Setting is and now we remember that is...
(1)
... so that with four decimal digit accuracy is...
Kind regards
1. prove that the infinite decimal .123456789101112....
is not eventually periodic, i.e. it represents an irrational number.
2. Let x be a real number having the infinite decimal expansion
x = 1.01001000100001... without the use of a calculator find the first four digits of x^(1/3)
Thanks!
thanks for your reply....
i want to answer the question in a way that references Cauchy sequences however. I know that the decimal 1.010010001... is a Cauchy sequence because you can make
a_1 = 1.01
a_2 = 1.01001
a_3 = 1.010010001
...
and you can make the difference between two successive terms of this sequence arbitrarily small by going out far enough
so is there some way to evaluate a Cauchy sequence to a certain degree of accuracy? I had trouble finding a sequence that represents this decimal.
my shot at #1 (would appreciate it if someone can tell me yes it's valid/no it's not and why):
1. prove that the infinite decimal .123456789101112....
is not eventually periodic, i.e. it represents an irrational number.
define a_n = n
then if this decimal is periodic, for some m and k,
a_m = a_(m+k)
a_(m+1) = a_(m+1+k)
...
a_(m+k-1)=a_(m+2k-1)
but a_n is strictly increasing, so a_m <> a_(m+k) for k > 0
so the decimal cannot be periodic.