How is

Buttons, paraphrased wrote:The reason I'm so confident: in the system I described, 1.000 = 0.123456.... Similarly, 1.0213500000... = 1.021346789A..., and in general the smallest number you can make starting with ***.*****k is equal to the largest number you can make starting with ***.*****(k-1). Therefore, there are no "gaps" in this system

not an argument? It's not a proof in and of itself, but it can be turned into one:

Lemma 1: In Buttons' system, the smallest number you can make starting with ***.*****k is equal to the largest number you can make starting with ***.*****(k-1).

Proof of Lemma 1: Since the smallest number starting ***.*****k continues with all 0s, and the largest number starting ***.*****(k-1) continues with the largest possible numeral in each position (i.e. n-1 in the 1/n! place), it suffices to show that 1/n!=n/(n+1)!+(n+1)/(n+2)!+...

We show by induction that n/(n+1)!+(n+1)/(n+2)!+...+(n+k)/(n+k+1)!=1/n!-1/(n+k+1)!

When k=0, we get n/(n+1)!=(n+1)/(n+1)!-1/(n+1)!=1/n!-1/(n+1)!, so the base case is true.

For k>0, we have, by the inductive hypothesis,

n/(n+1)!+(n+1)/(n+2)!+...+(n+k)/(n+k+1)!

=1/n!-1/(n+k)!+(n+k)/(n+k+1)!

=1/n!-(n+k+1)/(n+k)!+(n+k)/(n+k+1)!

=1/n!-1/(n+k+1)!

as desired.

This tells us the sum of the first k terms of the series is 1/n!-1/(n+k+1)!, so the sum of the series is 1/n! as claimed.

□

Lemma 2: If x is a real number 0<x<1, there are integers b

_{1}, b

_{2}, b

_{3}, ... with 0 ≤ b

_{k} ≤ k such that x=Σ

_{k=1}^{∞} b

_{k}/(k+1)!

Proof of Lemma 2: Let b

_{n} be the largest integer ≤ n so that Σ

_{k=1}^{n} b

_{k}/(k+1)! ≤ x. Then the sequence of partial sums is monotone increasing and bounded by x, so the series converges to a real number s less than or equal to x. Suppose s<x. Then for some N, 1/N! < x-s, so for n>N, Σ

_{k=1}^{n} b

_{k}/(k+1)! ≤ s < x-1/n!, and therefore b

_{n} is not the largest integer so that Σ

_{k=1}^{n} b

_{k}/(k+1)! ≤ x. We therefore have b

_{n}=n for all n>N. We therefore have a last m with b

_{m}<m, since we can't have b

_{m}=m for all m with x<1. But by lemma 1, the sum of terms after the mth is equal to 1/m!, hence we can replace b

_{m} by b

_{m}+1 and still get a partial sum ≤ x, contradicting our choice of the numbers b

_{n}. Therefore we must have x=s=Σ

_{k=1}^{∞} b

_{k}/(k+1)! as claimed.

□

Combining lemma 2 and the fact that all positive integers can be written as a sum of factorials where n! appears at most n times, we have the following theorem:

Theorem: All positive real numbers can be represented in the form Buttons described.

□

The full proof is of course more complicated than Buttons' argument, but Buttons' argument provided the basic framework that I fleshed out into a proof.

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson