@Secret what was heather's comment?

I had to show that, when given two numbers, $a\in\Bbb{R}\text{\\}\Bbb{Q}, b\in\Bbb{Q}$, $a+b$ is irrational. I tried Reductio ad absurdum, and assumed $a+b\in\Bbb{Q}$. Then, I assigned $a+b$ to $c$, whereas $c\in\Bbb{Q}$. $\implies a=c-b, c-b\in \Bbb{Q}$, $a\in\Bbb{R}\text{\\}\Bbb{Q}$, which obviously leads to a contradiction. This implies that $a+b\in\Bbb{R}\text{\\}\Bbb{Q}$. Is this proof correct?

@Mr.Xcoder yes

@LeakyNun Thanks! Do you know of any better method?

That's the method I'd use.

My presentation:

Let $a \in \Bbb R \setminus \Bbb Q$ and $b \in \Bbb Q$.

Aiming for a contradiction, let $a+b \in \Bbb Q$.

Since $b \in \Bbb Q$, we have $(a+b)-b = a \in \Bbb Q$.

This contradicts with the assumption that $a \in \Bbb R \setminus \Bbb Q$.

Therefore, $a+b \in \Bbb R \setminus \Bbb Q$.

Indeed, my wording wasn't the best. Yours is slightly simpler.

@LeakyNun I somewhat felt that another variable $c$ was a bit superfluous, thanks for showing a better "format".

no problem

@Mr.Xcoder this is what I'm saying you can use $\subseteq$ or $\subset$ interchangably in one way.

because $\subseteq$ implies that it may be equal.

not always necessarily equal.

@LeakyNun Hi :D

@KasmirKhaan hi

@Mr.Xcoder I prefer just saying contradiction, because Latin words sound so pretentious, lol.

@LeakyNun thought id sleep for an hour so sleepts 8 hours now haha =p

@LeakyNun waking up 8 pm is not nice ><

:o

:o

:D

@Jasper Reductio ad absurdum is a bit more natural to me, as it is the equivalent of Reducere la absurd in my language...

@Mr.Xcoder Oh OK, what is your language?

@Mr.Xcoder you're Romanian :O

@Jasper Romanian

(I read from your profile)

unfortunately I don't know much about Romanian

@LeakyNun Didn't you know already from the nineteenth byte?

@Mr.Xcoder Ah, the Romance languages are so similar that we might as well collapse them all into Italian, the closest to Latin, lol.

@Mr.Xcoder I forgot

@LeakyNun which languages can you speak fluently?

@Jasper Romanian is one of the furthest language from Latin, having Slavic influence

other than mathematics ;)

@Jasper Eugh, I hate Italian (no offence to Italians)

@Dodsy Cantonese, Mandarin, English

wow cool.

@Mr.Xcoder :C

@LeakyNun Different dialects of Chinese

?

they are mandatory in Hong Kong @Dodsy

You have to learn both Cantonese and Manadarin?

hm.

@AlessandroCodenotti no offence to Italians

@Mr.Xcoder yes, but you might like to think of them as different languages

@Dodsy right

From what I know of Mandarin/Cantonese, they are very different, correct?

@Dodsy correct

I've heard that Cantonese is much more difficult to learn.

@Mr.Xcoder If a+b is rational, then a=(a+b)-b is rational, a contradiction. Short and sweet, lol.

@Jasper lol

@Dodsy that's subjective :P

Fair enough. :)

Cantonese does have two more tones than Mandarin

i thought u left dodsy

and syllables ending in -p -t -k

I'll never leave Faust.

@LeakyNun I was chided by two people on the main site for providing an answer that pointed out the bad parts of a proof, while the other answer that simply said the proof was well done got many upvotes. This is one reason I will no longer have an account here.

@LeakyNun Like two different dialects of calculus, mathematics and Physics? :p

@Jasper are you going to leave us?

@Mr.Xcoder ...

@Jasper please don't leave dude.

