Proof of 1 Not equal 0.999...

0*infinity=(0+0+0+0...)=0 (x+0=x x=0)
1=0.999... (agreed) 1-0.999...=0
(type 1) 1-0.999...=(1/10)*(1/10)*(1/10)... (from division by 10, 1-0.9 etc)
ab=0 if a or b=0; 1/10=0 ;10*1/10=10*0 ;1=0 ;x*1=x*0 ;x=0
(type 2) (1/10)*(1/10)*(1/10) ...=(1/10)^infinity=1/infinity=0
1/infinity=0 then 0*infinity=1 bu 0*infinity=0 ;1=0 ;x*1=x*0; x=0
Every number equals zero .(by 1=0.999...)

I'm not >>40

>>37
if we take .999... to mean a decimal point followed by an "infinite" amount (perhaps it's better to say, uncountable amount) of nines afterwards then the answer to the question:
"what number is between .999... and 1?" is none. Or rather there exists no distinct number between .999... and 1 because they are the same number. Just like there are no numbers between 0 and .000... because they become the same number.

>>38
that number would actually be less than both 1 and .999... since both .999...0 and .999... have the "same" amount of digits (infinitly so for both). If one was subtract the former from the latter, one would get .000...9 which is greater than than the subtraction of 1.000...0 and .999...9 which yields .000... and therefore .999...0 is farther from one than .999... is. You would somehow have to create a number that has more digits than infinity in order to create a number between 1 and .999...