That's too many words

Brain hurt.

@JasonBourne The complete lack of credible evidence for is at least weak evidence against. One may also appeal to the principle of parsimony.

Definition of truth: to say of what is that it is, and of what is not that it is not, is true.

@JasonBourne One must first define "definition"

The existence of a deity is an untestable hypothesis; therefore lying outside the realm of what science can decide. Hence, it is irrelevant in any context except for one's personal belief and should be extended no further; and likewise should be protected and not diminished.

The argument over the validity of a religion is like arguing over whether red or blue is more awesome.

In other words, if everyone could just mind their own damn business about it (including you, Tim Tebow), that would be just swell.

I think blue is awesome because of Jason's gravatar :)

@EdGorcenski It would be nice if it actually were.

@BrianM.Scott It actually is; we just ignore that in our society.

Blue is awesome because blue is awesome.

@EdGorcenski No, it isn’t: people’s color preferences don’t have consequences of similar magnitude.

weak evidence against what? truth?

Oh, gack.

You're looking at it backwards. People's religious preferences don't have consequences of similar magnitude, until someone decides that they do.

The actual nature of a religious preference is as meaningless as "red vs. blue", but this meaninglessness offends people for some reason, and so they invent ways to make it meaningful

@BrianM.Scott I bet it gets closed.

I'll give 10 to 1 odds.

@EdGorcenski Clearly you’ve never read any history.

@skullpatrol Not taking that bet!

I've read much history; most of history is people saying "RED IS BETTER" "NO BLUE IS BETTER" "NOW YOU DIE"

well i think the question "is there a "why" to the universe?" COULD be meaningful....

and, moreover, is a distinct question than "how is the universe?"

But, there is no substance to their claims, and the only way to counter a policy-based argument predicated on faith (or lack thereof) is to remind everyone involved that any argument based on faith (or lack thereof) is isomorphic to "red vs. blue"

@BrianM.Scott -2 already.

@EdGorcenski Which in the real world is rather pointless. The human fact of the matter is that they aren’t isomorphic save in a dream world.

religious conflicts often have more to do with the alliance of religions with political factions than the underlying theology itself

@skullpatrol No close votes, though. I take that back: $2$.

The odds just went up to 50 to 1 :)

Yes but our society is built upon moving ever closer to that dream world (even if we'll never get there). I mean, that's the guiding idea behind that part of the 1st Amendment, for instance.

we are red! red is better. you are brown! now, you die!

The acceptance that the ideal exists, and even if it is not manifest, that we should strive ever towards it.

i think one can also ponder, is "goodness" or virtue" something that exists in a platonic sense, or is it merely a human invention which is purely relative?

@JasonBourne -4

@DavidWheeler I'm an existentialist. I don't give a shit about goodness or virtue ;)

@JasonBourne $4$ downvotes and $3$ close votes.

my guess is "too vague"?

100 to 1

Yes but if the asker wanted detailed logical analyses, the question would have been better written.

@DavidWheeler The three close votes are for not a real question.

@EdGorcenski what do you mean by that, exactly?

General reference...?

@DavidWheeler Which part?

@JasonBourne I think that enters the realm of philosophy at that point

"existentialist"

You know, Camus, Sartre, that whole deal

personally, i'm not a big believer in "absolute truth". i AM a big fan of "contextual truth".

Anyhow, I must run to a Dr appt.

bbl

L8r

i still don't know what that really MEANS.

Hi guys!

hi

@DavidWheeler I understand now, thank you very much for your answer on category theory question.

@BrianM.Scott Is it basically the same as this?

@skullpatrol Not really: the one that you found is much better focussed, with an emphasis on the practical problem of checking one’s work.

mathstackexchange has a certain preoccupation with "standards" which contribute to an atmosphere of non-user-friendliness

@BrianM.Scott Hmm...agreed.

people are socially ostracized for asking "bad" and/or "stupid" questions, creating an air of elitism

little constructive advice on how to refine and improve their search for knowledge and understanding is given...they are encouraged instead to go away and come back when they can confrom to the norm to a greater degree

A big part of math is conforming to convention, no?

The problem is that often times there's not even a way to fix some questions

or, sometimes the asker isn't interested in fixing the question - they want an answer to the original query in its original form, nothing else will do

well, narrowing the doorway is one way to keep a site from being overwhelmed. such may even be in the best long-term interests of the site itself.

@badp I’m of two minds about this one. I wouldn’t have closed it, but a real answer would take a pretty extensive essay.

@BrianM.Scott Are we talking about a specific question? (Sorry)

at the moment, yes

@badp Yes. Hold on a minute.

