@user193319 originally just thought that meant topology of operator norm, but it seems like wikipedia says that the strong operator topology has more open sets
> If a reasonably active room has no owner, or none of the owners have been in the room for a certain time, a new owner may be automatically appointed, to ensure that there's somebody who can administrate.
@user1732 it was rough, I think I did well, not sure if I'm gonna get 800+ (which is the score where I'd be comfortable not redoing it), hopefully at least 700+ though, so idk
I have registered for the October one just in case
It's annoying that it takes almost a month to get scores back though. It's a scantron, just put it in the computer and boom
There are 3 instances of this test per year, one in April, one in September (today), and one in October
Though since grad school apps are in the fall/early winter, you can only really take September and October unless you start while you're still in third year
> Are impredicative and nonconstructive objects like: Amorphous sets, Dedekind cardinals, Uncomputable functions, Bernstein sets, Banach Tarski paradox, Large cardinals, Hamel functions etc. have any application outside of itself (not just adding footnotes to existing theorems in core mathematics), set theory and mathematical foundations, or all they all useless artefacts of ZF and ZFC that can only exist within the world they create?
I won't care about how fantasy they are if I don't love them
Just like division by zero, you either demonstrate an example, or you crawl through the entirely of mathematical space to convince me that they never exist forever and ever
Impossible is a very strong word to me. It means that it cannot be realised when you look at every nook and cranny
This is why I like explosive generalisation, because it allows me to find the most general (and hence most unforgiving) proof of something such that not only it answers the question, but it answers it so well that in the far future, nobody will dared to raise it again
Inefficient things will beerased from existence, and in the mathematical world to ensure that is to have a proof so airtight that only nature can screw it by producing an experimental counterexample, and hence generating the 4th mathematical crisis
I must be have a personal Vedentta on the existence of questions lol :D
My attitude to the truth of something, assuming it is either true or false, is to prove that my belief is right and you shut up forever about its negation, or prove that my belief is wrong and I will shut up forever about its affirmative
In other news (assuming anyone had paid attention) the workings of the irrational slopes are wrong and they don't do anything because I implicitly assumed that the polynomials that has $s+t$ as root must be a combination of the individual polynomials of s and t when it is not
What I should have done instead is to pick one generic polynomial, and then plug int the values s, t and s+t, and then try to find a relationship if s+t is a root
will work on that shortly
Also: It appears the auto posters are not respected by the chat and the Rambles get frozen again
Hmmm, @Kasmir congratulated me. I was never consulted or asked if I was willing to be owner. Interesting. ... :P I guess robjohn and others are hardly ever around.
Question: Find $\alpha$ so that P is a measure, where:
$\Omega = \{1,2,3,4 \}, \mathscr{F} = P(\Omega), P(\{1 \})=\frac{1}{4}, P(\{2 \})=\frac{1}{6}, P(\{2,4 \})=\frac{1}{2}, P(\{3 \})= \alpha$
What is $P(\{2,3 \})$?
Here's my attempt:
Given
$\Omega = \{1,2,3,4 \}, \mathscr{F} = P(\Omega), ...
I'm very amused by the fact that, when introducing an extension of the reals with those curious properties like in the complex numbers, we can deduct incredible things about reals
I've just met a problem that goes like "prove that for any integers $a, b$ and $n > 0$, $(a^2+b^2)^n = x^2 + y^2$ for some integers $x, y$"
I had no idea on how to solve it. But, looking carefully, it's an (almost!) trivial result from $\Bbb Z[i]$
I find it curious that extensions of things can tell new facts about the "original" things.
When reading about logical systems, I always go for the "straighforward-est" way to approach any problem. So it would not make sense to extend a thing and, then, get any result from it
But since I stopped to think about this, $\Bbb C \twoheadrightarrow \Bbb R \twoheadrightarrow \Bbb Q \twoheadrightarrow \Bbb Z \twoheadrightarrow \Bbb N$
Either way, I can't see how a new structure with completely new operations can give any interesting result about the structured that generated it
Opinions, gentlemen?
Oh, for example: you can make quotient rings, mappings modulo R, etc. But creating new operations is so... insecure (?)
