wtf

On 9/6/2018 at 8:08 PM, Achilles said:

Anything that requires user input. Is transistors at the heart of it, acting as on and off switches which equate to values (most commonly 1's and 0's)?

https://www.youtube.com/watch?v=Xpk67YzOn5w

Didn't read any of the other posts so apologies if someone else mentioned this. You can build logic gates out of dominoes. You can build logic gates out of anything. Transistors are convenient because they are very fast. But that's only an implementation detail. Nothing in the definition of computing requires any particular implementation; and in fact it's a basic principle of computing that the particular implementation is irrelevant.

https://en.wikipedia.org/wiki/Domino_computer

On 9/7/2018 at 4:19 AM, Olfaction said:

The use of Cantor's diagonal slash applied to the output of TMH is then showed to lead to paradoxes ...

 

 

I haven't the bandwidth today to respond in detail. I noted this particular phrase, which is the point at which you stopped working hard to understand the proof and instead substituted handwaving. First, the proof is a diagonalization, but it's not the Cantor diagonal argument. Secondly. "lead to paradoxes," no. It leads to the conclusion that TMH can not in fact exist.

My suggested for you would be to go back to the proof of the unsolvability of the Halting problem; and no matter how long it takes you, go over ever single itty-bitty detail until you clearly understand the proof. Only then go forward with your new example, which frankly doesn't hold water at all. In short you are stopping well short of understanding right at the point where the work gets hard. You need to push through that and understand the proof.

You're basically reinventing linear algebra. You can form a matrix of the coefficients; and if the matrix is invertible, or equivalently has nonzero determinant, you can multiply right side of the equation (the constants) by the inverse of the coefficient matrix to get the answer. The benefit of this method is that it's easily programmed into a computer and requires no insight or cleverness as to what to do first. It's automatic.

The answer to your question about the algebraic method is that you can generalize this by talking about "elementary row operations," and algorithms by which you apply them to reduce your system of equations so that it yields the answer. 

I Googled around and this looks like a pretty good overview of what you're asking about.

https://en.wikipedia.org/wiki/System_of_linear_equations

On 8/27/2018 at 9:06 AM, Bean_Spiller said:

 

I hope you don't mind a couple of very general remarks, since I'm profoundly ill-equipped to offer substantive comment on matters of physics.

* Your title claims a disproof of something widely believed to be true. https://en.wikipedia.org/wiki/Bell's_theorem So the burden is on you to be clear in your exposition. Along these lines, it's not required but readers like myself would find it very helpful to have some context. Can you give us a little overview of the subject and perhaps an outline of why you think you have a disproof of something that's called, in the physics business, a theorem. A roadmap or high-level overview of why you think you know something that everyone else doesn't. 

* Nice pictures. As a math person I certainly appreciate the basic fractals like the Koch snowflake. I wonder if you can mention briefly how the study of mathematical fractals (which, after all, exist in continuous mathematical space) relate to physics, which, as I hope you know, resides in a space that's not known for sure to be continuous or discrete.

* As a stylistic note, it's not necessary to quote your entire lengthy post every time you reply to yourself. All that does is cause people to have to scroll endlessly to read through the thread.

Well anyway like I say I really can't help with the physics. But I have a lot of experience with people who claim to have disproved some famous theorem! The burden is on you to motivate readers to take you seriously. You can go a long way with context and clarity.

ps -- I skimmed a little, found this:

> You'll find that doing this will produce inside out spheres that are smaller than a Planck length:

Ok first, what is an inside-out sphere? Earlier you said photons are spheres. Do you mean mathematical spheres? Is that true? I don't believe physics says that, but I'm no expert on photons. But i know spheres. What is an inside-out sphere? A sphere is just a set of points in three-space equidistant from the center. If you turned it inside out that would be meaningless, it would still be the same set of points. The sphere itself is a two-dimensional manifold embedded in 3-space.

By the way you can reflect all of 3-space in the unit sphere, is that what you mean? Everything inside goes outside and vice versa, and the origin goes to infinity perhaps. Haven't thought about it. Another thing you can do is evert a sphere, turing it inside out by passing it through itself but (this is the clever bit) without leaving a crease. Surprising counterexample, nobody thought it was possible. https://en.wikipedia.org/wiki/Sphere_eversion

So what exactly do you mean by turning a sphere inside out?

Also, nobody can say ANYTHING sensible about anything smaller than a Planck length, because the Planck scale is the point at which physics breaks down. That doesn't mean there either is or isn't something interesting "down there," it just means that our present physics simply doesn't apply. We can't use to reason.

