The Proof of Collatz Conjecture - Explained

[Guest]

Youtube video "The Proof of Collatz Conjecture - Explained" -> https://www.youtube.com/watch?v=FIZjITBbi2Y

[Guest]

Your argument is cyclic (at 35:00). You say the limit exists, but it exists only for numbers reaching 1. So saying that the existence of the limit proves all numbers goes to 1 is wrong. It only proves that numbers reaching 1 reaches 1

[Nobody Knows]

At 35:00 is [tex]\lim_{i \to \infty }\frac{A_i}{2^{\nu(B_i)}} - \lim_{i \to \infty }\frac{B_i}{2^{\nu(B_i)}} = 1[/tex] how would you explain this equation to have on the left side 1 if not the way I did it ?

[Guest]

A few lines before, one of your limit was on the other side. [tex]\infty[/tex]+1=[tex]\infty[/tex]. So is [tex]\infty[/tex]+2. It does not mean that [tex]\infty- \infty[/tex]=1...or 2... or anything. You should be more careful with limits/infinities.

[Nobody Knows]

May I ask based on what you have assumed it is infinity ?

[Guest]

I could ask the same question to you. Based on what do you assume it to be finite? The burden of proof is on your shoulder.

For me it is simple: 19:00 you say [tex]A_{i+1}=4 A_i[/tex], so [tex]A_i \ge 4^i[/tex]
But it is well known that in a cycle, the exponent of 2 should be arround [tex]2^{\lceil i \log_2(3)\rceil}[/tex] which is [tex]\approx 3^i[/tex]
It is only when you reach 1 that you keep dividing by 4 without going down further allowing the limit to be finite

[Nobody Knows]

It is easy
If true is that
A = B + 1
Then logically
A - B = 1
Correct ?

[Guest]

Not when you play with limits and infinities
https://en.wikipedia.org/wiki/Indeterminate_form

[Guest]

Also your argument is really wobbly.
You have [tex]\infty=\infty+1[/tex], then you move things around [tex]\infty-\infty=1[/tex] to conclude that since you have 1 on the RHS, both values on LHS must be finite. That's not how it works

[Nobody Knows]

By saying [tex]\infty= \infty + 1[/tex] it looks that you are adding [tex]\infty[/tex] like a number, which it is not. For me [tex]\lim_{i \to \infty }\frac{A_i}{2^{\nu(B_i)}} = \lim_{i \to \infty }\frac{B_i}{2^{\nu(B_i)}} + 1[/tex] means immediately that both limits are finite or this equation is just wrong.

[Guest]

No. You are adding 1 to something infinite, and the result is also infinite. But I am a bit confused....this is math basics, you are contesting?
Do you agree that a limit to infinity can be infinite? If yes, do you agree that you can still add 1 to it? if Yes, do you agree the result does not become finite after the operation? If yes, than you agree that [tex]\infty+1= \infty[/tex] (where of course [tex]\infty[/tex] is just the representation of your limits).
Because you are implying that [tex]\infty+1= \infty[/tex] do not exist and therefore you cannot add 1 to something infinite (or add something infinite to something else), which makes no sense
You also seem to have no trouble using infinity in your limit but denying its existence in the result you get (a limit cannot be infinite?)

[Nobody Knows]

"No. You are adding 1 to something infinite, and the result is also infinite."
It is your assumption not mine. I started from [tex]\lim_{i \to \infty }\frac{A_i}{2^{\nu(B_i)}}[/tex] and after few transformations I have got it equals to [tex]\lim_{i \to \infty }\frac{B_i}{2^{\nu(B_i)}} + 1[/tex] from this I concluded both limits must be finite.

"Do you agree that a limit to infinity can be infinite? "
More or less fine for me. But if you want to know my opinion please read my other (short) paper
General Relativity Theory of Numbers -> https://www.researchgate.net/publication/365605224

"If yes, do you agree that you can still add 1 to it?"
No. If you can add 1 to it - means it was not infinite

"if Yes, do you agree the result does not become finite after the operation?"
Ther was no such operation in the first place.

[Guest]

Uhhh????
So for you this simple equation [tex]\lim_{x \to \infty}(x+1)=(\lim_{x \to \infty}x)+1[/tex] does not exists? or one of the limit is not infinite?
Man...seriously?

