NationStates Jolt Archive


2+2=5

Belschaft
23-12-2008, 00:20
2+2=5



Think about it.
Hydesland
23-12-2008, 00:21
I've thought about it, and discarded it. Now what?
Wilgrove
23-12-2008, 00:21
You need a calculator for Christmas?
Belschaft
23-12-2008, 00:21
I'm right. I really am.
Hydesland
23-12-2008, 00:22
Only if your numeracy system does not have a 4 in it.
Khadgar
23-12-2008, 00:22
Not in base 10 it's not.
Fartsniffage
23-12-2008, 00:23
For very large values of 2.

What is your point?
Saige Dragon
23-12-2008, 00:23
I'll give it to you. 4's a bit if ninny anyways, being all square rootable and shit.
Belschaft
23-12-2008, 00:23
If you lot really think about it the truth is obvious. 2+2=5. It's undeniable.
Hydesland
23-12-2008, 00:24
What are you playing at belschaft?
Wilgrove
23-12-2008, 00:25
If you lot really think about it the truth is obvious. 2+2=5. It's undeniable.

Can I have the stuff that you're on?
Belschaft
23-12-2008, 00:25
2+2=5. How can you lot not realise this? Did you skip maths from year two onwards or something?
FreeSatania
23-12-2008, 00:26
Is this British humor because I don't get it?
Hydesland
23-12-2008, 00:28
2+2=5. How can you lot not realise this? Did you skip maths from year two onwards or something?

Prove it. :p
Heikoku 2
23-12-2008, 00:28
...for high enough values of 2...

And...

SPAM.
Zainzibar Land
23-12-2008, 00:29
1+1=Fish
Chernobyl-Pripyat
23-12-2008, 00:30
There is no spoon.
New Mitanni
23-12-2008, 00:30
"Two and two are four."
"Sometimes, Winston. Sometimes they are five. Sometimes they are three. Sometimes they are all of them at once. You must try harder. It is not easy to become sane." -- George Orwell, 1984
Bouitazia
23-12-2008, 00:31
1+1=1
Fartsniffage
23-12-2008, 00:32
"Two and two are four."
"Sometimes, Winston. Sometimes they are five. Sometimes they are three. Sometimes they are all of them at once. You must try harder. It is not easy to become sane." -- George Orwell, 1984

Bad quote.

It should read "It's not easy to understand maths."
Hydesland
23-12-2008, 00:32
1+1=1

No, 1+1 = 2, here's proof:

The proof starts from the Peano Postulates, which define the natural
numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which
replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.
Wilgrove
23-12-2008, 00:36
No, 1+1 = 2, here's proof:
-snip-

*Brain esplode*
Belschaft
23-12-2008, 00:36
No, 1+1 = 2, here's proof:

The proof starts from the Peano Postulates, which define the natural
numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which
replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

But does that really prove it? How do we know what '1' and '2' are? Are they not meerly concepts of our minds? Do numbers really exist? What colour is imagination?
Hydesland
23-12-2008, 00:37
But does that really prove it?

Yes.


How do we know what '1' and '2' are?

Because we defined it.


Are they not meerly concepts of our minds?

Correct.


Do numbers really exist?

In what sense?


What colour is imagination?

I believe it is dark yellow.
Saige Dragon
23-12-2008, 00:37
But does that really prove it? How do we know what '1' and '2' are? Are they not meerly concepts of our minds? Do numbers really exist? What colour is imagination?

Black. Then trippy multi-coloured waves are overlaid. It's pretty cool.
Fartsniffage
23-12-2008, 00:39
What colour is imagination?

Octarine.
Belschaft
23-12-2008, 00:44
I thought it was flourecent indigo?
UnhealthyTruthseeker
23-12-2008, 00:45
Well, if one accepts that division by zero is a valid operation, in other words, that zero has a multiplicative inverse, then one can solve the following equation and arrive at one unique solution for a and b:

0*a=0*b

Unfortunately, this solution, and the acceptance of the existence of zero's multiplicative inverse, requires that all operators be defined over the trivial set. Namely, the set of one element, 0. In other words, given this set, 2+2 does in fact equal 5 because if we are talking about 2 and 5 being elements of this trivial set, then they must be equal to 0. However, this is not a desirable situation, as it requires the complete sacrifice of any usable mathematical structure.

