So if I could make a covariant functor you say it would have to be dual to the contravariant thingie?

I don't know yet what dual means, I don't really know any category theory despite my avatar.

I'm not claiming that. I honestly don't know.

I feel like you would have heard about it

@DylanMoreland So it's clear in advance what it would have to look like?

Sorry, in that message I was just trying to say that the contravariance is natural.

Ok, I don't know what that means yet : )

Don't try to fight it, in other words. Or do! Fight the man!

It just means that the functor reverses the order of things. $f\colon A \to B$ becomes $F(f)\colon F(B) \to F(A)$.

Oh, yes I know what contravariant means. But I don't know natural.

To me it feels as if I could define quite a variety of valid functors from a category to itself.

Ah, well natural has a technical meaning. I didn't mean it in that way.

You might be thinking of a "natural transformation", which is just a morphism of functors.

I think though the thing I'm trying to construct would probably not be natural.

It's not so important, I guess.

That's always the question. It's easy to get some arrow.

I assume you mean something like the constant functor that maps all the things to one thing?

That's doesn't quite work. You have to send identities to identities for it to be a functor.

But that works: I can send $id_i : M_i \to M_i$ all to $id: M \to M$, no?

I think we are imagining different things now.

Oops. : /

I am still imagining a functor which on objects sends $M$ to $\operatorname{Hom}(M, N)$.

Right. Silly me, I wandered off.

I guess then we can't send all $M^\prime$s to $Hom(M,N)$ for some $M$.

Actually, why not?

Ah, found it. Here is a nice example.

That would be different on objects, then. You could do it, I guess.

That would be a "constant functor".

I guess my point is that you shouldn't worry about this stuff too much unless you want it to be your life.

Which is fine, of course!

@DylanMoreland It completes me moves pinky to one corner of mouth

: )

Sorry to interrupt...:( But please can anyone try to solve my simultaneous equation question in linear algebra or homework tag Plzzzz

@user1396721 Did you ask it on the main site?

Such things usually get answered pretty quickly.

Ya i mean in the mathematics site

Hey then plz Dylan will u help me for that?

This doesn't really look like linear algebra.

I don't think I can say anything intelligent about this, sorry. There's some symmetry that you ought to be able to exploit, but I don't see how.

its okay:)

thanks

Do you just want real solutions?

Maybe even just integer solutions, given the elementary-number-theory tag?

So, regarding your answer here, does canonical mean that all diagrams would commute?

@MattN Canonical is definitely not a technical term. But I think "all the diagrams ought to commute" is a good definition in this case.

See this MO thread.

phew Thanks : )

@DylanMoreland Ah, thanks for that.

@user1396721: This one?

@user1396721 It seems like André has done the job.

His is a pseudonym, right?

I guess you'd tell me which question he's referring to.

@Gigili math.stackexchange.com/questions/152866/simultaneous-equations I think

@ Gigili the question of absolute value has also been asked by me

anyone know how to get 9 using only 5 fives?

i'm currently missing a $9$.

Can I divide $5$ by $5$

bbl

Me too.

yup. i'm trying to keep it to the basic 4 operrations.

What's wrong with $10-1$?

I mean, $55/5-5/5-5/5$

too many fives

I didn't read the question anyway!

if not i'd have just used 5+5 - 5/5

which is too few fives

You're supposed to use a definite number of $5$'s?

yup

basically you fix a $k$ and and $n$

then try to build $1$ through $12$ for it

i finished all but $9$

Oh, that makes sense.

Let me think then.

wow $9$ is a toughie

@Gigili btw, i like your avatar. however if you drew the smiley face on your nail it would literally be a thumbnail.

@user1396721: Robjohn has posted an answer to your question and he explained it in his comment, what's the problem?

Hey Gigili i am unable to get how to consider various cases for 'a'

@Eugene Oh you're right. Haven't noticed it, thank you.

I'm very interested in painting stuff on my nail.

@user1396721 Do you know how to simplify $|x-4|<a$?

I wonder how many times you're going to edit that @Eugene.

@Gigili OP suggested an edit to use dfrac

well frac rather but i chose dfrac

sorry if it's annoying that it keeps getting bumped up

Hey Gigili after simplifying we get x<4 and x>4-a

right?

