It makes perfect sense to speak of a base $10$ digit expansion. Why does it not make sense to speak of $10$-adic numbers or $10$-adic integers?


5 Answers 5


You can define the $n$-adic integers, even if $n$ is not prime. For example, every $10$-adic integer has a $10$-adic decimal representation, and you can add them and carry as usual.

One thing that I like about this system is an alternate way of writing negative numbers in base 10. Instead of having to put a negative sign in front, we have $-1 = \ldots999$, $-2 = \ldots998$, and so on. Here is an example of computing $37 - 50 = -23$ in the $10$-adic numbers: \begin{align*} \ldots0000037& \\ +\quad \ldots 9999950 &\\ \hline = \quad\ldots 9999987&\\ \end{align*}

And here is $37 \cdot -50 = -1850$: \begin{align*} \ldots0000037& \\ \cdot\quad \ldots 9999950 &\\ \hline = \quad \ldots 00000000 &\\ + \quad \ldots 0000185\phantom{0} &\\ + \quad \ldots 000333\phantom{00} &\\ + \quad \ldots 00333\phantom{000} &\\ + \quad \ldots 0333\phantom{0000} &\\ + \quad \ldots 333\phantom{00000} &\\ + \quad \phantom{00} \cdots \phantom{000000} &\\ \hline = \quad\ldots 99998150&\\ \end{align*}


Most of the other answers suggest that the reason we don't like the $10$-adic numbers, the reason they don't behave nicely is that they have zero divisors. But we are perfectly fine dealing with rings with zero divisors such as $\mathbb{Z} / 10\mathbb{Z}$. So let me elaborate on this.

A key part of any good introduction to $p$-adic numbers is a discussion of their topological properties, which come from the $p$-adic absolute value. For example, this is how we may obtain the $p$-adic numbers as a completion of the rational numbers $\mathbb{Q}$ with a different distance metric than the one that gives $\mathbb{R}$.

The problem with zero divisors in this context is that a ring absolute value cannot even be defined when there are zero divisors. Why? Because we really want it to be the case that $|xy| = |x| \cdot |y|$, so if $x$ and $y$ are zero divisors, then since $|0| = 0$, $|x| = 0$ or $|y| = 0$. If we think of the $10$-adic integers as the ring $\mathbb{Z}_2 \times \mathbb{Z}_{5}$, this essentially forces us to either give the $\mathbb{Z}_2$ part of every $10$-adic integer zero norm, or to give the $\mathbb{Z}_5$ part of every integer zero norm. So the $10$-adic numbers do not have a natural norm. This is probably a big reason that we don't usually study them.


I was going on at too great length in a comment to a discussion between Henning Makholm and Hurkyl, so let me put it all into an answer instead:

To show that $\Bbb Z_{10}\cong\Bbb Z_2\times\Bbb Z_5$, you need to find a pair of orthogonal idempotents, call them $E_2$ and $E_5$ such that $E_p^2=E_p$ for $p=2$ and $5$, $E_2E_5=0$, and $E_2+E_5=1$.

To do this, you find consistent sequences $\{e_{2,n}\}$, $\{e_{5,n}\}$ such that \begin{align} e_{p,n}^2&\equiv e_{p,n}\pmod{10^n}\\ e_{2,n}e_{5,n}&\equiv0\pmod{10^n}\\ e_{2,n}+e_{5,n}&\equiv1\pmod{10^n}\,. \end{align} And here’s how to do that: Use Chinese Remainder Theorem to get \begin{align} e_{2,n}&\equiv1\pmod{2^n},&e_{2,n}\equiv0\pmod{5^n}\\ e_{5,n}&\equiv0\pmod{2^n},&e_{5,n}\equiv1\pmod{5^n}. \end{align} And that would be the whole story, except that it’s easier to compute $E_2$ and $E_5$ than to describe the process. Start with the right numbers modulo $10$, namely $e_{2,1}=5$ and $e_{5,1}=6$. Raise the first of these successively to $2$-power powers, the second successively to $5$-power powers, and look at the last digits. But it’s even easier than that, ’cause you only need to start with $2$-powers of $5$ for your $e_{2,n}$’s, and subtract from $1$ for the other approximant.

So you start $5$, $25$, $625$, $390625\equiv0625$, etc., and easily get to ten digits with $8212890625\equiv E_2\pmod{10^{10}}$, and from that $E_5\equiv1787109376\pmod{10^{10}}$. I’ll leave it to the interested reader to do the three multiplications that show that these are indeed orthogonal idempotents modulo $10^{10}$.

EDIT — Addendum

user1952009 and Henning Makholm have asked what good all of the above is. It’s just that if you have a $2$-adic integer $S_2$ and a $5$-adic integer $S_5$, then you get a $10$-adic integer corresponding to $(S_2,S_5)$ this way: approximate $S_2$ by ordinary integers $\{s_{2,n}\}$ and $S_5$ by $\{s_{5,n}\}$, then at each stage, multiply the $2$-adic approximants by $E_2$ and the $5$-adic approximants by $E_5$. You get two $10$-adic integers, and you add them to get your $(S_2,S_5)$.

