Landmark math proof clears hurdle in top Erdős conjecture
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Re: Landmark math proof clears hurdle in top Erdős conjecture
#12Hindman's Theorem says that if you color the natural numbers with finitely many colors, there must be some infinite subset D of the natural numbers such that every finite sum of elements of D has the same color.
The proof, asontonishingly, is an application of free ultrafilters, and is actually simple enough that someone with an advanced undergrad background in pure math can understand it with just a few hours of reading (e.g. see [1])
Re: Landmark math proof clears hurdle in top Erdős conjecture
#13Re: Landmark math proof clears hurdle in top Erdős conjecture
#14I find the existence of number theoretic problems quite puzzling. I wonder what are the implications about the world we can make from them.
Maybe we don't know today if there's an application, but notice that proof mentioned using Fourier Transforms to study a set, the same way you might use it to study a radio signal. There was probably a time when people asked what use is FT?
Re: Landmark math proof clears hurdle in top Erdős conjecture
#15The conjecture by Erdős is the following: if A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges, then A contains arbitrarily long arithmetic progressions. Bloom and Sisask now proved that A contains infinitely many length-3 arithmetic progressions, following from their main result which is an improved upper bound for Roth's theorem. https://arxiv.org/abs/2007.03528
And just to expand a smidgen on that, the maths expression: A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges is one way of saying that the set A of integers is not too sparse. The set of powers of 2 does not satisfy this condition ... it's too sparse. The set of primes does satisfy this condition ... it's not too sparse, primes turn up "reasonably often".
Re: Landmark math proof clears hurdle in top Erdős conjecture
#16The conjecture by Erdős is the following: if A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges, then A contains arbitrarily long arithmetic progressions. Bloom and Sisask now proved that A contains infinitely many length-3 arithmetic progressions, following from their main result which is an improved upper bound for Roth's theorem. https://arxiv.org/abs/2007.03528
> The conjecture by Erdős is the following: if A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges, then A contains arbitrarily long arithmetic progressions This is quite hard to understand for people who don't already know what it means (including me). I started trying to translate it but there were a few parts I didn't understand, starting with: 1.) Is ℕ integers >0 or >=0? Wikipedia says it can be either. Maybe it doesn't m…
Re: Landmark math proof clears hurdle in top Erdős conjecture
#17I find the existence of number theoretic problems quite puzzling. I wonder what are the implications about the world we can make from them.
Re: Landmark math proof clears hurdle in top Erdős conjecture
#18Earlier quoted context omitted.
And just to expand a smidgen on that, the maths expression: A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges is one way of saying that the set A of integers is not too sparse. The set of powers of 2 does not satisfy this condition ... it's too sparse. The set of primes does satisfy this condition ... it's not too sparse, primes turn up "reasonably often".
Hmm is there an asymptotic statement underlying this? The powers of 2 are exponentially sparse (N integers contain at most 1/logN powers of 2), whereas primes are polynomially sparse (N integers contains N/LogN primes)
Would you care to clarify? If it's a question then I'll try to answer, but if it's a conjecture, can you make it more precise?
Re: Landmark math proof clears hurdle in top Erdős conjecture
#19Does it get us closer tot he twin prime conjecture?
Re: Landmark math proof clears hurdle in top Erdős conjecture
#20The conjecture by Erdős is the following: if A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges, then A contains arbitrarily long arithmetic progressions. Bloom and Sisask now proved that A contains infinitely many length-3 arithmetic progressions, following from their main result which is an improved upper bound for Roth's theorem. https://arxiv.org/abs/2007.03528
> The conjecture by Erdős is the following: if A ⊂ ℕ is such that Σ_{n ∈ A} 1/n diverges, then A contains arbitrarily long arithmetic progressions This is quite hard to understand for people who don't already know what it means (including me). I started trying to translate it but there were a few parts I didn't understand, starting with: 1.) Is ℕ integers >0 or >=0? Wikipedia says it can be either. Maybe it doesn't m…