Ronan Tegerdine

[Ronan Tegerdine @rcteg]

@San_adex001 @bengrossbg 0 and 1 are both elements of the ring. 1 is the multiplicative identity, meaning that a times 1 = a for all values of a, including a=0. 0 is not the same thing as nothing, which you can consider more like the empty set.

[Ronan Tegerdine @rcteg]

Almost all of these explanations are incorrect.

It's 1 because it's the *empty product* of the ring. In any ring the empty product is always the multiplicative identity (1), just like how the empty sum is always the additive identity (0).

It's the same reason that 5^0 is 1 as well; you're multiplying *nothing*, together, not "5" 0 times. That's why the argument people make of "but 0 times anything is 0" doesn't make sense here. There's not actually a 0 being used anywhere, it's purely a notational thing.

You are multiplying nothing, there's no symbol there, and so by definition you HAVE to get 1 or else you could do bizarre things like 4 times (empty product) and get something other than 4, which of course makes no sense.

To be clear, you have to keep in mind the difference between 0 and "empty"/null here. 0 is a number and an actual element of the ring, whereas empty is literally nothing, no elements of the ring at all.

[Ronan Tegerdine @rcteg]

@CAaronRodgerus @arithmoquine That's kind of exactly the point, it's not possible to enumerate all reals. There's no way to *actually* randomly generate a real number. There's too many of them. The mathematician's way of talking about these things is purely abstract, away from any implementation.

[Ronan Tegerdine @rcteg]

@CAaronRodgerus @arithmoquine https://t.co/DXGWjk8ZSE

No, it is possible for a subset to not be empty but to be so small relative to its superset that there is probably 0. A geometric interpretation that may help is thinking of the set as being "infinitely thin", so you couldn't throw a dart and land on it

[Ronan Tegerdine @rcteg]

@slayerofsnails @Ortho_Dixie @lordmiles I found it pretty hard to get through. There's about 30-50 pages of interesting analysis and then the rest is him repeating himself ad nauseam with claims that aren't actually really supported by history. It's worth reading those 30-50 pages to understand his hypothesis, though.

[Ronan Tegerdine @rcteg]

@mutawasiti1 @miniapeur doesn't happen in the integers, but in different spaces you can have zero divisors or torsion elements; in the integers mod 6 (so 0, 1, 2, 3, 4, 5) you have that 2*3 = 6 mod 6 = 0

[Ronan Tegerdine @rcteg]

@jesawyer I took a class with him last year, this is the absolute last place I expected to see him pop up

[Ronan Tegerdine @rcteg]

@latursk @BoltsJolts @Steve_Dangle No, that's not how that works. LTIR NEVER gives you extra space. If you have someone on LTIR for 6 million, your cap ceiling is raised by 6 million, but you also still have their 6 million cap hit. Acquiring an LTIR player has no effect on how many players you can field.

[Ronan Tegerdine @rcteg]

@GeopoliticalBum @cmvalentineb lol if you're curious i searched "deuterium ratio in hydrocarbons"

[Ronan Tegerdine @rcteg]

@GeopoliticalBum as you might expect the answer is wildly more complicated than a yes or no https://t.co/8p9xZ0R8tj