For instance, suppose you want something in $\Bbb Z_{10}$ that’s the $2$-adic integer $\sqrt{-7}$ and the $5$-adic integer $23/128$. To ten places, these are $b5;$ in hex notation for the $2$-adic integer and $4201334331;$ in $5$-ary notation for the element of $\Bbb Z_5$. As integers modulo $10^{10}$, these are $181$ and $8621216$, respectively. Now, $181$ is congruent to $\sqrt{-7}$ modulo $2^{10}$, but it’s something unwanted modulo $5^{10}$, so we multiply it by $8212890625$, which is $1$ modulo $2^{10}$ and zero modulo $5^{10}$. Result is good at $2$ and zero at $5$. Do the same thing for $8621216$, which is $\equiv23/128\pmod{5^{10}}$, but something useless modulo $2^{10}$, But $8621216\cdot1787109376$ is good at $5$ and zero at $2$. Add these two products modulo $10$, get $$181\cdot8212890625+8621216\cdot1787109376\equiv2479324341\pmod{10^{10}}\,.$$ This is a number whose square $\equiv-7\pmod{2^{10}}$, and such that when you multiply it by $128$, the result is $\equiv23\pmod{5^{10}}$, which is what I promised.

FURTHER EDIT — in response to Henning Makholm’s very valid objections.

I apologize for the logical messiness of all this. It would have been better to toss out all my previous posting and replace with what’s below, but that would have invalidated all the comments.

  1. A sequence is $10$-adically convergent (resp. Cauchy) if and only if it is both $2$-adically and $5$-adically convergent (resp. Cauchy).
  2. The idempotent $E_2$ is made from the sequence $\{5^{2^n}\}$ that is $2$-adically convergent to $1$ and $5$-adically convergent to $0$. The orthogonal idempotent $E_5=1-E_2$ is similarly made from a sequence $2$-adically convergent to $0$ and $5$-adically convergent to $1$.
  3. $\Bbb Z_2\cong\Bbb Z_{10}E_2$, because for every $\alpha\in\Bbb Z/2^n\Bbb Z$, there is $\alpha'\in\Bbb Z/{10}^n\Bbb Z$ such that $\alpha\equiv\alpha'\pmod{2^n}$ and $\alpha'\equiv0\pmod{5^n}$, namely $\alpha'=5^{2^n}\alpha$. Similarly, $\Bbb Z_5\cong\Bbb Z_{10}E_5$, because for every $\beta\in\Bbb Z/5^n\Bbb Z$, there is $\beta'\in\Bbb Z/10^n\Bbb Z$ with $\beta'\equiv\beta\pmod{5^n}$ and $\beta'\equiv0\pmod{2^n}$.
  4. There is a surjective ring homomorphism $\Bbb Z_{10}\to\Bbb Z_2$, by $w\mapsto wE_2$, similarly for $w\mapsto wE_5$ giving a surjective homomorphism to $\Bbb Z_5$.
    This should be enough to show that $\Bbb Z_{10}\cong\Bbb Z_2\oplus\Bbb Z_5$. But Henning asks why there is no other such decomposition. The answer has to do with the fact that the only idempotents in $\Bbb Z_p$ are $0$ and $1$.
  5. Let $e$ be any idempotent in $\Bbb Z_{10}$. I say that $e$ is either $0$, $E_2$, $E_5$, or $1$. Indeed, $e=e\cdot1=e(E_2+E_5)=eE_2+eE_5$. But $eE_2$ is an idempotent in $\Bbb Z_{10}E_2\cong\Bbb Z_2$, so $0$ or the unity there, namely $E_2$. Similarly $eE_5$ is either $0$ or $E_5$, and these four cases exhaust the four claimed possibilities for idempotents in $\Bbb Z_{10}$.
  • $\begingroup$ Seconding @user1952009: How does the existence of $E_2$ and $E_5$ with these properties imply that $\mathbb Z_{10}\cong\mathbb Z_2\times\mathbb Z_5$? $\endgroup$ Sep 8, 2016 at 20:27
  • $\begingroup$ @user1952009 and Henning Makholm, see my addendum above. $\endgroup$
    – Lubin
    Sep 9, 2016 at 2:51
  • $\begingroup$ This procedure still seems pretty mysterious. When you write "... which is $1$ modulo $2^{10}$ and zero modulo $5^{10}$", how does that property follow from your defining equations $E_2^2=E^2$, $E_5^2=E^5$, $E_2E_5=0$ and $E_2+E_5=1$? Are there additional secret properties of $E_2$ and $E_5$ you depend on but did not speak about originally, and if so what are they? $\endgroup$ Sep 9, 2016 at 6:35
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    $\begingroup$ It looks like it might have been "easier" to give a proof after all $\ddot{\smile}$. If it helps, Lubin's $E_2$ and $E_5$ are the numbers $M$ and $N$ in my answer that map to $(1, 0)$ and $(0, 1)$ respectively under the natural mapping of $\Bbb{Z}_{10}$ to $(\Bbb{Z}/2^i\Bbb{Z}) \times (\Bbb{Z}/5^i\Bbb{Z})$ for each $i$ (and in the latter ring it does make sense to call them orthogonal idempotents). $\endgroup$
    – Rob Arthan
    Sep 9, 2016 at 9:33
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    $\begingroup$ Guys, why are you faffing around? My answer supplies the proof that you are flailing around for. If it's unclear, please let me know and I will endeavour to clarify it for you. $\endgroup$
    – Rob Arthan
    Sep 9, 2016 at 23:06

