@Alucard I was curious about a question like this recently. Obviously it's not dividable into perfect sections, but I believe there's on the order of 10 acres of arable land per person. Maybe it was per household.
@Fargle hehe, yeah i don't think 1 house for every person is the right solution. some want to live together, in a small village or in a big city for various reasons, so the area per person varies...
@EricSilva At the moment I'm learning about algebraic curves (Riemann-roch etc). My idea of a proper algebraic geometry is what people reading Hartshorne are learning :\
I think mathematics might be the only discipline where you have experts freely sharing their knowledge and advice with everyone on so many platforms. It's blessed!
@Holo that was initally a joke but it's not all of lie. The construction of natural numbers is extremely simple in terms of set theory and it's connected to cardinality of infinite sets and that kind of stuff. Obviously everything is set theory, but - as I've told you initially - it's just a joke. xD
@Holo yes. But the "notion" goes before the construction.
Not long ago I was reading a paper about symbolic/mathematical logic and it's kinda weird that we have "set theory 1, natural numbers 1$ and then we construct everything.
@Holo I thought I was sort of familiar with the chapters that preceded it, but the set theory you learn in chapter 0 of analysis texts is not the same, I learned xD
Also: the "set theory 1, natural numbers 1" thing is not really about defining sets and natural numbers. It's about estabilishing that expressions like $\forall x_1 \dots \forall x_n$ are comprehensible.
@LucasHenrique well, the cardinal arithmetic is so; if a,b are finite cardinals then a+b,a•b is just like natural numbers. If one is infinite and the other non zero then a+b=a•b=max(a,b) and lastly a+0=a, a•0=0
One could rephrase that by, instead of defining the cardinality of a set, defining 'the cardinalities of A and B are the same' - this notion is just 'there is a bijection between A and B'.
That is what a mathematician usually means when they say 'the cardinalities are the same', as opposed to anything more esoteric.
@MikeMiller this is the usual way it is taught. We are defining |a|=|b| first. But we still need to know what the definition is, because sometimes we will prove things for sets and not proper classes and then we can't prove things for all sets of the same cardinality
The definition of cardinals without AC is more of something we need to go back to check ✓ on rather than something we need to focus and use
BTW the way you wrote it there seems like an unusual notation: $p_i+p_j \in \mathbb P \operatorname{iff} i=1 \lor j=1$. (And you did not say what you denote by $p_i$.)
I am computing the witt group of rational field. I come up with th map delta pi residue homomorphisam. So we get map W(Z_(p)) -> W(Q) -> W(Z/pZ) the composition of this map is 0. But then how to get W(Z) -> W(Q) -> W(Z/pZ) is 0 for this to be true W(Z)-> W(Z_(p)) -> W(Q) -> W(Z/pZ) we have find sequence like this but why W(Z) factors through W(Z_(p)) at localization of Z at p?
Consider a group $G$ which is the semidirect product $\mathbb{Z}_3 \ltimes (\mathbb{Z}_7 \times \mathbb{Z}_7)$ (internal semidirect product). Let $\phi:\mathbb{Z}_3->Aut(\mathbb{Z}_7 \times \mathbb{Z}_7)$. There will be $\phi_0 ,\phi_1 ,\phi_2$ corresponding to $\bar{0},\bar{1},\bar{2}$ of $\mathbb{Z}_3$. Then does that mean there will be 3 distinct non isomorphic Semi direct products, one if we choose phi_0, another if we choose phi_1 and another if we choose phi_2?
0. Jot down some notes of today in a report 1. Check a couple of chemistry calculations 2. Start putting the transition state template file together 3. Check out the 3D volume function when it is done 4. Start doing some investigation for simple d-finite and amorphous sets. Aim is to get some general but elementary theorems proofs so it will be used as tools to further manipulate and dissect these sets If time: Perhaps try some forcing stuff
Which of these CANNOT be the class equations for a group of the appropriate order? i)10 = 2 + 2 + 2 + 2 + 2 ii)15 = 1 + 3 + 5 + 6 iii)4 = 1 + 1 + 2 iv)6 = 1 + 1 + 1 + 1 + 1 + 1 v)6 = 1 + 2 + 3 I think for problem ii) we can't find such class equation, as $6$ can't divide $15$, so a group of order $15$ can't produce a orbit with cardinality $6$. Rest problems I think can be class equations. Can anyone check it?
If you write up the actual class equation, you will always group the $1$s since those actually form a subgroup, and grouping them is the most useful. But that still requires that you keep that group separate from any conjugacy class of that size
So sure, for this sort of problem probably there should be $1$s
Let $G$ be an abelian group acting transitively on a set $S$. Consider an element $s_0\in S$, if the $\text{stab}(s_0)=H$ then, can we say $\text{stab}(s)=H$ for any $s\in S$?
@LeakyNun Do you happen to know about full-adders while learning logic? I need to implement some kind of divider architecture with only a 4bit adder and I'm kinda stuck trying to change all my Boolean algebra into XOR
@Ted Given a homotopy of homogeneous polynomials $f_t$, you have corresponding algebraic varieties $f_t = 0$, each of which has a fundamental class. Is there a fundamental class for the whole mess, aka points $(t,x)$ with $f_t(x)=0$, so that $f_0=0$ and $f_1=0$ are homologous?
@Daminark I have a new blogpost where I try to motivate characters a bit and give a proof that for a finite group $G$ and a field of characteristic $0$, finite-dimensional representations of $G$ are uniquely determined by their characters
Alright so, if we have some order $\mathcal{O}\subset K$ (where $K$ is a number field), we define $\mathscr{D}^{-1}_{\mathcal{O}} = \{a\in K \mid tr_{K/\mathbb{Q}}(a\mathcal{O}) \subset \mathbb{Z}\}$
@BalarkaSen f are complex polynomials, varying continuously in a real parameter t. I guess I don't see the harm in assuming it's also real algebraic in t, but I don't think it's necessary.
Someone asked a nice question on main a while ago, how to show that eg an elliptic curve in P^2 is degree 3, in the sense of homology. I suggested taking a homotopy to a product of linear factors, where you get a union of lines, where this is clear.
And this should identify the inverse different with the dual module of $\mathcal{O}$. So now it turns out to be the case that if $\mathcal{O}_1 \subset \mathcal{O}_2$, then $\mathscr{D}^{-1}_{\mathcal{O}_2} \subset \mathscr{D}^{-1}_{\mathcal{O}_1}$. I guess because it's easier to map a smaller ring into $\mathbb{Z}$.
Now it seems the key bit is that if $\mathcal{O}_1 \subset \mathcal{O}_2$, and the latter is the ring of integers, then $\mathcal{O}_2 \subset \mathscr{D}^{-1}_{\mathcal{O}_2} \subset \mathscr{D}^{-1}_{\mathcal{O}_1}$
