Mathematics

6:16 AM

@SohamChowdhury Doesn't seem to have any geometric motivation behind inert primes there in the answer though.

user227867

@BalarkaSen Have you decided which university to go to for your studies?

Nope.

user227867

I can't believe I got 2 stars for that stupid comment, lol.

It's a comment I also agree with.

user227867

I like some Indian authors, like V S Sunder, R Narasimhan, M Ramachandran, and S Varadarajan.

user227867

6:24 AM

The problem is that many of their names are too long for me to remember. =)

Indeed.

user227867

At least B Sen is easy. =)

"BS" in short.

user227867

Like Brian Scott.

I had a different de-concatenation in mind.

Hi @Huy

user227867

6:30 AM

Like Bull Shit.

hey @Balarka

@JasperLoy Exactly

user227867

I think math degrees should not be called Bachelor of Science but of Arts instead, since math is not a science.

@Huy What are you working on?

@BalarkaSen: reseasoning my cast iron skillet. just woke up, having coffee, afterwards brunch at my parents' place

and in the afternoon, I'll prepare some more geometry for my students, then probably take a break

so mathematics-wise, not much at the moment, sorry to disappoint

maybe I'll study a bit about quaternions to see if there's something interesting that I can show the kid that I didn't know yet

user227867

6:33 AM

@Huy Also mention octonions.

8:07 AM

hi

8:29 AM

hiiiiiiiiiiii

user227867

Hi @usukidoll, lol

can someone look over one of my proofs? I think I got them but the way I worded it is crap

user227867

If the way you worded it is wrong, then maybe the proof is wrong.

I got this
If $a \vert b$ and $c \vert d$, prove that $ac \vert bd$\\
By the Divisibility definition, a divides b if $b=ax$ for some integer x and c divides d if $d=cy$ for some integer y. Multiplying b and d, we have, \\
$(ax)(cy)=bd$\\
$(ac)(xy)=bd$\\
Hence, $ ac \vert bd$\\

user227867

Anyway, I am only a banana.

user227867

8:34 AM

@usukidoll You mean ac in the second line.

huh?

user227867

ac divides bd in the second line you typed

oh shit typo

user227867

Yeah, shit, lol

now it sounds legit

user227867

8:36 AM

@usukidoll Just mention that xy is integer. That is it.

like since xy is an integer, multiplying b and d results in
(ax)(cy) = bd
(ac)(xy) = bd
Hence ac divides bd?

user227867

@usukidoll You should mention it just before the conclusion.

ef

errr
UGH!

user227867

@usukidoll I will show you how I would write it.

k

user227867

8:41 AM

@usukidoll Since $a$ divides $b$, $b=ax$ for some integer $x$. Since $c$ divides $d$, $d=cy$ for some integer $y$. Now $bd=(ax)(cy)=(ac)(xy)$ and $xy$ is an integer. Hence $ac$ divides $bd$.

user227867

Is that alright @usukidoll?

yeah

user227867

Do you think you should thank me now?

yeah gives chocolate frog

Mats Granvik

9:57 AM

10:13 AM

Hey

r9m

I am not particularly good with these stuff ... a friend asked :) so asking for help here: math.stackexchange.com/questions/1906134/… thanks! :)

hey @usukidoll .. I see you are a weird frog avatar! :P

I think it's a Pokémon.

bulbasuar

bulbasaur

u there?

Manolis Lyviakis

10:30 AM

hey

fellow numberfriends

What if you only like topological spaces?

what's the gcd(3,5)?

1

really?! well I was dividing a lot and got 1

but when I did the first part I got 2

Isn't it quite obvious? They're both primes.

10:34 AM

like 3 divided by 5 but the remainder was 2...
then I had gcd(2,3) did this again and got a remainder 1

well I was using EUclidean ALgorithm

Show by example that if $au+bv=d>1$, then $(a,b)$ may not be d.\\
I let a =3, u=4, b =5, v = 6
(3)(4)+(5)(6)=12+30=42
but gcd(3,5) is 1

And?

that's it?!

42 is greater than 1 but I highly doubt that gcd(3,5) produces 42

What do you mean with (a, b)?

ya

._.

?

10:39 AM

my d turned out to be 42 when I let a =3, u = 4, b =5 v = 6

But what do you mean with the notation (a, b)?

but gcd(3,5) is 1 so (a,b) isn't d which was 42.. right?!

like gcd(a,b)

it's the notation from abstract algebra an introduction 3rd edition

they left out the gcd

Not everyone is familiar with the notation used in your book. It's good to be unambiguous.

