
#1
Jan205, 08:00 PM

P: 114

.9i^2 means the square of .9999999999999999999999999999 INFINITE!
.3i^2 = .9i, so .9i should be able to be calculated too. This is rather an imaginary number, but it should be able to be calculated. If you don't know the exact answer, just tell me if .9i^2 is not .9i If .9i^2 is something other than .9i, then I think I have found a flaw in number theory, which I will post later. And don't say that .9i^2 = 1, that is another topic Thanks in advance 



#2
Jan205, 08:04 PM

Sci Advisor
PF Gold
P: 2,226

I assume that 'i' is NOT meant to mean sqrt(1).
0.999... = 1 therefore (0.9999...)^2 = 1^2 = 1 



#3
Jan205, 08:41 PM

P: 291

How is [itex](.\overline{3})^2 = .\overline{9}[/itex]?
[itex](.\overline{3})^2 = .\overline{1}[/itex], no? J 



#4
Jan205, 08:44 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

What is 0.9i^2? 



#5
Jan205, 11:03 PM

P: 114

1. ((1/3)*3)^2 2. 1^2 So, .9...i=.9...i=1? 



#6
Jan305, 04:49 AM

Sci Advisor
HW Helper
P: 9,398

Are we really going to have another 0.9recurring isn't one discussion? By definitoin of the terms involved they are equal, that is the system we have chosen to work in. This isn't a number theoretical fact, it is an analytic choice, one made so that we have a system in which we can do analysis.
As you're not giving a new definition for the symbols, nor are you defining a new algebraic structure, then we msut assume you're using the accepted mathematical ones. Maths isn't flawed (well, not because of this argument, anyway). 



#7
Jan305, 11:31 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,898

The only thing "new" is that unfortunate notion "i" for "infinitely recurring" which then allows him to write "i^2" as if it meant something!
The fact that he then says "This is rather an imaginary number" makes me wonder if this isn't someone's idea of a joke. 


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