Mathematics - 2014-04-25 (page 1 of 5)

Pedro Tamaroff

12:33 AM

@KarlKronenfeld @Mike

Karl Kronenfeld

Damn, I was just about to break the eerie silence, @Pedro.

Pedro Tamaroff

@KarlKronenfeld I read your mind. Don't worry.

=)

I have a problem.

I think I am doing things correctly.

Suppose $0\to M'\to M\to M''\to 0$ is exact. If $M',M''$ are finitely generated, so is $M$.

Karl Kronenfeld

ok

Pedro Tamaroff

I had first the idea of using that the above is equivalent to having an exact sequence $A^n\to M'\to 0$ and $A^m\to M''\to 0$, (we can take $m=n$) and obtain one $A^k\to M\to 0$, I am thinking I can get $k=n$:

But now I think I can just work with generators, and obtain a set of generators from $M$; like with vector spaces.

I will do it and see what I get.

Karl Kronenfeld

The second is fairly straightforward, but a variant of your first idea also works.

(I wouldn't try to establish k=n)

Pedro Tamaroff

12:40 AM

@KarlKronenfeld OK.

Enjoys Math

that's so hot

Pedro Tamaroff

What did you have in mind?

Karl Kronenfeld

Let $k=m+n$

Pedro Tamaroff

Oh, I wasn't talking about that. But yes, I get what you're saying.

Suppose $\langle x_1,\ldots,x_n\rangle=M'$ and $\langle y_1,\ldots,y_m\rangle =M''$.

Let $x_j'\in M$ be such that $x_j'\mapsto y_j$.

Then the claim would be $\langle f(x_1),\ldots,f(x_n),x_1',\ldots,x_m'\rangle$ generate $M$.

$$M'\mathop{\longrightarrow}^f M$$

user127001

Hey @JasperLoy

Pedro Tamaroff

12:50 AM

@KarlKronenfeld Done. =)

Anthony

Someone! I asked this this morning, and I think I got a sufficient answer, but to show that $[0,1]$ is not of measure zero, can I do the following?:

Actually I don't know how to formalize this.

It's covering compact, so every cover has some finite subcover.

MickLH

Guys! the arithmetic - geometric mean of $1$ and $x$ is $$\pi \div \int_{0}^{\frac{\pi}{2}}\frac{2}{\sqrt{1-\sin(t)^2\cdot(1-x^2)}}\,\mathrm{d}t$$

Anthony

In order for a finite subcover to cover something, it would seem intuitively that the combined length of the subcovers would need to be at least as large as the interval, but I don't actually know how to say that.

Fernando Martin

An interval $[a,b]$ is, by definition, of measure $b-a$

Anthony

Ugh

I don't think I have that...

I just know that it has a finite cover...

Fernando Martin

1:01 AM

How did you define measure if not?

Karl Kronenfeld

I'd guess Anthony uses open intervals.

Anthony

I'm in intro to real analysis

We've only talked about measure zero

And it's just as Karl said.

I think? Hold on.

In other words, for each $\epsilon$> 0, there some open set U containing Z whose total length is less than $\epsilon$

Fernando Martin

ok, but what's length then?

Anthony

I suppose I shouldn't have said length?

The question is

Show that [0; 1] is not measure zero. [Hint: recall that you know [0; 1] is covering compact.
Then show that for any finite cover of [0; 1], the sum of the lengths of the intervals is bounded
below by 1.]

Karl Kronenfeld

@Anthony So that's your definition of ($Z$ having) measure 0?

Anthony

1:06 AM

A countable cover by open intervals

Fernando Martin

So you know how to measure open intervals then, right?

Anthony

What do you mean measure an open interval?

Karl Kronenfeld

Aka the "length".

Anthony

I mean, I guess. Here this is the problem.

Karl Kronenfeld

ok, so you have finitely many open intervals covering $[0,1]$. There's actually a cute way to do this by replacing $1$ with an arbitrary positive real number, and inducting on the number of open intervals in your covering.

Anthony

1:11 AM

Oh the question wasn't even including in the screenshot

I don't even know what I'm asking anymore

Karl Kronenfeld

I was just going from where you left off when you initially asked your question in chat

Anthony

Yeah-so I have [0,1]

Every cover has a finite subcover

Then what? What's the best way to proceed?

