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1 + 1 = 3

Name: Anonymous 2006-03-06 1:26

Discuss.

Name: Anonymous 2006-03-06 2:23

∵1+1=3
∴3-1=1

Name: Anonymous 2006-03-06 4:38

anything + anything = everything

anything = everything

Name: Anonymous 2006-03-06 6:20

Let what is cheap be equal to 1.
What is expensive be equal to 2.
Because cheap < expensive.

We all know genuinely cheap things are rare.
But what is rare is expensive.
Thus expensive things are cheap, and 1 = 2 (from the above).
So 1 + 1 = 1 + 2 = 3

SCIENCE MOTHERFUCKERS.
The above is just as valid as the 0.9999~ = 1 bullshit.

Name: Anonymous 2006-03-06 6:22

>>4
Um, no. Expensive things are rare. Maybe you live in Africa where everything is expensive though.

Name: Anonymous 2006-03-06 7:49

>>5
Saying "Rare things is expensive" is different from saying "Expensive things are rare" motherfucker. They're not mutually exclusive either.

Go back to Logic 101.

Name: Anonymous 2006-03-06 8:16

f: A => ℂ
A: unknown class of number
f( 1 + 1 ) = f ( 3 )
f( 2 )     = f ( 3 )
Q.E.D

Name: Anonymous 2006-03-11 2:07

is tis base 10:???????

Name: Anonymous 2006-03-11 13:59

>>8
Even if is lower or greater than base 10 this equation would be wrong if in real or complex class;

Name: Anonymous 2006-03-13 16:54

1+1 = 1.88888888888888888888888888888

Name: Anonymous 2006-03-13 20:13

>>10
fail at funny and at troll

Name: Anonymous 2006-03-14 4:33

>>10
1.888888888888 = 1.9?

Name: Anonymous 2006-03-22 0:23

1.888888888888 = 1.9999999999

Name: Anonymous 2006-03-22 0:48

1.99999999999=1?

Name: Anonymous 2006-03-22 0:59 (sage)

>>6
go back to grade school, equal signs are equivalence relations when used between sets; this entails reflexivity.
the problem is the group of assumptions made on the sets.
C = set of cheap things
E = set of expensive things
if we assume C and E are well defined:
the intersection of C and E is the null set, because things that are expensive are by definition not cheap.
G = set of genuinely cheap things
you could argue that g is not well defined, or that it is in fact equal to C, however assuming some cheap things are not genuinely cheap:
since G is a subset of C, the intersection of G and E is the null set.
apart from that:
R = the set of rare things.
this set is not a subset of E.
if C and E are not well defined because of their subjectivity, we cant use them to make arguments anyway.

Name: Anonymous 2006-03-22 1:04

1+1=2

GET YOUR FACTS RIGHT

Name: Anonymous 2006-03-22 16:07

1.111... = 1

FACT.

Thus: 0.999... = 1.111...

Name: Anonymous 2006-03-22 16:07

>>17
Also, 0.999 + 1.111 = 2.

Name: Styrofoam !DWDMFPPpRw 2006-03-22 20:58

>>17

What the hell are you talking about?  1.111~ != 1.
1.111~ = 10/9.

>>18

Also wrong. 0.999~ + 1.111~ = 2.111~

Name: Anonymous 2006-03-23 8:36 (sage)

>>19 == Pseudoscientific mathematician.

Name: Anonymous 2006-03-23 11:24

0.999... is basically 1, though, because of it's silly representation. This is because if we have a number 0.999... then the number between that and 1 is smaller than any real number. So basically they are the same.

0.999... * 10 = 9.999...
9.999... - 0.999... = 9
9 / 9 = 1

=> 0.999... = 1

(though, this fails in the fact that 0.999... is not real mathematical representation)

Name: Anonymous 2006-03-23 23:37

0.999... = Vagina
1.111... = Penis

0.999 + 1.111 = 3:
Mother
Father
Child

Name: Anonymous 2006-03-24 18:28

consider this:

a = b != 0
a^2 = a*b
(a^2)-(b^2) = (a*b)-(b^2)
(a+b)*(a-b) = b*(a-b)
a b = b
b+b = b
2b = b
2 = 1
1 = 0

therefore any number = 0

Name: Anonymous 2006-03-24 22:54

(a+b)*(a-b) = b*(a-b)
a+b = b

DIVIDE BY ZERO!

Name: Anonymous 2006-03-25 9:11

0.999~ = 0.999~
1 + 1 = 2

Shut the fuck up all of you and get out of my internets.

Name: Anonymous 2006-03-25 13:14

0.999~ + 0.999~ = 2?

Name: Anonymous 2006-03-25 13:29 (sage)

penis + penis = gay ~= this thread

Name: Anonymous 2006-03-25 13:30

age

Name: Anonymous 2006-03-25 13:48

>>1

i lolled

Name: Anonymous 2006-03-27 16:34

>>27
Only if balls touch

Is 1.00 + 1.00 gay? Depends. If it results in 001100 it's not, but if it's 100001 then it is.

Name: Anonymous 2006-03-28 2:51

Synergy

Name: Anonymous 2006-03-28 18:43

Coadjuvancy

Name: Anonymous 2006-03-29 16:50

This is why I stick to arithmetic.

Name: Anonymous 2006-03-30 10:55

>>30
so if the swords cross that's not gay?

Name: Anonymous 2006-03-30 11:26

heresy. hipocrisy. tragedy. That is contemporany United States of North America.

Name: Anonymous 2006-04-03 1:46

The only way this can be right is if your corrupt the meaning of the symbols used to describe it.

Name: lys 2006-04-03 1:48

1+1=3 and pi=3
rejoice....

Name: Anonymous 2006-04-04 5:35

>>23
You, sir, are a fucking moron.

Name: Anonymous 2006-04-04 11:17

Prove that 1+1=3:

1+1=2; by >>23's theorem, 2=0 and 3=0; therefore 2=3.

Name: Anonymous 2006-04-04 15:45

Theorem: 2 = 3
Proof:
1^2 = 1
thus ln(1)/ln(1) = 2
1^3 = 1
thus ln(1)/ln(1) = 3
Clearly they are equivalent.
Q.E.D.
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