You can speak about $10$-adic numbers just fine, but they don't behave as nicely as $p$-adic numbers. For example, the $10$-adic integers have zero divisors, so no matter how you complete or massage them afterwards, you won't be able to embed them in a field.

  • $\begingroup$ I don't really understand why $\mathbb{Z}/6\mathbb{Z}$, $\mathbb{Z}/10\mathbb{Z}$ etc. are canonical rings that we study in abstract algebra class, but suddenly we are terrified of $\mathbb{Z}_{10}$ because it has zero divisors. $\endgroup$ Sep 9, 2016 at 10:41
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    $\begingroup$ @6005: Who's "terrified"? $\endgroup$ Sep 9, 2016 at 10:42
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    $\begingroup$ Not you, I guess. I just mean it isn't usually brought up in a discussion of $p$-adic numbers. $\endgroup$ Sep 9, 2016 at 10:43
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    $\begingroup$ Maybe zero divisors screw up the $p$-adic norm. $\endgroup$ Sep 9, 2016 at 10:43

It does, but the result is not as nice. And we have $\mathbb{Z}_{10} = \mathbb{Z}_2 \times \mathbb{Z}_5$, so we can understand 10-adic numbers in by breaking them apart into a $2$-adic and a $5$-adic piece, each of which does behave nicely.

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    $\begingroup$ Elaborating on why (or at least in what ways) the result is "not as nice" might be helpful. $\endgroup$
    – Wojowu
    Sep 8, 2016 at 14:00
  • $\begingroup$ @Wojowu: Exactly! $\endgroup$
    – Mario
    Sep 8, 2016 at 14:03
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    $\begingroup$ Are you sure $\mathbb Z_{10}=\mathbb Z_2\times\mathbb Z_5$ holds for $n$-adic numbers (rather than for molular arithmetic)? It doesn't work like that digit-for-digit -- for example $...005_{10}\times ...006_{10}$ would make $...001_2\times ...000_2$ or $...000_5\times ...001_5$ both of which are $0$, but the original 10-adic multiplication yields $...030_{10}$. $\endgroup$ Sep 8, 2016 at 14:11
  • $\begingroup$ @HenningMakholm I think it works. Rather than digit-by-digit, it's probably something like inverse limit of the last $n$ digits. So the 2-adic part would be $\ldots00101_2 \times \ldots00110_2$ and the 5-adic part would be $\ldots 0010_5 \times \ldots 0011_5$. $\endgroup$ Sep 8, 2016 at 14:18
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    $\begingroup$ It’s really easy to find a pair of $n$-digit base-ten numbers, call them $A$ and $B$, with $A^2\equiv A$, $B^2\equiv B$, $AB\equiv0$, and $A+B\equiv1$, all congruences modulo $10^n$. So you get $10$-adic numbers having the same relations. You use these for inducing your isomorphism between $\Bbb Z_{10}$ and $\Bbb Z_2\times\Bbb Z_5$. $\endgroup$
    – Lubin
    Sep 8, 2016 at 16:03

As has been said in other answers the 10-adic integers have zero-divisors. To sketch a proof of this, let me write $[m]_n$ for the equivalence class of $m$ modulo $n$, so $\Bbb{Z}/n\Bbb{Z} = \{[0]_n, [1]_n, \ldots[n-1]_n\}$. Then $[m]_{2^i5^i} \mapsto ([m]_{2^i}, [m]_{5^i})$ is well-defined and defines an isomorphism $\phi_i$ say between the ring $\Bbb{Z}/10^i\Bbb{Z}$ and the product ring $(\Bbb{Z}/2^i\Bbb{Z}) \times (\Bbb{Z}/5^i\Bbb{Z})$. There is a unique pair of classes $x_i = [m_i]_{10^i}$ and $y_i = [n_i]_{10^i}$, such that $\phi_i(x_i) = (1, 0)$ and $\phi_i(y_i) = (0, 1)$. By uniqueness $x_i = [m_{i+1}]_{10^i}$ and $y_i = [n_{i+1}]_{10^i}$. So you can form 10-adic integers $M$ and $N$ say such that for all $i$ $$M \equiv m_i \mod 10^i\\ N \equiv n_i \mod 10^i$$ and then as $m_in_i \equiv 0 \mod 10^i$ for all $i$, you will have $MN = 0$ in the ring of 10-adic integers.


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