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@Fast-Nop always up for a correction for my half baked maths. Thanks man.
Is it still true that that the fibonacci prime density would be a lower bound on the sensory of primes in general then? -
Yeah the density of the subset of prime numbers compared to natural numbers approaches 0 but doesn’t equal zero. I don’t think this will prove that the subset of Fibonacci primes is finite though. This article seems cool: https://vixra.org/pdf/...
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@Wisecrack Since fib primes are a subset of all primes, the number and hence density of fib primes is a lower bound for all primes, yes.
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@Fast-Nop sometimes we all need reminding that you can’t just assume 0*inf = 0. Thanks sir.
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@stackodev Mathematical discussions as a whole would normally have a "conclussion" at the end of the post, essay, book etc. In this case, the conclussion is that prime numbers are infinite(this is a fact as far as we know) then fibonacci PRIME numbers must be finite. At this point we would just look at what a prime number is vs a fibonacci number and then we would better understand what a fibonacci prime is. Once the jargon is taken out of the equation we can look at the general idea that @Wisecrack posted to prove or what hinted at his/her idea of the posting originally.
We then try and make sense of it only to realize that @Fast-Nop said.
But yeah man, when you don't understand something in math, read it slow and look at the jargon definitions. It is not much different than programming, and imho way more fun sometimes. -
@AleCx04 Better high on math than on meth. :)
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@Fast-Nop you said it brother. And the realization that we cannot assume that 0*inf = 0 was both brilliant and the key.
I raise my glass to you my good sir. This is why peer review is a thing in the real world and academia.
A tu salud! -
@AleCx04 "And the realization that we cannot assume that 0*inf = 0 was both brilliant and the key."
Looking back I've come to appreciate this post a lot more. I considered it good back then but in retrospect it was genuinely brilliant because typically 0*inf IS undefined under normal circumstances.
Math is hard.
When you wanted to know deep learning immediately
Maybe I'm severely misunderstanding set theory. Hear me out though.
Let f equal the set of all fibonacci numbers, and p equal the set of all primes.
If the density of primes is a function of the number of *multiples* of all primes under n,
then the *number of primes* or density should shrink as n increases, at an ever increasing rate
greater than the density of the number of fibonacci numbers below n.
That means as n grows, the relative density of f to p should grow as well.
At sufficiently large n, the density of p is zero (prime number theorem), not just absolutely, but relative to f as well. The density of f is therefore an upper limit of the density of p.
And the density of p given some sufficiently large n, is therefore also a lower limit on the density of f.
And that therefore the density of p must also be the upper limit on the density of the subset of primes that are Fibonacci numbers.
WHICH MEANS at sufficiently large values of n, there are either NO Fibonacci primes (the functions diverge), and therefore the set of Fibonacci primes is *finite*, OR the density of primes given n in the prime number theorem
*never* truly reaches zero, meaning the primes are in fact infinite.
Proving the Fibonacci primes are infinite, therefore would prove that the prime number line ends (fat chance). While proving the primes are infinite, proves the Fibonacci primes are finite in quantity.
And because the number of primes has been proven time and again to be infinite, as far back as 300BC,the Fibonacci primes MUST be finite.
QED.
If I've made a mistake, I'd like to know.
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