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-   -   1=2? (http://www.flashflashrevolution.com/vbz/showthread.php?t=11326)

Brass 06-6-2004 06:09 PM

1=2?
 
a=b
a^2=ab
a^2-b^2=ab-b^2
(a+b)(a-b)=b(a-b)
a+b=b
2b=b
2=1

my head hurts?

lightdarkness 06-6-2004 06:28 PM

In a word, No.

in variables it works, but try real numbers.

classof04 06-6-2004 06:46 PM

NO, in the last part

2b=b
you have to get rid of the coeficiant.

b=1/2b
but what darklightness said try real numbers

does 5*2=5?

DracIV 06-6-2004 08:47 PM

It took me a moment to realize it, but that is totally logical.

If a=b
a-b = 0
0(a+b)= b(0)
0 = 0

No matter what A and B are, both sides of that are always 0. That is why 2B is the same as B, because the answer is still the same.

Kilgamayan 06-6-2004 08:56 PM

Re: 1=2?
 
To get from (a+b)(a-b)=b(a-b) to a+b=b, you have to divide both sides by (a-b). But a is b, so a-b is 0, and you can't divide by 0.

This is the oldest trick in the book.

classof04 06-7-2004 05:46 PM

Eureka I got it E=mc2

IronMonk 06-7-2004 05:46 PM

i used to know a way that you could do it with regular math. no variables involved.

pxbluesoul 06-9-2004 04:13 PM

Quote:

Originally Posted by IronMonk
i used to know a way that you could do it with regular math. no variables involved.

No, Kilg nailed it, it involves dividing by zero. There are similar equations with similar trickery though, you might be thinking of one of those.

Yattasparagus28 06-9-2004 07:36 PM

a=b
a^2=ab
a^2-b^2=ab-b^2
(a+b)(a-b)=b(a-b)
a+b=b
2b=b
2=1

So, remember, both variables are the same. So....

1. x=x.
2. x^2=xx.
3. x^2-x^2=xx-x^2.
4. (x+x)(x-x)=x(x-x).
5. x+x=x. (Wrong, should be written as x+x=2x)
6. 2x=x. (See?)
7. 2=1. (Flat out wrong.)

It was said before that to get from step four to five, you must divide. This is true. Simplified, four is actually x squared minus x squared equals x squared minus x squared. It is still true here. If you simplify further, x^2-x^2=x^2-x^2 becomes 0=0. In fact, all of them do. The same problem is always on both sides, until it changes in step five. This means that in step four no matter what the variables a and b are set to, it will always simplify to zero. And when it simplifies to zero, dividing by zero is, obviously, impossible.

Brass 06-11-2004 02:54 AM

i sitll think its cool

pxbluesoul 06-11-2004 03:02 AM

Quote:

Originally Posted by Yattasparagus28
4. (x+x)(x-x)=x(x-x).
5. x+x=x. (Wrong, should be written as x+x=2x)

Run that by me again? You divided both sides by quantity x-x, so you're left with..

(x+x)(1)=(x)(1)
x+x=x

Anyway you look at it, x+x=2x should not show up, even though it is, in this case, correct.

Omeganitros 06-11-2004 03:07 AM

I love this kinda stuff, but for the love of god...

pxbluesoul 06-11-2004 04:14 AM

I haven't watched an episode of freakazoid in 4 years. That makes me sad.

jimerax 06-11-2004 07:10 AM

Already mentioned, you cannot divide them by (a-b), because it's definitely 0.
also, 2b=b => 2=1 is false, 2=1 or b=0 but 2=1 is constantly impossible.

so this proof is totally incorrect, that's all. but I think it's interesting logic.

RajginKisaragi 06-11-2004 08:27 AM

ARG! I hate math!

Omeganitros 06-11-2004 10:46 PM



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