How can we show that it is an integer 5-adic number?

Question

Show that the number $\frac{3}{8}$ is an integer $5$-adic and calculate the first five positions of its power series in $\mathbb{Q}_5$.

Could you explain me how we can conclude that $\frac{3}{8}$ is an integer $5$-adic number?

Also, is there a methodology to find the power series of a rational number in $\mathbb{Q}_p$?

Answer 1

By the sum formula of converging geometric series $$ -\frac18=\frac{3}{1-5^2}=3\cdot1+3\cdot5^2+3\cdot5^4+\cdots=\sum_{k=0}^\infty 3\cdot5^{2k}. $$ This series is clearly an element of $\Bbb{Z}_5$, so we have shown that $-8$ is a unit of $\Bbb{Z}_5$. We can get everything you need out of this:

• $-3$ is a 5-adic integer, $-1/8$ is a 5-adic integer. The 5-adic integers form a ring $\implies (-3)\cdot(-1/8)=3/8$ is a 5-adic integer.
• Using the above series expansion of $-1/8$ and the fact that $3/8=1+5\cdot(-1/8)$ we get that $$ \frac38=1+3\cdot5+0\cdot5^2+3\cdot5^3+0\cdot5^4+\cdots. $$

Let's check the beginning by elementary means. The sums of the initial segment of that series are $1,16,16,391,391,\ldots$. We verify this by observing that $$ \begin{aligned} &\frac38-1&&=-\frac58&&\text{is divisible by $5$}\\ &\frac38-16&&=-\frac{125}8&&\text{is divisible by $5^3=125$}\\ &\frac38-391&&=-\frac{3125}8&&\text{is divisible by $5^5=3125$} \end{aligned} $$ and so forth.

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A general method for finding a 5-adic series for a rational number is roughly the following. I first show how to represent $-1/m$ as a series, when $\gcd(m,5)=1$. We can view $5$ as an element of $(\Bbb{Z}/m\Bbb{Z})^*$. It is of a finite order, say $a$. This means that $5^a\equiv1\pmod m$, or that $5^a-1=km$ for some integer $k, 0<k<5^a$. Then $$ -\frac1m=-\frac k{km}=\frac{k}{1-5^a}=k\left(\sum_{i=0}^\infty 5^{ai}\right). $$ Here $k=\sum_{i=0}^{a-1}k_i5^i$ for some integers $k_i\in\{0,1,2,3,4\}$. Putting these together gives us the series with a periodic sequence of coefficients. The length of the period is $a$, and digits are the $k_i$s. The minus sign is a bit of a nuisance. The way to negate such a series is to subtract it from $$ 0=5+4\cdot5+4\cdot5^2+4\cdot 5^3+\cdots. $$ So for example with $m=7$ we see that $5^6-1=15624=7\cdot2232$. Here $$2232=2\cdot5^0+1\cdot5+4\cdot5^2+2\cdot5^3+3\cdot5^4,$$ so we get a series with a period of length six $$ \begin{aligned} -\frac17&=\frac{2232}{1-5^6}= (2\cdot5^0+1\cdot5+4\cdot5^2+2\cdot5^3+3\cdot5^4+0\cdot5^5)\sum_{i=0}^\infty5^{6i}\\ &=\cdots\overline{032412}032412032412. \end{aligned} $$ The negated number is thus $$ \frac17=\cdots\overline{412032}412032412033. $$ If the denominator is divisible by a power of five, we simply shift the series an appropriate number of positions. So for example $$ \frac1{35}=\cdots\overline{241203}241203241203.3. $$ Above the numerator was $\pm1$. A larger numerator poses no problem, because we simply multiply the series.