Karl Kronenfeld

I don't know the best way to proceed. I offered a suggestion just a moment ago though. :P

Anthony

Inducting on the number of open intervals?

Karl Kronenfeld

There will always be an interval which covers 1. Remove that, you have one less interval.

The set covered will be smaller, hence I suggested replacing 1 with some $x$.

Ram

1:18 AM

I have a question regarding Connected sums. T1#T2/T1-B^0 how to prove that this is homeomorphic to T2?

Fernando Martin

ping @Pedro

Karl Kronenfeld

lol I looked up $p$-divisible group in nLab.

Anthony

@KarlKronenfeld , I'm having trouble following. First, our finite subcover isn't necessarily disjoint, right? Second, if I remove a subcover, I will have one less interval, and the set will be smaller, but why does that help?

Pedro Tamaroff

HAI @FernandoMartin

@KarlKronenfeld

Karl Kronenfeld

Instead of proving that any finite cover by open intervals of $[0,1]$ has cumulative length (that $\sum (a_i-b_i)$ thingy) of at least $1$, generalize a little and show that $[0,x]$ any finite cover by open intervals has cumulative length of at least $x$.
Let $(a_1,b_1),\dots,(a_n,b_n)$ be your open cover of $[0,x]$ and suppose without loss of generality that $b_i\ge a_i$ (remove extra intervals as necessary). Then, $(a_1,b_1),\dots,(a_{n-1},b_{n-1})$ covers $[0,b_{n-1}]$ and $(a_n,b_n)$ covers $[a_n,x]$. There is some overlap and you can find a lower bound of the cumulative length using yo…

Fernando Martin

1:31 AM

I was trying to prove the exercise from last class @Pedro

The corollary of that result by Serre

have you thought about it?

Pedro Tamaroff

I didn't pay much attention. =P

Fernando Martin

smacks Pedro with a commutative algebra textbook

Pedro Tamaroff

$\mathcal O_0^{\rm an}$ where analytic functions in a nbhd of $0$.

Fernando Martin

germs of

Pedro Tamaroff

Yeah, yadda yadda...

Fernando Martin

1:33 AM

and let's denote it $A$ please

Pedro Tamaroff

And $\mathcal O_0$ where rational functions defined in a nbhd of $0$?

Fernando Martin

defined in 0

i.e. the denominator doesn't vanish at 0

Pedro Tamaroff

Ah, OK.

Fernando Martin

though yeah they don't vanish in a whole nbhd

Pedro Tamaroff

The claim was the former was a faithfully flat the other-module.

Fernando Martin

1:34 AM

faithfully

That's the result we are going to use

without proof

Pedro Tamaroff

Yes.

Fernando Martin

to prove the exercise

Pedro Tamaroff

And the exercise was?

Anonymous

@KarlKronenfeld :(

Fernando Martin

let's say $f, f_1, \dots, f_m$ are rational functions such that there exist germs $h_1,\dots,h_m$ such that $f=\sum h_i f_i$ (as germs)

Pedro Tamaroff

1:35 AM

OK.

Fernando Martin

then there exist rational functions $g_1,\dots,g_m$ such that $f=\sum g_i f_i$

Pedro Tamaroff

@Anonymous What's the matter?

Fernando Martin

I have proved it in the case where $f=0$

Anonymous

Egasp my name changed.

Fernando Martin

maybe it follows from affinizing (is that even a word?) the case $f=0$

but I don't see how

Pedro Tamaroff

1:40 AM

Let's write out what it means that $\mathcal O_0^{\rm an}$ is a faithfully flat $\mathcal O_0$ module.

Anthony

It is fixed!

Fernando Martin

It means that the "tensoring by germs over rational functions" functor is exact and reflects exactness

Ted Shifrin

Hi @Fernando @Pedro ... Damn, them is some fancy words you'ze slinging around, mr @Pedro.

Fernando Martin

Hi @Ted

Pedro Tamaroff

@FernandoMartin I know what it means. I'm just saying that it might help to write things down and look at them carefully.

@TedShifrin It's a big theorem by Serre.

Fernando Martin

1:45 AM

Oh I though you didn't write it down

Pedro Tamaroff

Our professor gave it out, and said we're going to use it.

Ted Shifrin

Yes, I've known a few of his big theorems :P

Pedro Tamaroff

@FernandoMartin I didn't.

