I posted and emailed where I was stuck 3 years ago…
They’re submitting it as a proof…
I don’t actually think they’re smart enough to catch up… I was just waiting for a reply, and then I’d dump the actual proof 
I planned this ahead… Stare at the numbers shitheads… And think you know what I know !
This is true for electrons do not rotate around a nucleus in an elliptical or circular orbit where their position and speed can
always be monitored. They are sub atomic and only partial information can be determined about them at any specific time
I learned that ‘electrons orbiting like planets’ was an abstraction for an energy field we don’t fully understand back in high school. It was the 90’s.
The Quantum Magi have been promoting that for decades. Quantum Mechanics is merely a fancy name for the statistics of mechanical motion. The term “quantum flux” merely means “we can’t figure this one out. It won’t give us a consistent number = ‘flux’.” Quantum Mechanics is not about physical entities, but rather their mathematical representations. When they can’t mathematically represent something as a fixed quantity, they merely declare that it doesn’t exist.
Electrons DO orbit nuclei. Quantum pseudo-physicists just can’t figure out how to mathematically track and predict their exact location (because they refuse to see reality). So they say, there is no orbit. It is magic." They seriously love magic and mystery. Quantum physics has nothing to do with actual science. They are merely mysterians just riding contemporary horse into fame and darkness.
An electron is a tiny congestion of ultra minuscule electromagnetic noise (I call “affectance”). It is only a “particle” because the congestion persists. And it orbits a nucleus because it cannot fall into the larger congestion (aka “mass”) of the opposite charge (“anti-charge”). Why it can’t is a complicated story that I explained some time ago (to Eugene, I think). It can’t merely sit still, it can’t move forward into the large anti-charged mass, and it can’t move away from the anti-charged mass, so it orbits like a fly around a cow’s ass.
Where is 1/3 = .333… on your list? Where is pi? Where is sqrt(2)?
The problem with mirroring across the decimal point is that a natural number only has finitely many digits. So mirroring leaves out all the infinite decimals.
For the binary tree expansion, I have no clue where pi abs sqr 2 are… I proved they are on the list though.
.3… Happens when the number 30 is listed …
30, 0.3, 0.(3 repeating)
My list for the rationals is absolutely complete .
Yes, and where is .333…?
Where is pi on your list?
This reply actually made me sigh…
I meant .333…
When I said .3…
Pi isn’t a rational, but it’s on the list…
I don’t see why it should. I reread your initial post. As best I can understand your description, the only numbers on your list have terminating decimal expansions. I don’t believe that you have any numbers on your list, rational or irrational, that have infinite expressions like 1/3 = .3333… and pi = 3.14159…
If I am misunderstanding your explanation, then I’m open to understanding your idea better. Perhaps you can explain it to me slowly and clearly.
Can you convince me, through a clear explanation of your idea, that you have .333… and pi on your list?
I do of course agree that you have each of .3, .33, .333, .3333, .33333, etc. You have every one of those. You have a million 3’s, a trillion 3’s. What you do NOT have, as far as I understand, is the decimal with infinitely many 3’s. Because if you mirror .333… across the decimal point you will get an integer with infinitely many 3’s. But all integers must have only finitely many digits. That’s where this idea breaks down.
One of my sequences is to use over line notation to designate infinite repetition , since I don’t know how to do that on a message board I simply use
0.(3 repeating) scroll up to see it.
The binary tree counts every possible infinite decimal expansion by slowly moving out and down forever.
I’m asking you to explain your construction in detail, since I don’t agree with what you said at all.
If you are doing this as a binary tree, then every node on the tree corresponds to a terminating decimal.
You can see this if you do the whole experiment in binary, so that each node of a binary tree represents a terminating sequence from the beginning to that point. So you would have say 0, and 01, and 010, and 0101, but you will never have an infinite-length bitstring ending in a node on the tree.
