Will mathematicians ever define zero as denominator? As you all know, Zero as a denominator is a no-go in maths. I always wonder if this could ever change.
For example you can’t use negative integers in the root but mathematicians defined complex numbers to “bypass” this hurdle, although the “bypassing” is a somehow exaggeration but I hope, you get what I mean!
I wonder if we could define the “super complex numbers “ , such as one devided by zero, equals to 1 over i. I am not sure if 1 over i would get in trouble with the notations of complex numbers!
 A: In $p$-adics we can consider divisions where the divisor is not multiplicatively invertible. Such divisions arise in the context of $p$- adic logarithms.
Suppose we seek a solution to the equation $2^x=3$ in $5$-adics. Strictly speaking, there is no solution because the exponent, to be determined below, gives powers of $2$ that do not converge uniformly to $3$; they actually firm four subsequently that separately converge to $3\zeta$ for each fourth of unity $\zeta$. To remove this complication we take fourth powers, so that $16^x=81$.
We then render
$x=\dfrac{\ln 81}{\ln 16}$
and apply the Maclaurin series for $\ln(1+x)$, which converges $p$-adically whenever $x$ is a multiple of $p$ for any odd prime $p$. This gives, using base 5 arithmetic:
$\ln(311)=...02010$
$\ln(31)=...10330$
Now, despite the denominator having a terminal zero and thus it lacks a multiplicative inverse, the $5$-adic equation
$...10330x=...02010$
does have a unique solution, which is obtained by canceling terminal zeroes and using ordinary division:
$...1033x=...0201$
$x=...1412.$
(Note the price we pay: we have list a "digit" from our logarithm calculations.) Switching back to base 10, this derivation says that $16^x\equiv81\bmod5^5$ when $x\equiv232\bmod5^4$. To retrieve our original equation $2^x=3$, we must separately apply the correct residue modulo 4 ($x\equiv3$), so we find that $2^x\equiv3\bmod5^5$ when $x\equiv2107\bmod(4×5^4)$.
