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ezmogee's Poker Blog Let a and b both be equal to 1. Since a and b are equal:
b² = ab
Because a equals itself:
a² = a²
Subtract equation 1 from equation 2:
a² - b² = a² - ab
Factor both sides of the equation:
(a + b)(a – b) = a(a – b)
Divide both sides of the equation by (a – b)
a + b = a
Subtract a from both sides and we get:
b = 0
But we set b to 1 at the beginning, so:
1 = 0
May 19, 09 23:56:22
I used to be a math major, but some of the rules are fuzzy to me now. I'm pretty sure there's an error in this proof. When a^2 - b^2 = a^2 - ab, if you fill in 1 you get 0 = 0. Same for the next step. But in the next part you get 2 = 1. So I'm thinking something is flawed in this part?
I also think you can't just subtract one equation from another in step 3. This is not a valid part of the proof, that's my guess.
May 20, 09 01:59:50
At the beginning, you assumed that a and b were both equal to 1. You then divided by the quantity (a-b), which is zero. You cannot divide by 0.
It is cool though.
May 20, 09 13:08:35
Here is proof that girls = Evil:
http://www.stacken.kth.se/lists/best-forestry/2001-05/jpg00000.jpg
Prove me wrong.
Subtracting equation 1 from equation 2 is where you start to go wrong I think. Why subtract at all, lets add, multiply, divide, you are creating the rules as you go aren't you!
May 20, 09 15:21:48
"Factor both sides of the equation:
(a + b)(a – b) = a(a – b)
"
This made me insta-WAT?
May 20, 09 18:41:28
[url]http://shhac.info/x/b/divide_by_zero.jpg[/url]
[url]http://halshop.files.wordpress.com/2008/04/dividedbyzero2.jpg[/url]
May 21, 09 06:00:42
can't divide by (a-b) cuz a-b = 0
this was a math question i got for homework at harvard summer school
May 21, 09 10:56:53
You can subtract equation 1 from equation 2 because both sides of e1 are equal and so you are essentially subtracting the same thing from both sides of equation 2.
http://www.angelzfunnyz.com/Portals/0/Gallery/Album/8/DivideByZero.jpg
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