@Eugene It's okay, I'm afraid it'll turn into CW if you edit a lot.

@user1396721 Unfortunately not, how did you get that?

@Gigili thanks for the info. i'm more interested in puzzle solving than rep points though. the only thing that would annoy me is that low rep users can edit

@Gigili i took a as positive and then considered x-4<a and -(x-4)<a

@Gigili also since i'm not very good at reading situations, i hope my "nail painting" remark wasn't overboard.

@Eugene Umm, I thought $5+5-5/5$ was okay to get $9$.

@Gigili it's missing one more $5$

@Eugene It wasn't.

@DavidWallace it was about granny, so it is indeed hardly, isn't it?

@Eugene I know, that's a bit difficult.

@Gigili thanks then.

@RajeshD I even don't what what to tell you. That is never polite, to say that somebody's English is not that good.

@user1396721 That's right. Or $-a<x-4<a$.

ok well i'll have to cheat

$5!/(5+5)$ - 5/5

wow

that was so off

haha

@ Gigili ya got it so is that the answer?

@user1396721 No, now you should consider different cases for $a$ , positive, negative and zero.

@Ilya : seriously? hmm...then I apologize. my bad...

@Eugene Mine was far better than that!

Ok then if 'a' is negative then shall i take it as '-a'

@Gigili indeed =D

@user1396721 You have upper and lower bounds to be $a+4$ and $-a+4$, see what happens to them for $a<0$, $a>0$ and $a=0$.

@Gigili wow, you seem to wear many hats.

Hum?

as in your diverse stack exchange interests

Haha.

unfortunately i'm only able to do math (which i'm not terribly good at really).

@Ilya : and also thanks for letting me know of such a thing, otherwise i could have done more damage to myself.

@Eugene You seem terribly good at it. Also, registering on different SE sites doesn't mean anything other than I was bored or I had a serious problem.

I can't see my answer being neglected. Someone please either upvote or downvote or comment.

@Gigili thanks for the compliment. however most of my answers are boring and most of my questions are rather silly.

@Ilya : I have just now seen that my comment on english being starred by someone.

@Gigili Jesus! someone did it!!

I didn't expect it at all, whoever has starred it, I request them to remove it please. Its very inappropriate

@Eugene No way!

Oh I see, he uses factorial.

yeah. it's a little bit of a cheat but oh well

@Gigili you don't want to try to make a clock?

A clock on my nail?

@Gigili while that would be pretty cool (as in a ton of micro painting) i meant this.

Looking at all those nasty stuff that user has done ... Not really!

I still think you ruined your answer with that $9$! I'd find a way to get it using the four basic operations.

well it's just a place holder for now

@Gigili i wonder if it's possible to prove that there cannot be an expression using only 5 fives for $9$

@Gigili JESUS CHRIST!!! he did it!!!

@tb In your answer here, is there a reason why you use $\rightarrowtail$ instead of $\hookrightarrow$?

Oh, I thought it was an inclusion.

Thanks, Zhen.

bbl

@Eugene Pft, I was so close to it. I found out that I can use parenthesis.

@Gigili sorry

So, a left exact functor is one that maps a short exact sequence $0 \to \dots \to 0$ to a short left exact sequence $0 \to \dots$. Now I'm a bit confused. In AM on page 22 we prove that the $Hom$ functors map short right exact sequences to short left exact sequences and short left exact sequences to short left exact sequences respectively. So we don't prove that the $Hom$ functors are (left/right) exact. But they are, aren't they?

Wow, look. He has three answers there @Eugene.

We should show these things for short exact sequences, not just for short (left/right) exact sequences, no?

@Gigili i know it's a bit annoying. i thought the beauty is to use the least amount of numbers

Hm, perhaps not. Otherwise they would probably do it in the book.

Yes, they are. Teddy shows it in his answer.

@Eugene No one would see a bunch of parenthesis on the clock instead of simple $1$, $2$, $\dots$, even a mathematician.

@Gigili especially $12$ freaking digits. i wonder how big the clock would have to be.

@Gigili besides $9$ is already done in the example. why reinvent a less efficient wheel?

