Divide by a number without dividing. Can anyone come up with a way to divide any given x by any given y without actually dividing?
For example to add any given x to any given y without adding you would just do:

$x-(-y)$
And to subtract any given x from any given y (that is, y-x) you could do:

$y+xe^{iπ}$
*edit: well since (i) is ($\sqrt{-1}$) and that is technically subtracting this one might not work perfectly but for the sake of the riddle and for the sake of example, I'm using that equation :)
How can you divide without dividing? Can anyone come up with equations that work for all $x$ and $y$ values? (For all intents and purposes we will leave out dividing by zero issues and what-not... don't worry about that...
 A: Look at the equation $\frac{1}{x}=a$. We use Newton's Method to approximate the solution. 
Let $f(x)=\frac{1}{x}-a$. The standard Newton iteration gives
$$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}=x_n -\frac{\frac{1}{x_n}-a}{-\frac{1}{x_n^2}}.$$
This simplifies to
$$x_{n+1}=x_n(2-ax_n).$$
Remark: Note that only subtraction and multiplication are used. If we start with $x_0$ close enough to $\frac{1}{a}$, the method converges rapidly. It was once used to implement reciprocal in software. 
A: For $y \neq 0$ $$\large x\div y = \dfrac 1y\times x = y^{-1}\times x = \large y^{\left(e^{i\pi}\right)}\times x = y^{\left(i^2\right)}\times x$$
A: Logs turn reciprocals into minus signs: $\ln(1/y)=-\ln(y)$.  Thus, $$x/y=xe^{-\ln y}.$$
(This is assuming that $y$ is positive.  If $y$ is negative, then $x/y=-xe^{-\ln(-y)}$.)
A: Take the logarithm that maps multiplication/division into addition/subtraction:
$$\frac{x}{y}=e^{\log{x/y}}=e^{\log x- \log y}.$$
$x,y >0$. 
Also, see my answer for multiplying natural numbers here:
Advocating base 12 number system