More concrete example:

0*4 = 0*5 (This is true over any ordinary set of numbers, but the next step isn't)

Multiply both sides through by 1/0:

4 = 5

This is a more concrete example of what must be true if we allow for zero to be a denominator.
Flimflammapip
23-12-2008, 00:45
If you beat someone with a club everyday, then starved them, then locked them in a five by five cell, then asked them questions and shocked them when they don't give you the answer you want to hear, and then hold a carnivorous rat up to their face, then I'm pretty sure they'd agree that 2 + 2 = 5.

I know some members of the Thought Police who could do that for you...
The blessed Chris
23-12-2008, 00:46
No, 1+1 = 2, here's proof:

The proof starts from the Peano Postulates, which define the natural
numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which
replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

This makes my brain bleed. And the OP is a fuckwit.
Belschaft
23-12-2008, 00:48
If you beat someone with a club everyday, then starved them, then locked them in a five by five cell, then asked them questions and shocked them when they don't give you the answer you want to hear, and then hold a carnivorous rat up to their face, then I'm pretty sure they'd agree that 2 + 2 = 5.

I know some members of the Thought Police who could do that for you...

There is no thought police - your meerly paranoid! Don't make me send you to room 101!
Trollgaard
23-12-2008, 00:50
If you are defining 4 as 5, then 2+2 would equal 5.
Saige Dragon
23-12-2008, 00:50
I thought it was flourecent indigo?

While 2+2 can equal 5 if 4 is busy being 3's bitch, imagination can never be fluorescent indigo. Why? Because it wouldn't match the tie.
Aceopolis
23-12-2008, 00:52
Simple non-brain-explodey proof that 2+2=4

Two letters: Sp
Two letters: am

Sp+am=Spam which is four letters
Heikoku 2
23-12-2008, 00:54
There is no thought police - your meerly paranoid! Don't make me send you to room 101!

*A programmer passing by* Why? What's in room 5?
Belschaft
23-12-2008, 00:55
While 2+2 can equal 5 if 4 is busy being 3's bitch, imagination can never be fluorescent indigo. Why? Because it wouldn't match the tie.

But I wanted it to have a turquoise tie......
Bouitazia
23-12-2008, 00:59
Math is always math, and the only tangible Truth we can be certain of.
It is however also a word, and an abstract concept.
No one would refute that one pile of salt added to another pile of salt equals anything other than one (albeit bigger) pile of salt.
And lets not even talk of different bases and higher math.
So the OP is wrong, but he may also be right, depending on his own definition.

Unless I am completely off here?
UnhealthyTruthseeker
23-12-2008, 01:02
Math is always math, and the only tangible Truth we can be certain of.
It is however also a word, and an abstract concept.
No one would refute that one pile of salt added to another pile of salt equals anything other than one (albeit bigger) pile of salt.
And lets not even talk of different bases and higher math.
So the OP is wrong, but he may also be right, depending on his own definition.

Unless I am completely off here?

Well, you're confusing quantitative and qualitative analysis. The quantity of sand in the final pile is the sum of the quantity of sand in the two piles. Pile is not a quantitative measurement, so it needn't obey any particular algebra.
Holy Cheese and Shoes
23-12-2008, 01:21
No, 1+1 = 2, here's proof:

The proof starts from the Peano Postulates, which define the natural
numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which
replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

That is not a self-contained proof, if your theory is based on defining natural numbers it's based on an unprovable axiom (or infinite recursive axioms in this case, I believe?) You're going to have to work out whether you want it to be true OR provable, according to Godel's incompleteness theorem. It is therefore no more true or provable than 2+2=5

Hmmm, on further investigation it still appears to be an open question (http://en.wikipedia.org/wiki/Peano_arithmetic#Consistency). :tongue:
Tamuff
23-12-2008, 01:30
What if there were covert decimal points making it 2.5 + 2.5 = 5.0?
Belschaft
23-12-2008, 01:32
What if there were covert decimal points making it 2.5 + 2.5 = 5.0?

