There's more than one way to do things  
PerlMonks 
Re: .999999... == 1? (Somewhat OT)by arashi (Priest) 
on Jul 25, 2001 at 01:27 UTC ( #99495=note: print w/replies, xml )  Need Help?? 
I'm not a math major, and I haven't really taken much math in college, but here goes anyways: The proof of this problem hinges on the assumption that 1/3 = .3 repeating. Is this a correct assumption? I can punch 1/3 into my calculator and it says .3 repeating, but is that really equal to 1/3, or is it just infinitely close to 1/3? 1/3 of something means it is broken into three equal parts that add up to 1, but since .3 repeating + .3 repeating + .3 repeating adds up to .9 repeating, wouldn't .9 repeating just be infinitely close to 1? Because of the infinity problem, there isn't a number between .9 repeating and 1, is this because .9 repeating isn't really a number, I mean you can't just write it out, because is keeps going, wouldn't that make it a limit function, more of a number concept than an actual number? Arashi I'm sure Edison turned himself a lot of colors before he invented the lightbulb.  H.S.
In Section
Meditations

