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00:25
hi
01:04
I am having trouble deciding whether the following series converges: 1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6... (two positive terms followed by negative term)
The quasi-alternating signs leads me to think I should find some way to use the alternating series test
01:27
doesn't
split partial sums into A+B-C, each of which is ~(1/3)ln(n)+const
IF....you keep "everything in order".....
indeed
(though terms within partial sums can be reordered at will ofc)
sometimes, anon, I think you're nearly half of mathstackexchange
I've barely done anything here the last few months
oh, you mean my accounts?
heh
well, it's the way of things....waves build, they crash, new waves form
oh...my...
I said I BELIEVE, I believe....my baby's got a black cat bone
02:13
@BalarkaSen, which book you use for multivariable calculus?
user174558
02:44
I use Kaplan and Lewis's Calculus and Linear Algebra 1 and 2. It's the best.
thanks
user174558
It's downloadable legally if you do a search. Also cheaply sold on Amazon.
user174558
It is sad that good old books are going out of print. Silly new books are so popular as they have lots of pictures and colour.
user174558
It is ridiculous for a calculus textbook to cost 200 USD.
03:05
@JasonBourne so are you studying math now?
I see that MIT won the Putnam competition again @anon
03:30
@rorty Nope. Grouping it as 1+(1/2-1/3)+1/4+(1/5-1/6)… probably works. If this diverges, so does the original.
03:44
@Silent Ted Shifrin, "Multivariable Mathematics"
03:56
Hi what does it mean to take the supremum under the unit disk in R^2?
$\|X\| = \sup_{\| x\|_2 = 1} \|X \bm{x}\|_2$
if $D$ is a set and $f$ is a real-valued function defined on it, $\sup\limits_{x\in D}f(x)=\sup\{f(x):x\in D\}$
also, use dollar signs for latex
So, does the supremum occur on the boundary of the unit disk?
Or not necessarily?
Hi @PedroTamaroff.
What are you working on?
04:00
Now? Trying to connect to some repository.
I meant math-wise :)
@anon Wait, so the set of the boundary of the unit disk meaning there exists a point $(x_1,x_2)$ such that the supremum is attained, i.e. not all points on the boundary yield the supremum?
@JesterTran have any thoughts? (btw ||x||=1 defines the unit sphere not the unit disk)
@BalarkaSen Looking at the contents of my topology course, I sit for the final tomorrow.
I mean today, Thrusday.
Best of the luck!
04:02
@anon Why? There are only two dimensions
@JesterTran in R^2, ||x||=1 is the unit circle
(circle is the 1-dimensional sphere)
@anon ok
@JesterTran (a) the supremum is attained because the unit (whatever) is compact. (b) "a supremum is attained" is not logically equivalent to "not all points yield the supremum" (only use "ie" for logically equivalent things)
@anon I don't understand
what don't you understand
04:05
apparently we're taking for granted that it's the 2-norm... :D
There is a subset of points in the unit circle such that we get the supremum. Which means not all points in the unit circle gives us the supremum?
@J.M. I'm taking for granted that JesterTran wrote $\|x\|_2$
@anon Yes, just found the relevant line. Carry on... ;)
$\{\bm{x} \in \mathbb{R}^2: \| \bm{x} \|_2 = 1 \}$
@JesterTran no, those English sentences don't mean the same thing. (because the phrase "subset of points in the unit circle" does not by itself preclude the subset being the whole unit circle). is English your first language? although I suppose these are mathematical things you're getting hung up on and not English
(what's \bm btw?)
04:08
Now, if they said proper subset, OTOH...
Yeah, I meant proper subset
Shoulda mentioned it to begin with.
@anon bolded vector
\bf (boldface)
@J.M. I wasn't precise
04:13
@MikeMiller Let me clarify a few things. So $MCG(T^2) \to GL_2(\Bbb Z)$ be given by sending an isotopy class of a self-diffeomorphism $T^2 \to T^2$ to the matrix corresponding to the induced automofrphism on $H_1(M_g)$. It's quite clear why it's surjective, because matrices there give rise to $\Bbb Z^2$-equivariant self-diffeomorphisms of $\Bbb R^2$ by multiplication, which descend down to self-diffeomorphisms of $T^2$.
