Mathematics - 2013-03-25 (page 2 of 3)

Thomas

17:04

Does anyone know: Is there a limit on how much reputation one can gain in a day?

CBenni

There was a discussion on Math.SE meta once

Jayesh Badwaik

@Thomas 200 from upvotes, accepted answers and bounties do not have any limits I guess. It is fully documented somewhere onthe site.

CBenni

you might want to look there

but it is unbounded by the number of bounties

Thomas

Ok, so what happens to the points? So they just never get awarded?

CBenni

yeah

you just lose them

Thomas

17:06

Thats not cool ...

CBenni

well

Thomas

How many hours left of today?

CBenni

nobody gets 200 upvotes in 1 day

Jayesh Badwaik

UTC

CBenni

7h

Thomas

17:06

I just hit 245 points for today on accepted and upvotes

CBenni

woah

Thomas

And I have missed out on 3 upvotes since then

CBenni

I would love to get 14 upvotes for a simple answer about quantors

wtf

I really do not understand Math.SEs vote distribution

Thomas

Yes, I don't get why there is a limit.

CBenni

probably just proportional to the users' current rep

Tobias Kildetoft

17:09

@CBenni generally, short answer that most people can understand, to questions most people can understand, get the most upvotes

Thomas

Oh well there probably is some reason

Yes, sometimes you see these long answers that actually never around to answering the specific question that the OP has.

CBenni

Well the reason is that people who create fake accounts and spambots do not modify their rep to infinity

Thomas

Ok

CBenni

because at some point you get really massive priviledges

Tobias Kildetoft

@Thomas no matter how active you are and how many questions you answer and ask, it would not be desirable for people to gain various moderator privileges until they have been using the site for an extended amount of time

CBenni

17:10

just imagine a 20k spambot

or someone who deletes thousands of posts

Thomas

@TobiasKildetoft: Yes, I understand completely. But why not just put a timelimit on each level up?

Tobias Kildetoft

The answers I have gotten the most rep from are generally the simplest ones

4

@Thomas no idea

CBenni

usually I am too slow on answering because I try to write the answer as detailed as possible

Thomas

Oh well. It is all good. It had just never happened to me before.

Tobias Kildetoft

@CBenni don't let that stop you from answering

Jayesh Badwaik

17:12

@Thomas The factors combine, the desire to not have spambots + desire for extended time periods etc. If you can achieve stuff with as general rules as possible, then it is probably better. Actually, I am in support of weighted rep, for the first ten votes, you should get 10 points each, then for next 10 votes, 9 rep points and so on.

CBenni

and then some 10k rep user comes by, writes some 2 line stuff, and gets 22 upvotes and accept

Tobias Kildetoft

@CBenni this is why I tend to mostly answer question I know not a lot of other people will be able to answer, even though it usually means at most a few upvotes if any

Thomas

@CBenni: I am definitely not fast. Sometimes I do get upvotes from simply answering the question as short and sweet as possible. Other times I notice a lot of short answers with little explanation and hten it makes sense to add a longer answer

CBenni

well I am only a maths student, Idhafc about most of what is written on this page ;)

well, actually I dont really care about votes+rep

as soon as I reach 1k I am happy ;)

Thomas

Yeah, you can't/shouldn't get too focused on the points :)

CBenni

17:14

I really wouldnt care if it werent for the privileges

Thomas

but at 1k you want 2k, then you want 5k then you ....

CBenni

lol :D Well no, I wouldnt start moderating the crap out of this page

however, I like the established user thing

Thomas

Oh well I will go again.

CBenni

cu

Tobias Kildetoft

@Thomas at 2k you want 3k, as that is where you get to start voting to close/reopen

user2175923

17:21

Hello, can anyone help me with a chinese remainder theorem question?

Tobias Kildetoft

@user2175923 probably

user2175923

I have X-k mod a = 0 and X+k mod b = 0 and I know a, k, and b. I wish to solve for X such that (X-k) and (X+k) are both positive

I've been at this for a very long time now and have had extraordinary difficulty getting a direct answer

Tobias Kildetoft

@user2175923 well, given any solution $X$, you can add any multiple of $ab$ and get a new solution

so add a large enough multiple to make it positive

user2175923

okay, but how to solve for X to begin with?

