Binary representation of 2-adic integers
I would like some examples of the binary representation of 2-adic integers that are not standard integers. What is the 2-adic expansion of $1/3$? Of $-1/3$? What number does $...010101$ represent?
$\endgroup$2 Answers
$\begingroup$We can write arbitrary rationals in their $p$-adic expansions by utilizing modular arithmetic and geometric series. To begin with, let's reverse-engineer the form of $p$-adics with periodic digits.
Suppose $0\le a <p^s$. Then
$$n+p^r\sum_{n\ge1}ap^{sn}=n+\frac{ap^r}{1-p^s} $$
Bottom line: we want a denominator of the form $1-p^s$ and a numerator with $p$-free part that is less than the power $p^s$ appearing in the denominator.
We will see it suffices to consider negative rationals $-1<a/b<0$ with $p\nmid b<0$.
To get $1-p^s$ from the $b$ in $a/b$, we employ modular arithmetic. Since $p\nmid b$ we know $p$ is a unit modulo $b$ hence $p^s\equiv1\bmod b$ for some $s$. Let $1-p^s=bc$. Then
$$\frac{a}{b}=\frac{ac}{bc}=\frac{ac}{1-p^s}=ac+acp^s+acp^{2s}+\cdots $$
Furthermore $ac<p^s\Leftarrow ac-1<-bc\Leftarrow (a+b)c<1\Leftarrow a<-b$ so this gives a full expansion.
Now let's re-examine our sufficiency hypothesis. If $x=x_0+x_1p+\cdots$ is a negative rational,
$$-x=(p-x_0)+(p-x_1-1)p+(p-x_2-1)p^2+\cdots $$
is the complete $p$-adic expansion of the positive rational $-x$. Thus we're safe considering negative rationals. Furthermore given arbitrary $x\in\Bbb Q\setminus \Bbb Z$, we can use a combination of subtracting positive integers and negation in order to get $x$ in the interval $(0,1)$, after which we find the $p$-adic expansion and then add the integers back on (which will either perturb a finite number of digits or telescope in an obvious way, e.g. $1+(1+2+2^2+\cdots)$ in $\Bbb Q_2$).
Finally, powers of $p$ can be factored out of a rational and put back in after the expansion of the $p$-prime part has a $p$-adic expansion; powers of $p$ simply shift the digits left or right.
Let's apply this to a concrete example, say $-26/21$ in $\Bbb Q_3$. Factor the $3$ out to obtain $-26/7$, then negate to obtain $26/7$, then subtract $3$ to obtain $5/7$, then negate again for $-5/7$. Calculate powers to obtain $3^6\equiv1$ mod $7$ and $1-3^6=-7\cdot104$. Therefore compute
$$-\frac{5}{7}=\frac{\phantom{-}5\cdot104}{-7\cdot104}=\frac{520}{1-3^6}=520+520\cdot3^6+520\cdot3^{12}+\cdots.$$
As $520_{10}=201021_3$ the above is $\cdots\overline{201021}_3$. Hence
$$\begin{array} -\frac{26}{21} & =-\frac{1}{3}\left(3-\cdots\overline{201021}_3\right) \\ & =-\frac{1}{3}\left(11_3+\overline{021201}021202_3\right) \\ & =-\overline{021201}02122_3 \\ & =\overline{201021}20101_3. \end{array}$$
$\endgroup$ 6 $\begingroup$To answer your last question first, $$...010101 = 1 + 4 + 16 + \cdots + 4^n + \cdots = 1/(1-4) = -1/3.$$ This also answers your second question. As for your first question, $$1/3 = \dfrac{1}{2}( 1 - 1/3) = ...0101011.$$
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