For injectivity, I am a bit unsure, but thoughts: Suppose you take a map $T^2 \to T^2$ which is identity on $H_1$. Since $H_1 \cong \pi_1$, that guy is also isom on $\pi_1$. $T^2$ is a $K(G, 1)$, so that map better be homotopic to identity. But why's it going to be isotopic to identity? Are homotopic diffeomorphisms of closed orientable surfaces always isotopic?
If I sub (1/sqrt2,1/2sqrt2) to remove the supremum into $\|X\| = \sup_{\| \bf{x}\|_2 = 1} \|X \bf{x}\|_2 $, does that give $\leq$?
you mean does $\|X\|\le \|X(1/2,1/2)^T\|$? or the other way around?
(seem to be using the letter X a lot..)
@anon nvm I got it
thanks
 
1 hour later…
05:39
√(p^2+q^2) + √(r^2+s^2) >= √(p^2+q^2+r^2+s^2)

for all real p, q, r, s.
True or false?
@Jester: Is $\sqrt 1 + \sqrt 2 = \sqrt 3$?
@TedShifrin Why does equality have to occur there?
That's a particular case of your question.
@TedShifrin $\sqrt 1 + \sqrt 2 > \sqrt 3$
So what you said doesn't work.
05:52
@TedShifrin I'm not following... why is that a particular case of my question?
$p=0$, $q=r=s=1$.
Why must it be an equality? It's greater than or equal
Oh, sorry, I missed the >.
Yes, so it's concavity of the square root function.
I apologize.
@TedShifrin No problem. Thanks for a concise answer
It's $(a+b)^2 \ge a^2+b^2$.
05:57
I see
(For $a,b\ge 0$)
06:26
@Jester: On further reflection I lied. Concavity gives us an inequalty going the other way — $\sqrt{2(x+y)} \ge \sqrt x+\sqrt y$.
Hi @Ted!
06:37
@TedShifrin Hi, still here? Must be getting late around your end of the world
I'm trying to prove whether or not the series 1 - 1/2^2 + 1/3 - 1/4^2 + ... converges. I think it diverges because there is more positive "fuel" than negative fuel, but am uncertain how to approach the proof.
@rorty Are you familiar with the alternating series test?
yes, but that does not apply here because the absolute values of the terms of the sum is not decreasing.
(at least, i think it does not apply in an obvious way...)
ahh, right
@rorty turn your thinking into a proof
06:42
but.........HOW
scratches head
split the partial sums into two parts. one part is bounded, the other isn't.
well the series sum(-1/(2n)^2) converges to 0
and the series sum(1/(2n-1)) diverges
but how do i use these facts?
@rorty no, it's bounded
@rorty you're trying to show the partial sums are not bounded. one part of the partial sums is bounded, the other isn't.
you are nearly asking me to do the whole problem for you
put 2 and 2 together
07:01
Hullo lovely mathematicians.
A set of vectors $\in V$ whose cardinality $< dimV$ can never be a basis of $V$, right?
because a basis of a Vector space has to have the same dimension...
@BalarkaSen how you like my picture :) math.stackexchange.com/questions/1741630/…
@ForeverMozart I like it. Nice haircut.
lol not that one
thats not me I'm talking about the picture in the answer
@BenjaminR Indeed
@TobiasKildetoft Thanks, just have a really dumb homework question that was making me question my sanity.
07:10
@BenjaminR Except the basis has a size (or cardinality), not a dimension
yep, sorry, that's right.
@ForeverMozart Nice. +1 from me.
me too :-)
thanks !
08:10
Hi @AlexClark.
08:47
Hey guys - quick question if anyone's involved in discrete math - could anyone prove this? Given Ri is partial order on Xi, where i = 1,2
Show R is a partial order on X1 x X2. (X1,X2) R (X1', X2') if (X1)R1(X1') and (X2)R2(X2')
I dont know if my proof is too "obvious" - I feel kind of dumb by just stating that in the ordered pair the first element is true and the second element is true because of the R1 and R2 already defined.
user174558
09:01
I got a brand new Springer GTM from Amazon that has handwriting in it, LOL.
user174558
I am going to return the item.
user174558
The amazing thing is how the previous user kept the book in such perfect condition except for the handwriting.
user174558
@skillpatrol Graduate Text sin Mathematics series
user174558
I thought I was having hallucinations when I saw the handwriting, LOL.
