**Part 1: **Normal numbers in base ten

**Part 2: Numbers in base φ and π**

**Part 3: **One is normal in base π

(This post assumes familiarity with the concept of number bases: decimal, binary, hexadecimal, and so on.)

Let’s say we want to write numbers in base φ≃1.618_ten (the golden ratio), or base π≃3.14_ten. So, for example, **1000**_π = π^3.

The allowed digits are {0, 1} for base φ, and {0, 1, 2, 3} for base π.

Why not just e.g. {0, 1, 2} for π? That wouldn’t be enough digits to express everything: there’s a gap between **0.222…**_π = (2π/(π-1) – 2) ≃ **0.934**_ten and **1.**_π = **1.**_ten.

## Lots of representations, out of order

In integer bases, decimal representations are almost unique (see previously): you can write 5/2 as **2.5**_ten or **2.4999…**_ten, but that’s the only kind of ambiguity.

In base φ, **100.**_φ = φ^2 = φ+1 = **11.**_φ. Uh oh.

Worse still, **110.**_φ = **1000.**_φ, which means that **111.**_φ > **1000.**_φ. Numbers aren’t lexicographically ordered anymore! Looking at two numbers, it’s not immediately obvious which is bigger.

(But there’s a limit to how bad this gets: the largest number starting with **0. **is **0.111…**_φ, which is equal to **10.**_φ. So if two numbers start off with the same digits, and then the first has a **1 **where the second has a **0**, the second number has to start “catching up” within the next two digit positions or it will end up being smaller.)

Wikipedia has lots of facts about base φ. It describes the “standard form” of a number as the representation that avoids the digit sequence **11**, which is always equivalent to the representation given by the greedy algorithm in the previous post. (In short: keep multiplying by the base and chopping off the integer part of the result.)

Mercifully, the greedy/standard representation in any base is unique and sorts lexicographically.

## Second-greediest representation

In any base, the greedy algorithm applied to one gives **1.**

The second-greediest representation (see previously) in an integer base is always something like 1 = **0.222…**_three or 1 = **0.999…**_ten.

In base φ, the second-greediest-rep is 1 = **0.11**_φ, which terminates.

In base π, it’s 1 = **0.301102111002022113…**_π, nonterminating, with no discernable pattern.

(Note that flipping any of the **0**s to **1 **or **1**s to **2** makes the number bigger than one.)

This kind of makes sense — the golden ratio is a much nicer number than pi (algebraic instead of transcendental, etc.).

But in base 5/2, the second-greediest-rep is 1 = **0.2101110000110121…**_(5/2), nonterminating, no discernable pattern.

And in base X≃2.732_ten, the positive root of x^2-2x-2, it’s 1 = **0.22**_X.

So which numbers have terminating second-greediest-reps for 1?

We can rewrite these representations as expressions like **11.**_φ = φ^2 and **22.**_X = X^2. The left-hand sides are polynomials in the base (φ+1 and 2X+2, in these cases), and we end up with the somewhat weird theorem:

**The second-greediest-rep for 1 terminates in base B iff B is a root of a monic polynomial whose other coefficients are integers between -floor(B) and 0, inclusive.**

What about repeating second-greediest-reps? Integer bases, sure, but any others?

Well, in base φ+1, we get 1 = **0.2111…**_(φ+1).

In general, when do we get something repeating?

…Left as exercise for the reader.

## When do we have a choice?

As we write down digits for some number in some base B, we can sometimes deviate from the greedy algorithm by writing down something smaller than necessary, leaving a remainder that isn’t in the half-open interval [0, 1).

For the second-greediest representation of a number, we do this once, at the beginning. If the number we’re representing is 1, we have a remainder of 1, which just barely falls outside [0, 1).

But sometimes we can be bolder. How large can the remainder r_i be? Presumably there’s some cap, r_i <= r_max.

Well, the largest contribution we can get from the remaining digits is **0.ddd…**_B, where d=ceil(B-1) is the largest digit. And that’s equal to:

r_max = ceil(B-1) / (B-1)

Sanity check: For integer B, this is just 1, and we are only allowed to have remainders in [0, 1]. This is consistent with the fact that tails of 9’s are the only non-greedy representation possible in base ten.

## Normal numbers in weird bases

Recall our statements about normal numbers in base ten:

- For almost all numbers, for any k, the distribution of length-k substrings of the base-ten representation approaches a uniform distribution.
- For almost all numbers, when following the greedy algorithm, the distribution of remainders approaches a uniform distribution on [0, 1)

I think the following is true in any base B, and fairly easy to prove:

- For almost all numbers, for any k, the distribution of length-k substrings of the base-B greedy representation approaches the same distribution D
_{k}(B). - For almost all numbers, when following the greedy algorithm, the distribution of remainders approaches the same distribution D(B).

The sets of numbers for which these two statements are true should be the same; let’s call them “normal numbers”, extending the usual integer-base definition. Furthermore, non-normal numbers may still have “normal representations” (whose remainders follow D(B)) under some non-greedy algorithm.

I think the following is true but probably *very difficult* to prove:

- In base pi, the second-greediest representation of one is normal.

Note that this is definitely not true in any integer base, or in base phi, where the second-greediest rep of one terminates or repeats.

What does D(π) look like, and what are its properties? See here.