Mathematics - 2015-02-08 (page 2 of 3)

teadawg1337

15:04

@Chris'ssis I've solved it!!!

Chris's sis

@teadawg1337 Awesome. What is the solution? :-)

teadawg1337

$\sqrt{2}-1$

Chris's sis

@teadawg1337 Congratulations! :-)

teadawg1337

@Chris'ssis Yay, I'm the first to solve it :D

Chris's sis

@teadawg1337 Indeed. Do you have a full solution or you only found the answer by some means?

teadawg1337

15:07

That was a fantastic problem

I have a full solution

Chris's sis

@teadawg1337 Yeah, it's created by me.

@teadawg1337 OK, that's great!

@teadawg1337 You're the first that told me the right answer so far. Anyway, knowing the answers things become easier.

teadawg1337

Shoot, I should've deleted it...

Chris's sis

@teadawg1337 What do you mean? Ah, yeah, the answer you posted above ...

teadawg1337

@Chris'ssis The answer, I should've deleted it

Yeah

:(

Chris's sis

@teadawg1337 No problem. :-)

@teadawg1337 You're really great!

teadawg1337

15:10

@Chris'ssis So I've been told :P

You said you've given this problem to professors as well?

Chris's sis

@teadawg1337 Yeah, I gave it to some students and professors and no one solved it so far.

@teadawg1337 If you talk to @robjohn he might delete that message with the answer.

@teadawg1337 That problem will be included in my book, it's one of nearly 300 problems there.

teadawg1337

@Chris'ssis I'll be sure to grab a copy when it's published :D

Chris's sis

@teadawg1337 :D

@teadawg1337 My plan is to publish a book with 300 problems as nice as the one you already solved, all containing that "wow" element.

teadawg1337

@Chris'ssis Knowing the problem was created by you, I was expecting a more complicated answer. This is a perfect example of why I love mathematics; the answer to something seemingly complicated often turns out to be quite simple

Chris's sis

@teadawg1337 Indeed. :-)

Committing to a challenge

15:19

@KajHansen I haven't got much gym in no sadly. I helped people move for days and then I got sick :S. I just got up to start my day 20 min ago and it's 1:20am, so I am pretty screwed up :P. I haven't even got much study in either sadly

Chris's sis

@teadawg1337 Here is a very interesting thing about this problem too: one can easily approach it wrongly but get the correct answer. :-)

teadawg1337

@Chris'ssis Really? I wonder if I used the incorrect method...

A Beautiful Mind

Hello @Chris'ssis @Committingtoachallenge

Chris's sis

@ABeautifulMind Hello

Committing to a challenge

@ABeautifulMind Hey J, how are you?

A Beautiful Mind

15:27

@Committingtoachallenge Not too good. Maybe I will email you soon.

Committing to a challenge

@ABeautifulMind That would be good :). My email box is so empty(other than the spam)

Chris's sis

@teadawg1337 Wait to publish my book :-)

teadawg1337

I'm assuming I did end up using the "bad" approach :/

Chris's sis

@teadawg1337 You solved it that fast because you probably have a good out-of-the box- thinking. Sometimes we do things that are apparently against rules, but they lead us to the desired results.

@teadawg1337 I often break the mathematical rules.

teadawg1337

@Chris'ssis If the answer turns out to be correct, I'd say you're merely bending them in your favor

Chris's sis

15:35

@teadawg1337 After doing that I try to explain why things worked that way.

@teadawg1337 It's not the case for the problem we discussed about. The way there is very clear.

A Beautiful Mind

My aunt made some pizza for me.

Chris's sis

@ABeautifulMind I wanna take a slice!

teadawg1337

@Chris'ssis Not at first, but the pieces slowly start coming together

Chris's sis

@teadawg1337 Yeap.

A Beautiful Mind

@Committingtoachallenge Do you have a place to stay now?

Committing to a challenge

15:49

@ABeautifulMind Yes, fortunately I am not homeless now :). I have full internet again also. After I get healthy it will be back to hardcore studies

Sayan Chattopadhyay

16:08

Hi Jasper

evinda

16:25

Hey @DanielFischer!!! Could I ask you something?
How can we determine if the diophantine equation $z^2=x^4+p^2y^4$ has a non-trivial solution with $xy \neq 0$ ?

Sayan Chattopadhyay

Hey guys a quick question:
Can I make a new axis in the coordinate system so that I can plot the points of a tesseract

Can someone help me in answering it

user105491

@Sayan Couldn't you just take the space X and consider the trivial fiber bundle E\times \mathbf{R}?