So I found this one sentence to be troubling. 

32 minutes ago, Zolar V said:

f(x) gives no useful information
G    gives useful information.

If I failed to see the distinction, what simple phrase or couple of sentences could you say to me that would give me a glimmer of an idea as to why you see a difference between G and f? G isn't related to Collatz, it's just some function you made up. That's how it seems to me. Can you articulate why you think these are different?

12 minutes ago, Zolar V said:

Taeto's and your point here are both irrelevant.  Of course there exists some function f(x) that satisfies the conditions on G, stating such a fact is mostly irrelevant to the topic at hand. Since we assume the function already exists,  our purpose here is to elucidate that function and show how the Collatz function is a natural byproduct of it. 

Why isn't Collatz a "natural byproduct" of f(x) = 2? Can you explain the difference in a few words? Why is G related to Collatz and f isn't?

1 hour ago, Zolar V said:

Lemmas ...
 

@Zolar, Did you understand @taeo's excellent point?

Let f(n) = 2 for all natural numbers 2. Then f satisfies your condition on G. It always outputs a power of 2, namely 2^1.

Now given some number n, let's apply Collatz. If it's even, then eventually it either hits ("converges to" in your terminology") 1; or else it hits an odd number. If it's odd, then f(n) = 2. 

So f is "inside the Collatz function" in your concept. Yet I hope you can see that f is just some arbitrary function that has nothing to do with Collatz. It doesn't prove anything. 

Do you understand this example?

4 minutes ago, Zolar V said:

I don't see your perspective.  

I simply can't help anymore. If I ever did.

> For any natural number N, there exists a function G such that any N converges to a power of 2 over M iterations.

So now there's a DIFFERENT G for each n? 

11 minutes ago, Zolar V said:

I know I dont,  i was outlining what I am going to show.  I suppose I posted it to see if it made sense

 

It makes no sense at all to me. You've reverted to the "spinning out of control" narrative that I thought I talked you down from the past couple of days. I was mistaken. I did no good at all. I can't do anything else here. 

18 minutes ago, Zolar V said:

2^4  sorry,  i was up for about 25hrs or by that point

 

I don't think I've served you by encouraging you in these last few pages. You haven't got a proof and your exposition is incoherent. I'm sorry.

11 hours ago, Zolar V said:

potentially 21 may be expressed as   2^3 + 5

Jeez man.

1 hour ago, Zolar V said:

I understand you're frustration. It is my hope to show you I am right, though it may take some time in making it a proof given your commentary here.

So lets get some terminology right:

Convergence:  I want to say a function converges to 2, if for any input the output is 2 after some amount of iterations.
Definition: " property (exhibited by certain infinite series and functions) of approaching a limit more and more closely as an argument (variable) of the function increases or decreases or as the number of terms of the series increases"

Would it be better to define convergence like this: For M iterations a function converges to 2 ?   
 

Theorem: A function that we are proving
Lemma: A supporting argument for the theorem
Corollary:  An unintended result
 

None of that is important. I understood what you meant by convergence very early on. You don't have to define what a theorem is. You have to figure out what you're trying to say about Collatz. 

1 hour ago, Zolar V said:

Suppose the prime reduction function is true
Take any odd natural number,   It is either prime or composite.

If it is prime, then adding 1 would result in all its primes being less than itself.  (we've already agreed on this)
if its a composite,  then adding 1 would still result in at least 1 of its primes being less than the primes in the composite.  (this needs to be shown.  This will be the result of step 2)

-----

if you split 3n+1 into 2n + (n+1),  then you have an odd integer being added to 1.  which fits our function.  the drill down part of our function is the same iteration in the collatz function.   Collatz has everything to do with the adding 1 mechanism.    I believe the (3) part in 3n+1 is irrelevant.  Since if this works it works for An+1 where A is any odd natural number.
 

Yes but the collapsing to 1 is completely different in the two cases. You're convincing me you haven't a proof. In any event this convo is now meaningless unless you supply a proof of your assertions. You can't keep tossing out vague plausibility arguments, especially ones I don't consider plausible. You claimed a proof. You don't actually have a proof. I'd say that's a problem for your claim.

I didn't think about your idea in detail, but if that's your research program, you should definitely get to it. Take your time, work your program, report back. 

