Page 187 - 'Blast_Into_Math

Blast into Math! Analatic nummer theora: ants, ghosts and giants

Since x 1 +1 >x 1 , and x 1 +1 ∈ Z , and x 1 is the largest element of S , this means that

x 1 +1 /∈ S.

So either

x 1 +1 =10=⇒ x 1 =9,

or

x 1 +1 x 1 1 x 1 1
>x =⇒ + >x =⇒ x − < .
10 10 10 10 10

If x 1 =9, then we can prove that we cannot have x n =9 for all n ≥ 1 by contradiction. So, let’s
assume x n =9 for all n ≥ 1, and so

∞
9
x = .
10 n
n=1
By the LAMP, with

N
1

S N = , T N =9S N ,
10 n
n=1
lim T N =9 lim S N ,
N→∞ N→∞

and we have already computed the limit of the sequence of partial sums S N for the geometric series
1
with ratio , so
10
1 1
lim T N =9 lim S N =9 10 =9 =1.
N→∞ N→∞ 1 − 1 9
10

But then

∞
9
x = =1,
10 n
n=1

and we assumed that 0 ≤ x< 1. So if x 1 =9, then we cannot have x n =9 for all n ≥ 1. In this case,

9 1
x 1
x< 1=⇒ x − = x − < .
10 10 10

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