are you talking about Ed Witten?

>>	Anonymous Mon Jul 1 23:34:36 2013 No.5873573 What the fuck? Why are you guys complicating it so much? It's just a simple addition and subtraction question 10 year old kids can do. http://www.wolframalpha.com/input/?i=1-1%2B1-1%2B1-1%2B1-1%2B1-1%2B1-1 You guys are retarded

>>	Anonymous Mon Jul 1 23:57:23 2013 No.5873598 it's cesaro convergent to 1/2, but diverges using "standard" techniques.

>>	Anonymous Tue Jul 2 02:11:52 2013 No.5873796 >>5873573 >10 year old Addition and subtraction are taught at a much younger age.

>>	Anonymous Tue Jul 2 18:49:23 2013 No.5875364 >>5873796 not in america

>>	Anonymous Wed Jul 3 02:27:55 2013 No.5876183 >>5875364 When is it taught there?

>>	Anonymous Wed Jul 3 02:33:26 2013 No.5876196 >>5876183 it's taught when you're 7 or 8

>>	Anonymous Wed Jul 3 02:42:56 2013 No.5876213 Series does not converge? Only took up to calc 2, what am i doing here?

>>	Anonymous Wed Jul 3 18:33:07 2013 No.5877598 >>5876196 I don't think such advanced math is taught in elementary school.

>>	Anonymous Wed Jul 3 18:36:46 2013 No.5877612 >>5877598 >amurica shit like this is taught at ages 5-7 everywhere else

>>	Anonymous Wed Jul 3 18:45:35 2013 No.5877639 More interesting: 1-2+3-4+5-6+....=1/4

>>	Anonymous Thu Jul 4 07:49:35 2013 No.5878696 >>5860772 the series ended in OP, it was nt continuing to infinity

>>	Anonymous Thu Jul 4 10:12:03 2013 No.5878851 >>5877639 Is this some "p-adic number" trick?

>>	Anonymous Thu Jul 4 19:15:03 2013 No.5880011 >>5877612 I think you mistook it for calculus.

>>	Anonymous Thu Jul 4 19:25:34 2013 No.5880044 >>5878851 Haha, no. You can prove that (1-1+1-1+1-1...)^2=1-2+3-4+5-6..=(1/2)^2. Of course, the series is divergent, but if it were convergent, it would be 1/4.

>>	Anonymous Fri Jul 5 02:28:04 2013 No.5880683 >>5873527 You have not heard of it?

>>	Anonymous Fri Jul 5 18:36:40 2013 No.5882092 >>5880683 No, I haven't. I don't know of a maths nobel prize.

>>	Anonymous Sat Jul 6 02:24:40 2013 No.5882716 >>5880011 What is calculus and how is it different from advanced calculus?

>>	Anonymous Sat Jul 6 02:26:10 2013 No.5882718 >>5858468 divergent series gtfo

>>	Anonymous Sat Jul 6 20:11:39 2013 No.5884192 >>5882716 Advanced calculus is more advanced than calculus. You wouldn't learn about para-quaternionic Kahler manifolds in calculus. But in advanced calculus you do.

>>	Anonymous Sat Jul 6 20:14:10 2013 No.5884196 File: 110 KB, 500x688, linklaugh2.jpg [View same] [iqdb] [saucenao] [google] >>5884192 >para-quaternionic >mfw people still use associative division algebras over global fields

Anonymous Sat Jul 6 20:38:10 2013 No.5884237

>>5882718
Yes the series is divergent, and yes, its sum can be defined in a very clean and unique way, and its value is 1/2. More surprisingly, 1 + 2 + 3 + 4 + · · · = −1/12. See https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_·_·_·

Remember that at some point a math teacher told you you couldn't subtract a number to a smaller number. Later you learned about relative integers. Same for non-integer powers, then for non-integer factorials and so on. Maybe your latest math teacher told you you couldn't define the infinite sum of (-1)^k, but in fact with more advanced maths you can.

>>	Anonymous Sat Jul 6 20:43:10 2013 No.5884247 >>5884237 My maths teacher also told me you cannot square root negative numbers. Are you gonna tell me that's wrong too? How much did that bastard lie to me?

Anonymous Sat Jul 6 20:56:40 2013 No.5884269

>>5884247
You can using complex numbers, but most of the time it's preferable to stick to the classic square root, defined for positive reals. Otherwise you cannot use the natural and useful property sqrt(x*y) = sqrt(x)*sqrt(y).

Indeed sqrt(-1) * sqrt(-1) = sqrt(-1)2 = -1
sqrt((-1)*(-1)) = sqrt(1) = 1

Anonymous Sat Jul 6 21:36:41 2013 No.5884314
File: 89 KB, 1000x1625, 1266374431326.png [View same] [iqdb] [saucenao] [google]

>>5858487
>The problem when people write things as you have is that you are treating "1-1+1-1+1-1+1..." as if without further clarification it is an innately meaningful, physical thing. It is not.

No, but there is a unique* simplest interpretation, and that's good enough.

>>5877639
>>5878851
>>5880044
>pic related

* (up to isomorphism, natch)

>>	Anonymous Sat Jul 6 22:00:39 2013 No.5884359 >>5884237 >Yes the series is divergent, and yes, its sum can be defined in a very clean and unique way, and its value is 1/2 Do you even know what divergent means ?

>>	Anonymous Sun Jul 7 02:36:11 2013 No.5884799 >>5884247 Your mathematics teacher is right. It is best to stay away from the masturbatory vendetta that is modern mathematics.

>>	Anonymous Sun Jul 7 03:00:40 2013 No.5884846 >>5858468 Ah hah, I love Numberphile. But I believe that that would converge to -1/2, because it's wobbling between 0 and -1.

OP Sun Jul 7 03:21:39 2013 No.5884869

Okay guys, OP here and self-saging.

seriously, I posted this shit a week ago hoping to see if anyone agreed whether or not the series genuinely converged.

It devolved into typical /sci/ bullshit about a day after, and yet it hasn't died yet.

Seriously, stop posting.

Anonymous Sun Jul 7 06:52:46 2013 No.5885083
File: 86 KB, 2100x1211, 5858468.png [View same] [iqdb] [saucenao] [google]

>>5884869
The thread is being bumped by a spammer who has been fucking with /sci/ in this manner on and off for about a year now.

>>5862012
>>5862917
>>5864141
>>5864808
>>5865935
>>5866834
>>5867893
>>5868517
>>5870654
>>5871385
>>5873168
>>5873796
>>5875364
>>5876183
>>5877598
>>5878851
>>5880011
>>5880683
>>5882092
>>5882716
>>5884192
>>5884799

Compare
>>5884369
>>5882390
>>5882654
>>5882360

If you want the thread gone before the bump limit, you will have to delete it yourself. Please talk to moot (moot@4chan.org) and the mods (irc.rizon.net, #4chan) about this so that they do something to end the spam.

Anonymous Sun Jul 7 06:58:10 2013 No.5885087

>>5885083
Holy shit, how many threads are you gonna spam with that paranoid delusional crap? Just because you don't like the threads, that doesn't mean there is an evil spammer. People are having science and math discussion on a science and math board. Take your meds and get treatment, you deranged psycho.

>>	Anonymous Sun Jul 7 07:07:09 2013 No.5885093 >>5884359 Yes, I do. Have you even read the wikipedia link?

>>	Anonymous Sun Jul 7 07:08:39 2013 No.5885095 >>5885087 Fuck off, spammer.

>>	Anonymous Sun Jul 7 07:17:39 2013 No.5885105 >>5885095 Jackson 5 GET!

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