@Eugene Indeed. I didn't notice it until now.

oh well

sorry he beat you to $5 = 9$.

@Gigili do you know german? i've always wanted to learn german

@Eugene Umm, yes. But I recently realized that I'm not good at listening.

@Gigili in general or to german? =P

@Gigili, @Eugene: you're talking here for the whole Saturday

@Ilya ?

i've only been awake for 3 hours

@Ilya Okay, how much should I pay for that?

@Eugene In general, in general. Why do you want to learn German, BTW?

@Gigili while it is not the common perception i think german is a beautiful language

it is indeed. Much more beautiful than French.

that's exactly what i think! unfortunately people think the converse.

people always tell how germans sound like they're fighting

@ZhenLin Are you still around?

I'm still reading 熊ちゃん's answer.

"Added: As witnessed by the argument above, left exactness of Hom is essentially the definition of left exactness in the abelian category of R-modules."

I think I don't understand what he's saying there.

@Eugene It sounds a little aggressive but is challenging, as opposed to French which has over 93029329032320932 words common with English.

@Gigili i heard the opera "die zauberflöte" and i've like german ever since. i've had no time to learn it though as i'm constantly working on mathematics.

which in arithemetic geometry goes the way of french

He answers my previous question at the end of his answer. So $Hom$ is not always exact. Which answers why AM don't show exactness of it. : )

@PeterTamaroff have you gotten anywhere on this? I have a solution using Bezout's Identity, as I half-jokingly mentioned to Bill.

@Ilya good day!

@PeterTamaroff That should be 1 or 3, not 1 or 2.

@anon good day to you, too!

I guess we lost Ilya.

yo

@anon yo

@robjohn He made that correction later in the chat.

I used (a+b)^2-(a^2-ab+b^2) = 3ab and then assumed a prime p divided a+b and ab...

@anon ah. it's interesting since I found that statement searching for "1 or 3"

@anon that is the point I used Bezout :-)

assume we have $ax+by=1$ and come up with $(a+b)(ax^2+by^2)-ab(x-y)^2=1$

Eh, I would just deduce p|a or p|b, either of which is only consonant with p|(a+b) if p divides both, a contradiction.

@robjohn bezout's for this problem?

@anon That works, but I prefer to avoid using Unique Factorization when proving things that might be used to prove Unique Factorization :-)

Is UF needed in that route?

(The definition of prime in alg num thry is that p|ab implies p|a or p|b at least.)

can't you just use that $a^2 - ab + b^2 - (a+b)^2 = -3ab$?

@Eugene yes, we used that earlier.

@robjohn yup yesterday in fact. did you come up with an alternate proof?

@Eugene Than what?

than than the $a^2 - ab + b^2 - (a+b)^2 = -3ab$?

@Eugene no. That I said we used. Then you need to show that $(a+b,ab)=1$

ah i see

Good evening!

this shouldn't be too difficult. let's see

Is there somebody familiar with control theory? detectible systems...

@robjohn did you prove it yet?

@anon not in this case perhaps, but you still need to show that there is a prime that divides $ab$ :-)

@Eugene Yes. I just showed it above and so did anon. We used different methods.

@robjohn here goes. if there is a prime $p \mid gcd(ab, a+b)$, then $p\mid a$ or $p\mid b$.

We assume it by hypothesis in order to show it's impossible, hence gcd(ab,a+b)=1 on the assumption gcd(a,b)=1.

@Eugene That was anon's method.

ah i see

ok then

i want to finish for completeness anyway

heh

then since $p \mid a+b$ we have that $p \mid a$ and $p \mid b$. a contradiction.

so $gcd(ab, a^2 - ab + b^2) = 3$ if and only if $3 \mid a+b$.

$\square$

nuts

anyone know the tex for qed?

@anon no, you've shown that if there is a prime that divides $ab$ and $a+b$ it must be 1. I think the problem arises in showing that the only unit is 1.

It's a picky point, but when dealing with elementary facts I worry a bit :-)

no it's sound. there are no primes that divide both $ab$ and $a+b$ if $gcd(a,b)= 1$

@Eugene You mean \blacksquare ?

no i prefer square

@Eugene yes, I know that.

is the proof incomplete then? i'm sorry i'm not understanding. i feel like i've jumped into the middle of a conversation i don't fully understand

@anon \square is what is used in the Proof environment I use.