> 5: Since our function converges to 2^n , so does the odd collatz function

Now that I understand the prime reduction function, I can't see at all how it relates to Collatz. It's a totally different procedure. You take a prime, add 1, drill each of its prime factors (to multiplicity) down to a power of 2. Well ok you have me believing that. 

If you have a reduction from 3n+1 to that, perhaps you should work on a very clear exposition of how this works. Because Collatz doesn't have anything to do with taking the prime factors of anything. I think there must be something wrong with your idea right here. 

1 hour ago, Zolar V said:

Hmm,  thats a tough question.  I want to say it isnt going to matter for the proof, but I think it will be injective,  i doubt it would be surjective.

"So, onward. How does this prove Collatz?"
- I believe our function is embedded in the odd function of the collatz conjecture.  That is to say we can manipulate 3x+1  to be  2x + x+1  where x is our p.  
  then 2x + (x+1)   and (x+1) is our function.  (yes i know,  its not defined here)
since G^n(p)= 2^n  then
  2x^n + 2^n    ->   2^n(x+1)  ->   2^n(G(x+1))  ->  2^n (2^m)

then apply the even collatz function n*m times. 
2^{n*m}  / 2^{n*m} = 1
 

So I know there is a lot missing/wrong.  But the underlaying mechanism to the collatz function is the prime reduction done by adding 1 to a prime. 

the odd collatz function will always converge at some point to a 2^n number,  and as soon as it does the even function will reduce it to 1

 

> I think it will be injective,  i doubt it would be surjective. 

Of course it's not surjective, its output is restricted to powers of 2.

> So I know there is a lot missing

Correct. Burden's on you. 

> the odd collatz function will always converge at some point to a 2^n number,

Of course. If you can prove that you become famous. 

8 minutes ago, Zolar V said:

all the terminal nodes are multiplied together to give a final result of 2^n and youre done

 

Ok. Is it important to you that the mapping from p to the final 2^n is injective? Meaning that two different p's give different n's? I don't think that's true but I haven't looked for a counterexample and won't bother if it's not important.

So, onward. How does this prove Collatz?

42 minutes ago, Zolar V said:

Possibly both operations, but the final result is going to be a 2^n number. (line 34 in my poorly written pdf)
   If all we need to do is multiply then that's all we need to do. Once we have that result,  n will equal the number of steps/iterations we needed to get to a 2^n number.
if we also divide by 2^n, then 2*n will equal the number of steps needed to get to 1.

 

Ok we can table that for later. So we build the tree, all the terminal nodes are powers of 2, and we do "something" to those powers of 2.

Now what. Please go slowly, one step at a time. Short and clear please.

11 hours ago, Zolar V said:

Edit:

I think it is sufficient. *  

Ok so what happens at the end? You fill out the tree and you have a bunch of terminal nodes that are powers of 2. Do you add them up? Multiply them? Does it matter? Once we nail this down we have a complete description of G and we can move on.

2 minutes ago, Zolar V said:

Oh thats really neat, i didnt know that.  I thought we only had expressions for special primes.

 

I edited it a little, give it another glance if you would. The whole subject of what functions are computable by algorithms and which aren't is very interesting. 

So please be clear. Is my tree concept sufficient for your needs? Because it seems damn simple to me, which means that I understand it. And I like ideas I understand better than ones I don't!

But if it's not sufficient for your purposes, do let me know. 

21 minutes ago, Zolar V said:

I added a little extra to my last post that i thought of. 

Lets accept and work with the tree method first.  If this gets nowhere then we can explore the adding idea.  Though even if the tree method does work, I think it's also worth exploring the adding idea.   

I'm not surprised,  being inductive would indicate you can systematically build primes.   I mean technically you can,  but this would be for all primes not special primes.

Here is an odd thought:
Suppose this idea was true. Some inverse equation/algorithm would build only primes.  Each equation we have now that creates some subset of primes would somehow be a part of or equivalent to that equation/algorithm

There IS a function that produces exactly the primes, in order. It's the function f that maps the natural number n to the n-th prime p_n.

This function exists. There's an algorithm for it:

For each natural number n:

    If n is prime, print it, otherwise don't.

It's slow but we don't care about that, all we care is that the function exists. It can be coded as an algorithm, that is, as a Turing machine. As such, it inputs a natural number and outputs the corresponding prime. Note that we'll need a subroutine that determines if a number is prime. For that we just use trial divisors. 

Also note that if we just want, say, f(45545434343), we just run it that far and print out only the value we want. 

There is no question that such a function exists, that it's computable, and that a brand new programming student would be able to write within a few weeks. 

