I'm going to be publishing my proof soon, but I've solved the conjecture about 78 years ago. Ive been sitting on the solution for a long while. Since the introduction of LLM's ive taken the opportunity to test the consistency of my answer.
Following the solution, ive found a couple interesting patterns in primes reciently as well. It seems to have a connection to the reinmann hypothesis. Namely the 1/2 part.

The density of primes is contained in the first half of all numbers. For m, m composite, m1 =2n. The primes that constitute 2n are all from 0 to n and none from n+1 to 2n. Meaning the relative density of prime numbers has a sort of upper bound where primes are sparse at higher than n.
Take p,q consecutive primes and m distance between them. Then m is a cyclical function, this relates fairly directly to the zetta function by the fact that the complex function is a description of rotation, and all solutions are within 0 and 1 where the real part is 1/2.
It means primes are predictable and there is a cyclical function in linear algebra that predicts primes.
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I was told on here that the infinite sum of 1/2^n = 1, and not just gets close but actually equals 1.
I can't help but notice that we are giving infinity a definite beginning point at 1/2 and a definite end point at 1. What could n possible equal to get to this point?
If this last point really is a solution to the equation, then wouldn't it have to be 1/infinity, or in other words, the "infinityith" point? If so, how can it be said that the natural numbers can numerate all points of a set of size alephnull?
]]>Ann:
111001...
000111...
101100...
110011...
000110...
010110...
transforms diagonal D1 to alternating '01' sequence T1,
010101...
which can't appear in any list per the cda.
Bill:
000110...
010110...
110001...
000111...
101100...
110011...
transforms diagonal D2 to alternating '10' sequence T2,
101010...
which can't appear in any list per the cda.
If D1 and D2 appear in any list, they must be members of the complete list.
T1=D2.
I.e. a missing sequence is only relative to an individual list.
]]>Each pixel contains 256 * 256 mosaic information, resulting in an image resolution of 2^{16} (65 536).
]]>or part of precalculus?
]]>But if we put them onetoone in a specific way, such as the naturals to the naturals from I, we see that the naturals of I get used up leaving 0 and the negative integers.
This seems to show that a correspondence from N to I can also not be onetoone.
The curiosity I get from this is just too much. It almost seems like this is an example of something that can be proved to be true and can be proved to be false.
I would have to think that my problem is that I am not allowed to correspond the naturals to only the naturals of the integers, but why not?
]]>
Here is the basic definition: If the largest prime factor of a number is less than or equal to the nth root of that number, then the number is nth root smooth.
I am curious about pairs of consecutive numbers that are both nth root smooth for large values of n. Let’s call them twin nth root smooth numbers. For example, 2400 and 2401 are consecutive numbers. The largest prime factor of 2400 is 5, and 5 is less than the 4^{th} root of 2400, so 2400 is 4^{th} root smooth. The largest prime factor of 2401 is 7, and 7 is equal to the 4^{th} root of 2401, so 2401 is 4^{th} root smooth. Thus, 2400 and 2401 are twin 4^{th} root smooth numbers.
I have found an example of twin 10^{th} root smooth numbers. Let A=2^2101 and B=2^210. A is equal to the following 64digit number:
1645 504557 321206 042154 969182 557350 504982 735865 633579 863348 609023
The largest prime factor of A is 1,564,921. I found this by using the calculator at the following website:
https://www.alpertron.com.ar/ECM.HTM
Since the largest prime factor of A is smaller than the 10^{th} root of A, A is 10^{th} root smooth.
The largest prime factor of B is 2, and since 2 is smaller than the 10^{th} root of B, B is 10^{th} root smooth.
So A and B are twin 10^{th} root smooth numbers.
My question is this: Is there a largest value of n for which twin nth root smooth numbers exist? I haven’t been able to find any examples twin nth root smooth numbers for values of n larger than 10, but I don’t suppose that means a lot.
]]>I'm really struggling with understanding how exponents make sense. According to a basic google definition:
"An exponent refers to the number of times a number is multiplied by itself."
This would indicate to me that 2 to the power of one would mean that 2 needs to be multiplied by itself... once. So 2 x 2 is 4. But I know that 2 to the power of 1 is actually one.
Why is 2 squared not 2x2x2? I can only see two "x"s in there...
Thanks for your advice.
