$3$-adic expansion of $- \frac{9}{16}$

Question

I get the $3$-adic expansion to be $1+1 \cdot 3+2 \cdot 3^2 +2 \cdot 3^3 + 0 \cdot 3^4+\cdots$. I'm trying to work out a pattern of the coefficients and think it is $1, 1, 2, 2, 0, 0, 1, 1, 2, 2, 0, 0,...$. To show this I need to show that $1+1 \cdot 3+2 \cdot 3^2 +2 \cdot 3^3 + 0 \cdot 3^4 + \cdots$ sums to $- \frac{9}{16}$, but I don't know how to manipulate this series in order to compute it.

Any help would be appreciated!

Answer 1

I will explain the case of $3$ - adic of $\frac{21}{50}$ and leave it to you to solve for $\frac{-9}{16}$

For $\frac{21}{50}$ you have expression $\sum_{i\in \mathbb{Z}} a_i p^i$ : $0\leq a_k<p$

$$\frac{21}{50}=a_0+a_1p+a_2p^2+a_3p^3+\cdots$$

So, how do you get $a_0$ ???

$a_0$ is chosen such that

$$\frac{21}{50}=a_0+a_1p+a_2p^2+a_3p^3+\cdots$$

i.e., $$\frac{21}{50}-a_0=a_1p+a_2p^2+a_3p^3+\cdots$$

see that $p$ divides $a_1p+a_2p^2+a_3p^3+\cdots$ so $p$ should divide $\dfrac{21}{50}-a_0$

i.e., $\dfrac{21}{50}-a_0\equiv 0 ~\text{mod} ~3$ i.e., $3$ divides $21-50a_0$

You have only three possibilities for $a_0$ namely $0,1,2$... Check it..

Now,$a_1$... $a_1$ is chosen such that

$$\frac{21}{50}=a_0+a_1p+a_2p^2+a_3p^3+\cdots\Rightarrow\frac{21}{50}-a_0-a_1p=a_2p^2+a_3p^3+\cdots$$

see that $p^2$ divides $a_2p^2+a_3p^3+\cdots$ so $p^2$ should divide $\dfrac{21}{50}-a_0-a_1p$

You know what is $a_0$. substitute it in above expression..

$\dfrac{21}{50}-a_0-3a_1\equiv 0 ~\text{mod} ~3^2$

Just repeat this accordingly and you will find what other $a_k$ are...

Answer 2

I guess I’m on a long-term rant to urge people to write their $p$-adic numbers as ordinary $p$-ary expansions extending (potentially) infinitely to the left. In your case, that would be ternary expansion, so, just as you learned in elementary school, sixteen comes out as $121;\,$. I like to use a semicolon for the radix point to remind myself that I’m not dealing with real numbers.

Instead of explaining everything from the beginning, let me refer you to two recent answers of mine, here and here. Let me take up from there:

Three-adically, $-1$ is $\cdots2222;$, so that $-9$ is expanded as $\cdots22200;$. Now, when you do long division, dividing $-9$ by $16$, you’ll need to know not only the expansion of sixteen as $121;\,$, but also twice that, which is $1012;\,$. When you start your division, of course the first (rightmost) two digits are both zero, and then you want to divide something ending with $2$ by something ending with $1$, so the first nonzero digit is $2$. Subtract twice$121$ from $\cdots2222$ and get $\cdots2221210;$. The rightmost digit is zero, as it has to be. Next step is to see that the required digit in the quotient is just $1$, so you subtract $121$ from $\cdots2222121$ and get $2222000;\,$. The rightmost digit is zero, as it has to be, but in addition you get two more zeros without cost, which means that the next two digits in your answer are zero. And lo and behold, you’re presented with the same problem you started out with, so you’ve achieved your repeating expansion, as $\cdots001200120012001200;$

Only remains to check that you have the right expansion, by using the formula for a convergent geometric series. Just looking at your expansion, you see that the common ratio is $10000;=3^4$, and your first term is $5\cdot9=45$ in decimal notation. Checking $a/(1-r)=45/(1-81)=-9/16$, yay.

Answer 3

As i understood, your question was how to find fraction form for
$...002211002211\in \mathbb Q_3$ (but not how fo find $p$-adic expansion for rational number).

Algorithm is the same, as for $p$-ary expression: $$ ...002211002211=2211(...+3^{12}+3^6+1)=\frac{54+18+3+1}{1-3^6}=-\frac{76}{728}=-\frac{19}{182}. $$