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Ok, this does not make sense....
I have proven that .9... = 1. (If you want the proof, just tell me)
but what about the floor() function? wouldn't the floor(.9...) = 0 and the floor(1) = 1? Therefore disproving my proof? |
Re: Ok, this does not make sense....
Yes, that does disprove you.
I'd love to see the proof, but it's undoubtedly like the proofs that prove 1=2. Hate to say it, but .999... isn't equal to 1, unless you're an microprocessor. --EDIT-- Yeah, so there's a proof. Big whoop. Sigma is overrated. ;) |
Re: Ok, this does not make sense....
Maybe this would help. . .
Linky |
Re: Ok, this does not make sense....
Summation was the way I proved it. Any thoughts on if floor(.9...) = 1 or 0?
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Re: Ok, this does not make sense....
Quote:
3/9= .3333333.... so 3/9 + 3/9 + 3/9 = 9/9 and .3333333.... + .3333333.... + .3333333.... = .999999.... Isn't infinity wonderfull? |
Re: Ok, this does not make sense....
You'd need a more formal definition of floor before you can prove or disprove anything. The formal definition of floor is: floor(x) is equal to the largest integer less than or equal to x.
You've already proven that .999... = 1, so 1 is the largest integer less than or equal to .999... and everything works out. Confusion only comes when you use the informal definition of ignoring everything after the decimal point. |
Re: Ok, this does not make sense....
x = 0.9999...
10x = 9.9999... 10x - x = 9.9999... - 0.9999... 9x = 9 x = 1 |
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