]]>Which law says a point has absolutely no space and a minimal length? How is it reasoned that a point of space has no physical extent?
When you're talking about volumes and hypervolumes, your set needs to be equipped with a measure function. The technicalities are too much to go into in a single post, but see the wiki link if you're interested.
The standard measure used on the Reals is the Lebesgue measure where the length (in one dimension) is just the difference of the endpoints.
So, if we're looking at the length of a point p, we need to just take pp, which is, of course, 0.
]]>]]>
Why 2 .(3) = 6
Why is it positive?
Why we accept that (minus sign) times (minus sign) is positive ? What is its origin?
Thanks in advance
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We apply the inductive method to prove that this conjecture is correct with all even numbers 2n with n ∈ IN. The process includes two steps
1. Prove that the conjecture is correct with n=2.
2. Assume that the conjecture is correct with a certain value n, prove that the conjecture is also correct with (n+1).
We proceed with our process in sequence as below
1. With n = 2 => 2n= 4 = 2+2.
The conjecture has been ascertained as correct in this case.
2. Now we assume that the conjecture is correct with n, the next step is we have to prove that it is also correct with (n+1).
Here is our solution
Let’s define A as the set of the odd and prime numbers x which qualify the expression x < 2n.
B is also the set of the odd and prime numbers y which qualify the expression y < 2n
S(x,y) is the statement to express the equation (x+y) = 2n.
By the axiom schema of Replacement, there is the existence of the set B to ensure that S(x,y) is correct with at least one element x ∈ A.
Since the conjecture is correct with n, we have B ≠ Ø.
Let’s define A+ as the set of the odd and prime numbers x which qualify the expression x < 2(n+1).
B+ is the set of the odd and prime numbers y with y < 2(n+1) and S(x,y) is correct with at least one element x ∈ A+.
By the axiom of Infinity, A+ and B+ are the next items of A and B relevantly in the infinite sets.
Since both A and B are ≠ Ø, then we can utilize the axiom of infinity to declare the existing of A+ and B+. That means A+ ≠ Ø and B+ ≠ Ø.
The conjecture has been finally proven as correct with (n+1). By the inductive method, it is determined as correct with every n ∈ IN.
Thomas Nghiem (Ontario – Canada)
Reference:
https://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Lian.pdf
]]>When you have square root of (2)^10 which is the result?
Because if I use the fraction notation results (2)^(10/2) = (2)^5 .... a negative number.
On the other hand, if the power is calculated first the result is positive.
Thanks in advance
]]>
Anyways, the engine showed there exists no soln . for this.
I tried this on few others and still got the same. I can't understand this.
1 and 1 both give 1 when multiplied by itself. No square of a number can be in negative form.
So sqrt(x)=1 should have the solution as 1.
Can anyone explain this.
Also, I searched this on certain sites and they explained with graphs of complex numbers containing parabola , hyperbola etc. I have not learnt calculus yet.
If anyone can explain this without calculus elaborately , I will be grateful
Best wishes
]]>
Both alpha_i and y_i belong to W. Therefore their sum ‘e’ should belong to W . This contradicts our postulate that e belongs to VW. Therefore each alpha_i belongs to VW.
Next we consider the equation
https://drive.google.com/file/d/10z63Xidgs3m8p04_C6ZiGh8Q6KTwPsh/view?usp=sharing
Incidentally I tried the Latex with the code button.But I am not getting the correct preview.
Example
\begin{equation}\bar{A}^{\mu\nu}=\frac {\partial \bar{x}^{\mu}}{\partial x^{\alpha}}\frac {\partial \bar{x}^{\nu}}{\partial x^{\beta}}A^{\alpha \beta}\end {equation}
\[{\ bar{A}}^{\mu\nu}=\frac {\partial {\bar{x}}^{\mu}}{\partial x^{\alpha}}\ frac{\partial {\bar{x}}^{\nu}}{\partial x^{\beta}}\]
Currently for a long time I am not a frequent user of Latex thanks to the equation bar of MS Word. But I do appreciate,like many others, the application of Latex in various forums.Help is being requested from the forum regarding Latex.Thanking in advance for help provided..
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Does that mean we can assume $\exists x \in \mathbb{R}(x \neq x)$?
If so, would this provide us with the basis for a field with one element?
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