Questions about p-adic expressions.

Question

In the notes, page 3, it is said that in $2$-adic and $3$-adic expansion, we have $$ \frac{21}{50} = \frac{1}{2} + 2 + 2^2 + \cdots \tag 1 $$ $$ \frac{21}{50} = 2\cdot 3 + 3^2 + 3^6 + \cdots \tag 2 $$ I am confused since it seems that the left hand side of (1) (resp. (2)) is different from the right hand side of (1) (resp. (2)). Why left hand side and right hand side are equal? Thank you very much.

Answer 1

I will explain the case of $3$ - adic and leave it you to the case of $2$ - adic...

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

If you know what p-adic norm of a rational number is then you can see that :

$3$-adic valuation of $\frac{21}{50}$ is $1$ i.e., you have only non negative powers in its $3$-adic expansion

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

Now the question is how do you find $a_0$ ???

$a_0$ 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 $p$ 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 $21-50a_0$

As there are only $3$ choices for $a_0$ you can easily check which $a_0$ is suitable..

Now, how do you find $a_1$ ??

$a_0$ 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 $p$ should divide $\dfrac{21}{50}-a_0-a_1p$

As you have already calculated what $a_0$ is you can just substitute it in above expression which results the point that :

$\dfrac{21}{50}-a_0-3a_1\equiv 0 ~\text{mod} ~3^2$ i.e., $3$ divides $??$

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

Can you conclude now?