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View Full Version : Ok, this does not make sense....

Michael Hill
10-18-2005, 08:41 PM
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?

Ryan M.
10-18-2005, 08:45 PM
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. ;)

Bcahn836
10-18-2005, 08:48 PM
Maybe this would help. . .

Michael Hill
10-18-2005, 09:00 PM
Summation was the way I proved it. Any thoughts on if floor(.9...) = 1 or 0?

mechanicalbrain
10-18-2005, 09:25 PM
Maybe this would help. . .

3/9= .3333333.... so 3/9 + 3/9 + 3/9 = 9/9 and .3333333.... + .3333333.... + .3333333.... = .999999....

Isn't infinity wonderfull?

Kevin Sevcik
10-19-2005, 09:52 AM
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.

coastertux
10-19-2005, 05:42 PM
x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1