# $5$-adic expansion of $−2$

Let $\alpha = \sum_{n=0}^{\infty} a_n\ p^n$ be the expansion of a p-adic unit. So $0<a_0<p$ and $0\leq a_n<p$ for $n \geq 1$.

Show that $\beta = -\alpha$ has the expansion $\beta = \sum_{n=0}^{\infty} b_n\ p^n$ with $b_0=p-a_0$ and $b_n=p-1-a_n$ for $n\geq 1$.

I have $b_n=-a_n$. And $0>b_0>p\$ i.e $\ -p>b_0-p>0$ so $-a_0=b_0-p$ . What about the other inequality.

Use this to find the $5$-adic expansion of $−2$.

How do I do this? $\alpha=2$ then?

What if $\alpha$ is not a unit and $a_0 =0$?

• Are you familiar with the fact that $$-1=\sum_{n=0}^{\infty}(p-1)p^n?$$ Either by using the geometric series formula or by observing that if you add one to the r.h.s. you cause an infinite sequence of carries - ending up with all zeros. You can prove the first claim similarly by checking that you get the endlessly cascading carry. – Jyrki Lahtonen Mar 5 '16 at 20:33
• $\alpha + \beta = 0$, reduce modulo $p$, then modulo $p^2$ etc. To find $-2=-1-1$ write the first $-1$ like the prevoius comment and add $-1$ to the first term of the series. – Maffred Mar 5 '16 at 21:09

This is my standard sermon on the topic. I recommend Strongly that you write your $5$-adic expansions in standard $5$-ary notation, so that $\cdots dcba$ stands for $a + b\cdot5+c\cdot5^2+d\cdot5^3+\cdots$. (And I like to put semicolon to the right of the units digit instead of a period to remind us that this is $p$-adic expansion.)

Now, if you subtract $1$ from the $5$-ary number $1000000$ you get $444444$, and remember how you got it: calculating right to left, doing the same borrow-and-carry procedure that you learned in elementary school. But $5$-adically, that $1$ in the sixth place is really small, standing for $5^6$, so let’s just subtract $1$ from $0$ and get the result $\cdots44444;\,$, with $4$’s extending all the way to the left. This is your expansion of $-1$. And you can check it with the formula $a/(1-r)$ for geometric series: here $a=4$ and $r=5$, evaluating sure enough to $-1$.

Since $-1=\cdots44444;\,$, surely $-2$ is what you get by subtracting $1$ from that, namely $\cdots44443;\,$.

What about the negative of an arbitrary $p$-adic integer $z=\sum_{i\ge0}a_ip^i$? Subtract it first from $-1$ to get $\sum_{i\ge0}b_ip^i$ where $a_i+b_i=p-1$, and then add $1$ to the result. Easy, right?

It seems to me that once you get in the habit of thinking of $p$-adic numbers in the way I’ve recommended, you’ll find your text’s explanation unnecessarily complicated.

• How would you set out the answer to the first part. And what would happen if $\alpha$ is not a unit and $a_0=0$? – Polp Mar 9 '16 at 20:42
• You can answer the first part by appealing to my algorithm; and if $\alpha$ is not a unit, the algorithm still applies. The nonunit twenty has $5$-adic expansion $40;$ and hence its negative is $\cdots44410;$ — no different from the negative of $2$. – Lubin Mar 9 '16 at 21:13
• So calculate $-1-\alpha$? – Polp Mar 10 '16 at 0:35
• Exactly, and then add $1$ to make it right. – Lubin Mar 10 '16 at 4:45
• Thanks, but I do not understand what the 5-adic expansion of $-2$ would be if $\alpha$ is not a unit and $a_0=0$? – Polp Mar 10 '16 at 22:51

I have $b_n=-a_n$.

No you don't: you have $b_n=-1-a_n$.

Perhaps the simplest method is to compute $\alpha+\beta$ term by term and show that each term is $0$.

The second part is easy: you have $\alpha=2$, so $a_0=2$, and $a_n=0$ for $n \ge 1$. Now just use the formula from the first part.

• How do I get the expansion with the inequalities for the first bit? – Tony Mar 5 '16 at 20:54
• I'm afraid I don't understand your question @Tony. – TonyK Mar 6 '16 at 1:20