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1 = 0

Name: Anonymous 2007-01-26 15:14

Proof:
1 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ....
  = (1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + ......
  = 0

kekekeke

Name: Anonymous 2007-01-26 15:14

BTW, the second line is the associative property.

Name: Anonymous 2007-01-26 15:15

The second line to the third line, I mean.

Name: Anonymous 2007-01-26 16:00

1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + .... = indeterminant

Name: Anonymous 2007-01-26 16:02

>>4
Indeterminate*

Name: Anonymous 2007-01-26 17:07

Just because you moved the parenthesis around doesn't make the result different.

Name: Anonymous 2007-01-26 17:27

>>6
Truth, as per the associative property of addition. Thus, 1 = 0.

Name: Anonymous 2007-01-26 17:49

>>7
Associative property of addition says a + (b + c) = (a + b) + c. This can be extended by induction in various ways to any finite sum, but not necessarily to infinite sums.

Name: Anonymous 2007-01-26 20:55

>>8
For the simple reason that the OP provided for us.

>>4
1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... = 1. What we are learning ITT is that
1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... != (1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + ...
.

By the way, there are only three dots in an ellipsis.

Name: Anonymous 2007-01-26 21:08

If you can come up with a field where the additive identity element and the multiplicative identity element are equal, then yes..  1 = 0 IN THAT FIELD.

Note this field isn't the real numbers (is that even a field?)

Name: Anonymous 2007-01-26 21:39

>>10
Er, wouldn't you only have that [the additive identity element in that field] = [the multiplicative identity element in that field] in that field? Even if they were called 1 and 0 respectively, they wouldn't be the same numbers we know.

Name: Anonymous 2007-01-26 21:40

>>10
It is by definition not a field. 0 and 1 are required to be different elements as part of the field axioms. Furthermore, even if you ignore that axiom, you end up with the trivial field {0} being the only one in which the additive and multiplicative identities are equal.

Name: Anonymous 2007-01-26 22:29

>>4
Not necessarily.  The Cesaro summation of that series is 1/2.

Name: Anonymous 2007-01-26 22:43

>>13
more liek 0

Name: Anonymous 2007-01-26 23:09 (sage)

>>13
Only if you leave out the brackets, dumbshit.

Name: Anonymous 2007-01-27 0:56

(1+-1)=a
(-1+1)=b
a=b=1-1=0
a+a+a+...=0+0+0+...=b+b+b+...=0

"1 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ....
  = (1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + ......
  = 0"
1=1+a+a+a+...=1+0
1=b+b+b+...+1=0+1
1=1

Name: Anonymous 2007-01-27 0:56

>>15
You fail at basic math.  Go back and do the result.

Name: Anonymous 2007-01-27 1:16

>>16
(1+-1)=a
(-1+1)=b
a=b=1-1=0
a+a+a+...=0+0+0+...=b+b+b+...=0

"1 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ....
  = (1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + ......
  = 0"
1=1+b+b+b+...=1+0
1=a+a+a+...+1=0+1
1=1
*

fix'd

Name: Anonymous 2007-01-27 2:04 (sage)

>>17
This involves infinity, so it's not 'basic' math. Additionally, I don't fail.

1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ...
   means
1 + sum 0..inf (-1 + 1) == 1 + sum 0..inf (0) == 1
   and not
sum n=0..inf (-1^n)

Name: Anonymous 2007-01-27 4:31

>>19
You do fail at basic mathematics dumbass because your expression is not a Cesàro summation for this series.

Name: Anonymous 2007-01-27 17:23

0 * infinity lawl

Name: Anonymous 2007-01-27 19:05 (sage)

>>20
Which expression? And how exactly is that relevant?
stop failing plox

Name: Anonymous 2007-01-28 14:46

law of L. Jenkins; always devide your outcome by 0

Name: Anonymous 2007-01-29 7:08

Perhaps this thread is just an elaborate troll to get people to talk about Guido Grandi?

Name: Anonymous 2007-01-29 15:51

uh, no...

1 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ....
  = (1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + ... + 1
  = 1

Name: Anonymous 2007-01-29 17:43

>>25

no.

Name: Anonymous 2007-01-29 18:06

>>24
I was thinking of the whole making something out of nothing concept but no, this is just a simple troll.

Name: dfgtfd 2007-01-29 20:53 (sage)

1*0=1*6
1=6
lolz

Name: Anonymous 2007-01-30 3:18

Name: Anonymous 2007-01-30 10:26

>>26
That's not a compelling argument.

Name: Anonymous 2007-01-30 10:27

Name: Anonymous 2007-01-30 11:50

>>18
>>25
same thing

Name: Anonymous 2007-01-30 18:27

The problem is that the partial sums , sum (i=0 to k) (-1)^i
doesn't really converge to a limit, so the infinite sum cannot be defined.

Name: Anonymous 2007-01-30 19:27

Simple Calc

gb2/grammerskool

Name: Anonymous 2007-02-01 17:52

I haven't been here for months and still there are dipshits debating the most retarded math problems - this one being a case in point.

>>25
>>18
:)
>>26
retard
>>1
retard in training

Name: Anonymous 2007-02-01 18:28

>>35
Fail nigger, because >>25 and >>18 makes no sense.

Name: Anonymous 2007-02-01 21:12

>>36
You're really retarded, then. 18 makes perfect sense.

Name: Anonymous 2007-02-01 23:20

Tell me >>25, what does the ellipses in:
(1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + ... + 1
supposed to mean?

Name: Anonymous 2007-02-02 10:54

>>1
Fail

1 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ....
you moved the one to the other side leaving you with->(0)  =      0->(1 + -1) + (1 + -1) + (1 + -1) + (1 + -1) + ......
 0 = 0

FAIL return to school retard

Name: Anonymous 2007-02-02 11:34

DOES ANYONE THINK >>1 IS NOT A RETARD?

IF SO YOU ARE A RETARD!

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