Tobias Kildetoft

by using the chinese remainder theorem (I assume that $a$ and $b$ are coprime since you mentioned it)

user2175923

17:30

they may not necessarily be coprime but it can be assumed there is a solution

i know how to solve it when they're coprime

Tobias Kildetoft

@user2175923 then start with something that solves the first one and keep adding multiples of a until you get something that solves the second one

user2175923

i believe my case is the same as finding the CRT of k and -k modulo a and b correct?

Tobias Kildetoft

if you start with something positive, you will then end up with something positive

CRT?

user2175923

chinese remainder theorem

X = k mod a and X = -k mod b

Tobias Kildetoft

CRT is not something you find

user19161

17:32

@Karl'sstudents That is a nice version sung by the composer himself it seems.

user2175923

I meant finding X via CRT

but still those systems are equivalent yes?

Tobias Kildetoft

yes

user2175923

okay, so given that system, I am after a systematic way to solve for X

i know i can always eyeball it etc but I am writing a program

κρανίοπεριπολία

Hi @Charlie @Karl'sstudents @JasperLoy

Tobias Kildetoft

@user2175923 what is your systematic way when the numbers are coprime?

user2175923

17:34

Chinese remainder theorem

The Chinese Remainder Theorem is a result about congruences in number theory and its generalizations in abstract algebra. It was first published in the 3rd to 5th centuries by Chinese mathematician Sun Tzu. In its basic form, the Chinese remainder theorem will determine a number n that when divided by some given divisors leaves given remainders. For example, what is the lowest number n that when divided by 3 leaves a remainder of 2, when divided by 5 leaves a remainder of 3, and when divided by 7 leaves a remainder of 2? A common introductory example is a woman who tells a policeman that...

see: case of two equations

user19161

@κρανίοπεριπολία Hey hey.

κρανίοπεριπολία

wazzup, pal?

Tobias Kildetoft

@user2175923 hmm, I guess one could try to generalize that based on the assumption that some solution exists

user19161

The Bridge on River Kwai is up.

Tobias Kildetoft

but it is probably about as fast to just add multiples of $a$ to a solution to the first until you get a solution to the second

κρανίοπεριπολία

17:37

@JasperLoy Did you watch it?

user19161

@κρανίοπεριπολία Only the video you linked to.

user2175923

So I would just solve for (k*b*inverse(b,a) + (-k)*a*inverse(a,b)) % (a*b)

but I can't do that if a and b are not coprime

how would I modify this?

Tobias Kildetoft

@user2175923 not sure, apart from doing it the long way I said

user2175923

what is the "long way"?

Tobias Kildetoft

the way I just described

user2175923

17:40

you said "then start with something that solves the first one and keep adding multiples of a until you get something that solves the second one"

i don't understand what you mean by this

Tobias Kildetoft

@user2175923 you know how to find a solution to the first one?

Chris's sister and pals

Greetings

user2175923

X = k modulo b? it seems like I could just set X = b+k, and then you are saying "keep adding b to X until it satisfies the secocnd congruence" right?

Tobias Kildetoft

@user2175923 well, until it satisfies the first congruence then, but yes

also, starting with $X = k$ seems more practical

caveman

17:57

hi @Chris'ssisterandpals

user2175923

18:42

still trying to figure out this problem, can anyone assist?

jtm22

Hey does anyone here know Discrete?

I'm trying to see if I'm getting this answer right

anyone here?

caveman

@user2175923, wha is the problem

@jtm22, what isthe problem

user2175923

1

Q: Using Chinese Remainder Theorem for two equations with non-coprime moduli

I have \begin{align} x & \equiv a\mod m \\ x & \equiv b \mod n \end{align} Normally I would solve for $x$ by doing $(an\cdot \operatorname{inverse}(n,m) + bm\cdot \operatorname{inverse}(m,n)) \bmod mn$ but I am not sure how to modify this for when $m$ and $n$ are not coprime. Assume a s...

number-theory

caveman

@user2175923 it is answered

user2175923

I can't get anyone to explain this thing in a simple, easily understood way that I can actually use

It's not answered in a way I can use

caveman

18:47

@user2175923, oh right, you want to know about CRT when not coprime.

user2175923

It's totally different notation with a bunch of extraneous detail that's only confusing me

jtm22

How many ways are there to distribute 5 balls into 3 boxes if the balls are labeled but the boxes are not? Would this just be 1 way?

caveman

@user2175923, do you know how to derive the usual CRT?

