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u/FormulaDriven Actuary / ex-Maths teacher 19d ago
For x>2, the proportion of natural numbers less than 2x that are not equal to 7 is (2x - 1) / 2x = 1 - 1/2x which has a limit of 1 as x tends to infinity.
So, because 100% of all numbers must not be equal to 7, there must be no number equal to 7.
This is the problem. The set of numbers that do not eventually iterate to 1 might (and probably does) have a density of zero over any subset of the natural numbers, but that doesn't prove the Collatz conjecture that there aren't any at all.
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u/gmalivuk New User 19d ago edited 19d ago
The proportion of integers that are not powers of 2 is (2n - n)/(2n), which is also one in the limit. Even though there are infinitely many powers of 2.
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u/FormulaDriven Actuary / ex-Maths teacher 19d ago
Indeed. Given the prime number theorem, you can use the argument to show that there are no prime numbers!
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u/Apart_Composer3952 New User 19d ago edited 19d ago
If it's calculating includes prime1 meaning 2 and prime2, meaning 3. Because neither any even number (2+30n) is prime except 2. Same for (3+30n) except 3, and (5+30n), only 5. This leaves us either of (1, 7, 11, 13, 17, 19, 23 and 29) + 30n to cover all possible primes exept 2, 3 and 5.
Using ((prime 4)-1)/prime 4 = 6/7 × ((prime 5)-1)/prime 5 = 60/77 × ((prime 6)-1)/prime 6 = 720/1001 ×...×((prime k)-1)/prime k showing a diminishing proportion.
((2x)-1)/2x includes multiples of 2, 3 and 5 in it's assessment.
If the proportion is getting smaller and smaller, and it becomes an overestimation is more proof of proportion approaching 0, because it doesn't contain the destructive effect of 2, 3 and 5 on continuous relative accuracy.
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u/FormulaDriven Actuary / ex-Maths teacher 19d ago
I'm afraid I'm going to have to bow out. I'm finding your arguments difficult to follow and they don't seem to be using conventional language to set out the logic - "destructive effect of 2,3 and 5 on continuous accuracy" is not terminology that I'm familiar with from analysis or number theory. I hope someone else can help you.
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u/gmalivuk New User 19d ago
Yes, most residue classes mod 30 do not contain many primes, but the others each contain infinitely many so what's your point?
The sequences 30n+1, 30n+7, 30n+11, 30n+13, 30n+17, 30n+19, 30n+23, and 30n+29 all contain infinitely many primes.
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u/Apart_Composer3952 New User 19d ago
Yes, they contain all of them, apart from 2, 3 and 5. So starting calculations from there, as l do in my high school style formulation, makes perfect sense. As you add primes creating composites on any of those 8, you are left with a diminishing overestimated proportion, never equal to (2x-1)/2x.
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u/gmalivuk New User 19d ago
But there are still infinitely many of them, so why does the diminishing proportion matter? If you're trying to show there are no exceptions to the Collatz conjecture, it is not enough to simply show that the portion of possible exceptions approaches 0.
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u/Apart_Composer3952 New User 19d ago edited 19d ago
Actively approaching 0 without leaving out any numbers is not the same as the example you use. Yours leaves out 1 when determining reals before 1. Where's 1's consideration? And all the other whole numbers for that sake. If l write x/y it isn't that rational. Why does it stop being considered rational by nature if x happens to = y? Schroedinger's fraction? You're wrong to call x/y rational until you know the numbers being used?
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u/gmalivuk New User 19d ago
Who cares? The singleton {1} is completely irrelevant when discussing the asymptotic density of infinite sets.
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u/Apart_Composer3952 New User 19d ago
Well it should be when saying that the reals reach density of 1. Calculating with blinkers on, l'd call it. There must be at least 2 different meanings behind 'reaching 1'. Actaully my x/y example doesn't hold water, but l think you might have got the point l was trying to make, even if you disagreed, which is fine.
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u/gmalivuk New User 19d ago
more proof of proportion approaching 0
But proving the proportion approaches 0 is irrelevant. Plenty of infinite sets of integers make up a proportion of all integers that approaches zero, such as the primes.
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u/Apart_Composer3952 New User 19d ago
Is it the same thing to say '..is also one.' as it is to say 'also reaches one.'
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u/Appropriate-Ad-3219 New User 19d ago
I remember in fact having heard that Tao proved almost every natural numbers satisfy the conjecture. At this point, the only problem is to prove that it is true for every integers.
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u/Apart_Composer3952 New User 19d ago
I think he uses randomised choices, therefore not involving every number. I don't know for sure.
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u/GonzoMath Math PhD 17d ago
Well, no. That would be ridiculous. "Almost all" isn't about randomized choices; it's about density.
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u/gmalivuk New User 19d ago
"Almost every" in what sense?
Because almost every integer is composite, in the sense that the asymptotic density if primes is zero, and yet there are still infinitely many primes.
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u/Apart_Composer3952 New User 19d ago
I imagine because his results didn't consider and exclude every number 'coming'. Only 99.99....% because of this. I believe there is randomness. I could be wrong, but l wouldn't understand it if l did have a look.
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u/gmalivuk New User 19d ago
Well like with primes, even if the density is 100% that doesn't stop there from being infinitely many exceptions.
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u/Apart_Composer3952 New User 19d ago
I can't imagine how that works.
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u/gmalivuk New User 19d ago
There are infinitely many primes but as you consider the interval [1,N], the percentage of integers in that interval that are composite approaches 100%.
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u/unic0de000 NaN 19d ago
It's also true of, say, the powers of 2. Maybe it's more obvious because of the faster-shrinking function.
