Recent content by Fala483

They should both work right? Neither should break math because of this trivial arithmetic problem.

That seems to result in the same problem.

The number won't be a real number. It might still be a more general kind of number. Now there is the corresponding problem in surreal numbers. If we allow also transfinite integers as the integers in rational numbers of the form a/b does some "middle infinite" decimal expansion correspond to...

Then how do you multiply infinities

So since 0.999... Squared ends in a decimal value of 1 and 1 squared ends in a decimal value of 0, 0.999... = 1 is a contradiction

I am trying to prove that all numbers of the form 0.999... Squared end in a decimal value of 1. For example 0.99sq = 0.9801 0.999sq = 0.998001 Etc. Is it possible to prove for all 0.999... ?

They didn't even shut down chernobyl

Let f(n) be of the form F(1) = 0.999999 F(2) = 0.9999999 F(3) = 0.99999999 Etc For arbitrary N >= 1 Let g(n) = f(n)^2 G(1) and g(2) end in 8 By linear relation all g(n) end in 8 Assume 0.999... = 1 0.999... X 0.999... must end in 8 per (4) 1 x 1 does not end in 8 By (5),(6),(7)...