Now this is a very interesting thing about the infinite binary tree. The number of nodes is countable; but the number of paths through the tree is uncountable. So you can in fact represent all real numbers with a binary tree; but a real number is an infinite-length path through the tree. Each particular node, however, only represents a terminating decimal (or terminating bitstring, same thing).
Wtf, you can’t even understand the basics of what is on the page, but I’ll give you a piece of advice about math and life in general… Without the right sequence, nobody will believe you, everything is about sequence.
Now the issue with the binary tree is that there are only two symbols a number can start with
0 and 1
So to start you need:
00 and 01
10 and 11
Now, for done reason, your brain thinks these are only finite strings, and to this regard it is malfunctioning completely…
If you have an algorithm from the binary tree that runs all the combinations, every infinite sequence is there and no finite sequences are there. Interpolating all the finite sequences is actually it’s own step!
Yes, I already acknowledged that. I don’t understand your argument enough to believe it. That’s why I’ve invited you to clearly explain it to me.
You’ve stated that you’ve sent your idea to many eminent mathematicians and physicists. I assume you haven’t gotten any replies.
But I’m replying to you. I happen to be knowledgeable about infinite sets, the real numbers, infinite decimal expressions, the infinite binary tree and related topics. I like to help people understand Cantor’s beautiful results. In short, with all of the letters and emails you sent out, I am the person you were looking for. I’m the one replying. So why not see if you can convince me with a clear argument.
The attempts to refute Cantor by mirroring across the decimal point or by constructing an infinite tree are common. You haven’t got anything new. You have muddy rehashings of known ideas that don’t work. The simple answer is that you have succeeded in enumerating the countable set of terminating decimal expressions. This is a known result. And you can’t possibly find any way of enumerating the reals. That’s also a known result. It’s 140 years old. If it was wrong, someone would have noticed by now.
I am offering to help you to articulate an argument clear enough so that one of two things will happen. Either
(a) You will understand why you are mistaken; or
(b) I will understand that you have shown an inconsistency in math, by demonstrating an enumeration of the real numbers.
I regard (b) as unlikely, but I have an open mind. That’s why I’ve asked you to explicate a clear argument. If you’ve refuted an established result, I’ll be the first one to sing your praises to the heavens.
Yes, agreed. I’m puzzled though. I’m not the one claiming I have sequenced the reals. You are. So I will simply quote you. “Without the right sequence, nobody will believe you.”
Amen.
Well you’ve got me there. Can you explain what you mean? I agree, I would definitely stake my mathematical reputation, such as it is, on the fact that the binary sequence 10 is finite. It has length 2, and 2 is a finite number.
You claim otherwise? How so?
I’d be the first to admit that my brain often malfunctions. Sometimes I can’t find my keys. Other times I go to the store and forget what I came for. That’s why I make lists. However, you can’t make a list of all the real numbers!! That’s the point at issue. The condition of my brain is fortunately not at issue, else I’d be in trouble. After all, here I am attempting to take you seriously, show you respect, and talk math. Please tell me this isn’t another one of my brain malfunctions!
I agree. If you had an algorithm to generate the reals, you’d generate all the reals. However, there is no such algorithm. Perhaps by “running all the combinations” you mean taking the limit of the paths. But then there are uncountably many such limits of paths. You are counting the nodes, which are indeed countable; but imagining the paths, which are uncountable. That’s your core error.
Well that doesn’t actually make any sense. If we do the decimal tree and assume an implied leading decimal point, then at level 1 we have .3; and at level 2 we have .33; and at level n we have .333…333 (n 3’s).
But there is no node in the tree that contains 1/3 = 333…
That’s the problem with the tree idea. There is no node on the tree corresponding to any infinite decimal expression. Rather, it’s the paths through the tree that can be identified with the real numbers. To get to the reals from the tree, you have to take the limit of every path. And although there are only countably many nodes in the tree, there are uncountably many paths.
I do not see how you have 1/3 on your list. Let alone any irrationals.
By the way if I take a while to reply to you, it’s because my handle is new and the mods age my posts. Perhaps in the hope that like wine, they’ll improve.
