@MatheinBoulomenos Regardless of the definition of projective one uses? It's obvious for some but it seems dubious to me for the "lifting property" one
I think it shouldn't be hard to get right inverses for surjective functions assuming that free modules are projective, but I'm busy with other stuff right now
Hello, how can we prove that it is impossible to express a surd $\sqrt{p_i}$ as a linear combination of one or more $\sqrt{z_i}$ where none of the $z_i$ are equal to $p_i$?
Is this well known proof, I cannot find it by googling
@LeakyNun one of the linear algebra problems I am solving in my undergraduate computer science course involves me having to assume that the above statement is true. so given that kummer theory seems so advanced, can i assume that the statement is true (Without having to include this proof)
I don't know how one would go about proving this elementarily: I'm not saying that an elementary proof doesn't exist, but that I don't have time to think about that
Hello guys, sorry to intrerrupt your discussion; I am looking for papers in complex analysis, accesible to someone who has only taken 1 graduate course of complex analysis. Does anybody know about such papers, or journals where I might find such papers? Any guidance is appreciated
oh...and then is it possible to express $\sqrt{\text{prime number}}$ as a linear combination of one or more real numbers $z_i$ where all $z_i$ are $\neq q\cdot\sqrt{p_i}$ for rational $q$?
@MatheinBoulomenos actually what I meant to ask was how do we know that $\sqrt{\text{prime number}}$ cannot be expressed as sum of all types of real numbers, so that includes real numbers like $\pi$ and $e$ :(
Hmmm, that's fair enough. I haven't noticed it yet myself, but I imagine if I start recompiling things repeatedly, I'll get irritated with it being laggy.
In my HW group for set theory we used a few weird symbols instead of the white square at the end of proofs during the semester. I remember snowflakes and bicycles
Consider the infinite extension $L=\Bbb Q(\sqrt{p_1}, \sqrt{p_2},\dots)$. We have a surjection $\mathrm{Gal}(\Bbb Q^{ab}/\Bbb Q) \to \mathrm{Gal}(L/\Bbb Q)$ where by Kronecker-Weber $\mathrm{Gal}(\Bbb Q^{ab}/\Bbb Q) \cong \widehat{\Bbb Z}^\times$ with $n \in \widehat{\Bbb{Z}}^\times$ acting on $\Bbb{Q}^{ab}=\Bbb{Q}(\zeta_{\infty})$ on roots of unity via $\zeta \mapsto \zeta^n$. Now let $p$ be an odd prime, then we have a quadratic Gauss sum $$\sqrt{p} = \varepsilon_p \sum_{n=0}^{p-1} \left(\frac{n}{p}\right)\zeta_p^n$$ where $\varepsilon_p = -i$ if $\left(\frac{-1}{p}\right)=-1$ and $\varep…
I find it interesting that the way biber works is actually an incentive to have a gigantic bibliography file you use for everything rather than making one for every project you type
@Eran the absolute continuity of $\nu$ wrt $\mu$ implies that $\nu$ is $0$ on any subset of a set of measure $0$ wrt to $\mu$ which is enough to imply that $|\nu|$ is $0$ on any subset of a set of measure $0$ wrt to $\mu$ by Hahn-decomposition
@Eran If $\nu^+(A) + \nu^-(A) = 0$, where both terms are non-negative, then $\nu^+(A) = \nu^-(A) = 0$. Therefore, for such a set, $\nu(A) = \nu^+(A) - \nu^-(A)$ is zero as well, and by assumption this means $\mu(A) = 0$.
What you were asking about was the implication $\nu = 0 \implies |\nu| = 0$, but you wanted the converse.
OK, then your question is straightforward by Hahn decomposition. If $A$ is $\nu$-null, then $A \cap P$ and $A \cap N$ have $\nu(A \cap P) = \nu(A \cap N) = 0$, and these are $\nu^+(A)$ and $\nu^-(A)$, respectively.
Of course this is a much stronger condition than just $\nu(A) = 0$.
