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?

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?

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.

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

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?

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.

x = 0.9999…

10x = 9.9999…

10x - x = 9.9999… - 0.9999…

9x = 9

x = 1