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Can someone explain this for the cretins in the room?
by whearyou 4y ago
Can someone explain this for the cretins in the room?
- mauriciolange 4y agoStill not a sufficient explanation but I pretty much like the headline: Mathematicians are finding inevitable structures in sufficiently large sets of integers.
- _a_a_a_ 4y agoThe word structure is commonly used but never seems to be defined.
- somrand0 4y agoyou need to study lots of higher algebra to get a sense of what they mean... all I can tell you (because it's all I know) is that "the structure" is preserved by transformations. and what the subject focuses on and studies is the transformations, more than 'the structures'. so I suppose they don't precisely and explicitly know what 'the structure' is; but the incredible thing is how this does NOT matter. as far as I've figured out so far, the point is that it gets preserved across transformations, and that they can inter-relate it across different mathematical objects.
- rightbyte 4y agoWon't a sufficiently large set contain any pattern?
- tromp 4y agoIn 2018, some mathematicians proved that any non-negative integer subset with a positive limit density (for some eps>0 and all large enough n, the subset contains at least eps*n numbers below n) contains a sumset, i.e. a set C that can be written as the sum of two infinite sets C = A + B (the set of a+b where a in A and b in B). More recently, they extended this to sums of more than two infinite sets.
- _a_a_a_ 4y agoI'm afraid that does not illuminate at all.
- deleted 4y ago[deleted]
- brookst 4y agoCan someone explain this comment for the sub-cretins in the room?
- Someone 4y agoLet’s start with finite sets, and in reverse: given the two sets A and B, define set C = {a + b | a ∈ a, b ∈ B} For example if A = {1, 3, 7} and B = {1, 4, 8} , we have C = {1 + 1, 1 + 4, 1 + 8, 3 + 1, 3 + 4, 3 + 8, 7 + 1, 7 + 4, 7 + 8} = {2, 5, 9, 4, 7, 11, 8, 11, 15} = {2, 4, 5, 7, 8, 9, 11, 15} That’s easy. Now, if you didn’t know that, but were given set C, can you find two sets A and B of size 3 that would produce C under that logic? An easy (but cumbersome) way to answer that is: produce all sets of 3 positive integers less than the largest number in C (15 in the example), compute the respective C’s and check whether that’s equal to the targeted C. This problem is about targeting infinite sets of integers for A, B, and C, with an extra twist: for what infinite sets of integers C can you find matching infinitely sized sets A and B that both are subsets of C? Since all of these sets are infinite, the easy but cumbersome method won’t work; neither set has a largest element. In addition, the result isn’t about a specific set of integers.
- 4y ago