Talk:Erdős–Graham problem

Still a "Conjecture" and not a "Theorem"?

I notice that it's called a conjecture, even though the result has been proven. I suppose we are waiting for the literature in this field (i.e., reliable sources) to "assign" it a name? Jwesley₇₈ 19:29, 23 March 2010 (UTC)[reply]

Glibichuk calls it the "Erdős–Graham problem". Maybe we should follow his lead and move it to that title? —David Eppstein (talk) 19:43, 23 March 2010 (UTC)[reply]

Yeah, that seems like a more appropriate title for the article. Feel free to move the article. Jwesley₇₈ 22:42, 23 March 2010 (UTC)[reply]
I also found, via Google Books, a reference to an "Erdos–Graham theorem" in "Discrete mathematics and its applications" (2006) by Sethumadhavan. He's apparently referring to a result from "Old and new problems and results in combinatorial number theory", which is where the conjecture was made (or just published?). It's possible that Sethumadhavan is referring to the same result. But, I don't see any mention of Croot. Jwesley₇₈ 22:42, 23 March 2010 (UTC)[reply]

Rewrite for comprehensibility

In an encyclopaedia it would be appropriate to explain things in such a way that people who do not already know about them can understand what is at issue. This article fails dismally. Looking at the first sentence:

What is combinatorial number theory? It is linked, the link takes me to Number theory with an anchor to Combinatorial_number_theory. But combinatorial number theory is not mentioned on the (rather long) page. Let's hope I already know what it is.

Integer, partition of a set, Egyptian fraction: OK, if I don't know what they are, the links help.

So I should be able to follow the first sentence? No: it's ambiguous as to whether any partitioning is required to fulfill the condition, or whether such a partitioning is required to exist. I know and you know that the former is meant, but it's not clearly stated. And the condition "one of the subsets can be used to form an Egyptian fraction" is ambiguous in such a way as to suggest the latter: if "used" means you use all of the members, then obviously most partitions will fail.

But maybe the second sentence will help? No, that's where everything really does go irretrievably pear-shaped. What is an r-coloring? It's not linked, probably because the link would take you to an article on linguistics. What is a monochromatic subset? Probably something to do with an r-coloring...

Never mind, while taking apart the reasons I didn't understand the first paragraph, I think I've got what it should have said.

Next paragraph suggests I'm wrong: it's starts by saying "in more detail", and then states something else. No longer are we talking about the existence of a a subset of a set defined by an arbitrary finite partition of the integers greater than 1, but instead we are considering possible limitations on the size of the largest member of such a presumed subset.

Never mind. Next paragraph: first sentence at least has something do do with the last paragraph. Second sentence reintroduces Egyptian fractions, so maybe we'll find out what the second paragraph has to do with the first. But what are smooth numbers? The linked article isn't bad, but doesn't help here, since the numbers are not said to be "B-smooth," "powersmooth," or smooth over some set. Never mind ... sum to unity if reciprocals add up to 6 ... always find interval where they add up to 6r, so if you've partitioned the integers into r sets, one of them at least will give you what you want. Makes sense. Except any sensible reader gave up some time ago. And even this one still doesn't know how the smoothness fits in.

Looking at the most recent reference [1] suggests that the article might also be wrong in substance, since it presents the partitioning result as the Erdős–Graham problem, and the density result as a "stronger form." In fact, the density problem is the original, the partitioning result addressing a weaker form.

2001:9E8:F3F:DD00:96C6:91FF:FE17:6B19 (talk) 13:58, 10 May 2022 (UTC) 2001:9E8:F3F:DD00:96C6:91FF:FE17:6B19 (talk) 13:48, 10 May 2022 (UTC)[reply]