Down the Rabbit Hole
Decimal expansions
Every real number can be expressed by a decimal expansion.
Why is $0.99999\ldots = 1$?
Through the looking glass
We are quite comfortable with numbers that carry on forever to the right of the decimal point. But have you ever considered numbers that carry on forever to the left?
This seems crazy, but let's see where the idea takes us.
First, let's try to figure out the value of
What is $\ldots 99999$?
But that would imply…
It appears that $\ldots 99999 + 1 = 0$ works out after all!
What's going on here???
Common sense tells us that $\ldots 99999$ is not a number.
…but these “proofs” that $\ldots 99999 = -1$ make it feel like there's something there.
Dr. Hensel's wonderful numbers
The main objection to $x$ being a number is that this sum clearly does not converge to a real number.
In 1897, Kurt Hensel discovered that we can build a rigorous mathematical theory of an alternate numerical universe in which infinite sums like this do converge:
The 10-adic numbers $\ZZ_{10}$
A 10-adic number is an expression of the form
\[ \cdots + \vas{a_4}10^4 + \vas{a_3}10^3 + \vas{a_2}10^2 + \vas{a_1}10 + \vas{a_0} \]with digits $a_i$ satisfying $0 \leq a_i \leq 9$. We write $\ldots \vas{a_4a_3a_2a_1a_0}$ as a shorthand.
Addition and multiplication work just like you would expect.
If $\vas{a}$ and $\vas{b}$ are 10-adic numbers, then the $n$th digit of $\vas{a + b}$ or $\vas{a\cdot b}$ only depends on the $k$th digits of $\vas{a}$ and $\vas{b}$ with $k \leq n$.
What about subtraction?
Addition and multiplication work in the familiar way, but what about subtraction?
- Subtraction is just addition with negatives: \(\,\, a - b = a + (-b). \)
- So, how do negatives work? \( -(\ldots 5429831) = \vas{\textbf{???}} \)
Can you see the shortcut for calculating the negative of a 10-adic number?
$-1 = \ldots 99999$
What other familiar numbers are hiding in $\ZZ_{10}$?
Consider the following 10-adic number with repeating digits,
We can use the same trick we used on $\ldots 99999$ to solve for $x$.
So $-\tfrac{1}{11}$ is a 10-adic number.
Some fractions live in $\ZZ_{10}$ too!
For which 10-adic numbers $n$ is $\tfrac{1}{n}$ also 10-adic?
I claim $\tfrac{1}{5}$ does not belong to $\ZZ_{10}$.
Powers
Some sequences of integers which obviously diverge in the real numbers $\RR$ turn out to converge in $\ZZ_{10}$.
| $2^{5^n}$ | $n$ |
|---|
The digits are stabilizing from the right.
The sequence $2^{5^n}$ converges to a limit in $\ZZ_{10}$!
What is this limit $b = \lim_{n\to\infty} 2^{5^n}$?
so $b = \ldots 879186432$ is a solution of the equation
\[ \vas{x^5 = x}. \]Mysterious limit
Recall that $b = \lim_{n\to\infty} 2^{5^n} = \ldots 879186432$ is a solution of
Thus $b$ must be a root of one of these factors.
$b \neq 0, \pm 1$, so $b$ must be a root of $x^2 + 1$.
Huh?
Consider the product $\,b(b^2 + 1)$.
Powers
Let's look at the powers of $b = \ldots 879186432$.
What happens if we replace $2$ and $5$ by different numbers?
| $n$ | $b^n$ |
|---|
Zero divisors
If $mn = 0$, then $m = 0$ or $n = 0$.
$\ZZ_{10}$ does not have this property!
- We call numbers with this strange property zero divisors.
- Not obvious, but the existence of zero divisors in $\ZZ_{10}$ stems from the factorization $10 = 2\cdot 5$.
In other words, $\ZZ_{10}$ has zero divisors because $10$ is not prime.
