Page 188 - 'Blast_Into_Math

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

We can complete the proof by induction. We have proven that we can find the first digit x 1 uniquely and

that the remaining digits cannot all be 9. Next, let’s assume we have found the first n digits x 1 ,... x n
analogously, so that these digits are unique, and

1
x 1 x 2 x n
0 ≤ x − + + ... + < .
10 10 2 10 n 10 n
We need to find the next digit and show that it is unique as long as the remaining digits are not all 9.
To do this we can again use a set. This time, let

s x 1 x 2 x n

S = s ∈ Z such that 0 ≤ s ≤ 9, and ≤ x − + + ... + .
10 n+1 10 10 2 10 n

Since

x 1 x 2 x n
0 ≤ x − + + ... + ,
10 10 2 10 n
we know that 0 ∈ S , and so S is not empty. By its definition the set S is a set of integers which is
bounded above by 9. Therefore it has a least upper bound. Let’s call this x n+1 . Since x n+1 +1 ∈ Z
and x n+1 +1 >x n+1 , this means that either

x n+1 +1 =10=⇒ x n+1 =9,

or

x n+1 +1 x 1 x 2 x n
>x − + + ... + ,
10 n+1 10 10 2 10 n

which we can re-arrange to

1
x 1 x 2 x n x n+1
x − + + ... + + < .
10 10 2 10 n 10 n+1 10 n+1
If x n+1 =9, we need to prove that the digits x m for m> n +1 cannot all be 9. We can do this by
contradiction. Let’s assume x n+1 =9 , and x m =9 for all m> n +1. Then

∞
x 1 x 2 x n 9
x = + + ... + + .
10 10 2 10 n 10 m
m=n+1
We can use the Geometric Σ Theorem and the LAMP to compute

∞
9

,
10 m
m=n+1

188