The Proof of Collatz Conjecture - Explained

[Guest]

sure.. you said "No. If you can add 1 to it - means it was not infinite", but whatever.
I understand fully what you did, and apparently better than you.
Your whole process is based on formulas (like [tex]A_i-B_i=2^{\nu(B_i)}[/tex]) which only works for and from [tex]N_i=1[/tex], so I repeat, your conclusion that [tex]N_0[/tex] reaches 1 is only valid for numbers reaching 1. Not only that, but you misinterpreted the meaning of your limits which are not correct.
Now It become clear that you will still try to sell your proof in the next decade(s) whatever is said to you, so I understand that I am only wasting my time talking in an infinite void. Good luck

[Nobody Knows]

sure.. you said "No. If you can add 1 to it - means it was not infinite", but whatever.
I understand fully what you did, and apparently better than you.
Your whole process is based on formulas (like [tex]A_i-B_i=2^{\nu(B_i)}[/tex]) which only works for and from [tex]N_i=1[/tex], so I repeat, your conclusion that [tex]N_0[/tex] reaches 1 is only valid for numbers reaching 1. Not only that, but you misinterpreted the meaning of your limits which are not correct.
Now It become clear that you will still try to sell your proof in the next decade(s) whatever is said to you, so I understand that I am only wasting my time talking in an infinite void. Good luck

"Your whole process is based on formulas (like [tex]A_i-B_i=2^{\nu(B_i)}[/tex]) which only works for and from [tex]N_i=1[/tex]"
This is correct.

"your conclusion that [tex]N_0[/tex] reaches 1 is only valid for numbers reaching 1"
This is NOT my conclusion. My conclusion is that [tex]N_0[/tex] reaches 1 only when [tex]N_0[/tex] can be represented in the "special form" presented in my paper and in my video, then I have proven that by using my procedure every [tex]N_0[/tex] can be represented in such "special form" and therefore every initial [tex]N_0[/tex] reaches 1.

I didn't expect that everyone will understand it and I'm also not surprised that those who are not able to understand are/will be frustrated by this fact and can be aggressive. I gave you some responses that you have completely ignored so it means for me that your goal was not to understand but to reply with your "opinion". You can have your "own opinion" like everyone else can have. Good luck to you too.

[Nobody Knows]

I want to address here as some say "problem" of limits at infinity from my paper and video, based on the example given by "Guest" here on the forum and as tarmaljed1609 and g3ncollaz in the comments under the video on Youtube.

As it was presented here during the discussion and also within some other comments under the video. It is ok in mathematics to say

[tex]\lim_{x \to \infty}(x+1)=\lim_{x \to \infty}(x)+1[/tex] this is OK

but not ok to move one term to the other side to have

[tex]\lim_{x \to \infty}(x+1)-\lim_{x \to \infty}(x)=1[/tex] but this is NOT OK

And the reason is that [tex]\infty= \infty +1[/tex] is fine, but [tex]\infty - \infty =1[/tex] is not fine as an effect of transformation of our first expression into second one.

This is NOT correct !

We just need to understand what is going on here. Basically limit at infinity means what is going on with a term when we are increasing x to infinity. But, infinity is not a number it is a direction. So increasing x to infinity means that we are changing a value of x through all possible values in the direction to infinity. It is "continuous process", we can increase x to make it bigger and bigger forever. During this process values of certain term of x (in our case below [tex]f(x)[/tex] and [tex]g(x)[/tex] ) become closer and closer to an asymptote it can also happen that values of this term of x will be exactly on the asymptote. It is generally accepted that when this asymptote is horizontal we say limit exists, if asymptote is not horizontal - limit does not exist, but we still can and should understand what is going on, because we can get valuable information from this.

Lets say we have two functions from our example above.
[tex]f(x)=x[/tex] and [tex]g(x)=x+1[/tex]
We can present both on the graph like below.

2023-11-06_203117.png (151.38 KiB) Viewed 1793 times

As we can see when [tex]x[/tex] is increasing (black arrow) to infinity, values of [tex]f(x)[/tex] and [tex]g(x)[/tex] are moving along red and blue lines/arrows. Even though we can say that these values are growing to infinity, but we can also say that each arrow is pointing to something else. Therefore equation

[tex]\lim_{x \to \infty}(x+1)=\lim_{x \to \infty}(x)+1[/tex] is OK, because as we are increasing [tex]x \to \infty[/tex] on the LHS we are moving along the blue line while on the RHS we are moving along the red line, but every point is always shifted "plus 1 up" which will be exactly the same as moving along the blue line. For any "extremely big" x we can say that g(x)=f(x)+1 and it is true.

Can we say based on this equation that [tex]\infty= \infty +1[/tex] ? No, because first of all infinity is NOT a number so we can NOT add to it, but also we have here two "different" infinities (directions) represented on the graph by the red arrow and the blue arrow. Problem starts when we are forgetting about this continuous process and trying to substitute our term by "value" like [tex]\infty[/tex]. Then making mathematical operations on it. Equation

[tex]\lim_{x \to \infty}(x+1)-\lim_{x \to \infty}(x)=1[/tex] is also OK, because as we are increasing [tex]x \to \infty[/tex] on the LHS we are "measuring" the difference between the point one the blue arrow and the point one the red arrow 1 which is in this case always 1.

Can we say based on this equation that [tex]\infty - \infty = 1[/tex] ? No, because we have here still two "different" infinities (directions).