@ÍgjøgnumMeg hi, sure. (a premiere problem of number theory if you ask me...) you mentioned a dissertation awhile back? did you work on one? also seeking an Msc? what kind of IT are you doing?
@vzn I just work in the IT department of my university (providing technical support to staff members mostly). My undergrad dissertation was on algebraic number theory, I plan to do an MSc focussing on this area. :)
Let's say i were to write a program which generates all permutations of symbols allowed for my particular programming language up to 1 million symbols, saves those permutations and then executes them as a program each in its own separate thread. Some of those programs will not be valid, some may not halt of which some can be checked with a checker program on if they will halt or not while others might not. We are only interested in those programs which will output a finite sequence of symbols.
Each program that outputs a finite sequence of symbols has its output stored in memory. Our main program then continues to check all permutations of symbols past 1 million symbols and attempts to find the first program which outputs a finite sequence that was NOT to be found/stored in memory.
One would think that now we found a finite sequence that requires more than 1 million symbols to output that supposedly new sequence but that is not the case since our main program can simply output it while only being a fraction of a million symbols
That might be closely related to berry's paradox
@vzn Yes it is but i am not sure this should prevent us from thinking about it. There has to be a FIRST program with more than 1 million symbols in length which would output a new finite sequence of symbols that cannot be generated by any program with less than a million symbols in size, no?
In order to circumvent the halting problem, we could further adjust our main program to check only all those programs which produce an output of a finite sequence of symbols within 1 trillion computing cycles/operations.
Now we would be looking for the first program with more than 1 million symbols in length which produces a finite sequence of symbols as output within up to trillion cycles. A sequence which wasn't produced by any prior program that is less than a million symbols in length, hence wasn't stored in memory so far
But again, our main program which is way less than 1 million symbols in length as it just goes through all permutations and stores their finite result in memory, could just check for the output of that first 1 million+ symbols in length program and print it out
Guys, may someone help me. I'm not sure why additive inverse elements of the integers are also generated by <1>, doesn't <1> mean something like 1+1+.....1? how can it possibly generate the inverses as well?
oh I see, <1> includes negative exponents, which I take to mean the additive inverses, am I right?
I just learned that one can construct the complex numbers from the rational numbers by completing the space of sequences of pairs of rational numbers with respect to the Manhattan metric (and then appropriately defining arithmetic). What kind of space do we get if we apply the same procedure but with respect to the usual Euclidean metric? (other than that we of course get a complete metric space)
Ok, thanks. That wasn't mentioned, I knew that it had to be a definition, it's because i've been told that if G is a group with a binary operation then the generator <g> $=$ $\{$ $g^n : n \in \mathbb{Z}$ $\}$
I learned what a field is in my very first semester of undergraduate math, in the linear algebra course, I learned about groups in the second semester, in abstract algebra
Guys, If I wanted to prove something of the form $\exists x \in X : P(x)$ that is the same as saying $\exists x \in X and P(x)$. So what should I say other than let x $\in$ X therefore P(x), because usually it feels more natural (to me atleast) to write it in that form
because therefore usually means implication, which isn't the case
Can someone tell me if my proof is correct? Prove that the identity e of a group G is the only element of G satisfying $x^2=x$. Clearly $e$ as the identity, satisfies the equation, since: $e^2=ee=e$ which implies that $e^2=e$. Assume there is another p$\neq e$ that satisfies the equation, therefore $p^2=pp=p$ that means p must be the identity which is a contradiction.
i asked a question on SE which seems to give rise to a paradox but once i finished writing it up clearly, i realized the answer and the mistake i made. Should i delete the question or let people answer it?
hm, i might be able to salvage my question after all just by increasing the number of operations from 1 trillion to a size large than required for my main program to finish
@Eran let $K$ be the splitting field of $X^{p^3}-X$ over $\Bbb F_p$. Pick $\alpha \in K \setminus \Bbb F_p$ and take the minimal polynomial of $\alpha$.