Anthony

I'm still a little lost despite Karl's help. To show [0,1] has nonzero measure, I don't really know how to proceed. He said to generalize, but I don't see how that helps...

Ted Shifrin

I'm glad you're finally in a class that challenges you, @Pedro.

Pedro Tamaroff

1:46 AM

Ted Shifrin

I prefer not to stay in groups.

@Anthony: It's not a totally easy argument, as I recall.

So you're supposing you have a countable cover by open intervals of total length $<\epsilon$?

Anthony

The route I was recommended to use, for [0,1] is to use the fact that it's covering compact.

Pedro Tamaroff

@TedShifrin I can't get enough of them. =D

What is Anthony trying to prove?

Ted Shifrin

That $[0,1]$ is not of measure zero, I take it.

Anthony

Yep.

Pedro Tamaroff

1:49 AM

The above is a group by Hall.

Ted Shifrin

yes, @Pedro, I dun recognized it.

Ram

Hi, can someone help me with this connected sum problem, T1#T2/T1-B^0 how to prove that this is homeomorphic to T2?

Ted Shifrin

OK, @Anthony, so what happens if we do that?

Jasper Loy

Just prove that it has measure 1 by applying the theorem that the length of an interval is its measure, lol.

Pedro Tamaroff

@Anthony Oh, and you define measure by the exterior measure using open intervals?

Ted Shifrin

1:49 AM

not helpful, @Jasper

Pedro Tamaroff

@Ram NOT ENOUGH INFORMATION FOR AN ANSWER.

Ted Shifrin

I don't think he's doing Lebesgue theory yet, @Pedro

Pedro Tamaroff

How do you define measure Tony?

Jasper Loy

When did Anthony become Tony? LOL.

Ted Shifrin

For example, I define measure zero in my manifolds and differential topology classes, but not measure.

Anthony

1:51 AM

Ahhhh

All I have is measure zero, and it was defined for me as i.stack.imgur.com/oDawU.png

Ram

@PedroTamaroff, T1 and T2 (say closed and connected) are two surfaces, and their connected sum as usual. And by cut and paste, it is clear that T1#T2/T1-{int open Ball} is homeomorphic to T2, but how to prove it in a better way?

Ted Shifrin

So we have a finite cover of $[0,1]$ by open intervals whose total length is $<\epsilon$? So then we should be able to show easily that the length of the interval is $<\epsilon$, right, @Anthony?

Ram

Compact too incase

Anthony

@TedShifrin You mean BWOC?

Ted Shifrin

gesundheit! say what?

Jasper Loy

1:53 AM

WTF is BWOC?

Ted Shifrin

@Jasper, you're back to name #127?

Jasper Loy

@TedShifrin Yes. I managed to change it before 30 days was over because I created a new account on another SE site.

Anthony

By way of contradiction, lol.

Pedro Tamaroff

Claim If $[a,b]$ is covered by finitely many open intervals $I_1,\ldots,I_m$ their total length is $>b-a$.

Ted Shifrin

Ohhh ... never seen that before, @Anthony :D Yes.

Anthony

1:56 AM

So yes, then we would need to show each interval is $\lt \epsilon$. So....?

Pedro Tamaroff

Let $(a_i,b_i)=I_i$.
*Claim 1*. We may assume that $a_0<a, a_2<b_1,b_2<a_3,\&c$.

Ted Shifrin

Wait. Are you proceeding my way or Pedro's way, @Anthony?

Pedro Tamaroff

Using this, we're done. =)

Anthony

Waaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaat is going on. I was listening to you Ted, but now I'm reading Pedro's.

Pedro Tamaroff

Make a drawing.

Ted Shifrin

1:57 AM

@Pedro now gives Ted's advice. Impressive.

Ram

@PedroTamaroff, "Make a drawing" is that for me?

Anthony

@PedroTamaroff I understand that if there are finitely many covers then each should have non trivial length, but I don't know how to say that formally.

Ted Shifrin

@Anthony: I meant the length of the total interval they covered had to be $<\epsilon$ if all the individual ones had lengths adding up to $<\epsilon$. Pedro's doing the contrapositive, essentially.

An open interval $(a,b)$ has length $b-a$, @Anthony.

Jasper Loy

Always be sure your argument is independent of the drawing used to motivate it.