Based on this example I wanted to show that I really don't care if in (3.65) (34:51 in video) of my paper [tex]\lim_{i \to \infty }\frac{A_i}{2^{\nu(B_i)}}[/tex] is infinity or not and also if [tex]\lim_{i \to \infty }\frac{B_i}{2^{\nu(B_i)}}[/tex] is infinity or not. For me it is important that both these terms when [tex]i \to \infty[/tex] become closer and closer to certain asymptotes in the way that formula (3.65) is correct. Based on math rules I can say "if X=Y+1 is correct then X-Y=1 is also correct no matter what X and Y are", and therefore (3.66) is correct, which is enough for me to prove Collatz Conjecture.

paper here -> https://www.researchgate.net/publication/351347153
video with explanation here -> https://youtu.be/FIZjITBbi2Y?t=2092

[Guest]

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Good Luck!

[Nobody Knows]

thx

[Nobody Knows]

I want to address here as some say "problem" of limits at infinity from my paper and video, based on the example given by "Guest" here on the forum and as tarmaljed1609 and g3ncollaz in the comments under the video on Youtube.

As it was presented here during the discussion and also within some other comments under the video. It is ok in mathematics to say

[tex]\lim_{x \to \infty}(x+1)=\lim_{x \to \infty}(x)+1[/tex] this is OK

but not ok to move one term to the other side to have

[tex]\lim_{x \to \infty}(x+1)-\lim_{x \to \infty}(x)=1[/tex] but this is NOT OK

And the reason is that [tex]\infty= \infty +1[/tex] is fine, but [tex]\infty - \infty =1[/tex] is not fine as an effect of transformation of our first expression into second one.

This is NOT correct !

We just need to understand what is going on here. Basically limit at infinity means what is going on with a term when we are increasing x to infinity. But, infinity is not a number it is a direction. So increasing x to infinity means that we are changing a value of x through all possible values in the direction to infinity. It is "continuous process", we can increase x to make it bigger and bigger forever. During this process values of certain term of x (in our case below [tex]f(x)[/tex] and [tex]g(x)[/tex] ) become closer and closer to an asymptote it can also happen that values of this term of x will be exactly on the asymptote. It is generally accepted that when this asymptote is horizontal we say limit exists, if asymptote is not horizontal - limit does not exist, but we still can and should understand what is going on, because we can get valuable information from this.

Lets say we have two functions from our example above.
[tex]f(x)=x[/tex] and [tex]g(x)=x+1[/tex]
We can present both on the graph like below.

2023-11-06_203117.png (151.38 KiB) Viewed 1735 times

As we can see when [tex]x[/tex] is increasing (black arrow) to infinity, values of [tex]f(x)[/tex] and [tex]g(x)[/tex] are moving along red and blue lines/arrows. Even though we can say that these values are growing to infinity, but we can also say that each arrow is pointing to something else. Therefore equation

[tex]\lim_{x \to \infty}(x+1)=\lim_{x \to \infty}(x)+1[/tex] is OK, because as we are increasing [tex]x \to \infty[/tex] on the LHS we are moving along the blue line while on the RHS we are moving along the red line, but every point is always shifted "plus 1 up" which will be exactly the same as moving along the blue line. For any "extremely big" x we can say that g(x)=f(x)+1 and it is true.

Can we say based on this equation that [tex]\infty= \infty +1[/tex] ? No, because first of all infinity is NOT a number so we can NOT add to it, but also we have here two "different" infinities (directions) represented on the graph by the red arrow and the blue arrow. Problem starts when we are forgetting about this continuous process and trying to substitute our term by "value" like [tex]\infty[/tex]. Then making mathematical operations on it. Equation

[tex]\lim_{x \to \infty}(x+1)-\lim_{x \to \infty}(x)=1[/tex] is also OK, because as we are increasing [tex]x \to \infty[/tex] on the LHS we are "measuring" the difference between the point one the blue arrow and the point one the red arrow 1 which is in this case always 1.

Can we say based on this equation that [tex]\infty - \infty = 1[/tex] ? No, because we have here still two "different" infinities (directions).

Based on this example I wanted to show that I really don't care if in (3.65) (34:51 in video) of my paper [tex]\lim_{i \to \infty }\frac{A_i}{2^{\nu(B_i)}}[/tex] is infinity or not and also if [tex]\lim_{i \to \infty }\frac{B_i}{2^{\nu(B_i)}}[/tex] is infinity or not. For me it is important that both these terms when [tex]i \to \infty[/tex] become closer and closer to certain asymptotes in the way that formula (3.65) is correct. Based on math rules I can say "if X=Y+1 is correct then X-Y=1 is also correct no matter what X and Y are", and therefore (3.66) is correct, which is enough for me to prove Collatz Conjecture.

paper here -> https://www.researchgate.net/publication/351347153
video with explanation here -> https://youtu.be/FIZjITBbi2Y?t=2092

Any other comment to this ?

[Guest]

By the definition of a limit, [tex]\lim_{x \to \infty}f(x) \to \infty[/tex] means that, for any positive number [tex]A[/tex], you will always find a number [tex]m[/tex] such as when [tex]x \ge m[/tex] then [tex]f(x) \ge A[/tex].

Starting from [tex]f(x) = x[/tex] then adding 1 to both sides of the inequality will yield a similar result (Take [tex]A' = A + 1[/tex]). That's what it means when the limit of a function goes to infinity ; It doesn't litterally equals [tex]\infty[/tex], it just abides by the definition above. To put it more simply, as x grows larger f(x) grows arbitrarily larger. It doesn't "reach" a number named [tex]\infty[/tex]. That's why it doesn't make sense to add a number to it : We don't do arithmetics with the [tex]\lim_{x \to \infty}[/tex] symbol, it's all simplifications derived from the rigorous definition.

If you want to demonstrate something related to limits, then you should start from the definition and continue step by step from there.