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As opposed to real number expansions which extend to the right as sums of ever smaller, increasingly negative powers of the base $p$, $p$-adic numbers may expand to the left forever, a property that can often be true for the $p$-adic integers. For example, consider the $p$-adic expansion of 1/3 in base 5. It can be shown to be $…1313132_5$, i.e., the limit of a sequence $2_5$, $32_5$, $132_5$, $3132_5$, $13132_5$, $313132_5$, $1313132_5$, … :

$\dfrac{5^2-1}{3}=\dfrac{44_5}{3} = 13_5; \, \dfrac{5^4-1}{3}=\dfrac{4444_5}{3} = 1313_5$ $\Rightarrow-\dfrac{1}{3}=\dots 1313_5$ $\Rightarrow-\dfrac{2}{3}=\dots 1313_5 \times 2 = \dots 3131_5$ $\Rightarrow\dfrac{1}{3} = -\dfrac{2}{3}+1 = \dots 3132_5.$ (Wikipedia, p-adic number)

I am unable to comprehend this. How does $\dfrac{5^2-1}{3}=\dfrac{44_5}{3} = 13_5; \, \dfrac{5^4-1}{3}=\dfrac{4444_5}{3} = 1313_5$ result in $-\dfrac{1}{3}=\dots 1313_5$?

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2 Answers 2

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Because is says that if $x=\cdots1313_5$ then $3x+1\equiv 0 \text{ mod } 5^n$ for all $n$, which is precisely what it means to be $0$ in $\mathbb{Q}_5$. Thus, you see that $3x+1=0$ so that $\displaystyle x=\frac{-1}{3}$.

EDIT:

Now that I have more time, let me be less glib about this response.

Whenever possible, we want to turn problems about $\def\Qp{\mathbb{Q}_5}$ $\def\Zp{\mathbb{Z}_5}$ $\Qp$ into problems about $\Zp$ since they are easy to deal with. So, how can we interpret $\displaystyle \frac{-1}{3}\in\Qp$, well since $\Qp$ is $\text{Frac}(\Zp)$ the only clear interpretation is that it is the element $x$ of $\Zp$ which satisfies $3x+1=0$. So, instead let us try to look for a solution $3x+1=0$.

To begin, let us recall how we can think about $\Zp$. Intuitively, $\Zp$ is the set $\{z\}$ of solutions to systems of equations as follows:

$$\begin{cases}z &\equiv a_1 \mod 5\\ z &\equiv a_2 \mod 5^2\\ z &\equiv a_3 \mod 5^3\\ &\vdots\end{cases}$$

where the equations are "consistent" (i.e. $a_i\equiv a_j\mod p^i$ for $i\leqslant j$). So, now if $x$ satisfies $3x+1=0$ then this should translate to mean

$$\begin{cases}3x+1 &\equiv 3a_1+1 \equiv 0 \mod 5\\ 3x+1 &\equiv 3a_2+1 \equiv 0 \mod 5^2\\ 3x+1 &\equiv 3a_3+1\equiv 0 \mod 5^3\\ &\vdots\end{cases}$$

So, we can solve each of these equations piecewise and find that

$$(a_1,a_2,a_3,\ldots)=(3,8,83,\ldots)$$

Ok, so, this tells us that $x=(3,8,83,\ldots)$...this doesn't look right? How can we go from this to the desired $x=\ldots1313_5$? The key is that we have the same element of $\Qp$ expressed in different forms. Indeed, the notation $x=1313_5$ means that

$$x=3+1\cdot 5+3\cdot 5^2 +\cdots$$

To reconcile this ostensible difference, let us write $3+1\cdot 5+3\cdot 5^2+\cdots$ in the same notation that we already have $x$ in. Recall that the correspondence between $\mathbb{Z}$ and these sequences is

$$m\mapsto (m\mod 5,m\mod 5^2,m\mod 5^3,\ldots)$$

Thus, we see that

$$\begin{aligned} 3 & \mapsto (3,3,3,\cdots)\\ 5 &\mapsto (0,5,5,\ldots)\\ 5^2 & \mapsto (0,0,5^2,\ldots)\end{aligned}$$

Thus, we see that

$$ \begin{aligned}3+5+3\cdot 5^2 &=(3,3,3,\ldots)+(0,5,5,\ldots)+(0,0,75,\ldots)\\ &= (3,8,83,\ldots)\end{aligned}$$

and voilà!

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I'll answer a simpler question, since the logic here is easier to see and extends to your question as well.

Consider that $5-1 = 4_5, 5^2-1 = 44_5, 5^3-1 = 444_5$. However, $...4444_5 = -1_5$. Why is that?