Ted Shifrin

Hush up, @Jasper. :)

Pedro Tamaroff

2:00 AM

@Ram Not really, no.

Ted Shifrin

@Jasper: I'll give you a linear algebra problem that I'd never seen before (it was posted earlier as a disguised differential geometry problem).

Ram

ok, can you suggest me some thing?

Jasper Loy

@TedShifrin No, no, please don't.

Pedro Tamaroff

@Ram I have no idea about what you're talking about =)

Ram

ohh, dont you know connected sums from Algebraic Topology?

Ted Shifrin

2:01 AM

Suppose you have three unit vectors $x,y,z$ in $\Bbb R^2$, and $x\cdot y = \cos\theta$, $x\cdot z = A$, $y\cdot z = B$. Prove that $\sin^2\theta = A^2+B^2-2AB\cos\theta$. I've never seen this before.

Fernando Martin

@PedroTamaroff alicia just sent an email

Ted Shifrin

@Ram: That's one subject Pedro hasn't studied yet.

Jasper Loy

@FernandoMartin Alicia or Alyssa?

Anthony

@TedShifrin, Little confused as to what is being said.

Pedro Tamaroff

@FernandoMartin Didn't get any.

Ram

2:02 AM

@TedShifrin :-(

Fernando Martin

Ahh, did you send her your email?

Maybe you're not on the mailing list

I'll forward it to you

Ted Shifrin

Pedro is telling you to lay 5 intervals down on the real line, @Anthony, overlapping so that they cover $[0,1]$. Why must their total length be at least $1$?

Ram

@TedShifrin, can you help me to prove my question more formally instead of waving hands.

Ted Shifrin

I wasn't paying attention to your question, @Ram, but in a few minutes I must go back to grading my students' homeworks. What is your question?

Ram

T1#T2/T1-B^0 how to prove that this is homeomorphic to T2?

some thing other than cut and paste technique

@TedShifrin

Ted Shifrin

2:04 AM

Well, what other technique is there for such a thing?

Ram

T1 and T2 are connected and closed

Ted Shifrin

These are surfaces, I take it?

Ram

yes

Ted Shifrin

You're taking $(X\cup Y)/Y$, essentially, right?

Ram

if we collapse the boundary right?

Anthony

2:05 AM

@TedShifrin That makes sense, but how do I saaaaay that.

Ted Shifrin

@Anthony: You right down the inequality you get knowing that the intervals must overlap.

Anthony

And also, then why is it that I can cover the rational numbers with zero length... it's so confusing.

Pedro Tamaroff

@FernandoMartin Got it.

Ted Shifrin

Yes, @Anthony. The hint for that is: geometric series.

you mean the boundary of the ball, @Ram?

Anthony

It's so strange though.

Ram

2:06 AM

@TedShifrin, yes, boundary of the ball.

Ted Shifrin

Yes. Because you're modding out by $T1 - B^o$, so the part that's left inside $T2$ is the boundary of the ball in $T2$.

Pedro Tamaroff

@FernandoMartin The books looks cool, I can see myself reading it during winter.

Ted Shifrin

@Anthony: There just aren't many rational numbers :P

I thought you're getting old during winter, @Pedro :D

Pedro Tamaroff

@TedShifrin I am.

Ted Shifrin

Getting old and reading math? Tut tut :D

Ram

2:09 AM

@TedShifrin, yes, I figured it out, but I want some one to confirm this. Thanks for your help.

Ted Shifrin

Sure, @Ram.

Ram

Happy grading, I have to grade from next sem :-(, will be in grad school.

Ted Shifrin

Yes, @Ram, I've been doing it for about 40 years now.

So I'm ready to let you young'uns do it :P

Ram

:)

Ted Shifrin

Just take it seriously and read and give serious criticism/feedback. Don't just put checkmarks on the papers.

OK, back to work for me.

Ram

2:12 AM

@TedShifrin, thanks for suggestion. I never been to Math school before this sem.

Thats really helpful

Ted Shifrin

Really? You're learning algebraic topology and you just started? I'm confused.

user127001

2:24 AM

Hi Ted

I decided against taking differential geometry next semester

I'm rather taking 2 intro analysis courses

Mike

2:53 AM

Hi Ted

I decided against taking 2 intro analysis courses next semester :)

user127001

lol Mike

sea turtles

3:09 AM

best troll I've seen of late

Pedro Tamaroff

"Although the function ... effectively yields the prime numbers at coefficients z1"

Heh.