[Nobody Knows]

You will find the answer to this simple question here: General Relativity Theory of Numbers -> https://www.researchgate.net/publication/365605224

[Guest]

No need to read your paper, the limit of a sum is the sum of limits, and obviously the equation I wrote is perfectly correct, and obviously both limits are infinite, and the +1 is still correct. I don't know why you send me to some paper other than you don't know what to reply. Apparently you don't accept Mathematician rules, but you want Mathematician to accept yours which allows you to wrongly claim a proof you don't have.

And for your other remark, you just didn't use the right formula at 19:11. The right formula is the one from proposition 4 in Sontacchi. The one you took at 19:11 is the one from proposition 7 which is only valid for numbers reaching 1, which is obvious in the preceding minutes of the video where you constructed your formula from 1. You should use [tex]A_i-B_i=2^{v_m}u^n(x)[/tex] at 14:28, because when you set [tex]u^n(x)=1[/tex], you already limit all your subsequent formulas to numbers already reaching 1, as I already mentioned. And proving that numbers reaching 1 reaches 1 makes no sense.

[Guest]

So to be clear, the correct equation is [tex]A_i-B_i=2^{\nu(B_i)}N_i[/tex] unless you already decided that [tex]N_i=1[/tex] which renders all your claims useless

[Nobody Knows]

No need to read your paper, the limit of a sum is the sum of limits, and obviously the equation I wrote is perfectly correct, and obviously both limits are infinite, and the +1 is still correct. I don't know why you send me to some paper other than you don't know what to reply. Apparently you don't accept Mathematician rules, but you want Mathematician to accept yours which allows you to wrongly claim a proof you don't have.

And for your other remark, you just didn't use the right formula at 19:11. The right formula is the one from proposition 4 in Sontacchi. The one you took at 19:11 is the one from proposition 7 which is only valid for numbers reaching 1, which is obvious in the preceding minutes of the video where you constructed your formula from 1. You should use [tex]A_i-B_i=2^{v_m}u^n(x)[/tex] at 14:28, because when you set [tex]u^n(x)=1[/tex], you already limit all your subsequent formulas to numbers already reaching 1, as I already mentioned. And proving that numbers reaching 1 reaches 1 makes no sense.

I see that you become emotional.
Answering some of your emotional comments.
1. You said that there is no need to read this paper not knowing what is in it, even though i said (knowing what is in it) that you should.
2. "I don't know why you send me to some paper other than you don't know what to reply." I know but response is to long and you can find it in my paper. You will not understand "why" unless you will read this paper.
3. "Apparently you don't accept Mathematician rules" - fallacy appeal to authority
4. "but you want Mathematician to accept yours" I actually don't care
5. "The one you took at 19:11 is the one from proposition 7 which is only valid for numbers reaching 1" thats the point of my proof I see you didn't follow
I can only recommend you to watch again and please try to understand not to reply.

[Nobody Knows]

Uhhh????
So for you this simple equation [tex]\lim_{x \to \infty}(x+1)=(\lim_{x \to \infty}x)+1[/tex] does not exists? or one of the limit is not infinite?
Man...seriously?

I will try to give you short answer here.
1. It exists because you wrote it here ... therefore it exists.
2. Question if it is correct ? In my opinion it is correct because you can convert "based on mathematical rules" your left side to the right side. Than you will have the same expression on the left and on the right, which satisfies equesion sign.
3. If [tex](\lim_{x \to \infty}x)+1[/tex] makes sense... for me generally NOT.

[Guest]

Don't play d*mb and d*mber, you know what "exists" meant, and I see that you realize your blunder (saying that one of the limit is not infinite when you can add 1 to it).

Only one thing left for you to realize: you admit that your whole proof literally start with the assumption that [tex]N_i=1[/tex], to prove that [tex]N_i[/tex] must be 1, but you still don't see that this is a cyclic argument, and that it automatically discard any other possibility from the start (despite this being the important thing to prove)

[Nobody Knows]

Sorry I deleted one post willl refraze it here . It is late here

Can you help me to recall where I did say "one of the limit is not infinite when you can add 1 to it" i think i said something different.

You said "Only one thing left for you to realize: you admit that your whole proof literally start with the assumption that [tex]N_i=1[/tex], to prove that [tex]N_i[/tex] must be 1, but you still don't see that this is a cyclic argument, and that it automatically discard any other possibility from the start (despite this being the important thing to prove)[/quote]"

It looks like you haven't seen my video fully but just previewed it. Concept that certain Ni=1 is correct only when N0 can be represented in required form. Whole proof is to show that this representation for every initial N0 is possible.