It is frightening to me how little you see…
Being the respondent to all of this…
You cannot start the binary tree from zero or one because there are an infinite amount of numbers that start from both.
Step 1:
0
1
Get it!! ???
Step 2:
00
10
01
11
Now you’ll notice just in the first two steps that it is expanding out and down… This is an algorithmic sequence that expands out and down FOREVER!!
In step 1 and step 2, every possible combination is there. As is true for step 3 and step 4 ad infinititum!!
You’re obsessed with the steps being finite and thinking it is only a finite sequence without realizing that it’s an algorithm that goes forever and never misses a single combination as it expands out and down FOREVER!!!
As for my counting of the rationals, I don’t use fractions, I only use decimals … One of the combinations is to run finite and repeating lines by defining them as such in extremely simple combinatorics… Go to my first post and read it again.
A wise man said …
Words to live by.
Thank you for responding. I found much to agree with in what you wrote. I do believe you are making the particular error that I outlined earlier. I hope you’ll have an open mind as I state my case.
Agreed. I always like to think of the tree as starting with a decimal point, since we have to have a single node as the root of the tree. Then to the left there’s a node representing 0 and to the right a node representing 1. A pair of finite bitstrings.
Yes exactly. Four finite bitstrings.
Yes I agree. The algorithm goes on forever. There are infinitely many levels and altogether infinitely many nodes. However, each node only represents a finite bitstring, as you just showed.
Here’s an example. Consider the counting numbers 1, 2, 3, 4, 5, 6, …
They go on forever too. But any particular number is finite! There are no “infinite numbers” in the list, ever. Even though the list of numbers goes on forever, each element of the list is a finite number. The list is infinite, but no particular element in it is.
It’s exactly the same with the binary tree. The tree has infinitely many nodes. But each individual node is a finite bitstring.
The next level in our tree is
000
001
010
011
100
101
110
111
There are 8 finite bitstrings.
In general, there will be 2^n strings of length n.
No matter how many levels out you go, you’ll always find that a node is a finite bitstring. Each node will eventually have a lot of bits, but still a finite number of bits.
So now if I ask you where is the infinite string 11111111…in the tree … you see that it is NOT on the list.
What IS on the list is each of 1, and 11, and 111, and 1111, and 11111, and 111111, and so forth. There’s a node corresponding to a million 1’s, a billion 1’s. For every natural number n there is a node on the tree corresponding to n 1’s.
But no node of the tree contains an infinite string. Even though there are infinitely many nodes and levels, every individual node contains a finite bitstring.
In order to represent the infinite bitstring 1111… we have to consider the entire path through the tree 1, 11, 111, 1111, …
But you aren’t enumerating those paths. You can’t, because they are not enumerable. That is the heart of the matter. You can’t enumerate the set of paths through the tree. You can enumerate the nodes, but not the paths.
A lot of people find this counterintuitive. I’ve even heard it referred to as a paradox, the fact that the infinite binary tree has countably many nodes but uncountably many paths.
Yes, every finite combination is there. And at step a zillion, every finite combination is there. There is NEVER an infinite combination in the tree. That’s what the algorithm does. It generates every FINITE bitstring.
Goes without saying.
As I explained, I’m quite clear about the fact that the number of levels in the tree is infinite. But the bitstring at each node is finite. There are no infinite bitstrings in the stree, by the very nature of the algorithm. The algorithm is not capable of producing an infinite bitstring.
Think of it this way. Suppose I have the infinite sequence 1, 11, 111, 1111, 11111, …
In fact I have an algorithm that keeps cranking out these strings of 1’s, one after another.
Even though there are infinitely many items cranked out; each item itself is a finite string of 1’s. There’s one 1, and two 1’s, and three 1’s, and a zillionty-zillion 1’s. But each item in the list consists of a finite number of 1’s.