From $10$ to $n$ to $p$
For any integer $n \geq 2$, positive integers can be written in base $n$ using the digits $0, 1, \ldots, n-1$:
\[ 7 = \vas{1}\cdot 2^2 + \vas{1}\cdot 2 + \vas{1} = \vas{111} \ \text{(Base 2)}, \qquad\quad 7 = \vas{1}\cdot 5 + \vas{2} = \vas{12} \ \text{(Base 5)}. \]The $n$-adic numbers $\ZZ_n$ consist of all expressions of the form
\[ \cdots + \vas{a_4}n^4 + \vas{a_3}n^3 + \vas{a_2}n^2 + \vas{a_1}n + \vas{a_0}, \qquad 0 \leq a_i \leq n-1. \]If $n = p$ is a prime number, then there are no zero divisors in the $p$-adic numbers $\ZZ_p$.
Just as every number $n$ factors into a product of primes, every $n$-adic number system decomposes into a product of $p$-adics according to the factorization of $n$. \[ 10 = 2 \cdot 5 \,\,\Longrightarrow\,\, \ZZ_{10} \cong \ZZ_2 \times \ZZ_5 \]
This decomposition is (a fancy version of) the Chinese Remainder Theorem.
Decomposing $10$-adics
Every $10$-adic number decomposes into a $2$-adic and $5$-adic number in a way that respects arithmetic.
| $\ZZ_{10}$ | $\ZZ_2$ | $\ZZ_5$ |
|---|---|---|
| 7 | 111 | 12 |
| 100 | 1100100 | 400 |
| …830261947 | …010111011 | …021340242 |
| …999999999 | …111111111 | …444444444 |
| …666666667 | …010101011 | …131313132 |
| …909090909 | …001011101 | …211402114 |
Recall $b = \lim_{n\to\infty} 2^{5^n} = \ldots 879186432$.
- Suppose $b$ decomposes as $b \mapsto (b_2, b_5)$.
- Since $b(b^2 + 1) = 0$ in $\ZZ_{10}$, we must also have \[\begin{aligned} b_2(b_2^2 + 1) &= 0 \ \text{ in } \ZZ_2, \\ b_5(b_5^2 + 1) &= 0 \ \text{ in } \ZZ_5. \end{aligned}\]
- But $2$ and $5$ are prime, so $\ZZ_2$ and $\ZZ_5$ have no zero divisors. Hence $b_2$ and $b_5$ are forced to pick a factor.
| $b$ | $b_2$ | $b_5$ |
|---|---|---|
| …879186432 | …000000000 | …032431212 |
So $b_5 = \sqrt{-1}$ in $\ZZ_5$!
Does every $\ZZ_p$ have $\sqrt{-1}$?
We saw that $\sqrt{-1} = \ldots 032431212$ in $\ZZ_5$. Does $\sqrt{-1}$ live in every $\ZZ_p$?
Suppose $(\ldots a_4 a_3 a_2 a_1 a_0)^2 = -1$ in $\ZZ_3$.
- $a_0$ can only be $0$, $1$, or $2$ in base $3$.
- Squaring these digits in base $3$: \[\begin{aligned} \vas{0}^2 &= \vas{0} \\ \vas{1}^2 &= \vas{1} \\ \vas{2}^2 &= 4 = \vas{1} \cdot 3 + \vas{1} = \vas{11} \ \text{(base 3)} \end{aligned}\]
Hence we can never get a $2$ as the ones digit of a square.
Thus $\ZZ_3$ cannot have a square root of $-1$!
So when does $\ZZ_p$ contain $\sqrt{-1}$?
What is the pattern?
$\ZZ_p$ contains $\sqrt{-1}$ exactly when $p$ leaves a remainder of $1$ when divided by $4$.
How often is $p \equiv 1 \bmod 4$?
Chebyshev's bias
The $p$-adic prism
Following the dubious identity $\ldots 999999 = -1$ down the rabbit hole led us to the wonderful world of $p$-adic numbers.
- Each $p$-adic number system is a parallel arithmetic universe to our familiar real numbers $\RR$.
We only get a complete picture by considering $\RR$ together with all the $p$-adics.
Slides