Daniel Fischer

16:49

@evinda Is $p$ a prime? It may be that a factorisation in $\mathbb{Z}[\sqrt{-p}]$ or $\mathbb{Z}[\sqrt{p}]$ helps. But that's just a guess.

A Beautiful Mind

16:59

Is cancer very painful physically? I don't know.

evinda

@DanielFischer Yes $p$ is a prime.

Sayan Chattopadhyay

Yes jasper

It hurts you physically as you see everything happening to you yourself

@evinda u seem to like number theory a lot

evinda

@DanielFischer $$x^4+p^2y^4=(x^2)^2+(py^2)^2$$
How can we continue?

Brian J

hi

anyone able to assist me on a question, relating to drawing a precondition decomposition diagram?

evinda

@DanielFischer Could we maybe check if there are $r, s$ such that:

$$x^2=r^2-s^2$$

$$py^2=2rs$$

$$z=r^2+s^2$$

Brian J

17:09

evinda

since our diophantine equation is of the form $x^2+y^2=z^2$? @DanielFischer

Sayan Chattopadhyay

Pythagorean equation @evinda

Brian J

I'm not sure where to start in drawing the diagram and can't find any tutorials on the web. Can someone point me in the right direction of a tutorial on how to create these diagrams?

Sayan Chattopadhyay

That means the equation will have infinitely many values.

evinda

Don't we have to check if there are such r and s?

Chris's sis

18:12

@r9m how are you doing? You seemed to be far more talkative in the past. Is there a reason you changed that? :-)

Mike Miller

Morning, @Ted

Semiclassical

hi @MikeMiller

Mike Miller

Hiya

Ilya_Gazman

18:50

Hello

hardmath

hello, indeed

Ilya_Gazman

19:08

Hey @hardmath

Can you help me understand some answers that I got?

3

Q: When $\sqrt{(x+a)^2 -b}$ is an integer?

While working on integer factorization problem, I came to this: How to find for which values of $x$ the next equation is an integer? $$\sqrt{(x+a)^2 -b}$$ $a,b$ are positive known integers In order to find a solution to this question, should I test all the values of $x$? Or can I simplify it...

hardmath

Sure, I'll try

Taking a look now.

Ilya_Gazman

tnx

hardmath

So it might have been expedient for you to include in the Question your own thoughts, so that the Readers who responded would have a better idea of your level of understanding. Can you give me a little of that context (your math background, why you are interested in this problem, what difficulty it posed for you).

Ilya_Gazman

@hardmath I have no math background at all. I am a developer and I am trying to solve the integer factorization problem. Why? Because I like challenge and this is what I been working on for the last several months.

hardmath

Okay, but the first thing missing in your problem is any restriction on what sort of values x can take. A casual Reader might assume that x is supposed to be an integer, perhaps even a natural number or positive integer. But in theory one might want to find real numbers x that satisfy the given requirement.

Ilya_Gazman

19:19

in my case x must be an integer.

In fact the way I solving it now, is testing all the possibilities starting from x =0 and until I find the solution.

hardmath

Okay. Now we also know that $\sqrt{(x+a)^ -b }$ is supposed to be an integer. That means we have two perfect squares $(x+a)^2$ and $(x+a)^2 -b$, with the gap between them exactly $b$.

Ilya_Gazman

yep

hardmath

If we do the search you are doing now, then we need to consider the possibility that there can be more than one solution.

However we don't want to keep searching for solutions if there are no more to be found.

anon

@Ilya_Gazman all solutions are obtained by finding same-parity (both odd or both even) divisors u,v|b such that b=uv and setting x=-a+(u+v)/2.

Ilya_Gazman

@anon sorry I don't understand this

@hardmath so far, so good

anon

19:27

you don't understand what I'm saying, or you don't understand why it's true? those are two different things.

Ilya_Gazman

what you are saying

anon

do you know what a divisor is?

do you know what even and odd mean?

Ilya_Gazman

yep

anon

then what don't you understand?

Ilya_Gazman

how do I find those parties?

anon

19:28

parties?

Ilya_Gazman

party*

parity*

anon

I said parity. Two integers have the same parity if they are both even or both odd. They have opposite parity if one is even and the other is odd. It's kind of like the "sign" of a number.