user2175923

Aaaghhhh!!!

caveman

@jtm22, i've got no idea you could probably just find that on google though

jtm22

18:49

I can't

user2175923

@jtm22 sounds similar to the en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) problem

caveman

@jtm22, since the boxes aren't labelled I'd label them A,B,C and make sure A>=B>=C

user2175923

try both of those combinatorics and see if one fits (k-1 choose n-1) or (n+k-1 choose k) or (n + k - 1 choose n - 1)

jtm22

but I'm being told it's either 1 or not

I'm confused.

caveman

ignore what you are being told

just solve the problem to your own satisfaction

jtm22

18:51

My satisfication is being right and that's what Im' trying to make sure of

caveman

do you think 1 is the answer?

user2175923

@caveman I am just trying to write a program that will return the right value given the two values and their moduli, which may or may not be coprime, and assumes a solution exists

jtm22

Caveman - do you?

caveman

if you're not going to answer my question I think I'll do something else

Tobias Kildetoft

@jtm22 can you not think of more than one way to distribute the balls?

jtm22

18:53

NO

It is not 1

Tobias Kildetoft

even if the balls are not labelled either, there is more than one way

jtm22

I think if it's (n+k-1 choose k) then that maeans it's (7 choose 5)

since N is 3, and k is 5

right?

How many ways are there to distribute 5 balls into 3 boxes = n=3, k=5?

user2175923

5 balls into 3 boxes means the order of the balls matter but not the boxes. so you have (1)(2)(345), (1)(2)(354), (1)(2)(435), (1)(2)(453), (1)(2)(534), (1)(2)(543), etc

caveman

@jtm22, what is your reason for asking "Caveman - do you?"

Jayesh Badwaik

Arrange the balls in $n!$ ways and then place the slashes as combinations?

jtm22

18:57

so would that mean it's 7 choose 5?

Jayesh Badwaik

yeah

multiplied by 5! that is I think.

jtm22

so, 21

wait are you sure it's multiplied by 5!?

Jayesh Badwaik

Its not multiplied by 5!. Hmm. Let me see.

user19161

Answered 3 lhf today, yay!

caveman

@jtm22 you are being rude ignoring me

jtm22

19:00

No, caveman. I answered you already.

There was no reasoning for it.

caveman

7 mins ago, by jtm22

Caveman - do you?

I'm asking you why you asked me this

jtm22

There was no reasoning for it, caveman.

No reason.

I don't know why I asked

caveman

that is a blatant lie

jtm22

I wanted your opinion on a question, I think. I don't remember.

ok

Dominic Michaelis

@jasper my students are so stupid i can't believe :(

caveman

19:01

@jtm22, it's not ok

user19161

@DominicMichaelis What level are they?

Dominic Michaelis

first semester

user19161

Can you stop calling people rude and calling them liars when they are not clearly so?

user19161

I am sick and tired of all this nonsense.

Dominic Michaelis

i did read today about 80 times: "A funktion is 2 times differentialbe iff it is a polynomial of degree 2"

Jayesh Badwaik

19:03

@DominicMichaelis why does every prof feel that his students are stupid, especially when everyone was a student once?

Dominic Michaelis

i am not a prof but still a student :D

Arkamis

@JayeshBadwaik The path behind always looks easier than the path ahead.

Jayesh Badwaik

@Arkamis That is what I wanted to say. :-)

Peter Tamaroff

youtube.com/watch?feature=endscreen&v=xPNZ7CRIb24&NR=1

Arkamis

It is much simpler to look back and say, "I did that and it was not hard" than to say "I'm about to do this and it will be easy."

user19161

19:04

@PeterTamaroff I saw your comment, hehe.

Peter Tamaroff

@JasperLoy "Look at me, I am awesome, I need no words to answer!" =D

user19161

@PeterTamaroff Did you give me an upvote? =)))

Peter Tamaroff

@JasperLoy Yeah,.

user19161

@PeterTamaroff Good, good.

Jayesh Badwaik

which answer?

user2175923

19:06

I get 120 btw

for the boxes question

user19161

@JayeshBadwaik The one on the Cauchy Schwarz inequality.

Jayesh Badwaik

:-)

user2175923

can anyone help me with math.stackexchange.com/questions/340866/…

user19161

@JayeshBadwaik Did you also give me an upvote? =)))

Peter Tamaroff

@JayeshBadwaik You so silly! You no gangsta.

caveman

19:08

@user2175923, I just answered it

user19161

I studied that proof in high school actually.