@Mr.Xcoder it's not really a dialect.

@LeakyNun I already have no MSE account, currently. But I will stay on ELL for a while, lol.

i dont even know what chided means\

@Jasper alright

@Dodsy @LeakyNun Of course that was a joke

oh I see.

@Faust where did you think I left to? when I said "gotta run"?

@Leaky 我說一点儿官话

I had to run to get pumpkin spice latte from Starbucks with gf :S

i dunno i assumed class

thats alot worse than class

yes.

And it's so hot here

:(

@LeakyNun You did not tell me what you think about assigment 2 :D

was it good at least?

@AlessandroCodenotti you typed them yourself? :O

No, not at all.

lol

@LeakyNun yeah, but I'm not sure how correct it is

@AlessandroCodenotti era perfetto

Hey @KasmirKhaan

I lived 10 months in China but I forgot almost everything I knew about Chinese

I think I like set theory stuff, though I've only learned basic notation.

@Faust Sup faust :D

@LeakyNun lol I thought you responded in Romanian at first :DDD

@KasmirKhaan doing my dam hw only really got AA left then i can have a drink

@Mr.Xcoder Leaky greets ppl using their native language many times =p

@AlessandroCodenotti come scribiste le parole?

@KasmirKhaan Well that wasn't a greeting

@Mr.Xcoder By the way, I have watched some Romanian movies before with werewolves, lol, but I can't remember the title.

@Mr.Xcoder how would you say it in Romanian?

@LeakyNun era perfect

@Faust Not a bad plan =P

@Mr.Xcoder sería "era perfecto" en español

Yeah, kinda

@Jasper I remember you told me you were an old man before.

@LeakyNun I have a xiaomi phone that came with a pinyin keyboard installed (scribiste is not written correctly, I'd use "hai scritto" here)

@Jasper is an old man?

@Dodsy Yes, I am very old compared to you folks, but I don't wanna say how old now.

@AlessandroCodenotti thanks :P

But your picture looks like you're $\leq 30$

yeah id guess around 30

@Dodsy Pictures can be deceiving. But note that I did not edit it at all. It's a virgin photo.

I'll be 27 when done my undergrad O_O

hm.

You look youthful, that's a plus.

@Jasper @AlessandroCodenotti nimen neng kan pinyin ma?

@LeakyNun did you ask them if they know pinyin?

@Dodsy which languages do you speak?

@Dodsy Well, I would look much younger if I were not sick. I can only hope that if I get well, I will look younger again.

@Dodsy yes I did

@LeakyNun Can you see phonetic code?

Translate is confusing :S

why nimen rather than just ni?

@LeakyNun very little mandarin, very little german, very very little french.

@Mr.Xcoder "pinyin" is the standard romanization system of chinese

@LeakyNun Wo keyi.

@AlessandroCodenotti perche sono duo persone

oh, I didn't see you tagged jasper as well, sorry

my understanding of Asian people is they look mid 20's until 30 years later then they magically look 50 and some time later they magically jump from looking 50 to looking 1000 +

@Jasper sorry to hear about your health. There's nothing worse than being ill, and we all take our health for granted at some point or another. I sincerely hope you overcome your sickness.

@Dodsy Last year when I went out, some girl asked me for donations for a charity. And she said 'Don't you have school today?' She thought I was 16, lol.

That's funny :)

i have never seen someone in transition from 50 to 1000 so i have to assume it happens over one night

@Dodsy So when are you getting married?

@Faust I have to say that is true for most Asians I know.

@LeakyNun but I knew you asked about pinyin because I saw "pinyin ma?" ma is just basically question mark isn't it? and I knew what pinyin was :D. I really have a limited understanding of Mandarin, probably better at German. My native language is English.

@Dodsy I see

@Jasper maybe after I get my PhD.

Am I right about ma?

it's like a question marker, right?

@Dodsy ja, du bist richtig

:D

@Dodsy Yes, something like that.

Danke!