Go through the list of divisors u of b (the negative ones too) and determine for which ones do u and b/u have the same parity - whenever they do, set v=u/b and then x=-a+(u+v)/2 is a solution.

hardmath

To go back to the searching idea, we can stop once the gap between the next two perfect squares is more than b, since all the squares beyond that point are further apart than b.

Before that point there may be more than one pair of squares that gives the difference b, and a little algebra helps us pinpoint them without random/brute force searching. That is what anon is getting at.

Gato

Hi

how can I count the basis on $F_q^n$? I am trying to find $\vert Gl(n,F_q^n)\vert$

hardmath

Let $y = \sqrt{(x+a)^2 -b}$, an integer, so that $b = (x+a)^2 - y^2$. Now we can use the factors of $b$ in combination with the difference of two squares to get a narrow range of possibilities for $x$.

anon

19:35

@Gato Do you mean "count the number of bases"?

Gato

@anon yes ^^

Ilya_Gazman

@anon I am not sure that I can factor $b$, its almost as hard as the original problem.

anon

@Gato Count the number of ways of selecting a basis. Selecting a basis has many subchoices: first you pick the first vector, then you pick the second vector, etc. Whenever you have a decision procedure like this, the total number of ways to go about it is the number of options in the first choice times the number of options in the second choices times etc. etc.

hardmath

@Ilya_Gazman: Are you trying to implement Fermat's method of factoring b?

Ilya_Gazman

nop

Gato

19:38

@anon As $\vert F_q^n\vert=q^n$, so for the first vector I have $q^n$ possibilities?

hardmath

Do you care if there is more than one solution?

anon

@Ilya_Gazman you're asking a number theory question on a math site without any reference to computational complexity or efficiency - you can expect to get a pure answer that's true in the abstract and reduces the problem to a better-known one (that's what mathematicians do !) but not necessarily any more practical than a naive approach to the original problem.

@Gato can any basis have the vector 0?

Gato

@anon oh right, $q^n-1$ so..

anon

also I suppose I should mention you need to divide the resulting product by n!, since there are n! ways of getting any given basis through a sequence of choices (corresponding to the different ways of listing out the elements of the basis in sequence).

Ilya_Gazman

@hardmath there could be more than one solution, but not likely. I didn't test it so... What I do is finding a solution, taking it to the next formula that gives me an actual factor of the number that I am factoring, and I just check if it is correct

anon

19:40

@Gato right. now how many options for the second vector?

hardmath

Well, you may have rediscovered Fermat's method of factoring. It seems at least closely related to what you are trying to do. In general the problem of finding a factor of a big integer can be approached with a hierarchy of methods.

anon

@Ilya_Gazman also, your proposed method seems to be asking to test an infinite number of values for x, and never knowing if one has found all solutions or not. at least in the answer I gave it's knowable how to bound where the solutions can come from.

Gato

@anon I need to choose $a_2$ outside of $span(a_1)$? (not sure how can I wrote this in english)

anon

@Gato yes

Ilya_Gazman

my actual equation is $\frac{\sqrt{(a+bc)^2 - 4bd} +bc + a}{2b}$

hardmath

19:43

The easiest/most elementary one is trial division. Then you probably want to try Pollard's rho for a while before turning to elliptic curves and quadratic sieve variants, the big guns of factoring.

Gato

@anon but how can I know the dimension of $span(a_1)$

anon

@Ilya_Gazman so you could have asked when (a+bc)^2-4bd is a perfect square.

Ilya_Gazman

where $a,b,c,d$ are integers and I increase the size of $a$ by one until I get an integer number, when I do, it will be the factor of $d$

anon

@Gato do you know the elements of <a_1>?

hardmath

I would recommend calling it an expression rather than an equation. Equation suggests that equality of two things is being asserted.

Ilya_Gazman

19:45

@anon that what I did, I just removed several constans

Gato

@anon the set of linear combination of $a_1$ so for a_2 it's $q^n-q$

hardmath

@Ilya_Gazman: So, how big is $d$ in your application?

anon

@Gato if there's only one vector you don't need the term "combination"; just say scalar multiples of a_1

but yes there are q^n-q choices for a_2

Gato

@anon okay thanks!

Ilya_Gazman

@hardmath it takes me miliseconds to factor up to $2^{40}$ numbers, minutes for $2^{60}$ and I want to improve it so I can crack RSA witch is $2^{512}$ Also I know that I probably will not get there ;)

anon

19:49

@Ilya_Gazman the question you asked on main seems slightly more general, since -4bd is a multiple of 4, and we aren't assuming x,a or b share any factors in the question you presented on main. turning a solution to the problem the posted into a way of finding tuples (a,b,c,d) for which your original radical is an integer may not be trivial, I'd have to think about it. (not sure which part of it you're calling x anyway)

Gato

@anon Now I would like to 'exhibe' a $q$-sylow subgroup of $Gl(n,F_q)$

anon

@Gato so q is a prime number?