No. 2+2=5
Katganistan
23-12-2008, 01:36
2+2=5



Think about it.
No it's not, O'Brien. You can never get into my mind and change me!
Belschaft
23-12-2008, 01:38
Ah.......

Three pages of mind games. Now i have things too do. Remember fols -


There is no spoon!
Katganistan
23-12-2008, 01:38
Octarine.
You must be a wizzard! :)
Fennijer
23-12-2008, 02:23
2+2=5

That sounds a little communist to me...

...or maybe something by Radiohead.
Hoyteca
23-12-2008, 02:33
Anpu has decreed that 2+2 may indeed equal 5. Anyone who disagrees is an enemy of all of Zepland and shall be crushed by Abraham Lincoln.
[NS]Kagetora
23-12-2008, 02:35
here's one, 1=-1

A=B, and B=C, then A must equal C, correct?

A=1
B=Square root of 1
C=-1

1=-/1=-1
Rotten bacon
23-12-2008, 02:35
well my friend actually proved it back in my junoir year of high school. using sig figs
2 + 2 acutally can equal five. dont ask me how to explain it. chemistry was never my strong point.
Lord Tothe
23-12-2008, 02:47
Under the rules of algebraic mathematics in the common base-10 numerical system, the OP is incorrect. Under other circumstances, he may be right. For example, in spherical trig, I can draw a triangle with 3 90-degree corners or draw arguably parallel lines that are not equidistant from each other at all points.
Miraclia
23-12-2008, 03:00
2+2=5

Think about it.

No, 3+3=7, because God created everything in 3+3=7 days!
Trollgaard
23-12-2008, 03:07
well my friend actually proved it back in my junoir year of high school. using sig figs
2 + 2 acutally can equal five. dont ask me how to explain it. chemistry was never my strong point.

Fuck sig figs!
Soheran
23-12-2008, 03:20
What if there were covert decimal points making it 2.5 + 2.5 = 5.0?

Then you'd get 3 + 3 = 5, not 2 + 2 = 5.
Nadkor
23-12-2008, 03:24
Good song. Not one of their best, though.
CthulhuFhtagn
23-12-2008, 03:25
Then you'd get 3 + 3 = 5, not 2 + 2 = 5.

Yeah, you have to do 2.4 + 2.4 = 4.8 or something along those lines.
IL Ruffino
23-12-2008, 03:56
26+6=1

:mp5:
Rotten bacon
23-12-2008, 05:44
Fuck sig figs!

believe me, i totally agree with you.
Amor Pulchritudo
23-12-2008, 05:50
Balls. 2+2= orgy!
Hoyteca
23-12-2008, 06:17
Balls. 2+2= orgy!

Bull. You'd never get away with saying such a falacy anywhere in Zepland, especially here in Carouselambra.
Lord Grey II
23-12-2008, 06:22
2+2=5



Think about it.

It is true. Allow me to prove it to everyone.

Choose any real number for A and B. Let A + B = T

A + B = T

(A + B) * (A - B) = T * (A - B)

A^2 - B^2 = TA - TB

A^2 - TA = B^2 - TB

A^2 - TA + ((T^2)/4) = B^2 - TB + ((T^2)/4)

(A - T/2)^2 = (B - T/2)^2

A - T/2 = B - T/2

A = B

See? I can make any number equal to any other number! So, 2+2 is easily equal to 5. Also 5,000,000,000,000, but that's not the point.

For those who figure out the trick, don't tell. It's more fun that way.
Hoyteca
23-12-2008, 06:27
It is true. Allow me to prove it to everyone.

Choose any real number for A and B. Let A + B = T

Blah blah blah numbers.

For those who figure out the trick, don't tell. It's more fun that way.

And if A+B=T, then A^2+AB+BT=T^2. How do you explain this?
[NS]Kagetora
23-12-2008, 06:29
OK, here. All numbers are either infinity or zero, which must be equal to each other

A=B, B=C, so A=C

A= any number times 0
B=0
C= any number times 0

A0=0=B0
Divide all by 0
What happens when you divide by zero? You get infinity (I know it's not actually a number)
so,
infinity=0/0=infinity
Vetalia
23-12-2008, 06:30
2+2=10...