Guys, is the statement : Let $a,b$ $\in G$. Then $ax=b$ has a unique solutions in G. of the form: $\forall a,b \in G$ $\exists !$ x$\in G$ such that $ax=b$?
Just out of curiosity, what would showing uniqueness look like in terms of proving the statement (in terms of logic), because to show uniqueness one must assume that another element satisfies the same property, but that is logically different
@LeakyNun Two metrics $d,d'$ on a set $X$ are strongly equivalent if there are two real constants $c,C$, such that $cd(x,y)\le d^\prime(x,y)\le Cd(x,y)$ for all $x,y\in X$. Generally, completing two metric spaces $(X,d)$ and $(X,d')$ where $d,d'$ are strongly equivalent gives two metric spaces $(\overline{M},D)$ and $(\overline{M},D')$ with the same underlying set and two also strongly equivalent metrics. By which way did you derive your answer?
@LeakyNun What do you think about this one: It suffices to prove that $x^3-a$ is irreducible for some $0 \neq a\in\mathbb{F}_p$. now, note that of $x^3=a$ has a solution then $a^{\frac{p}{3}} = a$ according to fermat's little theorem. BUT the polynomial $y^{\frac{p}{3}}-y = 0$ has at most $\frac{p}{3}$ solutions , that is there exists such $a\in \mathbb{F}_p$ that is not 3rd root of another element, thus $x^3 - a$ has no roots in $\mathbb{F}_p$ and is irreducible
@ÍgjøgnumMeg well the mathematics i've read usually contain a mix of symbols or words or both, when symbols are involved, things are way easier to understand, in my opinion.
@ÍgjøgnumMeg Usually when writing things out with symbols and words help with formulating a structure for a proof, for instance , the way @LeakyNun wrote it in terms of logic, was much helpful than what was written in plain english, in my opinion
Doing a directed reading project this quarter, adviser gave some suggestions that all sound appealing: modular forms, understanding BSD conjecture, or elliptic curves over finite fields
But yeah I guess I'm starting to become inclined toward modular forms, especially because I feel like I could use more AG before reapproaching elliptic curves
@pZombie not really sure what youre getting at, but there is something in what youre writing about understanding the distribution of outputs of TMs versus the TM enumerations, that is similar to thoughts from Kolmogorov complexity...
For undecidability idk, probably depends on what you want to know(like just a proof). I think Combinatorial group theory by Lyndon and Schupp has a proof
Although it isn't proving it for an explicit presentation
Yeah just a proof would be fine for me. We showed that hyperbolic groups have solvable word problem (via Dehn presentations) in the GGT course and the professor told us that it is undecidable for general groups, but we didn't prove that
@MikeMiller Nothing much, right now I am at Utah for the semester. I am procastinating on writing a test, and sort of thinking about this question, which I think I have a strategy for infinite type surface case, although quite a bit of work would be needed
I often have to integrate over solid angles $d\Omega$. But i have never seen (until today), an integral over a solid angle twice. Does anyone have an idea of what $d^2\Omega$ would mean?
@AlessandroCodenotti Have you been learning any forcing? There is is this neat result that it is consistent every set of reals is Baire measurable in ZF, and it has a cool implication that every group homomorphism $\mathbb {R} \to \mathbb{R}$ is actually continuous (after doing a bit of work)
And this further extends to every group homomorphism between Polish groups is continuous(in this model).
@AlessandroCodenotti hyperbolic groups are actually biautomatic even. There is a larger class(which WPiG discusses a bit) called combable and bicombable, which has similar definition except without some of the algorithmic stuff, and more geometric
Never mind about my question. It was just a difference in notation. $d^2\Omega$ in some papers means $d\Omega$ in others. The reason this is the case is that $\Omega$ stands for solid angle, and integration over it is integration over two variables, leading some authors to leave a $d^2$ at the start of their variables of integration.
You have this weird axiom about compact Hausdorff spaces which, if you ignore continuum hypothesis, gives you some cardinality strictly between $\aleph_0$ and $\mathfrak{c}$ whose power set has cardinality $\mathfrak{c}$