Gato

@anon yes.

anon

upper-triangular matrices, with 1s on the diagonal

Ilya_Gazman

@anon replace $a$ with $a+x$

anon

19:50

(I don't know how to give a hint for that one so I just gave the answer)

Gato

@anon no problem, thanks

hardmath

@Ilya_Gazman: As I mentioned at the beginning of our chat, it would be expedient to explain the context of what you are trying to do in the Question itself. It is well-known that factoring an odd integer in a nontrivial way is equivalent to expressing it as the difference of nonconsecutive squares. Consecutive squares give to trivial factorization (2k+1)*1 = (k+1)^2 - k^2.

Chris's sis

20:27

I was looking for a nice counterexample to the Fubini's theorem I saw some time ago on MSE, but I cannot find it anymore. Anyway, I can come up with some ...

There are some on MSE I see now but I like no one ...

Moshe

Hey, can someone help with some homework? I need to solve a nonlinear inequality.

anon

> just ask; don't ask to ask

16

Moshe

I'm having trouble with the nonlinear inequalities: 2b, 2c

do I need to complete the square or something?

hardmath

@Moshe: Keep in mind that for real inequalities a/b > c can be simplified by multiplying both sides by b, being careful however not to multiply by zero and to change the direction of the inequality if multiplying by a negative value.

The inequality a/b < 0 simply means that a,b are nonzero and have different signs (so their ratio is negative.

anon

@Moshe (2b): split into cases depending on the sign of x+1

same idea for (2c), just more involved

Moshe

20:41

@hardmath so for 2b, I can just move the denominator over by multiplying.

@anon how do you mean?

anon

@Moshe we don't like denominators in elementary algebra; we often get rid of them by multiplying equations or inequalities by them. what happens when you multiply an inequality by something? well, of course you multiply the left and the right side by that thing, but what happens to the inequality symbol depends on the sign of the thing you're multiplying by.

for instance if x+1>0 then (x^2+x-3)/(x+1)>1 implies x^2+x-3>x+1 implies x^2-4>0 implies x<-2 or x>2. but (x+1>0 and (x<-2 or x>2)) is equivalent to x>2. you can do the same idea for the case when x+1<0.

hardmath

@Moshe: As I said, worry about the sign of the denominator. x+1 could be negative, so if you "move it over in that case the inequality gets reversed.

Sorry, gang. Need to go attend to caffeine intake.

Chris's sis

Those inequalities work very nice by using simple sign tables. At least this is the way I did them when I was a kid.

anon

that's better yes

Moshe

What's a sign table?

(Context: doing this as Precalc review for a calculus 1 class.)

infinitesimal

20:57

A table with signs.

anon

@Chris'ssis well for (2c), not for (2b)

Chris's sis

@anon also for (2 b) works nice.

anon

oh?

it's >1, not 0

Chris's sis

@anon Just move 1 in the left side.

anon

yes that works

:-)

Chris's sis

20:59

:D

infinitesimal

:D)

Chris's sis

@Moshe Are those exercises coming from uni? I had to do a lot such ones when I was in the middle school. What nice days ... !

(they're gone)

@Moshe

►

This is the table I was talking about (it's in Romanian language, but this shouldn't be a problem).

Arkamis

Anyone around that can give some pointers on: math.stackexchange.com/questions/1139603/…

Chris's sis

@Moshe I found something in English

►

infinitesimal

2

Q: How does one construct a 'sign chart' when solving inequalities?

I'm working on solving inequalities for an assignment. The instructions also request that I draw a 'sign chart' along with each solution. I've never heard of a 'sign chart' before, and the internet also seems to have a limited amount of information. From what I can gather... PurpleMath prevents ...

inequality

user134177

21:22

hi

user134177

i need help

user134177

mathworld.wolfram.com/GausssCircleProblem.html you see N(r) and i want to prove the second equality. If i is odd, I set i=2j+1 and if i is even, let i=2k+2

user134177

so i get r^2/(4j+1) and r^2/4k+3

user134177

but how I change the index in detail?

user134177

21:43

oh, no problem

user134177

its easy^^

bolbteppa

Anyone here know their representation theory?