2, 4, 8, 16, 32, 64... Obviously we'll never run out of such numbers, there are infinitely many of them. But as n -> ∞, the fraction of numbers < n which are in the set, approaches 0.
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u/Apart_Composer3952 New User 19d ago edited 19d ago
This brings to me, the problem of induction in philosophy, sets that are members of themselves and all that tricky stuff. I might look at that to see if l can find some dividing line where your theorising example is on one side and mine is in the other. Because the 2 examples we're using definitely don't overlay each other perfectly to be able to say that the supposed truth or falsity of one strictly implies the truth or falsity of the other.
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u/gmalivuk New User 19d ago
Because the 2 examples we're using definitely don't overlay each other perfectly to be able to say that the supposed truth or falsity of one strictly implies the truth or falsity of the other.
No one is bringing up these other examples to say Collatz is definitely wrong. All we're trying to do is explain why your logic is flawed, because your logic could equally be applied to claims that are definitely wrong, and therefore there must be a problem with your logic.
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u/Apart_Composer3952 New User 19d ago
No problem. I accept the 'idea' that there is a one to one relationship between the reals between 0 and 1, and the whole numbers, but your example multiplies to an infinity within a boundary; 0 to 1. Mine multiplies into an infinity with no such boundary. Are there more whole numbers than the set of whole numbers? No! Therefore the set is exhaustive for whole numbers. Are there more reals than between 0 and 1? Yes! Therefore the set is not exhaustive for reals, indicating a difference.
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u/unic0de000 NaN 19d ago
All I'm doing is clarifying the meaning of "almost all/every" in this context. https://en.wikipedia.org/wiki/Almost_all#Meaning_in_number_theory
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u/Apart_Composer3952 New User 19d ago
Yes, l understand. All l'm doing is suggesting there must be more than one way of looking at this wording. We're at a crossroads of you accepting what has been decided as singularly true and me suggesting there must be some kind of dividing line. Thankyou for your involvement, at least you've carried on communicating even though it's obvious l'm no trained mathematician. Stay well, if you are well! If not, get better!
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u/Apart_Composer3952 New User 19d ago
For x>2, the proportion of natural numbers less than 2 x that are not equal to 7 is (2 x - 1) / 2 x = 1 - 1/2 x which has a limit of 1 as x tends to infinity.
The formulation never involves 7. Does 7 = 7? In mine no numbers are excluded for inspection.
Different!
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u/Apart_Composer3952 New User 19d ago
7/8 Did l write 7?
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u/FormulaDriven Actuary / ex-Maths teacher 19d ago
I don't know what that comment means. I was just showing how I could use your argument to "prove" that no natural number is equal to 7. From some of the other replies to me, you'll see we can "prove" that there are no numbers which are powers of 2, and so on.
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u/Appropriate-Ad-3219 New User 19d ago
With this kind of proof, you're saying that all real numbers are irrational.
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u/Apart_Composer3952 New User 19d ago
Please can you give me example of what you mean. I don't see it.
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u/Appropriate-Ad-3219 New User 19d ago
Rational numbers are numbers of the form p/q where p and q are both integers, with q being non zero. The definition of irrational numbers are the real numbers that aren't rational. As you know, both of these numbers exist (e.g 1/2 is rational, the square root of 2 is irrational).
There are even nice properties of these numbers, for example rational numbers are "everywhere" in R in the sense the rational numbers are dense in R. Being dense means that if I take a point x and and look at a segment centered at x, there is a rational number contained in this segment. This property is also true for irrational numbers.
Consider the uniform probability on [0, 1] which is the most natural probability measure, which means that for a segment [a, b] in [0, 1], the probability of falling into [a, b] is b-a which is also the length of the segment. You can prove that the probability of stumbling upon irrational numbers when picking a real number randomly following the uniform law is 100%. Thus, you have proved that 100% of the numbers in [0, 1] are irrational, but that doesn't mean there are no rational number since 1/2 is rational. So when you say : "Because 100% of all numbers must eventually reach a strictly smaller milestone, any arbitrary starting number is locked into an inescapable cascading chain of downward thresholds, forcing all trajectories to eventually collapse into the fundamental 2 → 1 trivial loop.", it is simply not enough to prove the Collatz conjecture. Of course, I could've misunderstood what you're saying when you say 100% of all numbers since I didn't read the whole post, but it seems you're using cumulative densities so it looks you're doing exactly that.
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u/Apart_Composer3952 New User 19d ago
What l'm saying is that as does the the 1N+1 example follow a geometric progression one series by one series accounting for all numbers reaching a smaller number as it accumulates, the 3N+1 follows the same exact geometric pattern of accounting, it just takes more series' to do so.
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u/gmalivuk New User 19d ago
But you seem to be arguing that the fraction of all integers that end up smaller approaches 1.
That's like how the fraction of reals that are irrational is 1.
There are still infinitely many exceptions.
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u/Apart_Composer3952 New User 19d ago
My numbers are being tested. How does the fraction of reals that are irrational being 1, inextricably link itself to what l'm saying? I can't help but think there is a distinction between the 2. Reals stand inbetween whole numbers. What l am doing leaves no whole number out and includes every whole number in it's analysis.
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u/gmalivuk New User 19d ago
The logic is the same either way. Such-and-such subset makes up a fraction of the whole set that is (or approaches) 0, and yet there are infinitely many elements of the subset.
If you don't like fractions and real numbers, stick with the many different examples of integers people have been trying to explain.
In the limit, 0% of integers are powers of 10, and yet there are infinitely many powers of 10.
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u/Fun-Cauliflower-8087 New User 19d ago
Hey this is similar to what I constructed, nice! You can see that growth and descent have binary structure. The ternary information causes it by interacting with the binary base, and giving a binary fractal.
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