Division By 0

Suppose we are dividing some number, lets say 5, by 0.
What kind of useful way could we represent this and how would we go about getting the inverse?

Lets see

5/0 = inf*0 rem 5
Assuming 0 goes into 5 infinite times.

while
0/0 = inf*0 rem 0

and
0/0 = k

where 'inf' represents a non-terminating process, or the set of all numbers in some set.

therefore

5/0 = 0*0 rem 5*1

but its a special kind of zero, because it too
represents the output of a process.
Consider it a production process.
That remainder gets carried around as a mathematical object

0/0 = inf*0 rem 5

5 * inf0 = 5/0

5/0 = inf*0 rem 5

Why?
Because 0 goes into 5 either 0 times, or an infinite number of times, with 5 as the remainder.
Actually its

5/0 = inf*0 rem 5*1

That one at the end is just the multiplicative identity, which I put in there
because I am obsessive.

Now think of 0/0 not as some heresy (you godless heathens), but like a point on the graph of a function with inputs x, and y.
Its a special dimension, sort of like the imaginary numbers, and i.
Or the matrix (lol).

We will represent it with the special symbol 'k'.

Now how do we take this, (inf*0 rem 5*1) and get 5/0 back out? And why are we using this structure? We'll see shortly.

We get back to 5/0 like so:

First we swap the 'inf' and the 5.

inf*0 rem 5*1 --> inf/0 = 5*0 rem inf*1

And then we make the 5 the numerator, and the infinity the denominator.

--> 5/inf = 0 --> 5/(5/inf) == 5/0

And this works because 5*0 is just 0, and 5/inf *approaches* zero, which only
works if 5/(inf) actually equals 5/(inf*1)
where the 1 is the multiplicative identity. Which seems like a tautology, but has a purpose.

Infinite division must approach 0. And we can represent n/0 as the *output* of some other division that resulted in n on the left side, and 0 on the right side, even if we can't compute it further than that because its indeterminant.

or, if we want more abstract representation:

inf*1 rem n*0 --> inf/0 = n*0 rem inf*1 --> n/inf = 0 --> n/(n/inf) == n/0

where inf/0 equals n*0 rem inf*1 for *all* n, though its not reversible if we
don't have the value of n in n/0 or inf*1 rem n*0


For example, take finding the limit of x and sin(0)
An indeterminate expression seems to have more than one answer, depending on how you evaluate it.

lim(x->0) (sin(x)/x)
Seems to be zero if you use direct substitution: sin(0)/0. However, sin x and x are almost equal to each other as you get close to zero, which actually makes the limit 1.

Lets work the problem a little bit.

where n = sin(0)

sin(0)/0 == inf*0 rem sin(0)*1 --> sin(0)*0 rem inf*1

sin(0)*0 rem inf*1 --> sin(0)/inf = 0 --> sin(0)/(sin(0/inf) == n/0 == n

NOW ASSUME something strange.

Assume our n is actually a number thats infinite on a given number line--only *relative* to
*some other* number, m.

It has a fixed value, and yet its infinite. How?

That number, m, is none other than our indeterminant point k=0/0

When we substitute 0 for our new dimension, k, then n/0 = n makes more sense (and I'll show you how in a moment).

And this works because

inf*1 rem n*0  -->   inf/0   = n*0 rem inf*1 --> n/inf = 0 --> n/(n/inf) == n/0

Thats n/inf = 0 --> n/(n/inf) == n/0

Think about it, k represents a special point on a function, where inf gets replaced with k
such that n/(n/(n/inf)) --> 0/(0/(0/0))

which after rearrangement...

0/0 = inf*0 rem 0*1 -- > (0*0 rem inf*1)

which becomes

(0*0 rem inf*1)

and when we plug that into our solution 0/(0/(0/0))

we get this:

(0*0 rem inf*1)/((0*0 rem inf*1)/((0*0 rem inf*1)/(0*0 rem inf*1)))

That 'remainder' inf*1 exists on the same imaginary number line as k.
Its not determinant on the *ordinary* number line, because it breaks some rules otherwise.
But it *is* determinant on this new number line

It translates to

inf/(inf/(inf/inf))

And what does that come out to?

It comes out to 1.

Eat it haters.
I'll take my noble prize at the special olympics now.

I want it delivered in the form of twenty two thousand ziplining tickets.
So all the kids in wheelchairs will get to experience not what its like to *walk*
for once in their life.

But what its like to FLY.