# Alien Math: The Reveal, Part 1

In part one, I will give strong hints as to what the Alien math system is, in part 2 I will reveal it all and why I think its a particularly coherent system. While working on the derivative from my first post, I started to get a sense of what thinking in this system is like.

Its all about function D, and its inverse, function C. D is the wrapper, and C is the unwrapper, and certain functions work best if their operands are wrapped, and some work best if they are unwrapped.

The expression for $\frac{dZ}{da}$ is huge so I will build it up, bit by bit.

First, we have K(D(b), b). It seems that K wants the first operand wrapped and the second not. We will name this sub expressions e1, e2, e3 and so on.

e1 = C(K(D(b),b))

So we wrap it up for K to work on, and the unwrap the result.

Reveal #1: C(K(D(a), b) = a * b

So e1 = C(K(D(b), b) = $b^2$

You can chain these values by stacking the repetitions on the left side like in:

e2 = C(K(K(K(D(b),b), a), a)) = $a^2 * b^2$
e3 =  C(K(K(QT, a), b) = 2 a*b

Its not as elegant because K is not associative or commutative, unlike ordinary multiplication.

Reveal #2: The numeral system uses RSTU as 0,1,2,3 in base 4. However, numbers start after the decimal point, so you have to use Q, which is the reciprocal function, to get numbers larger than 1.

For numbers like three that are difficult to get in base 4 using this method, you can use the alternate system. It uses E’s to count up small numbers. E = 1, EE = 2, EEE = 3 and so on.

QT is $\frac{1}{0.2 base 4}$ which is 2.

So we have our products and our literal numbers.  Now:

Reveal #3: C(Z(D(a), D(b))) = a + b

The wrapping and unwrapping is similar to K except both parameters are wrapped. It can be extended similarly to K.

e4 = C(Z(D(e2), Z(D(e3), D(E))) = e2+e3+1 = $a^2 * b^2 + 2 a b + 1$

Reveal #4: W(R, a) = -a

W(a,b) is not a – b, but when the first parameter is R (0), the result is the negative of the second param. This is the only way to introduce negative numbers in this system.

e5 = C(Z(D(W(R, e1)), D(E))) = -e1 + 1 = $-b^2 + 1$

Using Q as the reciprocal, and the construction in Reveal #1, we can construct the final expression, and expand it using the e’s

dz/da = C(K(D(e5), Q(e4)))
dz/da = C(K(D(C(Z(D(W(R, C(K(D(b),b)))), D(E)))), Q(C(Z(D(C(K(K(K(D(b),b), a), a))), Z(D(C(K(K(QT, a), b)), D(E))))))

You can simplify further by cancelling the C(D(?)) and D(C(?)) expressions.

dz/da = C(K(D(C(Z(D(W(R, C(K(D(b),b)))), D(E)))), Q(C(Z(D(C(K(K(K(D(b),b), a), a))), Z(D(C(K(K(QT, a), b)), D(E)))))) = $\frac{-b^2+1}{a^2 b^2 + 2 a b + 1}$