NationStates Jolt Archive

2+2=5



Think about it.

I've thought about it, and discarded it. Now what?

You need a calculator for Christmas?

I'm right. I really am.

Only if your numeracy system does not have a 4 in it.

Not in base 10 it's not.

For very large values of 2.

What is your point?

I'll give it to you. 4's a bit if ninny anyways, being all square rootable and shit.

If you lot really think about it the truth is obvious. 2+2=5. It's undeniable.

What are you playing at belschaft?

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?

2+2=5. How can you lot not realise this? Did you skip maths from year two onwards or something?

Is this British humor because I don't get it?

2+2=5. How can you lot not realise this? Did you skip maths from year two onwards or something?

Prove it. :p

...for high enough values of 2...

And...

SPAM.

1+1=Fish

There is no spoon.

"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

1+1=1

"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."

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.

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

*Brain esplode*

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?

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.

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.

What colour is imagination?

Octarine.

I thought it was flourecent indigo?

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.

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...

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.

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!

If you are defining 4 as 5, then 2+2 would equal 5.

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.

Simple non-brain-explodey proof that 2+2=4

Two letters: Sp
Two letters: am

Sp+am=Spam which is four letters

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?

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......

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?

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.

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:

What if there were covert decimal points making it 2.5 + 2.5 = 5.0?

What if there were covert decimal points making it 2.5 + 2.5 = 5.0?

No. 2+2=5

2+2=5



Think about it.
No it's not, O'Brien. You can never get into my mind and change me!

Ah.......

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


There is no spoon!

Octarine.
You must be a wizzard! :)

2+2=5

That sounds a little communist to me...

...or maybe something by Radiohead.

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.

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

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.

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.

2+2=5

Think about it.

No, 3+3=7, because God created everything in 3+3=7 days!

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!

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.

Good song. Not one of their best, though.

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.

26+6=1

:mp5:

Fuck sig figs!

believe me, i totally agree with you.

Balls. 2+2= orgy!

Balls. 2+2= orgy!

Bull. You'd never get away with saying such a falacy anywhere in Zepland, especially here in Carouselambra.

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.

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?

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

2+2=10...

...in base four, I'm fine!

Keh. As if math could be changed on a whim.

And I get some of my threads locked as spam...

And I get some of my threads locked as spam...

That's because Anpu wanted some of your threads to be locked.

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.

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.

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

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!

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.

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?

Here are some of my favorites.

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

How?

Duh

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 ,)

"So basically any number plus/minus/whatever any number can equal any number with enough manipulation?"


DUH

Good. You understand geometry.

write 9 like on a clock

._
|_|
.. |

I'll end the suspense early.

TWO+TWO= SIX

three letters +three letters= six letters..DUH!

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.

In other words show me it's not just smoke and mirrors .. with numbers.

2+2=5



Think about it.

Are you a first year philosophy student, by any chance?

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.

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.

"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.

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.

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

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 .

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

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:

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.

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.

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.

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

Five is right out. Three is the number of the counting, and the number of the counting is three. Amen, Brother Maynard.

1..2..5! Three sir. 3!

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

<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...

Ah, so that's why the economy is the way it is...

The economy is the way it is because money isn't real.

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.

2 + 2 - i^2 = 5

>.>

Yay imaginaries!

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.

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.

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.

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.

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.

Now you've gone and spoiled it! We could have amused ourselves for days....

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.

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!"

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.

2+2=5



think about it.
nö!

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

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.

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.

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.

Bull. You'd never get away with saying such a falacy anywhere in Zepland, especially here in Carouselambra.

What, you prefer threesomes to foursomes?