Dominic Michaelis

@jasper i did after i made an accidental downvote :D

user19161

@DominicMichaelis Ah, good good.

Dominic Michaelis

still need 60 reputation today :(

Peter Tamaroff

@JasperLoy The proof I like the best is the one considering $P(\lambda)=||x+\lambda y||^2$ when $x\neq 0$ or $x\neq \eta y$ for any $\eta \in \Bbb R$.

Kicks every other proof's asses.

user19161

19:10

@PeterTamaroff Where did you see that?

Peter Tamaroff

@JasperLoy I think 'twas Rudin.

gulan

hi there , any body know the answer of this question?

why the positive-definiteness of a riemannian metric implies it is non-degeneracy?

user19161

I like how Bill comments on answers and says how it is lacking and then tells the answerer to see his solution. Quite interesting!

user2175923

@caveman Let's say you had x = 21 mod 4 and x=-21 mod 10

caveman

@user2175923, there

@user2175923, you can use a stack of (value,modulus) pairs and keep trying to add new ones onto it by checking whether they are coprime with what's already in the stack

Peter Tamaroff

19:16

FUCK YEAH!

user2175923

I have a method that solves for gcd(x,y) already

Dominic Michaelis

i like this one here

user2175923

it should be 1 for coprime pairs

so you're saying divide the moduli by g

Peter Tamaroff

@JasperLoy Bill is a marketing machine of himself!

caveman

@user2175923, I added a worked example

Peter Tamaroff

19:21

He answers and he's like see THIS link here. And it's a related answer of his, where he links to yet another answer, ad infinitum!

user19161

@PeterTamaroff Well, he called himself Professor Bill Dubuque in that meta post! I wonder where he taught at... I find him an interesting person actually.

Peter Tamaroff

@JasperLoy Me too. I like his answers. He's succinct and efficient!

user19161

@PeterTamaroff Yes, like me. =)))

Peter Tamaroff

He also introduced me to an essay that deals with differential equations, special functions and said it is related to Lie Algebra.

I am much interested about that essay.

But it is too highbrow for me ATM.

user19161

Michael Taylor's 3 volumes of PDE is BOSS!

Peter Tamaroff

19:23

I studied ODEs and stuff from Spiegel

user19161

There is so much machinery used in them that the appendices can form complete books in functional analysis, representation theory and riemannian geometry!

user2175923

@caveman In my case, that a=b mod g thing will always hold true

but i am getting odd answers

are you saying that i can just treat it as solving for x = a mod m/g and x = b mod n/g?

caveman

@user2175923, you should just assume that your procedure takes arbitrary inputs and give an error token if there is no solution

user19161

Michael Taylor recently published a book on ODE as well.

user19161

His lecture notes on his home page are BOSS!

Dominic Michaelis

19:26

@jasper do you want to se the histogramm of the exam i corrected today ?

user2175923

@caveman sure; so I will check if that equivalence holds and if not, no solution. so when it does, then that thing holds?

user19161

@DominicMichaelis Sure.

caveman

what thing?

Dominic Michaelis

oh they failed so hard :(

user2175923

@caveman solving for x = a mod m/g and x = b mod n/g (only when a = b mod g)

caveman

19:28

no you threw away the condition that x = a mod g

user2175923

i am confused now

the usual program (which i use when a solution is assumed) is

function crt(a, m, b, n)
   return ( a*n*inverse(n,m) + b*m*inverse(m,n) ) % (m*n)

caveman

no

user19161

These number sequences, who knows what is the next number? Could be anything...

caveman

this requires n,m to be coprime

user2175923

what I did now is call crt(a, m/g, b, n/g) where g=gcd(m,n)

Dominic Michaelis

19:31

@jasper nope i said it is $\pi$

user19161

@DominicMichaelis lol

Dominic Michaelis

well in fact i said $\pi$, $\pi$ or $42$

caveman

I don't understand how mathgems is doing this without recursion

I see a way but it's not the same as his

user2175923

caveman, how would you modify the program?