anon

1 hour ago, by anon

> just ask; don't ask to ask

bolbteppa

21:59

Okay cool, so if you take a matrix (rank 2 tensor) you can decompose it into an a) anti-symmetric matrix, b) a traceless symmetric matrix, and c) a trace. This corresponds to taking a 9 component matrix and decomposing it into a smaller 5, 3, & 1 component matrices. Furthermore these three tensors somehow get mapped onto the first 3 spherical harmonics of order 0, 1 & 2 via 1 = 2*0 + 1, 3 = 2*1 + 1, 5 = 2*2 + 1, and this corresponds to the 3 quantum spin values -1, 0 & 1 somehow.

anon

c) a diagonal matrix?

bolbteppa

Yeah, you decompose it first into a symmetric and anti-symmetric matrix, then the symmetric matrix gets decomposed further because trace is an invariant right?

anon

so 3, 3 and 3 component matrices

@bolbteppa what do you mean by trace is an invariant?

bolbteppa

It's basically $A = A_a + A_b + A_c$, I mean trace is an invariant quantity for linear operators irrespective of the basis you choose to represent this operator in

anon

I don't see how that's relevant to the decomposition

you just split the symmetric matrix into its diagonal part and whatever's left

bolbteppa

22:05

No you don't just set the diagonals to zero or split it up, you can include diagonals so long as the trace is zero

anon

huh?

bolbteppa

Page 9 of this pdf :) pmaweb.caltech.edu/Courses/ph136/yr2012/1211.1.K.pdf They strangely allow you to include the diagonal!

anon

I suppose by pg9 you mean the one listed pg6 but ninth in the pdf file

bolbteppa

There is a geometric explanation in terms of unit ellipses on page 10, but I think it just also makes sense because of the invariance of the trace

yeah haha sorry :D

I believe I have just decomposed a general matrix into irreducible representations of the rotation group or something, what am I actually doing? How does this relate to taking homogeneous polynomials and computing spherical harmonics that way? Can I apply this method to the matrix representations of the symmetric group? How do characters relate to this? Group reps is complicated :(

anon

@bolbteppa trace is a number. the thing the author is calling "trace" with scarequotes is obviously the diagonal part of the symmetric part of the matrix.

I have no idea why you think invariance of the trace (the scalar quantity) has anything to do with this

bolbteppa

22:11

It's a literal trace, equation (1) on pdf page 9 illustrates this, that is a double sum on the indices, the $\Theta$ is a constant in equation (4)

Ted Shifrin

heya anon

anon

heya

bolbteppa

and on page 10 pdf he says $\Theta$ just represents "one component" :)

anon

@bolbteppa okay, so am I to assume the author is decomposing a matrix into an antisymmetric part, a traceless symmetric part, and a scalar multiple of the identity (where the scalar is the trace of the original matrix divided by the dimension)? or what is the decomposition here?

bolbteppa

Yeah, the idea is to take an arbitrary matrix (rank 2 tensor) depending on 9 coordinates and represent it in terms of 'smaller' quantities

anon

22:17

@bolbteppa your questions seem to be all over the map here

@TedShifrin yo

bolbteppa

It's a question about representation theory :) Have you studied it?

anon

in the abstract for finite groups, not really for lie groups

bolbteppa

Cool, well I really want to understand it for finite groups, I'm doing a course in it, but this is the first thing I've seen that makes some sense haha :)

Ted Shifrin

yo, yo @anon :P

bolbteppa

Apparently decomposing a matrix $A = A_a + A_b + A_c$ into an a) anti-symmetric trace (representing a vector since 3 components), b) a traceless symmetric tensor (representing a unit ellipsoid) and c) a trace (I think representing the scaling factor of this ellipse) is a way of representing the matrix in terms of irreducible representations of the 3-D rotation group, I mean this is heavy machinery, I don't know what I'm doing tbh

Ted Shifrin

22:21

similar decompositions are used for curvature tensors in Riemannian geometry, too

anon

do geometers / quantum mechanics call matrices "traces" or something?

Ted Shifrin

not that I've heard ... but we certainly talk about trace-free parts of the tensor ...

bolbteppa

Wow so that's what I'm actually doing in general relativity :D

Ted Shifrin

I would guess so, yes, @bolbteppa

general relativity is a bunch of semi-Riemannian geometry :P

bolbteppa

So in that case I would be representing the Riemann curvature tensor matrix in terms of irreducible representations of the 4-D orthogonal group!?

