.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
How is [itex](.\overline{3})^2 = .\overline{9}[/itex]? [itex](.\overline{3})^2 = .\overline{1}[/itex], no? --J
Not really -- they both boil down to the meaning of the terms involved. It's a mathematical fact that, in the decimals, [itex]0.\bar{9}=1[/itex]. If you prefer working with a system where that is not true, then you're going to have to explicitly specify how that system works.
0.9...i=1 Yes, .9i = 1, I have though this out much, there is mathamaticly no difference from .9i and =, thus, they are equal. My point in creating this new thread was to try to find a flaw in number theory. See, if .9i^2 does not equal (! or <>) 1, then math is flawed, because these 2 problems won't have the same answer. 1. ((1/3)*3)^2 2. 1^2 So, .9...i=.9...i=1?
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).
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.