12:30 AM
@Rithaniel looking at this, I think you can avoid the axiom of choice: en.wikipedia.org/wiki/…
At least if the subset is supposed to be finite

Okay, but you would need the axiom of choice if the subset were infinite?

1:08 AM
"every infinite set has a countable subset" needs the axiom of choice
"every finite set has a subset of a given lesser cardinality" does not

2 hours later…
2:46 AM
@LeakyNun how about "every set has a finite subset"? (I'm guessing AoC indeed isn't needed)

@Semiclassical just take the empty set :)

3:07 AM
Why is there no ERO that corresponds to substitution? More generally, why is substitution left out of Linear Algebra entirely?

@LeakyNun pffffft
"every nonempty set has a finite subset"

yes, take the empty set
it's finite and it's a subset

"every infinite set contains a subset of cardinality $n$" for a given natural number $n$

that doesn't need AoC
finite choice is a theorem of ZF

yeah, I thought it didn't need AoC (that's what I was asking for)
but I don't know how to prove anything in ZF, soooo

4:00 AM
Hi I am high school student solving problems. On permutation and combinations, a problem is in party there are 5 Russian and 5 British couple, we have to find total no of shake hands if Russian women will. Not hand shake with any man, and any women will not shake hand with her own husband
What I do is total hand shakes are 190,i minus shakes hand between Russian women and British man that is 25 - shake hands of us Russian women with her own husband, - shake of British ladies with their husband am I right or I missed an case will anybody help me

2 hours later…
6:32 AM
not your usual zoom

6:44 AM
It amazing but I love power of tau

Hmm interesting, there seemed to be a lot of islands floating in a void
I wonder if it is because of those branch cuts mentioned in the description

6:58 AM
Last night dream, the following variational equation were seen, where $C$ is some nonlinear operator
$$C^2 - \lambda Cx = 0$$
The instruction is to pick a suitable $\lambda$ in order to minimise the operator norm of $C$
Reality check: From the look of it, it might seems it is a straightforward quadratic equation. However the equation is actually:
$$C \circ C - \lambda C \circ x = 0$$

Yes

And since $C$ is nonlinear, $C\circ (f-g) \neq C \circ f - C\circ g$, you cannot factorise it
Worse $C \circ \lambda f \neq \lambda C \circ f$ where $\lambda$ is some complex number

7:22 AM
Secret have you seen my question

7:56 AM
hello guys
if I have calculated divergence of a vector field in Cartesian coordinate then will the divergence change if I were to convert the vector field into cylindrical coordinate system and then calculate divergence?
I think it won't. But can anyone confirm?

1 hour later…
9:00 AM
@yuvrajsingh uh... I am not terribly good at solving handshake problems, you might need to find someone else to help you
All I can say it is a variation of the famous handshake problem
2

I need help with this problem. The problem goes like this: In some countries it is customary to shake hands with everybody in the meeting. If there are two people there is 1 handshake, if there are three people there are three handshakes and so on. I know that the formula is $\dfrac{n(n-1)}{2}$...

5 hours later…
2:12 PM
$$C^2 = \lambda C x$$
Let $x_0$ be a fixed point of $C$. Then:
$C^2(x_0) = \lambda C (x_0)x_0$
$x_0 = \lambda x_0^2$
$x_0 = 0 \lor \frac{1}{\lambda}$
Hence $\lambda \neq 0$
and hence $C \neq 0$ thus excluding trivality
Let $C (x_1) = C(x_1 + c)$ for some $c \in \Bbb{C}$. Then:
$C^2 (x_1) = \lambda C (x_1)x_1$
$C (x_1 + c) = \lambda C (x_1 + c)x_1$
$C (x_1 + c) + \lambda C (x_1 + c) c = \lambda C (x_1 + c) (x_1 + c)$
$C (x_1 + c) + \lambda C (x_1 + c)c = \lambda C^2(x_1 + c)$
$\lambda C^2 (x_1 + c) -C (x_1 + c) (1 + \lambda c) = 0$
ok nvm that is wrong
$C^2 (x_1+c) = \lambda C (x_1+c)x_1$
$C^2 (x_1+c) + \lambda C (x_1+c) c= \lambda C (x_1+c)(x_1+c)$
$[C^2 (x_1+c) - \lambda C (x_1 + c) (x_1 +c)] = - \lambda C (x_1+c)c$
$\lambda C (x_1 + c)c=0$
$\lambda,c=0$ tell us nothing, thus dividing $\lambda, c$ gives
$C(x_1+c)=0$
or there exists $x_1$ such that $C(x_1)=0$
meaning $C$ has at least one root
Now, I don't know of any more advanced techniques in nonlinear functional analysis to continue this without flooding the chat thus I will pick it up later
One interesting question, perhaps, is to ask what happens when $C$ is compact
I will be surprised if the literature have few examples of these quadratic looking nonlinear functional equations