> Are impredicative and nonconstructive objects like: Amorphous sets, Dedekind cardinals, Uncomputable functions, Bernstein sets, Banach Tarski paradox, Large cardinals, Hamel functions etc. have any application outside of itself (not just adding footnotes to existing theorems in core mathematics), set theory and mathematical foundations, or all they all useless artefacts of ZF and ZFC that can only exist within the world they create?
2
At least for complex numbers, the reason it works is because new routes become available to get from A to B by using the extended structure, such as contour integration of some real functions
which otherwise will be extremely complicated to get to via the direct route
In fact, going as far back to the axiom of choice, the reason it is useful even though ultimately everything relevant to real life can be formulated in finite terms is because many theorems and results become much nicer when infinity is used as a shortcut
(though I am being inaccurate here, cause using axiom of choice vs using only finitist foundations is not really an example of an extension)
Examples where the extended structure does not tell anything extra about the original structure are for example $\text{No} \twoheadrightarrow \Bbb{R}$
where $\text{No}$ are the surreals, though as it turns out the surreals are very useful to reason about certain game theoric games
Another example are the fractional wheels, which extends the field of rationals with involutions, but otherwise it does not tell anything more about rationals
So it is not always true that algebraic extension will give us more
I think it is mainly because while we lose total order when going from reals to complex, we gained algebraic closure, which give us the fundamental theorem of algebra, which give us a very powerful tool to query about reals.
But can we predict what we get before the extension is being created, I don't think we can do that without working out what happens when we add axioms to extend a structure
Alessandro and Tobias might be able to give you more detail on when is an algebraic extension will tell us more about the parent structure and when it does not
@LucasHenrique this is so incredibly powerful and so easy to prove. Can't understand how it is possible even though you've already explained the total-order/algebraic-closure relation
Ok for that particular question, think about it this way: In the complex plane, I am sure you are aware that it can be rewritten as $e^{ir^2n}=e^{is^2}$ and thus the result is apparent when the exp are rotations about the origin. Yet if you only have $\Bbb{Z}$, then all you have is the x axis projections of these rotation thus the result is less obvious that there has to be pairs of integers that satisfy the result
So in a way,some relationships of the reals is a shadow of the relationships in the complex plane
which is why extensions to complex plane helped solved many problems of the reals
In 1847, Lame gave a false proof of Fermat's Last Theorem by assuming that $\mathbb{Z}[r]$ is a UFD where $r$ is a primitive $p$th root of unity.
The best description I've found is in the book Fermat's Last Theorem A Genetic Introduction to Algebraic Number Theory. For the equation $x^n + y^n = ...
We have $$ a_{n+1}=(-1)^{n+1}\cdot \Bigg( (-1)^{n}\bigg((-1)^{n-1} \Big(\big(\cdots\big) + 2\cdot (n-2) \Big)+2\cdot (n-1) \bigg) +2\cdot n \Bigg)=\prod_{i=2}^{n+1}(-1)^i\cdot a_1+\sum_{i=0}^{n}(-1)^{i+1}\cdot 2i=\prod_{i=2}^{n+1}(-1)^i\cdot 0+\sum_{i=0}^{n}(-1)^{i+1}\cdot 2i=2\cdot \sum_{i=0}^{n}(-1)^{i+1}\cdot i$$ or not? How can we calculate that sum? @Rudi_Birnbaum
no, but observe that $$ a_{n+1}= (-1)^{n+1}\Bigg(\cdots \bigg((-1)^{2}\Big((-1)^{2}\big((-1)^{1}(2\cdot1)+2\cdot2\big)+2\cdot3\Big)\bigg)\cdots+2\cdot n\Bigg) $$
Not really a question cause it is incomplete. It is once again those episode of having an irresistible urge to solve $\pi+e$ (or more generally, the rationality of the sum of two transcendentals) whenever some users reminded me of its existence by e.g. mentioning about transcendental numbers
somehow the fact that this is an open question bothers me like there is no tomorrow
According to the plot of the set of irrationals, irrational numbers closest to 0,1 are the most irrational, followed by those closest to 1/2, then 1/3, 2/3 and so on
and this relation is periodic with a period of 1 as shown by the Thomae's function
ok... will try this one more time before going back to global structure of irrationals (cause that one is more interesting and less discouraging since it does not involve a famous open question)
Suppose $s, t \in \Bbb{I}$ $c_i \in \Bbb{A}$ for all $i \in \Bbb{N}$, $n \in \Bbb{N}$ and all other letters unlabelled are assumed to be real. Consider a generic element $P \in \Bbb{P}(\Bbb{A})$ as follows:
Since $P$ is a polynomial, which always terminate under some finite terms $k < \omega$ (that is there exists some $M$ such that for all $k > M$, $c_k = 0$), it follows rearrangement is possible. Rearranging give us:
@MaryStar in nutshell I would recommend to first guess the law (by making a table and looking sharply at the formula, you will observe that even and odd make their own series, and that the first members $n<4$ are "not in the trend") and then you can use induction.