Mike Miller

22:23

Morning, @Ted

Ted Shifrin

good night, @Mike

bolbteppa

Such a simple way to express what look like big ideas

Ted Shifrin

finally finished grading my diff geo homework papers ... and, o joy, exams on Thursday for my birthday

Mike Miller

Given an integrable function $f$ on $\Bbb R^n$, what (hopefully weak) conditions can I put on $f$ so that $\int_{|x| \leq \varepsilon} f \to 0$? For $f \in L^2$, I can use Holder's to show this is true, but I don't think it's true for arbitrary $f \in L^1$.

Ted Shifrin

yup @bolbteppa ...

why isn't it absolute continuity of the integral when $f\in L^1$?

Mike Miller

22:26

Ah, good point

Thanks

Ted Shifrin

bows

that actually came up in one of the graduate problems for my diff geo course I just graded :P

(trying to do $\int_C \kappa\,ds$ when the plane curve has a finite number of corners)

Mike Miller

So as a result, $\int f = \lim \int_{|x| \geq \varepsilon} f$.

Ted Shifrin

sure, @Mike

user43418

How can I solve this ? $y'+ln(y)+1=0$

y being a function with variable x

Ted Shifrin

I'm not sure there's an explicit formula, @user43418. Can you integrate $\int dy/\ln(y)$?

Mike Miller

22:31

@TedShifrin OK, then I'm not sure what's wrong with this argument I've got. I've got a differential operator $D$ and a function $f \in L^1_{\text{loc}$ with $Df = 0$ away from $0$, where $f$ blows up. It seems to me, then, that if $g$ is compactly supported, $\int fDg = \lim \int_{|x| \geq \varepsilon} fDg = \lim \int_{|x|=\varepsilon} fg$ by the above assumptions on $f$ and integration by parts, no?

user43418

@TedShifrin That's what I did for now

0

Q: Differential equation with logarithm

$y'+\ln y+1=0$ $y' +ln(y) +1 = 0$ $y' = -ln(y) - 1$ $dy = [ -ln(y) - 1 ] dx$ $-\frac{dy}{(ln(y)+1)} = dx$ But I am stuck here..

differential-equations

Ted Shifrin

whoa, slow way down, @Mike. $D$ is a general differential operator? Do the integration by parts more carefully.

Mike Miller

Sorry. It's even order.

Ted Shifrin

Is this in $\Bbb R^n$, @Mike?

Mike Miller

Yeah.

Ted Shifrin

22:33

I still don't see how you got the boundary term.

Mike Miller

(I'm actually only doing $-\Delta + a^2$, but saying even order above seems like it works fine.)

Ah, I see, sorry.

Fixing that should probably fix the seeming contradiction I pull up later. Thanks.

user43418

@TedShifrin What do you think ?

Ted Shifrin

I think it's non-elementary @user43418

user43418

Yes

but the hint says it a incomplete Gamma function

Ted Shifrin

or the Li function ... whatever

user43418

22:38

but I can't seem to get it..

Ted Shifrin

Li is immediate — that's the integral I gave you

Ramanewbie

22:51

hi @ted

Ted Shifrin

bonsoir, @Ramanewb

Ramanewbie

what's up

Ted Shifrin

Just recovering from a day of grading ... You been having a fun day wandering the Paris streets?

Whoa ... it's past your bedtime again!

Ramanewbie

@ted not yet, I've about 10 minutes left...

Ted Shifrin

0:04 is your bedtime?

Ramanewbie

22:54

@ted You've been grading all day long ?? poor of you...

Ted Shifrin

Well, I started with tennis at 08:30

Ramanewbie

@ted it's not midnight at the minute -_-

Ted Shifrin

is @Hippa back at school?

Ramanewbie

@ted yes he is, but he might be quite tired today

Ted Shifrin

you wore him out?

Ramanewbie

22:56

@ted "wore out" ?

@ted what do you mean ?

Ted Shifrin

c'est toi qui l'as rendu épuisé?

Ramanewbie

@ted not really... yesterdy we both stayed up late for a ferrofluid experiment !

Ted Shifrin

oy vey

Ramanewbie

I helped him as I could with that...

"oy vey" ?

what's that

Ted Shifrin

that's Yiddish for "oy vey" :)

Ramanewbie

22:58

"Oy vey iz mir," "Oh, woe is me"

Ted Shifrin

c'est ça :P

The phrases I know in French aren't quite so emotive.

Ramanewbie

lol...

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$Mathematics$ Mathematics