But doesn't Bezout's theorem contradict what you are saying?

O_O!

so am I wrong?

Well take $a = 2, \; b = 4, \; u = -1, \; v = 1, \; d = 2$.
$$(2)(-1) + (4)(1) = gcd(2, 4) = 2$$

10:58 AM

Show by example that if $au+bv=d>1$, then $(a,b)$ may not be d.
MAY NOT BE D THOUGH!!!!

so I'm proving that by example

Ooh, misread that, my mistake.

my D was 42 but when gcd(3,5)= 1
$ 1 \neq 42$

You can easily construct a solution that fulfils your condition.

a=3,u=4,b=5,v=6

12+30=42 but gcd(3,5) = 1 whoops not equal

Let $a, b \in \mathbb{Z}$ and let $u, v$ be integers such that Bezout's relation $au + bv = gcd(a,b)$ is fulfilled. Let $d = 2 \times gcd(a,b)$ then $$2au + 2bv = 2 \times gcd(a,b) = d$$ Then your solution is the quadruple $(a, \; b, \; 2u, \; 2v)$.

11:04 AM

but I'm wondering if what I did was right?! since 42 is d but the gcd(3,5) was really not d

It is clear that $d > 1$.

yeah

Yes, it's correct.

yay ^_______^

11:18 AM

the sum of three squares
$n \equiv 3(mod4)$
I think $3^2+3^2+3^2=27$ which is 3 in mod 4

@Huy Ah, I see.

Busy with teaching, I see.

quite a bit, mostly because I've never taught 9th graders before, so I need to get some experience how much will be enough for how long

I kind of underestimated them at first, so now I'm trying to adapt

I have this strange experience in here that a generic 9th grader is more mature (in the sense of being more serious about learning, actively thinking more and asking questions, quick at understanding, etc) than a generic 11th grader.

11:34 AM

no, it's completely the opposite here

Might be a coincidence because I have only actively interacted with 3 batches (one being my own, when I was a 9th graders, the next one, and the one now - I am in 11th now)

mostly because the school where I work at is for 9th till 12th grade, so 9th graders are new and very self-confident and believe they are the smartest kids alive, but after 1-2 years at the school, they realise how much they still have to learn and become much more humble

Oh.

and since most teachers just do revision at the beginning to make sure everyone's on the same level, they are convinced they know everything already

Yikes, yes, dealing with overly excessive self-coincidence is a bit of a trouble.

11:36 AM

yes, I'm glad I don't only teach 9th grade this year

Is the quaternion guy from 9th?

yes

I have a bad feeling about his eagerness to learn quaternions then

same

I'll find out soon enough

pro-tip: just don't be the guy who's teaching us physics and says eagles fly at escape velocity

just a random suggestion. might be useful

11:43 AM

I'm not teaching them any physics any time soon anyways

I need a light bulb. Can't solve the problem I want to.

buy Philips Hue. they make you smarter.

you should send an ad-proposal to Philips on that theme.

user227867

12:10 PM

@Huy You should not teach quaternions to a 9th grader. If he is so inclined, he can learn it himself, lol.

@JasperLoy Why not?

If someone is genuinely interested, one should not discourage - rather encourage - based on intellectual maturity, IMO.

user227867

@BalarkaSen If a 9th grader is talented enough to learn quaternions, he can learn it himself. If he isn't talented enough, then he has no business learning it.

I do not agree with that point of view. One doesn't need to be talented to be interested in learning something.

user227867

If one cannot understand 1+1=2, he should not try to understand 2+3=5.

user227867

I feel that too many people are trying to learn things beyond them these days.

12:20 PM

Maybe he does understand complex numbers and so on. Maybe he does understand the definition of quaternions but wants to know what it's good for.

I don't think one should deny outright to help someone learn something beyond the textbooks. This is the kind of attitude which discourages most people to learn anything at all, especially in high school.

user227867

Then he can learn it himself, like I said.

No, I don't think this is true.

user227867

For me, I learnt everything on my own.

E.g., I am interested in learning quantum mechanics. I probably can't learn it myself without significant guidance.

user227867

Another problem is finding the wrong guidance.

user227867

12:25 PM

@BalarkaSen I think you can. You are genius.

I tried to put some effort in understanding the basics. I don't understand everything satisfactorily well enough in the little of the book I have read.

user227867

Yeah, we need to read a few books sometimes to see things. One needs a good book, and one needs a good teacher.

user227867

Not all books are good, and likewise not all teachers are good.