I checked the transcript but found none

math.stackexchange.com/questions/291525/what-a-true-proposition maybe?

yes

@badp Beat me to it. Yes, that’s it.

Betting is now [Closed]

in all fairness, it's more of a philosophical question then a mathematical one

@DavidWheeler That's not really true, there's a branch of mathematics about this

blog.stackoverflow.com/2013/01/…

If you guys are interested^

I could take my notes on first order logic and copy paste answers there

The OP’s last comment turns it into a very practical question; the original didn’t include enough context.

well, issues of "provability" and "self-contained definitions" of what "true" means in a logical system can be very deep and pertinent questions

The question's real problem is one of scope

In lessons we spent two weeks or so answering these questions

There's no need to ask us to write the book on first order logic

@BrianM.Scott What is your result here?

@badp That’s not quite what it’s doing.

it seems to me that if the question was re-posed, or suitably sharpened, there would be a more favorable response. but if i were the OP, i'd think to myself: "to heck with this, i'll go somewhere else".

@BrianM.Scott It's asking us first to define a mathematical model where you can have a thing that represents a proof and then define enough properties until you have a way to say "this is a correct proof"

@Novice A little over $57$%. If you dig around, you’ll find a query that gives the top several. Arturo, Qiaochu, and a couple others have higher rates.

@BrianM.Scott Wow, mine is under 6%.

Thanks to one of my own answers which I accepted, or I'd have been around 4%.

Or rather, made to do so.

What is your denominator?

68.

$31\%$ surprising

I am starting to pity upon myself.

@Novice Among users with at least $1000$ answers: Arturo (62.2), joriki (59.5), Davide Giraudo (58.5), me (57.1), and David Mitra (54.4). Drop the answer cutoff to $500$, and t.b. heads the list at $75$%, Willie Wong and Georges Elencwaig move into third and fourth, and David Mitra drops to eleventh.

The script is here.

@Novice Which means that your percentage is still pretty volatile. I have $4320$ answers, so mine doesn’t change much.

No one is answering my question about what ideles are

@BrianM.Scott Indeed.

@user58512 Something I’ve never learned about, I’m afraid.

t.b. reminds me of tuberculosis; I bet I am not the only one.

@Novice He originally used his actual name, but apparently he started getting unwanted e-mail.

@BrianM.Scott Is he still inactive?

@skullpatrol I’ve not seen him in a while.

@BrianM.Scott Too bad.

Arturo hasn't been seen either.

This is completely ridiculous.

@Novice Because I don’t upvote as often as I should. I tend to get involved in answering questions and forget about voting.

This script suggests that Marvis and BenjaLim won the Enthusiast badge three days after joining.

The same is with Fanatic.

There is a BIG difference in meaning.

Enthusiast << Fanatic

Of course.

But look into the script.

What do you want me to look at?

@skullpatrol data.stackexchange.com/mathematics/query/10418/… enter "Fanatic".

I don't seem to get what this answer says.

How did Katie Banks change her name to Jamie Banks?

@Novice If a positive integer $n$ has $d$ digits in base ten, then $10^{d-1}\le n<10^d$, so $d-1\le\log_{10}n<d$. In other words, $\log_{10}n\approx d$.

@BrianM.Scott hmm...

oh!

I'm tired, very busy day

@Novice Note also that when the number of digits is large, multiplying an $m$-digit number by an $n$-digit number yields about an $(m+n)$-digit number $-$ just like logs.

@BrianM.Scott Yes, I see that in the comments. Thanks!

@Novice You’re welcome!

@BrianM.Scott Thanks for dropping by :D

@skullpatrol My pleasure.

We appreciate your help.

logs of numbers correspond more closely to how we feel about size, rather than how we actually measure size. we probably have some sort of logarithmic circuit in our brains somewhere.

Heh, I got 10 points this morning, then a user was removed and I lost 10 points. Figures.

@robjohn @BrianM.Scott I wanted to share with you guys something I found out relatively recently and can best be described in the form of a question: "what is the most common noun in the English language?"

is it a word we use all the time?

Yes.

@skullpatrol Did David Wheeler just use the word? :-)

No googling allowed...

When you're ready the answer is here.

LOL, I knew that the italics had to do with something.

@robjohn That is a BIG hint :-)

lol the is not a noun

it's a definite article

If you're looking for nouns then "I"

kinda

<_<

"I" is not a noun, either, it's a pronoun.

@badp Pronoun...

pro-noun!

It's kinda almost the same!