09:04
what did they write?
user174558
@robjohn It should be 'Why?' and not 'WHY???', LOL.
user174558
@skillpatrol Notes in blue ink on many pages.
INK?!? >8(
user174558
IKR, WTF
WHY??? :P
user174558
09:06
Haha, how did this even happen?
user174558
Am I hallucinating?
nah, it's real
hi @robjohn
user174558
I tell you, the book is in perfect condition otherwise.
user174558
09:07
Not a single page is crumpled.
how much did it cost?
user174558
56 USD.
user174558
How can someone read and even write on the book without crumpling a single page?
user174558
I smell a fish...
do you lose shipping and handling if you return it?
user174558
09:09
I think maybe Amazon can refund the shipment too, but the shipping is not much, so not a problem.
user174558
@skillpatrol Any answer to my above question?
which one?
user174558
The last one with a question mark.
i never crumple pages
user174558
Maybe I should ask on Skeptics, LOL.
user174558
09:12
Or Lifehacks, LOL.
it depends on how carefully you turn the pages, no?
user174558
But the book just looks so new that it is quite impossible... This person must be a nutcase
perhaps
user174558
@skillpatrol It must be you. =)
what's the title?
user174558
09:15
Conway: Functions of One Complex Variable I
user174558
LOL
09:43
this is the proof for why it is linear:
can anyone explain this proof?
09:56
Morning!
Can someone explain to me Euler's number? Who cares if we use 2.7 to some power instead of 3?
@tylerl-uxai Because those would be different numbers
for approximations, 2.7 might be close enough, or it might leave you missing Mars by a few miles
And in math, we might not care about approximations but about actual values
whoa
I like the way that sounds... Tell me more :D
What's the purpose of exponents on Euler's number should be my question
what problem does that solve exactly?
@tylerl-uxai e is a mathematical constant that appears naturally. for instance, in the solution to exponential growth models, the standard normal distribution (bell curve), rotations using complex numbers, permutation statistics, etc.
@tylerl-uxai It has several properties. Are you familiar with derivatives?
since it shows up everywhere, we might as well have a name for it
10:00
yes, I am familiar with derivatives.
What's it doing there?
It is also the factor your saving would increase by each year if you had an annual interest of 100% accumulated infinitely often
100% is 1 in the denominator?
@tylerl-uxai that's related to $e^x=\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n$ (which is in turn a limit of successive approximations to solutions to a growth model)
and exponent is how often, right?
whoa whoa whoa...
how do I plug that fancy talk in to something easy to read?
see "LaTeX in chat" above the starboard ------------->
10:02
It is also the unique value of $a$ such that the graph of $a^x$ has slope $1$ at $x=0$.
@JasonBourne No...
you drag a thing to your bookmarks bar, come back to this tab and click the bookmark to render latex
no one's gonna help me? :(
how do I code this for you guys? I can do something for you in exchange for what you're helping with
code what? you want to type equations for us to read?
10:04
I can fix this bookmark bullshit. What am I supposed to do to convert it to latex? Copy and paste it in the text box?
yeah bro
convert what to latex? do you know how to type in latex?
if it takes less than 20 min of programming and i can do it on codepen.io
@tylerl-uxai Did you get the bookmark from the webpage?
Sorry, I sort of want to fix your latex converter
what do I do? I should ask before offering a solution. I don't fully understand the system and how it works..
@tylerl-uxai fix it? is it broken?
10:05
I just go to the url, and uh
do what?
do what the instructions say: (i) put the thing in your bookmarks bar, (ii) come back to this tab, (iii) then click the bookmark
user147690
Hey @BalarkaSen, sorry my laptop un-hibernated me into a chat, and I didn't realise I was here :P.
tell us which step you're at
does that work in google chrome? I'm frustrated already... This neesd to be re-written
so people can convert it online
yes it works in google chrome
what do you mean by convert? convert what?
10:06
@tylerl-uxai if you're too frustrated with the bookmark, there are some other options you could try:
16
A: Any chance of MathJax in chat?

