I only just learned about dozadic and p-adic numbers after watching a YouTube video. I found it interesting that negative numbers and real numbers could be represented by an infinitely long positive integer. Which made me wonder, could imaginary numbers also be represented as 10-adic or p-adic number? What would be the dozadic equivalent of i1?
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When $p\equiv 1\pmod 4,$ yes. – Thomas Andrews Jun 19 '23 at 01:05
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2And they aren't "infinitely long positive integers." It is unwise t9 think of them as positive. They look positive, but $p$-adic numbers are not naturally ordered. – Thomas Andrews Jun 19 '23 at 01:07
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1@ThomasAndrews To be careful: $i$ can be defined in the $p$-adics when $p\equiv1\pmod4$, but that doesn't imply that every purely imaginary number can be (much less every complex number). – Greg Martin Jun 19 '23 at 01:41
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1Most real numbers cannot be naturally represented as $p$-adic numbers (or the other way around). https://math.stackexchange.com/a/4007515/96384 What Thomas Andrews means above is that the equation $x^2+1=0$ has two solutions in $\mathbb Q_p$ for $p\equiv 1$ mod $4$. It is potentially unwise to call these solutions $\pm i$ though, or to conflate them with the imaginary unit and its conjugate in $\mathbb C$. They only behave like those algebraically. – Torsten Schoeneberg Jun 20 '23 at 02:25
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Right. The reals and the $p$-adics, for all the so-many primes $p$, not only don’t correspond with each other, each of them is completely ignorant of the existence of the others. – Lubin Jun 20 '23 at 19:04