Put even more generally, for any ring of p-adic integers and any p, $\sum_{n=0}^\infty (p-1) \cdot p^n = -1$. So $...111111_2 = -1_2,$ $...222222_3 = -1_3$ $...444444_5 = -1_5$, etc.

The above is always true, even though for any partial summation for n=0 up to some finite s $\sum_{n=0}^s (p-1) \cdot p^n = p^{s+1}-1$. So even though $2222_3 = 3^4-1$, and $44444_5 = 5^5-1$, $...222_3 = -1_3$ and $...444_5 = -1_5$.

Why is that?

Well, you might have noticed that some algebra on the equation $\sum_{n=0}^\infty (p-1) \cdot p^n = -1$ leads to $(p-1) \cdot \sum_{n=0}^\infty p^n = -1$, which finally leads to $\sum_{n=0}^\infty p^n = \dfrac{-1}{p-1} = \dfrac{1}{1-p}$, which is the usual formula for the summation of a geometric series, except $|p|$ no longer has to be less than 1. What's going on here?

To answer the question, let's stop considering any specific value of $p$ and instead look at the more general ring of formal power series $\mathbb{Z}[[p]]$, which consists of all finite or infinite series of the form $... + a_2 p^2 + a_1 p^1 + a_0 p^0$, where the $a_n \in \mathbb{Z}$. In effect, this ring is like a precursor to the p-adic integers: it's kind of like a p-adic integer ring where we haven't decided what value we're setting $p$ to yet.

Now, one element in this ring is the series $... + p^2 + p^1 + 1$, where the coefficients for each term are 1 (note that $p^0 = 1$). Now, observe what happens if we multiply by $p$ in this ring:

$(... + p^2 + p^1 + 1) \cdot p = ... + p^3 + p^2 + p^1$

Also, look at what happens if we subtract 1:

$(... + p^2 + p^1 + 1) - 1 = ... + p^3 + p^2 + p^1$

You'll see that we've gotten the same thing, and so $(... + p^2 + p^1 + 1) \cdot p = (... + p^2 + p^1 + 1) - 1$.

This is the quirky bit of behavior that, ultimately, is the source of your misery with the 5-adics. When you start with a formal power series ring, and you take the element corresponding to the infinite series $... + p^2 + p^1 + 1$, multiplying by $p$ is the same as subtracting 1.

This can contradict our intuition at first, because when we deal with actual geometric series of reals like we're used to, as opposed to "meaningless" formal expressions of symbols that we're just manipulating like we are here, that behavior only holds if $|p| < 1$. In that case, it checks out intuitively because multiplying by a value < 1, and subtracting 1, are both operations that typically giving you a "smaller" number than you started with, and in the case of the geometric series with $|p| < 1$ which converges in the reals, you get the same thing both ways.

On the other hand, naively if you think of the p-adics as being numbers where we're assigning to $p$ a value that's greater than 1, you might expect that multiplication by $p$ gives you a naively "larger" p-adic number than you started with, whereas subtraction by 1 ought to give you a naively "smaller" number than you started with, and hence you conclude that the two values couldn't ever possibly be the same for a $p$ that's greater than 1.

This intuition can be the source of much confusion, and in fact that's the genius of the ring of formal power series: the identity $(... + p^2 + p^1 + 1) \cdot p = (... + p^2 + p^1 + 1) - 1$ always holds, no matter what, intuition about ordering be damned! We're not actually "assigning a value to p," we're just performing operations on these formal series, and they just form a ring like anything else.

Now let's see what the implications of that are: firstly, with some algebra, we quickly find

$(... + p^2 + p^1 + 1) \cdot p = (... + p^2 + p^1 + 1) - 1$

$(... + p^2 + p^1 + 1) \cdot p - (... + p^2 + p^1 + 1) = -1$

$(... + p^2 + p^1 + 1) \cdot (p - 1) = -1$

And now we're back where we're started. So if we now DO want to do something like "setting $p$ to 2," we can do that by setting $p-2=0$, or in other words taking the quotient ring of $\mathbb{Z}[[p]]/(p-2)$, where $(p-2)$ is the ideal generated by p-2. The weird behavior from before sticks, and you get the ring $\mathbb{Z}_2$ of 2-adic integers. And since the above identity still holds, you get $...1111_2 = -1_2$. If you use the same quotient ring technique to "set" $p=5$, you get $...44444_5 = -1_5$. And so on.

So now we've come full circle. It's because of this same principle that you can get other strange results, such as that $...4444_5 = -1_5$, and then also other related things like that $\dfrac{...4444_5}{3} = -1/3_5 = ...1313_5$. And all of this "quirky behavior" ultimately derives from the quirkiness of the formal power series ring itself.

This should make you think a bit about what it means to say one p-adic number is "larger" than another, and if it's really possible to order the p-adics in the same way we order the naturals.

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