Landmark math proof clears hurdle in top Erdős conjecture
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Landmark math proof clears hurdle in top Erdős conjecture
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Re: Landmark math proof clears hurdle in top Erdős conjecture
#2The 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.
Re: Landmark math proof clears hurdle in top Erdős conjecture
#3The 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
#4The 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 matter? Maybe it must be >0 otherwise 1/n makes no sense?
2.) When you ask whether the sum of 1/n for all n in A diverges, how do you know what order to sum them in? Does it diverge regardless of the order? Since A only contains positive integers doesn't sum of 1/n for all n in A always tend towards infinity?
EDIT: I see now that sum of positive integers doesn't always tend towards infinity, thanks to ColinWright's comment about powers of 2 (for which the sum of 1/n tends towards 1).
Also, since the numbers are all positive, it doesn't matter what order you sum them in, you're basically just asking whether the sum of 1/n for all the numbers in A is finite or not.
Re: Landmark math proof clears hurdle in top Erdős conjecture
#5The 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…
Dense math can often be complex to decipher. Sometimes it feels like reading an esoteric codebase to me!
1) Yes in this case 0 is not included in N. I’ve seen N defined both ways depending on the context, so it can be confusing when it’s not given explicitly.
2) By definition, a series sum is based on the limit of partial prefix sums. E.g 1/a_1, 1/a_1+1/a_2, ... It is an interesting question mathematically, if the sum stays the same when you arbitrarily rearrange the terms of the series. In general the answer is no (see Riemann rearrangement theorem), but as this series is only made up of positive reals, it can be rearranged arbitrarily without change in how it converges (or doesn’t).
To the second part of your question, take any geometric series, i.e. A = {1, r, r^2, ...}, then it will converge. There are other classes of series that will converge in this case, and the conjecture is basically asking to characterize sets with diverging series as needing to be “large and dense” in a certain sense.
Re: Landmark math proof clears hurdle in top Erdős conjecture
#6The 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…
1. Yes, as you say it doesn't matter
2. Order of summation doesn't matter when all numbers are positive
> Since A only contains positive integers doesn't sum of 1/n for all n in A always tend towards infinity?
no, think for instance sum(1/n^2) EDIT: or sum(1/2^n) which solves Zeno's paradox (Achilles vs Tortoise)
Re: Landmark math proof clears hurdle in top Erdős conjecture
#7Earlier quoted context omitted.
> 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…
Dense math can often be complex to decipher. Sometimes it feels like reading an esoteric codebase to me! 1) Yes in this case 0 is not included in N. I’ve seen N defined both ways depending on the context, so it can be confusing when it’s not given explicitly. 2) By definition, a series sum is based on the limit of partial prefix sums. E.g 1/a_1, 1/a_1+1/a_2, ... It is an interesting question mathematically, if the su…
Note that a geometric series can also be a multiple, e.g. (a, a r, a r^2, ...). It converges if and only if |r| < 1.
Re: Landmark math proof clears hurdle in top Erdős conjecture
#8Earlier quoted context omitted.
Dense math can often be complex to decipher. Sometimes it feels like reading an esoteric codebase to me! 1) Yes in this case 0 is not included in N. I’ve seen N defined both ways depending on the context, so it can be confusing when it’s not given explicitly. 2) By definition, a series sum is based on the limit of partial prefix sums. E.g 1/a_1, 1/a_1+1/a_2, ... It is an interesting question mathematically, if the su…
Note that a geometric series can also be a multiple, e.g. (a, a r, a r^2, ...). It converges if and only if |r| < 1.
Good point, I was not being fully general there!
Re: Landmark math proof clears hurdle in top Erdős conjecture
#9Does it get us closer tot he twin prime conjecture?
Re: Landmark math proof clears hurdle in top Erdős conjecture
#10Very promising, although the jury is still out: "The new paper is 77 pages long, and it will take time for mathematicians to check it carefully. But many feel optimistic that it is correct. “It really looks the way a proof of this result should look,” said Katz, whose earlier work laid much of the groundwork for this new result."