Following on from a previous post (http://tglad.blogspot.com/2010/07/that-crazy-formula.html) it is an interesting exersize to push the number of operators back another notch. Here's a somewhat speculative idea.
Take the ultra-operators of the previous post and increase their position by one, so operator 1 is equality, 2 is addition, 3 multiplication etc. Then make the 0th operator the comma operator.
The previous post talked about having a number-of-values (rather than a set of), this concept is defined by the use of ',' to allow more than one. So the value of a,b is simply the two values a and b, i.e. well it is just a,b i.e. it doesn't really operate on the two value other than to allow them to exist side by side.
This could extend the previous crazy formula to:
e (4) i (3) pi (2) 1 (1) 0 (0) .
i.e. e^i*pi + 1 = 0,.
where . is the empty element, or void, or the contents of the empty set.
It leads to a new creation story for numbers (going from right to left in above equation)...
In the beginning there was nothing (void)
For there ever to be more than this, the concept of a plurality (',' operator) is added
The first value can now exist alongside void, and since there is no arithmetic yet it is and can only be nothing/zero
Being a proper value it equates to something, hence the '=' operator can exist
The existence of equality allows for inequality (something that doesn't equal 0), we name that something 1, the unit.
What is to 1 as 1 is to 0? this recursive question leads us to define the set of integers and consequently the addition operator. (and inversely the subtraction operator as seen by seeing the above equation as -1 + 1 = 0)
Repeated addition on the integers leads to a definition for the '*' operator (and inversely divide) which leads to the quotient numbers.
This somehow leads to a new type of number not represented so far, pi the architypal transcendental.
Repeated multiplication leads to exponentiation operator
This (and its inverse, logarithm) require the new number e
Exponentiation leads to the question x^2 = -1 what is x? which leads to the development of the imaginary number i.
Writing this made me think, most operators come in pairs, so what are the inverses of ultra operators 0 and 1?
I would guess that the inverse of equals is 'doesn't-equal' (!=), but what does it evaluate to? Well wrap the values inside sets and '=' becomes the intersection operator I think.. so '!=' would probably be the equivalent of union minus intersection, or a xor b.
Harder it the ',' operator, what is its inverse? (call it ';') what is a;b?
My guess at the correct answer is that it is its own inverse (or equally it has no inverse). What do you think?