In order to get an infinite string of ones, 111…, you need to take the limit of that sequence, if you think of it that way. And that’s what real numbers are. Real numbers are limits of infinite sequences of finite strings. The number pi is the limit of the sequence 3, 3.1, 3.14, 3.141, 3.1415, … You can put all the finite sequences in a tree; but the limits are represented by the paths. That’s what the real numbers are. They’re the paths through the infinite binary tree.
I read your first post a couple of times and can’t understand it. Without a more clear exposition I can’t comment. You did mention mirroring and I know mirroring doesn’t work, since a real number has infinitely many digits to the right of the decimal point, but only finitely many to the left. That’s why mirroring doesn’t work. You can have .1 and .11 and .111 and .1111 but you can’t have .111…
This is the issue I have with where you are stuck in your own contradiction.
You brought up 1/3 for example…
Now 0.333… Is also a “bit string” using an algorithm for division which we use an inferential proof to determine that it is an “infinite number” as you said.
The same thing is happening with thus binary tree sequence… Every combination is being enumerated …
Now we can split hairs here and say that 1,2,3,4,5,6,7,8,9… Isn’t actually enumerated because there are numbers there we will never count, BUT!! We know that sequence is the complete set of counting numbers using the +1 algorithm through inferential proof, inferential proof being every bit as powerful as an explicit proof.
The issue with the binary tree sequence is that the numbers at the beginning of the expansion always stay the same, so… As you build the tree, there will in fact be a 3.14 that is the real enumerated one that really corresponds to pi. All these numbers are eventually enumerated (again, not splitting hairs here). If we can get past this part, I’ll explain my mirroring technique more.
Let’s consider the binary tree. You say the infinite bitstring 111… is in it somewhere. I’d just like to know where.
Here’s level 1, consisting of all the length-1 bitstrings. 0, 1
And level 2, all the length-2 bitstrings: 00, 01, 10, 11
Level 3: 000, 001, 010, 011, 100, 101, 110, 111
In general, level n contains the 2^n bitstrings of length n.
Just tell me, where is 111…, the infinite bitstring? It’s not in level 1, it’s not in level 2, it’s not in level 3, etc. So where is it?
That’s the only number that doesn’t get enumerated in the binary tree sequence…
Since it is rational, it gets added with those.
I actually explained this in the emails I sent out, that it’s only an illusion that the bottom number gets enumerated, but since it is a rational, it still gets counted.
Ok, then you agree that 111… was not on the list after all.
Ok, so you build your tree, which you now agree only contains all the finite-length strings.
Now you add in all the infinite-length rationals that you forgot. That’s ok actually, because there are only countably many rationals, and adding a countable set (of numbers in the tree) to all the rationals you missed, still leaves you with a countable set.
But notice that you didn’t just leave off 111… You also left off the binary representation of 1/3 = .01010101010101… Where was that on the tree? It’s not. You had to add that in too. You left off ALL the infinite-length rationals, not just one.
But now where are all the infinite-length irrationals like the binary representation of pi, sqrt(2), etc.? You left those off too.
To put this more concretely, let’s think about the decimal tree. You have 3.1, 3.14, 3.141, 3.1415, … in the tree. But where is pi? Did you need to add that at the end too?
So now you have two lists. You have the list of numbers that are already on the tree – those are the finite-length strings. And now you have another list of numbers that must be added in at the end. But how many of those are there? Well, there are uncountably many. That’s the flaw in your argument. You didn’t just leave off the infinite-length rationals. You also left off all the irrationals.
I did not received one of those emails. You left me off your list. Just as you left all the irrationals off your list. Thus proving that I’m irrational
How can it be an illusion? You just agreed that 111… is not on the list. And I just pointed out that you left ALL the irrationals off your list. That’s not an illusion. That is the central fact of the infinite binary tree. It has only countably many nodes. It leaves off all the infinite-length rationals; and it leaves off all of the irrationals.
And what of the binary representation of all the irrationals? They all have infinite-length strings, so NONE of them are on your list. You have uncountably many irrationals left over that aren’t on the tree.