...in base four, I'm fine!
Nanatsu no Tsuki
23-12-2008, 06:58
Keh. As if math could be changed on a whim.
SaintB
23-12-2008, 07:21
And I get some of my threads locked as spam...
Hoyteca
23-12-2008, 07:23
And I get some of my threads locked as spam...

That's because Anpu wanted some of your threads to be locked.
Naturality
23-12-2008, 07:30
2+2=5



Think about it.

That what we think we know isn't set in stone? That there may very well be other ways to get to where we want to go, than what we know now?

Other than that , I don't know what you are talking about. Plain out 2+2=4 to me mathematically. But I'm no math genius.
UnhealthyTruthseeker
23-12-2008, 07:32
It is true. Allow me to prove it to everyone.

Choose any real number for A and B. Let A + B = T

A + B = T

(A + B) * (A - B) = T * (A - B)

A^2 - B^2 = TA - TB

A^2 - TA = B^2 - TB

A^2 - TA + ((T^2)/4) = B^2 - TB + ((T^2)/4)

(A - T/2)^2 = (B - T/2)^2

A - T/2 = B - T/2

A = B

See? I can make any number equal to any other number! So, 2+2 is easily equal to 5. Also 5,000,000,000,000, but that's not the point.

For those who figure out the trick, don't tell. It's more fun that way.

Square root of a number squared is only guaranteed to be that number if the number is positive or zero. If A=B, then both A and B are 1/2 of T so A - T/2 is 0 and B - T/2 is zero. However, if they aren't the same, then one of the two, either A - T/2 or B - T/2 is negative. The square root of a negative number squared is the additive inverse of that number. In other words, you need to put the "plus or minus" symbols in to denote the possibility of opposite signs. Both sides having the same sign => A = B. The two sides being differently signed => A + B = T, which we already knew. It was more clever than those hidden divide by zero errors, though.
Naturality
23-12-2008, 07:37
Square root of a number squared is only guaranteed to be that number if the number is positive.

Ok .. do that right now with 7 9 and 14
Intangelon
23-12-2008, 07:39
Well, if one accepts that division by zero is a valid operation, in other words, that zero has a multiplicative inverse, then one can solve the following equation and arrive at one unique solution for a and b:

0*a=0*b

Unfortunately, this solution, and the acceptance of the existence of zero's multiplicative inverse, requires that all operators be defined over the trivial set. Namely, the set of one element, 0. In other words, given this set, 2+2 does in fact equal 5 because if we are talking about 2 and 5 being elements of this trivial set, then they must be equal to 0. However, this is not a desirable situation, as it requires the complete sacrifice of any usable mathematical structure.

More concrete example:

0*4 = 0*5 (This is true over any ordinary set of numbers, but the next step isn't)

Multiply both sides through by 1/0:

4 = 5

This is a more concrete example of what must be true if we allow for zero to be a denominator.

But we don't. Next!

It is true. Allow me to prove it to everyone.

Choose any real number for A and B. Let A + B = T

A + B = T

(A + B) * (A - B) = T * (A - B)

A^2 - B^2 = TA - TB

A^2 - TA = B^2 - TB

A^2 - TA + ((T^2)/4) = B^2 - TB + ((T^2)/4)

(A - T/2)^2 = (B - T/2)^2

A - T/2 = B - T/2

A = B

See? I can make any number equal to any other number! So, 2+2 is easily equal to 5. Also 5,000,000,000,000, but that's not the point.

For those who figure out the trick, don't tell. It's more fun that way.

Bolded does not follow. Next!
Qazox
23-12-2008, 07:47
Kagetora;14328002']here's one, 1=-1

A=B, and B=C, then A must equal C, correct?

A=1
B=Square root of 1
C=-1

1=-/1=-1

One of my favorites.

But here's an easier one to figure out:

2+2=6.
Naturality
23-12-2008, 07:50
One of my favorites.

But here's an easier one to figure out:

2+2=6.

So basically any number plus/minus/whatever any number can equal any number with enough manipulation?
[NS]Kagetora
23-12-2008, 07:50
Here are some of my favorites.

Half of 8 is 3, and half of 9 is 4. Half of 13 is 3.

How?
Naturality
23-12-2008, 07:52
Duh
Bouitazia
23-12-2008, 07:53
Kagetora;14328850']Here are some of my favorites.