as i have misunderstood you apparently/somehow

caveman

    function crt(a, m, b, n) // input requires m,n coprime. outputs a value mod mn
        if gcd(m,n) != 1 then error "inputs not coprime in CRT"
       return ( a*n*inverse(n,m) + b*m*inverse(m,n) ) % (m*n)

user2175923

19:37

right

but then now to solve for when m and n are not coprime and there is a solution

caveman

@user2175923, you start with [(a,m),(b,n)] meaning you want to find x = a (m) and x = b (n)

user2175923

yes, solving for x

caveman

if m,n have gcd g you check a = b (g) then change this to [(a,m/g),(b,n/g),(a/g)]

Dominic Michaelis

@jasper you looked at that histogramm ?

user2175923

so i first check if a%g==b%g, yes?

caveman

19:40

@user2175923, yes - now you want to solve a system of 3 equations with CRT - but you still don't know that they are all coprime

we know that m/g and n/g are coprime

user19161

@DominicMichaelis So full marks is at the top?

caveman

so you could change it to [(crt(a,m/g,b,n/g),m*n/g^2),(a,g)]

Dominic Michaelis

on the x axis there are the points, and on the y axis the number of students getting the points

there are no marks fixed yet

user19161

@DominicMichaelis I see, no labels at all! -1. =)

Dominic Michaelis

oops sry

user2175923

19:42

sorry got lost again

so it becomes another congruence in the form of

x = crt(a,m/g,b,n/g) modulo mn/g^2, x=a modulo g ?

caveman

@user2175923, Here is the big picture explanation: You want to solve a system of congruences - the tools you have work to either solve coprime congruences or split a congruence into coprime congruences - so we have to use these together to reduce the problem to what you can solve

yes, x = crt(a,m/g,b,n/g) modulo mn/g^2, x=a modulo g is what I meant by [(crt(a,m/g,b,n/g),m*n/g^2),(a,g)]

user2175923

and then how would i solve that final congruence? are mn/g^2 and g coprime?

caveman

@user2175923, recursion

you don't know they are coprime (having checks like that built into your programs would help you a lot)

user2175923

ok so i just keep checking if gcd(modulus1, modulus2)==1

caveman

@user2175923, one thing you do know is that gcd(mn/g^2, g) is strictly smaller than g

This justifies the recursion (proving it terminates)

Dominic Michaelis

19:49

@jasper but they really failed didn't they?

user2175923

@caveman so

function crt(a,m,b,n)
    g=gcd(m,n)
    if g==1
        return (a*n*inverse(n,m) + b*m*inverse(m,n)) % (m*n)
    else
        return crt(crt(a,m/g,b,n/g),m*n/(g*g), a, g)

@caveman

caveman

I recommend not writing code like that

user2175923

it's recursive

caveman

I mean lack of error checking

user2175923

what's the error?

(is what I wrote wrong?)

caveman

19:51

it's only wrong sometimes

user2175923

?

caveman

you should use else if a%g==b%g ... else throw error

user2175923

function crt(a,m,b,n)
    g=gcd(m,n)
    if g==1
        return (a*n*inverse(n,m) + b*m*inverse(m,n)) % (m*n)
    else if a%g==b%g
        return crt(crt(a,m/g,b,n/g),m*n/(g*g), a, g)
    else
        throw error

Dominic Michaelis

this answer here is so wrong (the one i commented)

user19161

20:37

@DominicMichaelis Oh well, without the definition of "fail" and the labels, the histogram cannot be interpreted, lol.

Dominic Michaelis

on the horizontal line there are the points which they have

and on the y axis there is the number of students having those amount of points

27 points are the maximum by the way

user19161

@DominicMichaelis There must be something wrong about my PDF viewer then.

user19161

I see absolutely no numbers on the graph.

Dominic Michaelis

there are at least numbers

oh there are numbers

user19161

I am using Evince on Debian, lol.

Dominic Michaelis

20:42

argh i still need 40 points, i want to go sleeping :D

caveman

how can I view peoples previous names?

"recent names"

user19161

@caveman You cannot.

user19161

I really enjoy typing solutions on this site, helps me practise my LaTeX and also exposition.

3

user19161

Hey @amwhy I just sent you an email.

amWhy

@JasperLoy Oh, goodie...!

κρανίοπεριπολία

20:59

youtube.com/watch?feature=player_embedded&v=5IBRbzf3Fws#!

►

user19161

Do you guys notice that on the questions that I answer, the other answers always have many votes? That is cos I upvote all the other answers, lol.