2:41 PM
Can anyone help me in one chemistry question please?

@Sonal_sqrt The divergence operator is invariant under coordination transformations

Sep 17 at 14:49, by Math Geek
chemistry rooms are very inactive
Sep 17 at 14:42, by Mike Miller
This isn't a chemistry channel and we're not here to do your homework. What is wrong with you?
And I Must Spam
Never-ending cycles of help and being helped phenomena in chat rooms -> frustration

@Secret sorry i won't do that again. I was just seeking some help.it's a doubt not homework help.Anyways i'm really sorry. i won't disturb anyone of you.Goodbye

3:12 PM
@mathgeek there's only one chemist who frequents this chatroom and he's not often here so I'd say you're better off posting on a main page for chemistry
if there is one

^ @ loong
In other news, it turns out this particular question is quite interesting. I just don't get why some of the interesting questions in the world are asked like a homework format
I really want to (censored) all help vampires in this world for corroding the education sphere so these questions can be given justice

3:28 PM
@Secret whats new, playing any games lately? :)

Nothing much really, mostly PhD and politics
If anything, last night dream caused me to done some research on nonlinear functional equations, some of the analysis were done a few messages above

was at a VR startup last nite got ½ thru cyberpunk great game/ graphics, VR escape room, kinda like blade runner + matrix, but very hard. HTC vive pros, highly recommend it :) vrescape.arvilab.com/cyberpunk vrescape.arvilab.com

1 hour later…
4:51 PM
Hi, I have a small question. I have a theorem that says: you have n linear equations modulo 2 with three variables per equation and with at most k occurrences of each variable. So, I did not write all theorem since it is another thing. But what I want is just to give example for what it said. e.g. suppose k=3, n=4. Then I don't know how to make 4 linear equations such that each linear equation has only three variables and they appear at most 3 times?
I mean you cannot do it. I feel I miss something

1 hour later…
6:00 PM
@Ultradark what are you learning in algebra?

@ShineOnYouCrazyDiamond permutation groups

Nice
$S_n$
You'll probably know more than I do, once your done with this section

What rule says that you can rewrite $$\frac{\sqrt{37}}{\sqrt{18}}$$ to $$\frac{\sqrt{74}}{6}$$ ?

multiplying by a fraction equivalent to $1/1$
You have to reason and find which one you want to choose
I'm guessing $\sqrt{2}/\sqrt{2}$
Then they simplified the $3^2$ in the bottom

Ahh, I see now, thanks!

6:10 PM
nvm, I was going to give you a nice link but can't find it

I never thought about multiplying a fraction with a fraction containing squareroots in order to simplify the fraction in focus afterwards ... that's clever!

That's a cool site
I want to prove that one-way functions exist
:)

6:31 PM
henlo

6:44 PM
henlo

Is there a quick way to figure out the full list of elements in the group generated multiplicatively by $\left( \begin{array}{cc}2 & 0 \\0 & 1\end{array} \right)$ and $\left( \begin{array}{cc}1 & 1 \\0 & 1\end{array} \right)$?

Powers of those

if you prove one way functions exist you prove that $P \ne NP$

Yeah, but conjugation gives you something different. Like $\left( \begin{array}{cc}2 & 0 \\0 & 1\end{array} \right)^{-n}\left( \begin{array}{cc}1 & 1 \\0 & 1\end{array} \right)\left( \begin{array}{cc}2 & 0 \\0 & 1\end{array} \right)^{n}=\left( \begin{array}{cc}1 & 2^{-n} \\0 & 1\end{array} \right)$
Which is not a power of either.

rofl fair
I was being facetious

6:52 PM
Ah, gotcha :P
I suppose it's just going to require some exploration.