@Eugene No, that proof works, but there are time I would rather use Bezout since it is more fundamental than using prime factorization.

fair enough. i think the fundamental theorem of arithmetic is also fundamental though =p

There are times when prime factorization is used circularly.

i agree.

i was making a joke

@robjohn are you a number theorist?

@Eugene no. Bill will confirm that :-)

oh ok then.

i wanted to ask a question about elliptic curves

i think some of the answers completely miss the point that the motivation is a clock.

@anon: but mainly I wanted to use Bezout to yank Bill's chain :-)

:)

@anon thanks :-p

@anon I should pin it :-)

how the crap you would fit $12$ twelves on a clock is beyond me

haha

@Eugene What the heck is the $n,k=19$

@Gigili lol

overreaching?

I think he'll next post an answer for $n,k=10000000$

I'll buy the clock.

@Gigili you might need a truck to take it home

Most of the answers to this question use either the derivative of $\exp(x)$ or the derivative of $\log(x)$. It seems to me that if you know either of those, substitution and L'Hopital bypasses the point (which I think is finding the derivative of $\exp(x)$).

@Eugene No problem, I'd have it. I'm not sure if it can go through the door.

@Gigili you might also need a one story house that is as big as a two story house to have a wall to place it on

@PeterTamaroff hi!

@robjohn I cracked the $\gcd$ problems yester day, thanks anyways!

@Eugene Yelloo

@PeterTamaroff we cracked it again above

@Eugene Okay, I'm on it.

@Eugene Let me see.

@Gigili you're right it would be a pretty cool clock. useless for telling time though since it would have to be square

@PeterTamaroff i realized that i didn't think my solution through yesterday. i'm sorry about that

@MartinSleziak Thanks. I'd have to properly look at it but at the moment I'm too busy doing algebra.

@robjohn your angry square always looks so ominous

i believe that's the point though

@Eugene Why so? was it wrong?

well i don't think i solve the second part correctly if i remember

@Eugene We did the following:

@Eugene You can use it as closet or something.

nice

@anon Thanks.

Eric nuked this

@PeterTamaroff yup we did. i think i might have messed up the last line?

i don't know

can someone verify a mess up?

@Gigili are you german perhaps?

@Eugene In there? What's the mess up?

hm

maybe not?

i'm not sure

maybe you're right. maybe there's no mess up.

@Eugene What seems strange?

the last line seems funky to me

but i guess not. maybe i was overthinking it

@Gigili a clock closet would be nice as well.

@PeterTamaroff did you see the Bezout method I showed earlier?

$(a^2,b^2)=(a,b)$ is only true when $(a,b)=1$.

@Eugene No?

@anon Sure, but the hypothesis is $(a,b)=1$

I was hypothesizing why Eugene might find the last line funky :)

@Eugene It's not an angry square, it's a mean square

@robjohn I scanned it. I'll look at it now.

Maybe it's mean because it's angry.

@Gigili oh. i thought that's how you learnt german

@robjohn the mean square is definitely more ominous than an angry one

@anon it could be >8(

@Eugene Why would I learn German then? Unless I am masochist.

@Gigili or a sadist who loves to inflict german onto others.

@Gigili achtung gigili!

@robjohn I'd be tough to come up with $(a+b)(ax^2+by^2)-ab(x-y)^2=1$...

@PeterTamaroff It must not be, I came up with it :-)

that and das ist der schreibtisch is all i know

@robjohn True! =)

But what follows from that?

$\gcd(ab,a+b)=1$?

@PeterTamaroff That $(a+b,ab)=1$

yup!

we proved that just now

only when $(a,b)= 1$ though

@Eugene no, this is the bezout method

@Eugene Sure, not that simplifes to $ax+by=1$

yes i know. but i was referring to (ab, a+b) =1 which we proved =D

Or does it?

Wait =P

I have a new one

!

I think I nailed it.

@PeterTamaroff if you can write $1$ as a linear combination of $a+b$ and $ab$, then they are relatively prime.