What I am about to say, if true will really surprise a lot of people. We all have a tendency to believe that once something is locked in stone, then by virtue of that, it is unshakable. Well, that's mostly true, but mankind has a vast amount of learning left to do...

Picture a spreadsheet, but it's no ordinary spreadsheet. To the left and to the right it extends out infinity. Up and down it also extends out infinitely.

In each cell we place a real number. We start with one in the center and generate the rest of the spreadsheet.

Now, we start in the middle, and we make some program to add the following digits in each cell to the right by:

.1, .111..., .10101..., .1001001..., .100010001..., ...

and to the left by

..., 10001, 1001, 101, 11, 1

and we leave the center at 0.

Going upwards we shift the number to the left by 1 decimal place (divide by 10), and going downwards we shift it right by one digit (multiple by 10).

Now, we place an irrational in the center, lets say Pi, and we do the generation:

...

... 141.4159... 41.4159... 31.4159... 32.4159... 32.5270... ...

... 14.1415... 4.1415... 3.1415... 3.2415... 3.2527... ...

... 1.4141... 0.4141... 0.3141... 0.3241... 0.3252... ...

...

What exactly are we are looking at?

This is a spreadsheet, that when stretched out to infinity in the left, right, up and down directions will generate irrationals. Lots of them. Possibly all of them.

Since we can circle around the spreadsheet starting it the middle and making gradually larger and larger loops, we can map each and every cell onto N. This means that our spreadsheet is countable, and thus if this is all of the irrationals, then the irrationals themselves are countable.This would also mean that R itself, which is made up of the countable rationals and the now countable irrationals is countable.

How do we know this works. One interesting way is to start at Pi and generate a table large enough to get to sqrt(2). If this works, the sqrt(2) will be in the table, and we can navigate to it.

In fact, all other know irrationals are also in the table.

Why?

There are lots of reasons, which I have been circling around for a couple of weeks now. Lets start with some basic premises:

- All irrationals are a non-repeating path that is infinitely long.

- If you add a repeating (or finite) path to a non-repeating one, the result is a non-repeating one.

- If there is a path between any two irrationals, and it is countable, then the path is finite.

- Irrationals in base 10 are just a representation of the underlying number.

- the +1 or +.1 pattern permutes the number, but not the distance between the permutation, so we need to distance the changes from each other, thus +11, +.111..., +101, +.10101... etc.

- the infinitely long permutations permute the tail.

- we have an infinitely long number, so we need infinitely long permutations (this was the most painful part to figure out).

- the shift operators, left and right accomplish the goal of moving the permutations into infinity.

- the table, itself generates all of the permutations and combinations, stretching out to infinity in four different directions.

We can start with any irrational, and get to all of the others.

But there is more:

The bi-infinite spreadsheet (because normally a spreadsheet only goes to infinity in two directions, rows and columns), is actually a binary infinite node tree, as I discussed in an earlier post.

Iterating the tree or the spreadsheet produces a magic sequence, as my other post also showed.

We can see that the spreadsheet is trivially countable by just starting in the center and circling around,

Still, the fact that it is shows that Cantor's diagonalization doesn't apply to the enumeration precisely because there are two different sets of infinities make the result a magic sequence.

And just a little bit more:

In one of my comments in Good Math, Bad Math I talked about how I saw infinities. Pretty much, that perspective and what I wrote this morning in my own comments on the Magic Sequence post, confirm that my understanding is pretty solid.

So another set of results, that is related:

All binary trees are countable, whether or not they contain irrationals.

A "possible" result is that there is no such this as uncounability. Everything in a formal system is countable.

There are at least two categories of infinity: calculus infinities and set theory ones. They are different, but relatable.

There are two different types of calculus infinities, the type that is "going" to infinity, and the type that is already "there". The first type is "dynamic", while the second type is "static".

As set theory infinities, this is static:

3.1415...

And this is dynamic:

{ 1, 2, ... }

Infinities are pretty much interchangeable, but they do come in axises. Thus we can have 1D, 2D and 3D infinities, and probably more beyond that.

Axis are static or dynamic.

A bi-infinite spreadsheet by itself is 2D, filled with irrationals is 3D. Cantor's diagonal proof only works on 1D or 2D. sequence are 1D or 2D, magic sequences are 3D.

It may be true that two dynamic axises don't map to a static/dynamic pair, I'm not sure, I think about it.

You can map from a dynamic infinity to a static one, as I showed in my comment about counting binary trees, but you have to be careful.

To go from a dynamic to static, the concept of actually going "through" infinity with a path, and onto another infinity is crucial. This whole piece is crucial to make set theory consistent, or the fact that "uncountable binary trees" exist causes it be be ill-defined.

We can frame any set theory infinity, either static or dynamic, in terms of calculus infinities. We can step through calculus infinities, one step at a time, to view the behavior.

The set:

{ 3, 3.1, 3.14, 3.1415, ..., Pi }

In calculus infinities, this starts as a dynamic infinity, that "teleports" itself to the static infinity containing Pi at the end. In set theory infinities, this just is.

Another ambiguity that should be noted:

{ 3, 3.1, 3.14, 3.1415, ..., }

This "syntax" should only be used to specify all of the number in the set "approaching Pi", but not Pi itself. Doing this resolves another problem.

I think that's all of it for now.

A special thanks for all of the people who patiently put up with my questions and provided answers over the last two weeks, without your help I could never have debugged the program.