Jasper Loy

@user127001 2? What do they cover?

Anyone reads Shakespeare here?

sea turtles

did macbeth, othello, midsummer, and much ado in high school. it's on my bucket list to read it all.

Jasper Loy

@seaturtles It's not Old English right? So I should be able to read it right?

sea turtles

It's closer to today's english than old english. One difference is people say trivial things in elaborate ways to show off how clever shakespeare is and to let the audience feel good about understanding it. (Or at least that was my impression.)

Jasper Loy

3:17 AM

@seaturtles I will recommend you one book with all his works then. Ready for it?

sea turtles

I already have his entire works.

In print form.

Jasper Loy

Oh, OK. I am getting one.

sea turtles

not counting any apocrypha

Jasper Loy

I have retired from Math and Eng SE. I will now focus on ELL SE, lol.

Fernando Martin

ping @pedro

Jasper Loy

3:20 AM

Then I will delete my account one day, lol.

@seaturtles Still waiting for you to tell your name, lol.

sea turtles

le sigh

Pedro Tamaroff

@FernandoMartin ?

Fernando Martin

did you think about the problem?

Karl Kronenfeld

@JasperLoy Dissing people whose first name is Sea?

Pedro Tamaroff

@FernandoMartin Nope.

I am doing AM exercises though.

I should sleep.

Jasper Loy

3:22 AM

@KarlKronenfeld I am still confused between you and Arkamis!

Or... miss CA.

The teddie is back.

tb?

No, Ted Shifrin in chat.

Ted Shifrin

Hi @Mike. You mean at grad school?

sea turtles

3:24 AM

oh, the later teddie

Ted Shifrin

Hi @anon

Jasper Loy

Hi @ted, it's too late, lol.

Ted Shifrin

You're always too late for something, Jasper.

Jasper Loy

I wonder what would happen if SE decided to remove all chat rooms.

Ted Shifrin

Hi @user127001: You don't want to take my class? :) it's typically less abstract than analysis.

we'd get more work done, Jasper.

Pedro Tamaroff

3:28 AM

:15150707

Jasper Loy

I am really shocked that his mum threw away all his math books.

Karl Kronenfeld

@PedroTamaroff Nice number, there.

Pedro Tamaroff

@KarlKronenfeld Yes. =)

Ted Shifrin

Yes, Jasper, she sounds unfit to be a parent.

Jasper Loy

@KarlKronenfeld I forgot, are you a grad student?

Karl Kronenfeld

3:30 AM

@JasperLoy nah

Jasper Loy

@KarlKronenfeld Are you gonna be one in future?

Ted Shifrin

Jssper, don't you ask @Karl that every week?

Pedro Tamaroff

@KarlKronenfeld Suppose $M\simeq N$, $A-$modules. $M'$ another $A$-module.

Karl Kronenfeld

@JasperLoy I'd like to be one in the future.

Jasper Loy

Hehe, I have bad memory now.

@KarlKronenfeld OK, maybe we'll meet in future.

Pedro Tamaroff

3:31 AM

Then $M'\otimes M\simeq M'\otimes N$, @KarlKronenfeld.

anorton

Anyone here good with elementary combinatorics? I can't figure out why I'm wrong here, and I'm headed AFK for the night... this is going to bug me; I can tell. math.stackexchange.com/questions/768266/…

(sorry for jumping into the middle of a conversation...)

Ted Shifrin

Mine's definitely getting worse :) Seriously worried about Alzheimers ...

Karl Kronenfeld

@JasperLoy Until then, I'll see you in my dreams, right?

Jasper Loy

@KarlKronenfeld Yes, you learn quickly.

Ted Shifrin

Hi @anorton

Jasper Loy

3:32 AM

@TedShifrin I really hope there is a drug for it.

anorton

Hi

Karl Kronenfeld

@PedroTamaroff Yuppers (tensoring with $M'$ is a functor after all).

Ted Shifrin

Hi again @Karl

Pedro Tamaroff

@KarlKronenfeld I was actually going to give a proof, but nevermind.

Karl Kronenfeld

Hi, @Ted, not sure if I greeted you earlier.

Jasper Loy

3:33 AM

An old woman in the streets stopped there and started peeing. The urine flowed to my shoes.

