That second line makes me go [citation needed]

Ill look the command for the set Q

\mathbb{Q}

Okay thanks :D

i.e. Q in math blackboard font

okay =p

the problem i am working on is

for reference, there's also \mathbf (boldface), \mathcal (cursive) and \mathfrak (fraktur font)

x*y =x+y+xy

fraktur is weird so don't worry about it

@SteamyRoot In a test on Vector Calculus, the same lecturer asked us to sketch the field lines of $F = \nabla ln(x, y)$, in a question worth 15 marks, (which all of us were confused about after cause of amt of marks), when the marks came back we all got marked down because according to him, none of us "bothered to prove" $F$ was conservative (even though the question never asked us to prove $F$ was conservative in the first place), not to mention that its conservative by definition or anything

I showed the associatitvity and the identiy

2 sided identity and inverse

I need to show closure

well, if x,y are rational numbers, is x+y rational?

should I say that x*y cannot be equal to -1

yes

how about xy?

Yes =p

@Perturbative Yeah, that just plain sucks :/

yeah. and x+y+xy is therefore rational as well.

but I dont have all the rationals

don't need to

its \mathbb{Q} - {-1}

needs more backslashes

just need to make sure that, for any pair of x,y in Q-{-1} you pick, x+y+xy is in the group as well

oh hmm

Okay ill keep wokring on it :D

and you've shown that x+y+xy is definitely in Q. so all that remains is to argue that it's in Q-{-1}

i.e. if x,y != -1 are rationals then x+y+xy is rational (which we agreed is true) and != -1

So if you can justify why x+y+xy isn't -1 then you're done

So I need to show that x*y is in G for all x,y in G

Right

this part is easy

but the part of x*y cant be equal to -1

of course, if z=x*y then the inverse axiom tells you that y=z*(x')

I need to figure it out

so it's definitely possible to find y such that x*y produces a desired z.

(this is not what you need to prove for closure, but i thought it worth mentioning)

right. so you need to figure out why x+y+xy can't be -1

but if x+y+xy were -1, then 1+x+y+xy would be 0. do you see an issue with that?

Yes we used x and y to be -1 when -1 is not allowed

that's not a clear reason.

it's easy enough to see that, if x or y were -1, then that would indeed be zero.

x*y= -1 iff x=y=-1

why?

also, it's not "iff x=y=-1"

it's "iff (x=-1 OR y=-1)"

yes

:D

meant that =p

one of them can be 0 as well

maybe so, but it's clearly different than x=y=-1

yes =p

anyways. i still haven't heard a clear reason.

if 1+x+y+xy = 0, why must x=-1 or y=-1?

(1+x) (1+y)=0

Exactly!

:D

it looked not clear at first

but when i factored few terms

it was clear :d

yep. hence why I said that A-B=0 is sometimes easier to prove than A=B directly

So x+y+xy = -1 is equivalent upon rearranging and factoring to (1+x)(1+y)=0, and the only way for a product of rationals to equal zero is...?

one of the factors is 0

:D

right. so either x+1=0 or y+1=0, which is exactly what is forbidden by the assumption that x,y are in Q-{-1}

So x,y in Q-{-1} indeed implies that x*y=x+y+xy is in Q-{-1} as well. Hence Q-{-1} is closed under *.

one thing I don't remember is what the name of that group is

i know I've seen it before, but..

I suspect it's related to the following observation: $$\left(1+\frac{x}{xy}\right)+\left(1+\frac{y}{xy}\right)=\left(1+\frac{x+y+xy}{‌​xy}\right)$$

[Random]

abstrusegoose.com/63

oh. or maybe the fact that $\frac{a}{1+x}+\frac{b}{1+y}=\frac{a(1+y)+b(1+x)}{1+(x+y+xy)}$

in which case it's saying how adding reciprocals modifies the denominator

Nice thanks man @Semiclassical

How to type the set Q

and delete one element ?

I did this \in \mathbb{Q}-{-1}

gave error

keep in mind that TeX interprets {, } as enclosing brackets

e.g. \mathbb{Q} is Q done with mathbb font

so if you want to have actual {,} in your text, you'll need to use \{, \}

the slashes tell TeX that you want {,} as symbols rather than as TeX code

ah got it =P

thanks :D

so it works?

\mathbb{Q} -{-1\}

it still gives error damn it

@Semiclassical

you need \ in front of both { and }

(x,y) \mathbb{Q} -{-1\}

you only have it in front of }

okay one second

\mathbb\{Q\} -\{-1\}

like this?

no.

something not right

you only use \{, \} when you want it to display as {}

you still need regular {} when doing stuff like \mathbb

$10^2+11^2+12^2=13^2+14^2$

aaaaaaaaaaaaa

just show me please :D

ill learn from that example

$\mathbb{Q}-\{-1\}$.

thanks =p

remember, you can right-click the text in order to see the mathjax (if it's formatting it)

is there a way to make it type once you got few letters right?

when i do this \tex

I don't know what you mean.

it shows text commend

but I cant click on it with right mouse click

I have to type the whole thing each time

@Wildcard $\implies 10^2+11^2+5^2=14^2$

huh

triple-click it to select the text, then copy that?

or just select the text manually with your cursor and then copy

Got it ! :)

Also, a cute observation:

@Semiclassical Have you see this demo playthrough? This is super-scary.

under this group law, (x-1)*(y-1) = xy-1

ah, P.T.

haven't had the nerve to watch that yet :P

I watched it yesterday and I couldn't sleep

Legit horror

I usually watch versions of those things without commentary

Oh I would personally recommend against that in this case

b/c it's just that unnerving?