@Dodsy when u getting ur phd?

at this rate?

the only bits of german i know are from T. S. Eliot

when I'm 60.

hmm was going to offer to race you

but that doesnt seem like a challange

Lookin like 9 years, so I'll be 32.

@AlessandroCodenotti was "sono" right?

the only German i know is Arschloch

nice.

@LeakyNun yes, but "due" is the correct spelling

"Frisch weht der Wind, der heimat zu? Mein Irisch Kind, wo weilest du?" I barely know what the individual words mean, but I know what that sentence means/what it's a reference to

@AlessandroCodenotti oh, right

I see Kind = child

Balarka barely knows German but can utter German poetry :P

and wo = where

that's no german poetry

@BalarkaSen well it rhymes

@Dodsy yeah that much i can extract

@LeakyNun sure but it's a small part of The Waste Land

I like your taste in poetry.

@Dodsy There is the Kinder Bueno chocolate brand.

anyone speaks Bavarian German?

@BalarkaSen do you like keats?

I think it means something like, "Fresh wind blows towards my home, my Irish child where are you?"

@LeakyNun that the same refrence as Bavarian donuts?

no idea

I got "fresh wind" "my irish child where are you"

but wouldn't have gotten blows towards my home

@Mr.Xcoder I searched the history and found that you spoke a phrase in Bavarian German lol

if that's any indication of my level of german.

@Dodsy Somewhat, although I haven't read much by him. I don't like the Romaticism that much

I see.

well I'm off to do some readings.

have a good day folks.

seeya

i guess i should do some hw

@LeakyNun And what does that mean?

you deleted that message though

imgur.com/a/HGRUj

@LeakyNun I had a phone call and lost the pace of the conversation. What was this related to?

was wondering how you knew the phrase

@Abcd do you know slope from ax+by+c=0?

I speak german (somewhat) fluently.

@Mr.Xcoder interesting

@LeakyNun Yes, $-a/b$

und Bayern?

Wie geht es dir? :P

@Abcd so they are parallel, because they have the same slope?

@Mr.Xcoder dies ist nicht Hochdeutsch?

Standarddeutsch, ich weiss nicht die Name

@LeakyNun I understood that, what about that $\lamda$?

@Abcd well that's a parameter that you can change

@LeakyNun It means how are you (doing)? lol

@LeakyNun I realised that wherever there is $\lambda$ in this chapter on straight lines I am getting confused.

@Mr.Xcoder yes, but isn't that standard German?

@LeakyNun it is standard german. Do you want us to speak Bavarian instead?

@Mr.Xcoder wi di mag

@LeakyNun gut.

@Mr.Xcoder also...?

If so, I'd much rather abstain from a conversation in Bavarian. I am not really confident about my skills.

@Mr.Xcoder denn sprich in deutsch :P

i dont understand what ur asking @Abcd

$\lambda$*

@LeakyNun Mir geht es gut. Danke!

brb

@Faust why the characterisation of parallel lines to $ax+by+c=0$ is $ax+by+\lambda=0$

@Faust Hi faust! Just asking what's the need of a lambda in that equation/

Wagwan mandem

what is the definition of parameter in mathematics?

im confused by w.e that says but we know the angle is bisected so it is equal on both sides

for that to be true his result follows directly fromt he defn of that angle and traingle

@Faust we?

whatever

those two triangles have the same side and angle

so the ratio of those sides must be equal

w.e= whatever

its one of euclids proved things

@Faust Oh! I didn't know that the $^o$ sign indicated equality of angles here.

the line says it bisects the angle

they must be equal by defn of bisect

Oh yes! Dumb me! Thanks @Faust

gl ^^

Reposting for others to see too.

@Faust gl?

good luck lol

@Abcd In this case, you should just check an English dictionary.

@Abcd This may not be entirely inclusive, but usually "parameter" refers simply to an input to a function--something that is allowed to vary.

Hi @Fargle

For example, the "t" in parametric equations for curves would be an example of a parameter. It is the input to the function that maps the t values to n-dimensional coordinates.