Ilmari KaronenAs a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

I right click or like hold the link and drag up?
when I go on a new computer, I get it working in 30 seconds with just button clicks
user147690
@tylerl-uxai Yes. Make sure show bookmarts is on, drag the 'Start ChatJax' to the bookmark bar, come back here and click it.
@tylerl-uxai whichever method you prefer to get it in your bookmarks bar
user147690
I will test how long it should take now.
10:07
dude! Live demo! click live demo!
user147690
3 seconds
all you people need is to re--do your link with the instructions ctrl + f" Live Demo" (without the quotes)
and people can copy and paste it in! Ta da :) high five team
I'm also highly specialized in usability
Hi @AlexClark
user147690
Hey @TobiasKildetoft, how are you?
@AlexClark Good
gonna do a final check through of the paper we are working on, then I need to do some minor revisions on a paper I submitted almost a year ago, which I got the referee report for yesterday (accept with minor revisions :) )
10:09
Change the description as follows: Latex Conversion mathjax.org/ ctrl + f" Live Demo" (without the quotes)
I've got to go. Hopefully someone emails me about euler's number. My email address is [email protected] Good luck with your maths ;D
I'm not changing any description to some technical stuff when the current method is just clicking and dragging
user174558
Hey @AlexClark, LOL.
user147690
Oh awesome stuff :D @TobiasKildetoft, is the paper on Arxiv already?
look at anon trolling
he knows my solution is easy as pi
and he wants the same stupid system in place
user147690
@tylerl-uxai I assume you are trolling tbh
10:11
@AlexClark The accepted one has been there for a long time.
someone down-vote all his possts
go to mathjax.org click live demo for me
user174558
I think @tylerl-uxai is @anon
tell me that is easier please
Original version is from Nov 2013
user147690
I meant the final check one
10:12
I've got to go. Please don't let anon rob you of the work I did
what just happened
@AlexClark Ahh, no, that should be done soon though, unless we decide to add more stuff, which depends on whether we can make some arguments work in general
user147690
What'll the title be?
otherwise we will probably not proceed further in that direction, and leave the next few special cases to some students at some point
user174558
Oh dear, he got suspended.
10:14
@AlexClark Simple transitive $2$-representations of small quotients of Soergel bimodules
user147690
@BalarkaSen I am going to write my alg geo topic on "The group structure on a plane cubic curve", so I need to learn about divisor class groups and such
user174558
@AlexClark What book do you recommend me for algebraic geometry?
2 mins ago, by tylerl-uxai
I've got to go. Please don't let anon rob you of the work I did
???
Mazorchuk and I convinced each other today that the Coxeter types $I_2(8)$ and $I_2(9)$ should also be doable, but it would take a ton of tedious computations unless we can find a more general argument
@anon someone wants to sell their solution
user147690
10:15
Did you ever understand this?
user147690
Oct 9 '15 at 6:07, by Tobias Kildetoft
I am also working on understanding the 2-category of Soergel bimodules in type $B$, hoping to understand if all simple transitive representations come from left cells
user174558
It appears that my LOL is the ??? of @robjohn
user147690
@JasonBourne No idea sorry. I am working with Gathmann & Milne, mainly Gathmann, but there are many good books, and I don't recommend constricting yourself to just one anyhow
@AlexClark Yes, one of the results in the present paper is that all simple transitive $2$-reps are cell $2$-reps in type $B$, under some assumptions on the apex of the $2$-rep
user174558
@AlexClark Hey I think I recommended both those to you, LOL.
10:17
Plus similar results for all other Coxeter types (corresponding to finite Coxeter groups)
user147690
@JasonBourne I think you did recommend gathmann, but no idea on milne :P
user147690
@TobiasKildetoft That is really really awesome
user174558
@AlexClark I see. My name appears in Milne, ROFLMAO.