Otherwise you need to drop to about #61 with "people"

(This whole discussion is also self-reinforcing "the", "I" and "people" up in the classification. Yay for feedback loops.)

omg, 8-th grade english class is coming back to me now...split infinitives... adverbial clauses, dangling participles....make it stop....

@DavidWheeler I reject your sentence theory and replace it with my own, for the joy of natural language parsers!!

i cannot parse that. reinitialize.

Your sentence theory is rejected and replaced with my own. Your natural language parser will certainly find joy in working with the new grammar.

$n \to n+1$, if samevalue.old = samevalue.new goto begin end

Refrigerator apricot double mumble dobedoo truth wants.

@skullpatrol Yes.

compile error. continue (y/n)?

@JonasTeuwen Thanks for replying :)

@DavidWheeler as far as I know, most of our senses work with logarithmic scales. Luminosity and hearing certainly do.

See the decibel scale for sounds.

well guise, its bin swell but im off 2 my virtual transgender cosplay. l8er.

@DavidWheeler lol cu

Wow "number" comes in at 22 for nouns?

"one" at 35 and "two" at 84 overall

Pretty interesting.

Yep.

"time" almost has "numbers" inherently associated with it.

thanks to you people's free star week we surpassed 5k stars without anyone noticing the landmark

:bummed:

You will always be the landmark star of this room.

What do homomorphisms from (N,+,0) to other monoid M look like? I am trying to figure out why N represents forgetful functor U : Mon -> Set

@Daniil Every such homomorphism is determined by where it sends 1, and hence the homos are in bijective correspondence with the elements of M (since for every m in M, there is a unique homo sending 1 to m).

@anon yep, thanks

We could've taken B = {0,1} instead of N, right?

Since N is a free monoid of B

If I understand this correctly

no

N is the free monoid on one element (under addition), not two elements, and it is unclear what a monoid homomorphism B->M would be since B is just a set. (You can give it a monoid structure under multiplication, but this is irrelevant.)

N is a free monoid on one element? And which element is that? It can't be one, since we can not get zero by summing ones...

Let $f:\mathbb{R}^n\to \mathbb{R}$ be a linear function. One wants to find the minimum for $f$. Why is it clear that the minimum is unique? "Because it is linear" is not the answer that I'm looking for...

Unless I am completely wrong and just talking shit atm, which is easily imaginable, sorry.

@Daniil Whether or not N contains 0 is a convention. Which convention do you think makes the most sense here?

If it's a monoid it has to contain 0

Since there has to be a monoid zero

ah

well, then sure, N is the free monoid on one element, 1, and 0 can be obtained by summing 1 zero times

Note that since B={0,1} is just a set, in the free monoid F generated by B, 0,0+0,0+0+0,... will all be distinct by fiat and all different from the identity of F (which we can call 0_F if need be).

You cannot assume the symbol '0' satisfies any relations in the free monoid being generated, because otherwise freeness will be violated (even though the symbol 0 typically is chosen to denote the element satisfying such relations). And the zero of a free monoid is always the empty string with zero 'letters' from the original 'alphabet' (set).

@WouterZeldenthuis Why... not?

That's like the answer right.

Say $f$ is minimal in $x_0$.

Then the minimum is attained...?

why would a dual vector (as a function) even have a minimum?

@anon, ah, you are right

It is on a finite dimensional space and hence bounded.

The constraint region is usually (weak)-compact.

@WouterZeldenthuis Unless I'm mistaken, f(x,y) = x + y has no minimums.

I can't write a proof for this math.stackexchange.com/questions/291609/… but I know the answer intuitively

@badp that has many minimums

@JonasTeuwen I don't understand - a linear map ${\bf R}^n\to{\bf R}$ will be of the form $x\mapsto a\cdot x$ for some $a\in{\bf R}^n$ under the usual inner product. Pick $x=Na$ for arbitrarily large $N$ and get arbitrarily large outputs (unless $a=0$), no?

namely $(x,-x)$

a not 0 means |a|_p > 0, and the absollute value is discrete so just take epsilon small enough that it can't change value..

@anon Sure, but minimization problems usually have like constraints and so yes no yes?

That's the whole bloody business usually.

@WouterZeldenthuis Uh, no, for x → -infty, for y → -infty, x+y → -infty without bound.

But I agree it is not mentioned.

@JonasTeuwen In which case the domain wouldn't be ${\bf R}^n$, I would assume.

why am I writing \br instead of \bf, my brain is messed up today

The domain of your operator is.

@badp, yes you are right, my mistake

was minimizing absolute value

oh.

Mea culpa!, Mea culpa! shoulded the Greek dude.