And in the scenario when a help vampire is also on and one of the comment is a response to a help vampire, I have the irresistible urge to flood the chat because since the conservation does not flow anymore, might as well accelerate it to chaos and restart the chat anew
@Faust People often speak of one topology being bigger or smaller than other; what one means when saying this is that one topology is set theoretically contained in the other. You say that the weak operator topology is finer (bigger) than the strong operator topology, but this seems to conflict with Daminark's comment, which states that the strong operator topology contains more open sets than the weak operator topology (i.e., the SOT is bigger than the WOT).
I think I still don't have enough knowledge on polynomials to try to tackle $\pi + e$, so I will return to this later. Hopefully future me when got triggered to solve $\pi + e$ again, will remember that we have already tried the very general investigation above and obtain:
$$P(x+y) = P(x) + P(y) - P(0) + Q(x,y)$$
and that nothing further can be said unless there are other special values besides $Q(x,0)=Q(0,y)=0$ were found to simplify it further
Although if this is a JEE-style problem there's probably some way to find a relation between $A$ and $B$ so that you don't have to calculate them explicitly (nothing of this sort comes to mind at the moment, but you can try transforming one integral into another)
I have a set of N-vectors, in d-dimensional space such that, $\vec{r_{i+1}} -\vec{r_i} = l\vec{a_i}$ wherein $a^{(k)}_i = {-1,0,1}$ (i.e the k-th component of the i-th vector takes one of these three values.
And my goal is to compute the n-th moment of the $\langle |\vec{r_N}-\vec{r_i}|^n\rangle$
In general, one can write the ened to end distance $\vec{r_N}-\vec{r_i} \sum_{i=1}^{N} \vec{r_i} = \sum_{i=1}^{N} l \vec{a_i}$
say, for the second moment, the we need $|\vec{r_N}-\vec{r_i}|^2 = l^2 \sum_{i,j=1}^N \vec{a_i} \boldsymbol{\cdot} \vec{a_j} = l^2\sum_{i,j=1}^N \sum_{k=1}^d a^{(k)}_i a…
Hmm... I don't get what the irrationality measure is doing. There is no point in this stack that there are countably many solutions of $(p,q)$, other than those stacks seemed to follow a pattern of 7
user131753
@DavidReed There is an alternative interpretation of the situation.
@KarlKronenfeld That is a plot of the rationals (white lines) up to denominator 100, with 4 irrational numbers plotted on it. I am trying to make sense of what the irrationality measure is doing by running from q = 1 to 30 and then the purple lines showed the distances $|x-\frac{p}{q}|$ for each given q that satisfy the inequality
But so far, there is no indication that there are infinitely many solutions because at each integer value of q, I only ever get two solutions
so I have no idea how to reason with the irrationality measure
Yup, the purple lines are the distances $|\pi-3 - \frac{p}{q}|$ that satisfy the inequality and the first rational of each $q$s are labelled by the light yellow lines e.g. $\frac{1}{2}, \frac{1}{3}, \frac{1}{4},...$
It should be noted that only up to q=19 is visible and done by inspection, anything beyond are calculated
Question: Find $\alpha$ so that P is a measure, where:
$\Omega = \{1,2,3,4 \}, \mathscr{F} = P(\Omega), P(\{1 \})=\frac{1}{4}, P(\{2 \})=\frac{1}{6}, P(\{2,4 \})=\frac{1}{2}, P(\{3 \})= \alpha$
What is $P(\{2,3 \})$?
Here's my attempt:
Given
$\Omega = \{1,2,3,4 \}, \mathscr{F} = P(\Omega), ...
