On Sunday, June 30, 2013 2:48:17 AM UTC-5, Archimedes Plutonium wrote: > I am trying to think of a easy fast trick to determine if Floor pi *10^603 as integer 314159..32000 is a member of the sequence 2, 4, 8, 16, 32, 64, . . . > > > > AP
Rather a silly question on my part, and my only excuse is it is late at night.
That sequence cannot get away with the last digit being either a 2, 4, 6 or 8, so Floor pi * 10^603 is not a member. So I wonder if Floor pi*10^600 as integer 31415..32 is a member?
The reason this is bugging me, is that macroinfinity produces microinfinity, so that 1*10^603 produces 1*10^-603 so the first infinity number is 1*10^603 + 1*10^-603. And if that division into 1 is messy, the entire algebra is messy.
So far, I have explained this away by saying that pi digits determine the exponent in which we find the infinity borderline. Once we determine that it is 10^603, then we claim that 1*10^603 is the borderline and not Floor pi. When we work with 1 and then 603 zeros following, the division into 1 comes out perfectly even and not messy.
For example, in the 10 Grid system, as we pretend 10 is the borderline and last and largest finite number then 0.1 is the smallest nonzero finite number and the smallest infinity number is 10.1. So everything is smooth fresh and clean. But suppose instead we pretend 13 is the borderline, then 1/13 is messy with 0.076 and we cannot then say 13.076 is the smallest infinity.
Calculus is cleaning up the infinity borderline numbers.