If you're trying to prove one way functions exist you are probably wasting away the rest of your natural life (given you stick with the problem that long). This is a high risk high reward venture. Why not try something a little more tractable?
All I know is that I wake up one shoe at a time just like the rest of us

7:26 PM
Perhaps dumb question: If I have a sequence of numbers $x=(x_1,x_2,x_3)$ is it the same as writing an infinitely long sequence $y$ with infinitely many zeros appended to the end, $y=(x_1,x_2,x_3,0,0,...)$?

@jonem Not the same. But it many cases they can be considered equivalent

Thanks!
@TobiasKildetoft Do these sequences have a name? Or how is this modification referred to?

8:16 PM
(removed)

8:31 PM
in Jokes, Sep 7 at 5:47, by Martin Sleziak
Q: When did Bourbaki stop writing books? A: When they realized that Serge Lang was a single person.

### Jokes

fun
Everyone is welcome^

8:44 PM
So, I have a set $\{\sum\limits_{x_i\in X}n_ix_i|n_i\in\mathbb{Z}\}$ where $x_i\in\mathbb{Q}$ and $X$ is finite. What must I claim before claiming that this set has a least positive element?

Are you trying to prove that every finitely generated subgroup of $\Bbb Q$ is cyclic?

Well, no, I'm trying to prove that a particular set of matrices is not finitely generated.

Namely, matrices of the form $\left( \begin{array}{cc}1 & a \\0 & 1\end{array} \right)$

you can rescale your set such that it's contained in $\Bbb Z$: just multiply all elements in $X$ by a common denominator

8:49 PM
Where $a\in\mathbb{Z}[\frac{1}{2}]$
That works

thus reducing the statement to $x_i \in \Bbb Z$
then you just need to know that subgroups of cyclic groups are cyclic
or that $\Bbb N$ is well-ordered I guess
the group $\{\left( \begin{array}{cc}1 & a \\0 & 1\end{array} \right) \mid a \in \Bbb Z[\frac{1}{2}]\}$ is isomorphic to the additive group of $\Bbb Z[\frac{1}{2}]$
you can just note that for every f.g. subgroup of $\Bbb Q$, the denominators of the fully reduced forms are bounded
but they are unbounded for $\Bbb Z[\frac{1}{2}]$

Also, I probably shouldn't outright say that $a\in\mathbb{Z}[\frac{1}{2}]$ as I've only show inclusion in one direction: That $a$ is in a subset of the rationals that has $\mathbb{Z}[\frac{1}{2}]$ as a subset.

what exactly is your group of matrices?

Given that $G$ is the group generated by $\left( \begin{array}{cc}1 & 1 \\0 & 1\end{array} \right)$ and $\left( \begin{array}{cc}2 & 0 \\0 & 1\end{array} \right)$ under multiplication, the group I'm considering is the subgroup with the main diagonal both equal to 1.

I think that subgroup should be exactly $\{\left( \begin{array}{cc}1 & a \\0 & 1\end{array} \right) \mid a \in \Bbb Z[\frac{1}{2}]\}$. For the other inclusion, note that both generators and their inverses have entries in $\Bbb Z[\frac{1}{2}]$

9:01 PM
Is anyone here up for an integration question?

and $\Bbb Z[\frac{1}{2}]$ is a subring of $\Bbb Q$, so $M_{2 \times 2}(\Bbb Z[\frac{1}{2}])$ is a subring of $M_{2\times 2}(\Bbb Q)$

Well, ultimately the only thing I need to show is that this subgroup isn't finitely generated.
But that would be good practice.
Something along the lines of "any entry of a matrix in $G$ must be a linear combination of $\frac{1}{2}$, $1$, and $2$, thus any $a$ would have to be contained in $\mathbb{Z}[\frac{1}{2}]$?"

you also have $\frac{1}{4}$ in there
but if you say linear combination of (possibly negative) powers of $2$, then you're fine

Ah, yeah, "powers of two" would be the way to express it.
Danke for the help, by the way. Doing free abelian groups in abstract, this week.

9:42 PM
hiya again

Is there a name for the trivial rule of inference $\frac{A}{A}$?