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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? |
In a word, No.
in variables it works, but try real numbers. |
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? |
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. |
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. |
Eureka I got it E=mc2
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i used to know a way that you could do it with regular math. no variables involved.
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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. |
i sitll think its cool
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(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. |
I love this kinda stuff, but for the love of god...
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I haven't watched an episode of freakazoid in 4 years. That makes me sad.
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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. |
ARG! I hate math!
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