Ted Shifrin

No harm @Karl

Karl Kronenfeld

:)

Jasper Loy

I think maybe she has Alzheimers.

Anyway, Spivak has written a mechanics book I just realised.

Maybe he will write an entire series on physics for math people.

I lost interest in physics once I realised there was too much math even without it.

Ted Shifrin

His goal 40 years ago was to understand through general relativity. That's what started him writing/learning differential geometry.

Now he's getting old ...

Jasper Loy

If I were interested in physics, I would buy the ten volumes by Landau and Lifshitz.

Ted Shifrin

3:39 AM

I'm going to read Feynman when I retire.

Jasper Loy

It seems to me that Feynman isn't really good for learning physics. It's not a usual textbook.

Ted Shifrin

He's masterful at giving the ideas. I don't want to learn pedantically, inch by inch.

Jasper Loy

I am waiting for someone to write say ten volumes on all of mathematics, similar to Bourbaki but covering almost all topics.

I guess that will never happen, unless I try to do it myself, lol.

Ted Shifrin

Not likely. There is the Russian three-volume survey of math, written 40 years ago. But nothing like Bourbaki ....

Karl Kronenfeld

What would the purpose of that be?

Jasper Loy

3:43 AM

@KarlKronenfeld For an interested student to learn almost all topics in a concise and consistent manner

It's really my dream to have such a series of books.

Fernando Martin

I don't think that is even remotely possible

Ted Shifrin

I concur.

Have you tried learning from Bourbaki, Jasper?

Jasper Loy

By almost all topics I mean that many can be omitted. I think it is possible.

@TedShifrin The level of generality is too high for it to make sense to me. But the general topology was good.

Pedro Tamaroff

@FernandoMartin I'm trying to find an injective $A$ module morphism $A^n\to A^m$ with $n>m$.

Fernando Martin

Well, there's the GTM series

@Pedro: That's the hard problem

Ted Shifrin

3:46 AM

Precisely. You need to be very advanced to learn from it ... Basically a summary for experts.

Pedro Tamaroff

@FernandoMartin Oh?

Jasper Loy

But hey Landau and Lifshitz could do the ten volumes for Theoretical Physics!

Karl Kronenfeld

@PedroTamaroff You can pick your own $A$?

Pedro Tamaroff

@KarlKronenfeld Yeah, I have to find a counterexample.

I was trying $\Bbb Z$. =P

Fernando Martin

I could give a hint @Pedro

Karl Kronenfeld

3:47 AM

:P

Ted Shifrin

Tensoring with $\Bbb Q$ makes $\Bbb Z$ unlikely.

sea turtles

2 mins ago, by Pedro Tamaroff

I was trying $\Bbb Z$. =P

try that some more

Pedro Tamaroff

@seaturtles I think it should work, yes.

sips drink

sips saliva

3:50 AM

love that one

@PedroTamaroff That looks like you.

Pedro Tamaroff

@JasperLoy Of course, I'm bold and noseless.

@seaturtles I think I got it.

Jasper Loy

I think Thai ghost movies are the scariest kind.

sea turtles

that's nice

Jasper Loy

3:52 AM

English ghost movies are the least scary.

sea turtles

@PedroTamaroff with $\Bbb Z$? popcorn popping

Pedro Tamaroff

@seaturtles What is?

Ted Shifrin

Show me, @Pedro. I'm befuddled.

Karl Kronenfeld

I'm befuddled as well.

$\mathbb Q$ is flat, etc.

Jasper Loy

Anyone read Knapp's Basic Algebra, Advanced Algebra, Basic Real Analysis, Advanced Real Analysis? They cover so many topics...

I thought of buying them once, but did not for various reasons.

Ted Shifrin

3:55 AM

Nope.

user127001

Hi @Ted. I had to make a decision between 'number systems and foundations of analysis' and differential geometry and I thought the number systems course would serve me better for my math background

Jasper Loy

@user127001 It's good to do analysis before differential geometry which uses analysis.

Pedro Tamaroff

@KarlKronenfeld Maybe I should try a matrix ring?

Fernando Martin

@Pedro

Ted Shifrin

I'd need to see syllabi and know the teachers to give good advice, @user127001, so I won't. :)

Fernando Martin

3:57 AM

please let me give you a hint

user127001