Dominic Michaelis

@jasper and everyone who upvotes you does it too

arg need 3 more upvotes than i can go sleeping :D

user19161

@DominicMichaelis I think we should not just upvote the first correct answer or the best answer, but upvote all reasonably good answers posted within a reasonable period of time.

Dominic Michaelis

yeah i guess that would be better

user19161

@DominicMichaelis No need to cap every day bro!

Dominic Michaelis

21:09

but i want to get those super fancy super sexy superpowers :D

user19161

@DominicMichaelis You will burn out soon! 100 a day is good enough!

Dominic Michaelis

oh

user19161

Rumour has it that Debian 7 might be released this weekend, but I think it could take another 4 weeks.

Dominic Michaelis

soon i will have 7.500 repuation

user19161

21:38

So today I answered 5 lhf, yay!

Chris's sister and pals

Hi

Dominic Michaelis

capped :)

caveman

21:55

hi

Dominic Michaelis

@jasper i nearly got you in the quartal scoring :)

Chris's sister and pals

@caveman: you should use "wiseman" nickname. ;)

caveman

hehe

The Chaz 2.0

Does anyone know of any questions/answers on MSE that discuss (what I hope to be correct)...

... when you go from octonions to quaternions, you gain associativity

...when you go from quaternions to complex numbers, you gain commutativity

...when you go from the complex numbers to the reals...?
Does this seem familiar, or should I start a new question about it?

caveman

you get a total order

Dominic Michaelis

21:58

thats a funny direction

The Chaz 2.0

Right. Are there any resources on MSE already?

Dominic Michaelis

normally you go from the real numbers to the complex and then to the quaternions and so on

4

Q: Why are properties lost in the the Cayley-Dickson construction?

Motivating question: What lies beyond the Sedenions? I'm aware that one can construct a hierarchy of number systems via the Cayley-Dickson process: $$\mathbb{R} \subset \mathbb{C} \subset \mathbb{H} \subset \mathbb{O} \subset \mathbb{S} \subset \ldots $$ "Reals" $\subset$ "Complex" $\subset$ ...

abstract-algebra quaternions octonions sedenions

The Chaz 2.0

YES

Thank you, Dominic

Dominic Michaelis

maybe this even fits better math.stackexchange.com/questions/86434/…

The Chaz 2.0

Those are exactly what I was looking for. Thanks again.

Dominic Michaelis

22:01

glad i can help

Chris's sister and pals

@DominicMichaelis: nice from your side.

@caveman: today I was thinking to prove that $\frac{\displaystyle x+\frac{x^3}{1\cdot3}+\frac{x^5}{1\cdot3\cdot 5}+\frac{x^7}{1\cdot3\cdot 5\cdot7}+\cdots}{\displaystyle1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2 \cdot 4 \cdot 6}\cdots}=\int_0^x e^{-t^2/2} \ dt$

caveman

@Chris'ssisterandpals, thats nice, i think you can make a continued fraction for it

of the LHS

that wont help to prove it though

Dominic Michaelis

isn't it sinh/cosh ?

caveman

no

Dominic Michaelis

oh right the factorials are wrong

caveman

22:10

@Chris'ssisterandpals, I cannot imagine how to proceed

Chris's sister and pals

@caveman: no pb. I only shared one of the problems I was thinking of today.

caveman

@Chris'ssisterandpals, I think this is a really great interesting problem

I just wish I had the skills to give some input on them

I think it is related to hypergeometric functions

can we differentiate the LHS with respect to x?

Chris's sister and pals

@caveman: I think we can. Does it help?

caveman

@Chris'ssisterandpals,, I have no idea if it helps

Chris's sister and pals

ok

caveman

22:15

if the derivative looks easier to deal with, and we prove they have the same derivative - that might b a good start

but how would you differentiate the LHS? it looks impossible to deal with to me

well I guess you could just get it as a big product and multiply it out

Chris's sister and pals

@caveman: I'm not sure yet I want to do it. Maybe there is another way.

caveman

yeah that is definitely not going to make things easier

its just the first idea I had

Chris's sister and pals

@caveman: all ideas are precious. Thanks.

@caveman: I'll think of it tomorrow. Have a nice day! ;)

caveman

ok, and tell me if you get any further

@Chris'ssisterandpals bye

user19161

22:31

@DominicMichaelis Good! Soon you will reach 100k!

Transcript for

$Mathematics$ Mathematics