Half of 8 is 3, and half of 9 is 4. Half of 13 is 3.

How?

Split eight down the middle,
Dont know..
Split 13, 1 and 3, take away 1..left is 3.. easy ,)
Naturality
23-12-2008, 07:53
"So basically any number plus/minus/whatever any number can equal any number with enough manipulation?"


DUH
[NS]Kagetora
23-12-2008, 07:54
Good. You understand geometry.

write 9 like on a clock

._
|_|
.. |
Qazox
23-12-2008, 07:55
I'll end the suspense early.

TWO+TWO= SIX

three letters +three letters= six letters..DUH!
Naturality
23-12-2008, 07:55
Show me its actually math and not just take and give to make.

Not that I'd be able to decipher atm.. but any who.
Naturality
23-12-2008, 07:57
In other words show me it's not just smoke and mirrors .. with numbers.
Philosopy
23-12-2008, 08:41
2+2=5



Think about it.

Are you a first year philosophy student, by any chance?
Everywhar
23-12-2008, 09:22
2+2=5

As a computer science student, I am offended. Why did you just trash arithmetic?

In a standard decimal number system with the standard operations and full closure, 2+2 simplifies to 4. And by the reflexive property, 4=4 and 5=5. But 4 != 5. It's not a hard concept.
Naturality
23-12-2008, 09:46
2+2=5

As a computer science student, I am offended. Why did you just trash arithmetic?

In a standard decimal number system with the standard operations and full closure, 2+2 simplifies to 4. And by the reflexive property, 4=4 and 5=5. But 4 != 5. It's not a hard concept.

4 ! means 4 forever? On? That's what ! means in my wow macros. I'm sure they got that shit from math.
Risottia
23-12-2008, 09:49
"Two and two are four."
"Sometimes, Winston. Sometimes they are five. Sometimes they are three. Sometimes they are all of them at once. You must try harder. It is not easy to become sane." -- George Orwell, 1984

2+2=5 because there's the enthusiasm of the workers, as the poster for the 1st five-years plan went (because Stalin wanted CCCP to reach the objectives of the plan in 4 years only). That's where Orwell took it from.

Anyway, 2+2=5, why not? All we have to do is to define another abelian group with another sum on Z.
Hoyteca
23-12-2008, 09:58
Are you a first year philosophy student, by any chance?

Must be. He forgot to factor in the properties of Zeppelin, the awesomeness of Lincoln, and Anpu in general when coming up with his mathematical thesis. For shame.
Bokkiwokki
23-12-2008, 10:02
Simple:

a = b => a² = ab => a² - b² = ab - b² => (a+b)(a-b) = b(a-b) => a+b = b => b + b = b => 2 = 1
Therefore, 2 + 2 = 5 => 2 + (1 + 1) = 5 => 2 + (2 + 1) = 5 QED

Or sumfn along those lines. :D
Risottia
23-12-2008, 10:11
Simple:

a = b => a² = ab => a² - b² = ab - b² => (a+b)(a-b) = b(a-b) => a+b = b => b + b = b => 2 = 1
Therefore, 2 + 2 = 5 => 2 + (1 + 1) = 5 => 2 + (2 + 1) = 5 QED

Or sumfn along those lines. :D

Proof invalidated. Multiplication and then division by 0 (bolded) because your hypotesis says a=b, hence a-b=0 . Another possible division by 0 (in red) because you didn't state b != 0 .
Bokkiwokki
23-12-2008, 10:15
Proof invalidated. Multiplication and then division by 0 (bolded) because your hypotesis says a=b, hence a-b=0 . Another possible division by 0 (in red) because you didn't state b != 0 .

Sshhhhhhh! Don't spoil this perfectly valid proof! :tongue:
Or I'll have to bore you with any of a large number of others of this kind (http://en.wikipedia.org/wiki/Invalid_proof). :D
Risottia
23-12-2008, 10:48
Sshhhhhhh! Don't spoil this perfectly valid proof! :tongue:
Or I'll have to bore you with any of a large number of others of this kind (http://en.wikipedia.org/wiki/Invalid_proof). :D

Aha! You're attempting one of these (http://en.wikipedia.org/wiki/Proof_by_intimidation)! :tongue:
The Archregimancy
23-12-2008, 11:12
2+2=5



Think about it.