Hi @Balarka, what's shaking?

Currently nothing but old memes, Riemann surfaces and ODE's :P

@BalarkaSen what sort of ODE?

For now linear

@BalarkaSen then why on earth are you doing it?

why on earth am i doing what

linear ODEs

linear system of ODEs are slightly more nontrivial to understand than just "linear ODEs", but i have just started reading a textbook

so that's why

I see

At least you don't have to consider stochastic ODEs.

Friggin' gross.

if you're going to learn ODEs better start with linear systems than solving y'' + x^2y = 0 or some arbit shit :P

@Fargle Oh you're into those right?

@BalarkaSen Well, trying to be

I lack some fundamental definitions in stochastics

@BalarkaSen ooh, close to my favourite DE :P

its a good exercise problem, but solving random ode's are not really what the theory of ode's is about

there's a lot of nontrivial dynamical meaning to the ode's people actually look at

@Fargle What are those? I have no idea about that stuff

@BalarkaSen Typical ODEs are deterministic, in that there is no "noise" to consider; with appropriate initial/boundary conditions, your solution is determined by the equation.

right

Picard-Lindelof black magickery

In stochastic ODEs, part of your DE is a stochastic process--think Brownian motion for example--and its solution will also be a stochastic process. There's noise to consider.

ah

Hi chat

I have a quick logical question

What do we do when assuming something is true implies it is false and assuming it is false implies it is true?

@SimplyBeautifulArt contradiction.

well, for example?

@LeakyNun But what do we do with it?

@LeakyNun No idea :D

@SimplyBeautifulArt you've proved the theory inconsistent.

this is the fall of naive set theory.

Hm

Okay

or you've made some false assumption before that.

Okay

Cantor's theorem hinges on a similar construction

briefly:

Let $A$ be a set and $P$ its power set.

Assume that there is a surjective function from $A$ to $P$.

Let $B$ be the set $\{x \in A \mid x \notin f(x)\}$

Since the function is surjective, there exists $\zeta$ such that $f(\zeta)=B$

then $\zeta \in f(\zeta) \iff \zeta \notin f(\zeta)$

you haven't proved the theory wrong; you just proved that $\zeta$ doesn't exist.

Yup, okay

thus there is no surjective function from a set to its power set

Group theory puzzle: let $G$ be a group and $H,K \lneq G$. Prove that $H \cup K \ne G$.

(I do have a solution)

I'm forgetting what $\lneq$ is supposed to denote

Would a pithy gloss on the statement be: No group is a nontrivial intersection union.

intersection?

Friday afternoons are apparently bad for my brain

Are H, K intended to be groups themselves?

yes

that's what $<$ means

Yeah, the less than sign means specifically "subgroup of" as opposed to merely "subset of"

I have an argument for finite groups, but I need to think a little longer about the general case.

@Fargle I also have an argument for finite groups

and I also needed to think a little longer about the general case

(finite group is easy)

Lagrange's theorem turns out to be pretty nice.

@Fargle right.

Ohhhhhh.

Hi :)

Clearly there is some element $h \in H$ for which $h \notin K$, and similarly a $k \in K$ such that $k \notin H$.

@Fargle bingo

And $hk \notin H$ because if it were, then $h^{-1}hk = k \in H$, contradicting assumptions.

Similarly, $hk \notin K$. But this contradicts the assumption that $G = H \cup K$.

...spoiler alert.

y u parse it in reductio absurdum

Because I felt like it

>_>

@BalarkaSen how would you do it?

kh $\in G \setminus (H \cup K)$ lol

@Fargle how reductive and absurd

I like proofs by contradiction

if only because they make intuitionists sweat

#triggered

xD

Wagwan Mandem again

@Balarka and @Fargle yo

hi

heyo

There was a slick proof of the fact that holomorphic functions are open maps using Rouch\'e's theorem but I can't remember it

Anybody recall this?

ooh latex acute accent