The problem is, we don't have a nice simple closed-form expression for it. We have to brute-force it. So the question isn't the existence of such a function. It's that we don't have an expression for it.

10 minutes ago, Zolar V said:

Alright,  I can work with that.  I didn't know that is what you wanted.   You are exactly right then for step 3.  all you are doing is adding 1 to each prime, one prime at a time per iteration.
 

-very true that it does.  but can you think of any number that is even, prime, and greater than 2?
the largest prime in an even number must be half the size of the even number,
Example:  P is our first prime, 
if  P+1  = 2^r * k   
then  /[ \frac {P+1} {k} = 2^r /]

Now suppose there is a prime factor in k that is larger than P,  then if we were divide k we would not get a positive even number.   I believe we also know that r >= 0

The claim that every branch terminates in a power of 2 is obviously true but a formal proof eludes me. I haven't spent much time on it. Something about the induction bothers me but I can't think of how a counterexample would work. I've decided to put the issue aside for now.

So let's just say I believe every subtree terminates in a power of 2. But this is different than your "adding" idea. Is it sufficient to get your argument off the ground? Either you have to accept that my "tree" method serves your purposes; or else you have to explain to me your adding idea, which I confess makes my eyes glaze.

> I will renew my efforts to clearly define a set of functions

No, please, let's go back to the point where we were in agreement and work from there. I have a plan. We were in perfect agreement at a certain point then (I feel) you spun off in a million different directions. The expression that came to my mind was "thrashing," which is what a computer does when it's so busy managing virtual memory that it can't get any actual work done. It gets busier and busier but accomplishes nothing.

Please don't take that as pejorative or "condescending." God it's not the first time I've been called that online. It's just the way I write. Or as Jessica Rabbit said, "I'm not bad, it's just the way I'm drawn."

Believe it or not I'm trying to be helpful with those remarks. Because we're having TWO separate conversations. One is about the math. The other is about the exposition of the math. So I am not attacking your idea. I perfectly well stipulate that you might be Ramanujan and I'm damn well not Hardy. Maybe you're right that a smarter person than me could figure out what you're talking about.

But on exposition, I am simply playing the part of a typical, somewhat dim, reader of mathematics. I read a line, try to figure out it, write down some simple examples. So when you write some exposition and I say you're thrashing, I am simply telling you that you have lost your reader. You have to make it much simpler to get through to the likes of me. 

Now here is what I posted before.

Speaking of mathematical communication, we ALL have a lot to learn. I thought I was making a clear point, a meta-point, a big-picture point. But I forgot to make my point! So here it is. There isn't a convenient "function" for G. Just express it as an algorithm.

For that matter, this is exactly how Collatz works! I can't think of a convenient way to express Collatz as a function, even though there is a function in theory. But in practice it's far better to work with it as a procedure or recipe.

So if you would forget about calling G a function, and notating it as if it were a function, your exposition would improve dramatically and I wouldn't have to spend so much time yelling at you that G's not properly defined. I'm sure we're both tired of it by now.

G is not a function. It's a procedure. Think about it that way and talk about it that way. Else we're doomed.

Having said that, we are in agreement through step 3. You did at one point say that you "keep the power of 2 and only divide it out at the end." That is the first moment of murky divergence from our point of agreement. Not only that, in your recent posts you are clearly using the 2^n in some way along with the other primes. So this is important to you and I don't quite understand it. It seems to complicate things. 

Now ignoring the 2^n, whether you divide it out or include it in something later, we have the remaining primes. For example in the case of G(139), we have 139 + 1 = 140 = 2^2 * 5 * 7. Note that G(139) is our notation for "Input the prime 139 into the G recipe." It is not a function. But now that we've defined the notation, it's ok. As long as we only input primes!!

Now my idea, which is evidently not what you have in mind, is then to recurse through the procedure on each of 5 and 7. So 5 + 1 = 6 = 2 x 3 and 3 + 1 = 2^2. And 7 + 1 = 8 = 2^3. So we claim (although I have not yet formally proved) that each subtree terminates at a power of 2. But that is my idea. (*)

At that point, you tried to explain to me that you do something with five 7's, and that's where I lost the thread. And in an attempt to explain this to me, you're (FROM MY VIEWPOINT, this is not meant to be pejorative) spiraling out of control.

So if you see any hope of going back to our most recent point of perfect agreement, that might be helpful.