Yuuup

heh

i also have a tendency to watch them with the sound off at times

just to not get too creeped out

heh

there's not much jumpscares in there

(and because sometimes I can't find a version without commentary, and that's just annoying)

RIP :(

the whole setup is amazing. it's a loop; you walk along a corridor and everytime you go down a stair it's the same corridor but some things change

indeed. wtf konami

and it gets more and more horrid

@Semiclassical yeah silent hills ftw

ah. i've seen that setup in a few of the things I've seen

and it is a pretty effective tactic

(which they indeed probably stole from PT. it inspired a lot of imitators)

Have you watched any playthroughs of RE7?

I really hope Death Stranding will be good

According to Wikipedia: "The Schwarzschild radius and Dirac equation – relating to the theory of relativity and quantum field theory, respectively – appeared on dog tags worn by Reedus's character in the teaser."

$(i\not{\partial}-m)\psi=0$

bah

@Semiclassical Yup

Good game

@Semiclassical Can I send you my texstudio homework , so you can correct my commands? I m having really hard time with them

bleh, that's not great looking

no.

:(

not going to do your HW for you.

Not the homework

just the texstudio thing

I did the actual problems

But they dont look professional on pdf

if they want you to write up your work in TeX, then correcting commands is HW as well.

hmm it was optional

we can handwrite them as well , but hmm they prefer tex

and after all I need to learn that thing sooner or later

then struggle through it :/

you're not going to learn except by doing

^

Read the errors in the log, and learn what they mean and how to fix them.

All righty =p

been 2 days now on that thing

its a nightmare , my eyes hurt ><

are you looking at example text? that should help a lot

It takes a lot of time and effort at first. All pays off in the end, though.

Ehm I do that but its not allways clear what they mean

yes something like that

So I can see how some commands are written

Hello, lets say I have a number and I want to round it to the nearest tenth. 1.946

Would come out to 1.9 correct.

Yes

Smaller than 1.95 rounds to 1.9. Larger than 1.95 rounds to 2.0.

Is there any variation of rounding where 1.946 could possibly turn into 2.0? Like this -> 1.95 -> 2.0

@Trevor If there is, I haven't heard of it. It probably wouldn't be very useful.

Okay thank you

If you draw it out on a number line, you'll see that 1.946 is closer to 1.9 than to 2.0. @Trevor

I didn't think so, but I got confused because an internal system seems to be calculating that number as 2.0 instead of 1.9

Right that makes sense

Must be a bug

Appreciate the help. I'll look further into it then.

Hi again yall

How to write a good proof?

I need to show that the set of a+bsqrt(2) forms a group under addtion

I need to check associativity and closure those are easy

should I show that A*e= e *A for all A in G

and that AB =BA

I mean How much detail should I put

So, this is why office moves are awesome: free books

They had a cart of them downstairs, and I managed to come up with (among other things)

VI Arnold's "Math Methods of Classical Physics", and his book on geometrical methods in ODEs

Whittaker and Watson

Saunders Mac Lane's book on category theory

and Bishop-Crittenden

oh, and a copy of Gradshteyn and Ryzhik i.e. a huge book of integrals/series tables

I'm supposed to be reading geometric methods in ODEs

neat

Arnold's is this book, to be clear: books.google.com/…

His math methods book contains a chapter about halfway through on differential forms

Can anyone show me a proof that $i^i = e^{-\frac{\pi}{2} + 2\pi k}$? I feel like an idiot but I don't follow the $2\pi k$ part.

sup chat

in fact, it's the first chapter in the part of the book he does on Hamiltonian mechanics

@Semi that book looks neato

and the chapter after that is "Symplectic manifolds"

so that should give a sense of his priorities

it's a pretty good find, yeah @EricSilva

okay, time to go

@SirCumference $i = e^{\left(\frac{\pi}{2} + 2\pi k\right)i}$ for all $k \in \mathbb{Z}$

I tried to read the mathematical methods in classical mechanics at some point but I got sidetracked

I wanna pick it up at some point and try again

Hello!!! Let $b_1< b_2< \dots< b_{\phi(m)}$ be the integers between $1$ and $m$ that are relatively prime to $m$ (including 1), and let $B=b_1 b_2 b_3 \cdots b_{\phi(m)}$ be their product. How can we show that either $B \equiv 1 \pmod{m}$ or $B \equiv -1 \pmod{m}$ ? Could you give me a hint?

@SteamyRoot That...doesn't help

I'm stuck on where the $2\pi k$ part comes from

Because, if you rotate by $2\pi$ radians, you end up on the same spot again.

@SteamyRoot ...wow

Somehow that wasn't hitting me

@Trevor Maybe it's rounding in one place to the nearest hundredth and then later rounding to the nearest tenth. (A bug, as suggested.)

@Evinda for Googling purposes, those phi(m) numbers b_n are known as the totatives of m

So you want to show that the products of the totatives is one integer away from being a multiple of m

I think my hint would be: Every totative of m is a unit in the ring Z/mZ.

Sigh, MathJax isn't showing up

$\text{Test}$

Displays fine for me, so we can understand it at least

Wait, I might've fixed it: $\text{Test}$

Nope

One last try: $\text{Please work}$

Sigh...

@Semiclassical Can you try sending me a message with MathJax?

Gas is getting kinda expensive and tricky to find (and it's a long line when you do find it) after the hurricane

It'll reopen hopefully on Wednesday but like eek