And $diety spake and said "If they exist, let them be equal..."
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@blakeyrat said:
@boomzilla said:
Jokes are supposed to be funny.The boiling cats stuff gets a pass due to the fact that this is a trolling forum, too.
Also it's a, you know, JOKE.
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On the OP, it occurs to me that this is what happens when the civil rights movement gets carried away.
[b]ATTN BLAKEY[/b]: MATHEMATICS SECTION, DO NOT READ @derula said:
@Scarlet Manuka said:
I wasn't talking bullshit. You've misinterpreted my post. I wasn't trying to be mathematically precise, because this is not a mathematics forum. So I skipped over the details about defining an equivalence relation via a.compareTo(b) = 0, defining a partial order ≤ on the equivalence classes as [a] ≤ [b] iff a.compareTo(b) < 0 or a.compareTo(b) = 0, and the extra conditions required to make this all work (mostly extra transitivity conditions, e.g. if a.compareTo(b) < 0 and b.compareTo(c) = 0 then a.compareTo(c) < 0). Instead, I took the properties and translated them back into terms of how they'd work with the compareTo() function, since that's what's relevant on this forum. In that formulation, saying that a.compareTo(b) < 0 iff b.compareTo(a) > 0 is indeed a result of the antisymmetric property, along with trichotomy which seems to be a reasonable assumption since it just says that a.compareTo(b) always returns a result. Again, I didn't spell that out because it's at a level of detail that doesn't really belong on this forum.@zelmak said:
It's not being antisymmetric either. The antisymmetric property says that if a≤b and b≤a, it follows that a=b. The compareTo method doesn't actually only define one partial order, it defines two partial orders and one equivalence relation. I.e., it compareTo < 0, the two numbers are in relation 1, if compareTo = 0, they are equal, and if compareTo > 0, they are in relation 2. The rule he mentioned is just to make sure that relations 1 and 2 are "consistent", that is, a<b if and only if b>a. (Depending on how you define "negative", it might also say that equality has to be symmetric)The function should also be transitive; that is: if a.compareTo(b) is negative, then b.compareTo(a) should be positive.
That's not being transitive, that's being antisymmetric. The transitive property says that if a.compareTo(b) is positive and b.compareTo(c) is positive, then a.compareTo(c) must be positive (and similarly if you replace 'positive' with 'negative' throughout).
Also, "<" is not a partial order. It's not antisymmetric. Don't talk bullshit.
Now, who wanted to hear about Abelian groups?I condensed all this down into "more or less" in my original post. That was supposed to be an indicator that while I was well aware that there was more going on in the background than what I'd said, I didn't consider it to be relevant. I guess it didn't work. Though I did even add in a note about how we'd actually be defining the relationship on the equivalence classes generated by a.compareTo(b) = 0. How is it possible that you read my post enough to criticise my definitions, without reading the part that said they weren't the real definitions?
But enough of that. Let's talk about Abelian groups instead.
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@Scarlet Manuka said:
"more or less" in my original post. That was supposed to be an indicator that while I was well aware that there was more going on in the background than what I'd said, I didn't consider it to be relevant. I guess it didn't work.
"more or less" means "I dunno, but I'mma post anyway".
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@Scarlet Manuka said:
I wasn't talking bullshit. You've misinterpreted my post.
Doesn't matter, I was mostly trolling. Also TL;DR.
@Scarlet Manuka said:
But enough of that. Let's talk about Abelian groups instead.
Yay! Well... the classification of finitely generated Abelian groups is kinda boring. Come to think of it, I've never really worked with any infinitely generated Abelian group other than (ℚ,+), (ℚ,⋅) or (ℝ,+) / (ℝ,⋅)... are Lie groups of such sorts, I mean, infinitely generated and Abelian? Also... I mean, it's quite obvious that (ℝ,+) is isomorphic to (ℝ,⋅) via exponential function / logarithm. But the same doesn't hold for (ℚ,+) and (ℚ,⋅), because ex not necessarily rational when x rational... are they still isomorphic thanks to some other isomorphism?
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@derula said:
@Scarlet Manuka said:
I wasn't talking bullshit. You've misinterpreted my post.
Doesn't matter, I was mostly trolling. Also TL;DR.
@Scarlet Manuka said:
But enough of that. Let's talk about Abelian groups instead.
Yay! Well... the classification of finitely generated Abelian groups is kinda boring. Come to think of it, I've never really worked with any infinitely generated Abelian group other than (ℚ,+), (ℚ,⋅) or (ℝ,+) / (ℝ,⋅)... are Lie groups of such sorts, I mean, infinitely generated and Abelian?
Some are abelian, all are infinitely generated (as [abelian, if they are] groups).
Also... I mean, it's quite obvious that (ℝ,+) is isomorphic to (ℝ,⋅) via exponential function / logarithm. But the same doesn't hold for (ℚ,+) and (ℚ,⋅), because ex not necessarily rational when x rational... are they still isomorphic thanks to some other isomorphism?
No. Suppose f were an isomorphism. Then f(1) would have rational n-th roots for all n > 0. That, however, is impossible by the fundamental theorem of arithmetics.
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(I have to add that after posting this I discovered that of course I meant (ℚ+,⋅) / (ℝ+,⋅), and that of course I know many more infinitely generated Abelian groups... just never studied them as such alone, but instead as the rings / modules / vector spaces they also happen to be. Not that it matters, because the point of the post was mostly to annoy blakey...)
