The impossible math problem...

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Unimaxium said:
oh, and the proof that 1=2

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

actually there's a problem in that proof. now see if you can figure out what it is. and joe and company, no spoiling the answer for others! :p
You are dividing by a^2-ab but that equals 0 so you can't do that.

I don't like the proof that 0.9 recurring = 1

Because I know it doesn't but it proves it does. It goes like this.

For any number If you multiply by ten, take away the original number and then divide by 9 you get back to the original number so -

a =0.9999999.....
10a =9.999999999..........
10a-a=9a = 9
a=1

:grumpy: I don't like it.
 
nvr2low said:
how did you get 2 outside on 2(a^2-2ab). i follow up to that point but i cant think how you how you do it. if you simplify

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

you should get

(a^2-2ab) = (a^2-2ab)

right?

you bring a^2 over and subtract out?

so how do you go from that to

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

how did you get 2 out of 1 and 1 out of the other? im stumped right there.

because if you simplify the left side of the equation above you get 2a^2 - 2ab, which equals 2(a^2 - ab).

But daniel's point is right that you're dividing by zero in the last step.
 
ok, i see it now, guess i missed it the first time
 
Here's a simple maths problem.

You have a hotel with infinite rooms. All the rooms are occupied. Recently, a batch of infinite number of guests arrived. You have to put all these new guests into the hotel rooms, without kicking out any of the current guests.

How do you do that?
 
I don't want to be in hotel with so many people...
 

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