Ok, this does not make sense....

[Michael Hill]

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.]

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]

Maybe this would help. . .
Linky (http://mathforum.org/dr.math/faq/faq.0.9999.html)

[Michael Hill]

Summation was the way I proved it. Any thoughts on if floor(.9...) = 1 or 0?

[mechanicalbrain]

Maybe this would help. . .
Linky (http://mathforum.org/dr.math/faq/faq.0.9999.html)
Or.... if your too lazy to read through that (like me!)

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

Isn't infinity wonderfull?

[Kevin Sevcik]

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]

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1