Well then, f(x,y) = |x| has infinity minimum values :)

So I guess the formalization of an optimization problem includes restricting a given function to a smaller domain, instead of assuming the domain being optimized over is understood a priori.

Now, the question may be asking about the minimum value rather than the point of minimum.

@anon The domain can depend on ... many things.

in the example i was looking at the function was actually to $\mathbb{R}^{\ge 0}$

As you take the largest.

But uniqueness usually follows from strict convexity in this type of problems.

Oh, look, a restraint :)

(although now the only linear function from R² to R+ that I can think of is f(x) = 0 and friends)

ok, my question was unclear and stupidly formulated. what i should have said is that we have a function that is linear in n variables, subject to minimizing a constraint

err

what shall I do

rather, and we want to minimize an error function involving the function

I hope you've learnt your lesson now. Next time the evil bunnies will be released (redundantly I should add they will kill you and feast on your soul).

ok, everyone, forget what i said. i realize what i've done now

Which is?

i haven't stated the problem correctly to you guys. i was very confused. basically i was looking at an error function that is quadratic in n variables, and whose derivative is thus linear...and therefore such a function has a unique solution

I see, so you used us to phrase the question properly. Good on you.

yes, thanks for being my victims, err, sounding board

Rubber ducking is fun

occasionally

Those rubber duck vibrators?

Have never used those. Is it?

Joshua Bell is a quite 'in your face' violist. Opinions?

@JonasTeuwen violists use vibrato, but probably no rubber ducks

I wish I never dropped playing violin.

It is a wonderful device to create wonderful things.

Additionally I even find the male virtuous violist sexy. If that isn't strange.

Didn't Einstein play the violin?

Yes.

I wonder what age he started at...

So young.

No, that's the thing about it!

Yes.

how do I show that exp(iy) + exp(-iy) is real from the fact exp(z)' = exp(z)?

I want to say that it's because its symmetry with respect to i |--> -i, but that's for polynomials

Just full in the Euler formula.

@JasonBourne In those situations, I add an edit at the start of my answer explaining that "-this answer was for the original question before the OP changed the question to something totally different that makes my answer look ridiculous" :)

help

@JasonBourne Yep -I have found it works well!

This was just posted by casperOne in the moderator's chat room

@robjohn cos b i ?

@mick, en.wikipedia.org/wiki/…

It says prime numbers are evenly distributed over residue classes

@user58512 I know that.

why did you ask a question that implies you don't know this?

I did ?

Do people label/refer to enumerations while writing documents in LateX? If yes, what do they use?

> What is the analogue PNT for these type of primes ?

@user58512 Yes assuming it is indeed 1 mod 12 this is well known yes. i just needed " chinese remainder " I knew the rest.

I have the list of conditions that are needed for a function to be called say "lambda function". Then, how do I ""name those requirements, I am writing them in the enumerate environment?

I realised the answer a minute after I posted ... is that ok ?

@user58512 marlu's answer seems to do the job there, I think

Im new to ring theory ... hate to admit but its true

that part of the question trivial compared to the other part. It doesn't make sense but whatever.

@user58512 sorry - what doesn't make sense?

hahaha

@mick, you can view Gaussian integers like this. And the norm N(a+ib) = a^2+b^2. Because it forms a lattice (it's just a square grid) you get a Euclidean algorithm (forall gaussian integers x,y there exists q,r with N(r)<N(Y) such that x = qy+r) this implies that it is a UFD.

how does the euclidean algoritm show that a + b sqrt(-5) is not a UFD ?

sorry to reverse things but that might help me

Also I do not know why lattice implies UFD (if it does)

lattices easily have euclidean algorithms (but only Gaussian and Eisenstein integers are) - you'll see this when you prove the statement I just made.

Any euclidean algorithm implies unique factorization, that's an easy result too

I have much to learn.

but i have to go

to show Z[sqrt(-5)] is not a UFD find two different factorizations of the same integer

but if i cant find one for say another ring , how do i show its not an UFD ?

@JasonBourne yes - I saw that - wel done!

I don't know

Maybe worth a question

maybe you could compute Dirichlet's class number formula numerically to a few decimal places and invert it

I think I understand the lattice thing a bit better , it implies unique meaningfull remainder hence GCD or such ...

I do not know how to compute a class number

it's just a formula, I'm saying you might be able to write a computer program that does it

thanks

bye bye

bye

Not to sound condenscending, but both Gone and Math Gems appear in top 10. :-0

Also Calvin Lin appears at number 5 while being on SE for only 24 days or something.

In constrast, did has been here for all of the 2 years.