In another recent question, the 10-adic numbers came up (along with the usual issues of not really being a field due to 10 not being prime, etc). I had a thought: ordinary binary numbers (either integers or reals) can be trivially interconverted between binary, octal, hexadecimal, etc notations. Is this likewise true of the 2-adic numbers, and if not, why not?

Or, the converse - is there an e.g. "8-adic" ring that is distinct in its properties from the same notation from grouping 2-adic numbers into groups of three digits, etc.?

  • $\begingroup$ More generally, the $n$-adic integers are isomorphic to the $\text{rad}(n)$-adic integers where $\text{rad}(n)$ is the product of the prime divisors of $n$. This is in turn isomorphic to $\prod_{p \mid n} \mathbb{Z}_p$. $\endgroup$ Jan 23, 2016 at 22:26

1 Answer 1


No, the $8$-adic completion of the rationals is isomorphic to the $2$-adic. You will recognize that if you put $|8|_8=1/8$, then it must follow as the night the day that $|2|_8=1/2$.

  • $\begingroup$ Am I right in thinking here that distance is no longer the highest power of $n$ that divides $\lvert x-y\rvert$? As only $8^0=1$ divides $2$ so by my definition $\lvert 2\rvert_8=1$. $\endgroup$ Jun 3, 2019 at 16:20
  • $\begingroup$ ...because I would seem able to extend your argument to make the incorrect claim that $\lvert \sqrt2\rvert_2$ must equal $\frac1{\sqrt2}$. $\endgroup$ Jun 3, 2019 at 16:25
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    $\begingroup$ But it’s certainly true that $|\sqrt2\,|_2=\frac1{\sqrt2}$, assuming that you want to extend the standard $2$-adic absolute value to $\Bbb Q(\sqrt2\,)$. $\endgroup$
    – Lubin
    Jun 3, 2019 at 21:47

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