Minitrue writes original post doubleplusgood duckspeak goodthink.

Disagreement crimethink.

BB says 2+2=5 Inner Party goodthinkful as it explains plusungood chocolate ration figures for final quarter.
Risottia
23-12-2008, 11:18
Minitrue writes original post doubleplusgood duckspeak goodthink.

Disagreement crimethink.

BB says 2+2=5 Inner Party goodthinkful. Minitrue use it in news about chocolate ration figures for final quarter.

Previous post corrected, secbreaching and crimethink. Chocoration is always plusgood.
The Archregimancy
23-12-2008, 11:22
Previous post corrected, secbreaching and crimethink. Chocoration is always plusgood.

Stand corrected. Previous post crimethink. Yes, chocoration always fulfilled plusgoodwise.

Submit self to Miniluv for further correction.
UNIverseVERSE
23-12-2008, 11:51
Kagetora;14328002']here's one, 1=-1

A=B, and B=C, then A must equal C, correct?

A=1
B=Square root of 1
C=-1

1=-/1=-1

False. B = \sqrt{1} = \pm 1

Therefore all you have said are that the two square roots of one are one and minus one, not that one and minus one are equal.

Kagetora;14328610']OK, here. All numbers are either infinity or zero, which must be equal to each other

A=B, B=C, so A=C

A= any number times 0
B=0
C= any number times 0

A0=0=B0
Divide all by 0
What happens when you divide by zero? You get infinity (I know it's not actually a number)
so,
infinity=0/0=infinity

False, division by zero is undefined, and

ak = bk -> a = b

only holds for non-zero k
Extreme Ironing
23-12-2008, 12:04
Five is right out. Three is the number of the counting, and the number of the counting is three. Amen, Brother Maynard.
Zainzibar Land
23-12-2008, 12:19
1..2..5! Three sir. 3!
Risottia
23-12-2008, 12:32
1..2..5! Three sir. 3!

"five is right out. once the number three, being the third number be reached, then lobbest thou thy holy hand grenade of antioch towards thy foe, who being naughty in my sight, shall snuff it."

http://www.youtube.com/watch?v=xOrgLj9lOwk
Lord Grey II
23-12-2008, 16:29
<snip>
A + B = T

(A + B) (A - B) = T * (A - B)
<snip>

Bolded does not follow. Next!

Sorry to revive this nearly dead thread, but... what? It follows perfectly fine. All you're doing there is multiplying both sides by (A - B). That's perfectly allowed... the math is almost perfect for the whole thing.

Yes, there is a trick, but it's not that... in fact, it's been posted at least twice I think...
South Lorenya
23-12-2008, 16:42
Ah, so that's why the economy is the way it is...
Dondolastan
23-12-2008, 22:18
The economy is the way it is because money isn't real.
UNIverseVERSE
23-12-2008, 22:50
Sorry to revive this nearly dead thread, but... what? It follows perfectly fine. All you're doing there is multiplying both sides by (A - B). That's perfectly allowed... the math is almost perfect for the whole thing.

Yes, there is a trick, but it's not that... in fact, it's been posted at least twice I think...

That's not valid if A = B, and you didn't exclude the possibility. There are also further steps of dubious validity, primarily involving square roots.
Christmahanikwanzikah
23-12-2008, 23:18
2 + 2 - i^2 = 5

>.>

Yay imaginaries!
UnhealthyTruthseeker
23-12-2008, 23:27
That's not valid if A = B, and you didn't exclude the possibility. There are also further steps of dubious validity, primarily involving square roots.

It's perfectly valid to MULTIPLY through both sides by 0, as long as you remember that there may be an inherent loss of information involved in doing so. Just like you need to remember when taking a square root that you need to account for the possible addition of extraneous data.
UNIverseVERSE
23-12-2008, 23:32
It's perfectly valid to MULTIPLY through both sides by 0, as long as you remember that there may be an inherent loss of information involved in doing so. Just like you need to remember when taking a square root that you need to account for the possible addition of extraneous data.

Well, yes, it's valid. However multiplying both sides by zero does render the rest rather meaningless, which is what I'm referring to. And yes, I know my language was unclear.

