2=1

[dragonfly]

Proof that 2 = 1: (note: "^" denotes exponent)

 

a = b

a^2 = ab

a^2 - b^2 = ab - b^2

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

(a + = b

(b + = b

2b = b

2 = 1

 

All the other mathematicians will get this right away. Even if you're not one I'm sure you'll get it. But if you use it in highschool they'll think you're a genius.

Edited January 6, 2007 by DarkArchon

[dragonfly]

Someone must see the flaw of logic in here...

[TheGeek]

All of its flawed

Especially this part:

(a + B ) = b

(b + B ) = b

 

your assuming that a=b right? Making this equation your basing it that they are different. That would be able to convince some 9th or 10th graders.

Edited January 8, 2007 by TheGeek

[dragonfly]

Actually, there is only one flaw in there. Those B's are not supposed to be capitalized, my bad. But there is only one flaw and it is before those lines. a=b assumes that a has the same value as b. The steps are all valid (except one) to lead to this point, which is the purpose of a proof. It's all factoring, adding to both sides, subtracting from both sides, squaring both sides, etc.

 

Good try, but there's only one small flaw...

Edited January 8, 2007 by DarkArchon

[ZesteR]

( a + b ) = b

( b + b ) = b

 

you're saying the quantity a + b equals b

then you say the quantity b + b equal b

 

you cannot change one term inside a quantity

 

i'm seconded guessing myself, this part and the VERY first part confuses me.

 

edit: a lower case b next to a closed parenthesis created a smilie with big sunglasses on lol

Edited January 12, 2007 by ZesteR

[appalachian_fox]

It's an oldie but goodie.

 

I'd suggest doing some variable substitution to find the flaw. Remember, a = b. Reduce everything to a one-variable operation and it will become more clear.

 

Zester, you're on the right track. You CAN change the variable name, because a=b. However, look more closely at what those equations say about the values of those variables.

 

Geek, you're on the same track, too, but you are missing something important. There does, in fact, exist a number for which those equations are true.

[ZesteR]

I'm busy at work again.. but when things slow down, ill figure it out again..

thanks for the tip

[NOFX]

Proof that 2 = 1: (note: "^" denotes exponent)

 

a = b

a^2 = ab

a^2 - b^2 = ab - b^2

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

(a + = b

(b + = b

2b = b

2 = 1

 

The flaw starts at

 

(a + = b

 

lets say A=3

3=3

3^2 = 3*3...................................9=9

(3^2)-(3^2)=(3*3)-(3^2)..............0=0

 

Everything is correct so far

 

(3+3)(3-3) = (3-3)(3).....................0=0

(3+3) = 3....................................6 !=3 that is incorrect

[dragonfly]

soooo cloooose!!! what is incorrect about it??? We need a rule...

 

Fox, that's a sweet hint.

 

nice shades

Edited January 12, 2007 by DarkArchon

[NOFX]

what incorrect about it??

 

6 doesn't equal 3!!! thats what wrong. I don't know specific names for flawed mathematics, I'm a computer programmer and am somewhat accustomed to finding flawed logic, especially within my own code.

[dragonfly]

Those last two lines you wrote: what is wrong with them? How do you get from the first line to the second, and can you actually do that?

[EbilDustBunny]

OH Snap where am I?

[JackieChan]

OH Snap where am I?

Math class...

 

*runs*

[boiler]

all that crap works if both a and b are equal to zero, but that doesn't mean 2=1

[dragonfly]

It would still be falst if a and b both equalled zero. That's another hint. Wow, no one's gotten it yet really (except for fox). I'll post the answer tomorrow as to why the proof is false.

[Nick Soapdish]

i think you can't assume that they are communitive

 

i think you can't assume that they are communitive

 

a^2 =not= ab

[dragonfly]

if a=b then it follows that a^2 does infact = ab

 

This statement is correct.

[appalachian_fox]

The flaw starts at

 

(a + = b

 

lets say A=3

 

[...]

 

Yes, but the idea is to show that it doesn't work for ANY value of A...

 

all that crap works if both a and b are equal to zero, but that doesn't mean 2=1

 

Precisely! What if A = B = 0? What then?

 

What if I told you that x^2 < x? You might tell me I'm crazy. What's 0.5^2? It's true...but only for certain values of x.

 

I think everyone is convinced that A = B = 0, and that there are no other values that work for this particular problem. Plug zero in for A and B and go through the math, step by step. Is there anything peculiar about 0 which you can use to find a flaw? This is particularly important in computer programming, and depending on your compiler/interpreter can throw some very interesting errors in either compile or runtime.

 

You are all getting sooo close! One more step, you're almost there. Don't stop now!

Edited January 16, 2007 by appalachian_fox

[boiler]

a couple of those steps would thus be dividing by zero if a=b=0, and as we all know nothing likes being divided zero times

Edited January 16, 2007 by boilersax

[appalachian_fox]

a couple of those steps would thus be dividing by zero if a=b=0, and as we all know nothing likes being divided zero times

 

Gold star.

 

That, or we could just say the whole equation boils down to nullity * 2 = nullity...

[boiler]

woot! gold star for me. guess not ALL of my math skillzzzzz have died after college

[dragonfly]

ding ding ding! Cant divide by zero! Boilersax wins, and fox gets honourary mention times infinity

Edited January 16, 2007 by DarkArchon

[NOFX]

I don't see any division by 0.

 

I know you can do your fancy math and divide each side and draw smiley's faces and make the equations much simpler, but my head doesn't work like that. I see it as basically two statements on each side of the =. And both statements produce the same answer, until you get to a+b = b

[appalachian_fox]

I don't see any division by 0.

 

That's because the division by zero is hidden...

 

a = b

a^2 = ab

a^2 - b^2 = ab - b^2

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

(a + b ) = b

(b + b ) = b

2b = b

2 = 1

 

We're going to divide this result by (a-b ) on both sides to simplify. Of course, a = b = 0, so a - b = 0 - 0 = 0. Dividing both sides by (a-b ) is a division by 0.

 

This is the part I had the hardest time when I first saw this, because once you get that a = b = 0, you want to remove extraneous stuff, but in this case a-b=0 is critical to the problem, and it's easy to turn it into a 0 = 0 problem...which isn't helpful.

Edited January 17, 2007 by appalachian_fox

[dragonfly]

yep! simple answer: when you get a^2 - b^2 you get zero, and cannot factor by division from there, since you'd be dividing by zero.

Edited January 16, 2007 by DarkArchon