Return Styles: Pseud0ch, Terminal, Valhalla, NES, Geocities, Blue Moon.

Pages: 1-4041-

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.

Name: Anonymous 2006-04-04 16:03

>>40

ln(1), you know what it is, motherfucker?

Name: Anonymous 2006-04-04 16:04

Some people have no sense of humor...

Name: Anonymous 2006-04-04 17:47

>>42
Nobody wants that kind of humor

Name: Anonymous 2006-04-04 21:07

>>41
ln(1) is the power to which e would need to be represented in order to equal 1.
but
log base sqrt(1) of 1, i.e. the power to which sqrt(1) would need to be represented in order to equal 1, would be defined and not 0, and you can use any base log for that division.

Name: Anonymous 2006-04-04 22:51 (sage)

>>19

.999+1.111=2.11 not 2.111

.999+1.001=2

Plus, if 2 = 0 and 1 > 0 then 1 > 2 which is broken...

Have fun here...

Name: the pope 2006-04-05 1:56

if 1+1=3, then I am the pope.

Name: Anonymous 2006-04-05 2:49

>>44

yeah so you have

thus ln(1)/ln(1) = 2

thus ln(1)/ln(1) = 3

which gives you the expressions 0/0 which are not allowed.

Name: Anonymous 2006-04-05 2:51

>>44

also for any non zero real number x. x^y = 1 iff y = 0.
sqrt(1) is a non zero number, so
>would be defined and not 0
is clearly false.

Name: Anonymous 2006-04-05 12:01

Trolling /sci people with intentionally flawed logic, misconceptions and nonsense is hilarious.

Name: Mathonymous 2006-04-06 11:54

1,9999+1,1111=3
1,999+1,111=3
1,99+1,11=3
1,9+1,1=3
1(2)+1=3
........../lol

Name: Anonymous 2006-04-06 12:04

>>50
Incorrect numbering. it is meant to be 1.999... and 1.111...
the '...' represent the endless number series

Name: Anonymous 2006-04-06 18:17

Let A be an algebra where:
1 and 3 are constant symbols
+ is a binop symbol
1+1=3 is an axiom

An infinite number of such algebras can be constructed.

Name: Anonymous 2006-04-07 3:59

>>52
Yes, they can, but to anyone but the maker of the algebra would see it as gibberish and the effort would be moot.
If 1 + 1 = 3 then 1 = 1.5, and since 1 = 1, then 1 != 1.5. Therefore, 1 + 1 != 3

Name: Anonymous 2006-04-07 5:04

1=0, 0=1, there is no 2, without 2 there is no 3

Name: Anonymous 2006-04-07 6:42

but 1 obviously does not = 0, because 1 is something, and 0 is nothing

Name: Anonymous 2006-04-07 7:55

>>53
You fail at universal algebra. Badly.

Name: Anonymous 2006-04-07 10:27

>>53
he's saying something like, pretend 3 is the new 2, and 2 is the new three,
1 + 3 = 2
1 + 2 = 4
3 + 2 = 5
he's just pointing out the only way to really ever prove 1 = 2 would require changing the way we intepret the numbers.

Name: Anonymous 2006-04-07 12:26

it's called proof by contradiction, by making a stupid assumption, and showing how stupid the assumption is, therfore falsifying it.

Name: Anonymous 2006-04-07 14:23

>>14

1.99999999=2 dumbass

Name: Anonymous 2006-04-07 15:18

Dumbasses? In my world4ch?

Name: Anonymous 2006-04-14 11:19

1+1((+1)hidden)=3

Name: Anonymous 2006-04-14 19:54

Should I get my math teacher up in hurr?

Name: Anonymous 2006-04-15 7:33

Yes, but he will say this is correct because he has tenure. 

Name: Anonymous 2006-04-15 7:49

NINTENDO 64 GET

Name: Emperor 2006-04-15 8:21

What do you think?

Name: Anonymous 2006-04-15 8:56

公理
・先頭元0が存在する。
・任意の自然数aにはその後者suc(a)が存在する。

定義
suc(0)=1;suc(1)=suc(suc(a))=2
定理
suc(a)+b=a+suc(b)=suc(a+b)
証明
suc(1)=1+1.
また、1=suc(0)
∴suc(suc(0))=2
∴1+1=2 QED

Name: Anonymous 2006-04-15 11:11

>>66
all i see r lots of squares o_o

Name: Anonymous 2006-04-16 1:23

>>67

You don't have the "vision".

Name: Anonymous 2006-04-16 6:00

みんな日本語ではなそうよ

Name: Anonymous 2006-04-16 6:42

>>15
Your use of sets is clearly flawed.
You haven't studied the great fluicity of calculus where if
a fuction is in a specific set of numbers there is a limit to which this function gets close to a number outside of it's own set and thus since numbers are themselfs functions of information it is not flawed that the a number cannot achive a number outside of it sets and even if two numbers of the same function in the same number set cannot achive a number outside of it's set. Thus, you should study the physics of how numbers can formulate in this real plane that we live in.

Name: Anonymous 2006-04-25 1:37

         ∧_∧   ┌────────────
       ◯( ´∀` )◯ < 1 + 1 = 3
        \    /  └────────────
       _/ __ \_
      (_/   \_)
           lll 

Don't change these.
Name: Email:
Entire Thread Thread List