As for the rest, I'm on my holiday, I'm not going to be thinking about it at this point.
Intangelon
24-12-2008, 02:47
Sorry to revive this nearly dead thread, but... what? It follows perfectly fine. All you're doing there is multiplying both sides by (A - B). That's perfectly allowed... the math is almost perfect for the whole thing.

Yes, there is a trick, but it's not that... in fact, it's been posted at least twice I think...

Yes, but by doing the same thing to both sides, you've not changed the original equation at all. And since that doesn't work, nothing that follows works.
Ifreann
24-12-2008, 02:58
Kagetora;14328002']here's one, 1=-1

A=B, and B=C, then A must equal C, correct?

A=1
Therefore B = 1 and C = 1.
Belschaft
24-12-2008, 03:02
I serioulsy can't believe you guys are still arguing over this. The secret behind 2+2=5 -

I was bored and just typed up some random two sentence crap. There is no magic formula. I was justing messing with you.

Hope your not pissed off.
Holy Cheese and Shoes
24-12-2008, 03:04
Now you've gone and spoiled it! We could have amused ourselves for days....
Glen-Rhodes
24-12-2008, 03:14
So.. you want us to increase production so that we could achieve in four years what you asked us to achieve in five? You bastard. Where's your Christmas spirit?

Your nonchalant "it was a joke" doesn't fly anymore.
Holy Cheese and Shoes
24-12-2008, 03:19
So.. you want us to increase production so that we could achieve in four years what you asked us to achieve in five? You bastard. Where's your Christmas spirit?

Your nonchalant "it was a joke" doesn't fly anymore.

LOL!

Yes, I can see it now, Stalin's last words : "It was all just a joke! I never thought you would take me seriously!"
Belschaft
24-12-2008, 03:27
Damn - foiled again! My refernce to a joke was, as you have guessed, meerly a poorly planned atttempt to increase worker productivity. Now I'll have to send both of you to the Gulag.
German Nightmare
24-12-2008, 03:30
2+2=5



think about it.
nö!
Christmahanikwanzikah
24-12-2008, 04:53
I serioulsy can't believe you guys are still arguing over this. The secret behind 2+2=5 -

I was bored and just typed up some random two sentence crap. There is no magic formula. I was justing messing with you.

Hope your not pissed off.

Oh wow. What a surprise. I couldn't possibly believe that. I mean, I spent my entire day thinking of a way to solve this problem. I'm so distraught.

...

Naw. :D
Lord Grey II
24-12-2008, 05:03
Yes, but by doing the same thing to both sides, you've not changed the original equation at all. And since that doesn't work, nothing that follows works.

... what?

Since when can't you add things to equations in such a way that the original equation doesn't change? Damn, I can't count the number of times in Algebra through Calculus the teacher said "and now here you multiply by a special one in order to solve..." Unless you're implying that math is wrong... Nah. God, I hope you're not right. I'd hate to be going into Calculus III and still not have a damn clue what I'm doing. ;)

I mean, I realize the equation is bunk. I realize this. I posted it as a clever math joke. But, you are fixated on the part that is... correct.

And kudos to the person who mentioned that A and B can't be equal in the given. I never even thought of that. I'll have to add that in when I write it places.
Gaeltach
24-12-2008, 05:12
Simple! Take two pieces of rope and tie two knots in each. When the ropes are put together there are magically five knots. :D

In all seriousness, I recall doing this proof waaaaaay back in high school but I have no recollection of how it was shown back then.
UnhealthyTruthseeker
24-12-2008, 05:36
Yes, but by doing the same thing to both sides, you've not changed the original equation at all. And since that doesn't work, nothing that follows works.

You'd be surprised what multiplying both sides of an equation through by the same value can do. In fact, that's exactly what the method of integrating factors, useful for solving certain types of differential equations, is. Another seemingly useless but actually rather useful technique is multiplying something by one. It can be used to construct common denominators, get i's (the imaginary constant) out of denominators, turn impossible integrands into elementary integrals, and much more.
Amor Pulchritudo
24-12-2008, 06:03
Bull. You'd never get away with saying such a falacy anywhere in Zepland, especially here in Carouselambra.

What, you prefer threesomes to foursomes?