(*) I'm nervous about this. I can see it's a strong induction ... having proved it for all primes below a certain point, it's true for larger ones because p+1 has factors that are smaller than p. On the other hand, it's a funny induction. The induction step never depends on any of the values that have gone before. I am starting to wonder if it's true.

For example ... suppose we know that 3, 5, 7, 11 all go to 2^n. What if we then put in a p such that p+1 only has prime factors of 13 and above? There's no actual induction step, you have to look at every prime by hand. I am starting to wonder if there might in fact be a counterexample. Your argument depends on p_i < p so I believe this impacts your argument. 

6 minutes ago, Zolar V said:

Our function G(p) = p+1.
 

So by definition, G(103 + ... + 103) isn't defined because the argument to the function is not prime.

> p = 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103 

No, that number is not prime. 'p' has denoted a prime since the beginning. If that's no longer true, you sure didn't mention it. 

 I think instead of posting anymore, you should go back to the drawing board and try to work through your idea. Maybe someone else can offer suggestions. I'm out of ammo. 

2 minutes ago, Zolar V said:

"As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?"
Lets select 103.  then we have 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 copies of 103 added together.
G(103+103+103+103+103+103+103+103+.......+103) = 103 +103+103 ....   +    (103+1)

I can't make heads or tails. I can't be helpful anymore. I can see you're multiplying and adding a bunch of numbers, but there doesn't seem to be anything for me to work with. I need to let this go. On the left hand side you have G of a composite number. We're already in perfect agreement that this is not defined. You can input a prime to G and recurse down a bit till your exposition falls apart, but you can NOT input a composite number to G. I have to let this go, I have nowhere to go with this.

12 minutes ago, Zolar V said:

"No prob, take your time. I doubt anyone will beat you to Collatz and claim credit. You have a few more days at least, I'm sure."
- I would hope so, Ive been sitting on this for almost a year.

I am attempting to explain how I handled 5 and 7 in the computation of G(139).  We are following the paper,  line 40 to be exact. 

"What do you mean "extracting" 7 20 times? This really is very unmotivated, it's not possible for me to follow your logic. As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?"

extract, to take out one piece of a whole. Think about a number as an entire collection of collections of objects that are equivalent.   If we only consider the additive operation then the number of equivalent objects within a number is how many different numbers can add up to equal that number.   E.G. 7.   1+6 = 2+5 = 3+4 ... = 2+2+2+1 = ... 7
A whole is 7, a piece would be take out a 2 or a 4 or some other part of 7.  

In our case our object is 2^2*5*7 which means you can extract a 2,5, or 7 and there are  2*5*7 , 2^2 *7, or 2^2*5 copies respectively.
If we extract 7, then we are extracting all 2^2*5 copies of 7.  So 7 +7+7+7+7+7+7+7+7+7+7+7+7+7+....+7
And we iterate through that sum of primes.

"As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?"

- Currently we select a prime and reconfigure the product of primes into a sum of that prime.  Then iterating through that on each prime. 
The core concept of adding 1 to a prime and reducing the primes within it is the primary mechanism we want to chase.   On a very rudimentary level, the mathematics that I know, only allow you to do something like add to a particular value only if you can group it together.    Obviously you cannot just select random primes within a product to add 1, but you can do just that over a sum of primes.


 

> - I would hope so, Ive been sitting on this for almost a year.

Numerous scatalogical jokes omitted.

> extract, to take out one piece of a whole. Think about a number as an entire collection of collections of objects that are equivalent. 

I am certain I have no conceivable idea what you are talking about, even after imagining various set-theoretic constructions of various classes of numbers.

For the third or fourth time in a row you are pointedly ignoring my request for context. 

> Currently we select a prime and reconfigure the product of primes into a sum of that prime.  Then iterating through that on each prime. 

"reconfigure." What am I supposed to make of that? You still won't tell me how to iterate through 5 x 7.

I don't want to sound negative, but your last half dozen posts haven't made any sense and haven't addressed my concerns, nor to you seem to be making any progress in clarity. I am near the end of my ability to be helpful. I did some useful work in helping you to define G, but once you do the initial factorization of p+1, the rest of your idea is simply unexplained. I think we're at around three days with no progress at all. 

> The core concept of adding 1 to a prime and reducing the primes within it is the primary mechanism we want to chase.   On a very rudimentary level, the mathematics that I know, only allow you to do something like add to a particular value only if you can group it together.    Obviously you cannot just select random primes within a product to add 1, but you can do just that over a sum of primes.

Simply doesn't parse. Maybe I should stop posting. I have nothing to add.