@AlexClark Next step will be to try to remove that assumption in type $B_3$ which seems surprisingly hard
@robjohn my best guess is tyler was having fun thinking of his own solution, decided to to an entertaining performance of a breakdown in order to be ironic, but caught flags from using a naughty word
user174558
10:19
So it's not OK to say sick fucks but OK to say fuckin bombs? LOL.
user147690
@TobiasKildetoft So at no point in your career do tedious computations seem worth it :P
@AlexClark Well, if they can prove a general result, sure. If they only deal with a single Coxeter type then probably no (unless that type could potentially be the stepping stone to larger types)
user174558
@AlexClark Except in Calculus I.
user147690
I think I may have already asked you this, but do you know the basis of canonical type?(different from the canonical basis)
We did consider that the case $I_2(9)$ could be of some interest as it is the smallest odd non-prime
@AlexClark No, I know of neither
user147690
10:21
Oh
user147690
I thought you'd know the good bases and the canonical basis, but maybe math really does split into too many directions
user147690
Apparently they are really helpful for highest weight reps
@AlexClark I have heard about the canonical basis many times, but I have never actually learned what the term means
I don't think I have heard the term "of canonical type" before
user147690
Yeah I thought you wouldn't of
user174558
This reminds me that in linear algebra, there is the Jordan canonical form, the Frobenius canonical form, and the Smith canonical form.
user147690
10:24
I am looking at the basis of canonical type for $U(\frak{n}_+)$
user147690
1) elements of $B$ are weight vectors, 2) $1\in B$, 3) each right ideal $(\theta_j^p U(\frak{n}_+)\otimes \Bbb Q)\cap U(\frak{n}_+)$ is spanned by a subset of $B$, 4) In the basis induced by $B$, the left multiplication $\theta_j^{(p)}$ from $U(n_+)/\theta_j U(n_+)$ onto $\theta_j^p U(n_+) / \theta_j^{p+1} U(n_+)$ is given by a permutation matrix, 5) $B$ is stable under an antiautomorphism in $U(n_+)$ and 6) $B$ is stable under the bar involution.
user147690
Those the the 6 conditions for $B$ to be a basis of canonical type, really long actually :P
Hi, does somebody have some experience with something called "Thermodynamic formalism"?
@AlexClark What is $\theta_j^p$?
user147690
Oh sorry, $\theta_i$ are the generators of $n_+$
user147690
10:29
and $\theta_j^p$ is just $\theta_j \otimes \cdots \otimes \theta_j$
@AlexClark Hmm, unfortunate choice I think, but ok.
user147690
Is there another common meaning for $\theta$?
@AlexClark It is often used for projective functors on category $\mathcal{O}$
user147690
Oh okay haha
user147690
Fair enough
Could someone, please, explain me this proof?
@AlexClark Hmm, so you have already fixed a basis of $\mathfrak{n}_+$?
user147690
Yep
user147690
So associated to $\frak{sl}(2)$ it is just $\theta_1$ and then for $\frak{sl}(3)$ it is just $\theta_1,\theta_2$
10:31
Seems odd to have something be called of canonical form when it depends on the choice of basis of something else.
user147690
Oh sorry no I mean
user147690
I just have a choice of generators
user147690
I have a labeling for my generators specifically
user147690
Like $\theta_1 = \begin{bmatrix}0&1\\0&0\end{bmatrix}$ for $n_+(2)$
@TheProgrammer it's possible the reason you're not getting any responses is that it's not clear what you need to be explained about the proof, or what you don't understand about it.
user147690
10:33
Then $\theta_1 = \begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}$
@AlexClark Ahh, I suppose the basis is chosen to consist of vectors corresponding to positive roots?
user147690
No basis for $n_+$ sorry, I said the wrong thing, I just mean we have a labelling for generators
@AlexClark And the generators are chosen corresponding to the simple roots?
user147690
For the weight vector thing in condition 1 of the canonical type basis
user147690
@TobiasKildetoft If they are I hadn't noticed it :P
10:35
@AlexClark I mean, there are manu possible choices of generators for your Lie algebra
and those choices will change what property 2) says
user147690
Well there will not necessarily be a unique basis of canonical type btw(although there is for $A_1,A_2,A_3$)
user147690
Property 2 being $1\in B$?
@AlexClark Woops, I meant 3)
Right, but if it depends on a choice of generators, then the word canonical seems misplaced
user147690
Hmmm how would it change property $3$?
user147690
So we write $f=U(n_+(3))$ say
user147690
10:39
then we get a grading $f=\bigoplus f_{i,j}$
@AlexClark property 3) is about applying a specific element, namely $\theta_j$, which could have been chosen differently
I guess it might turn out not to depend on that choice, but I would be surprised
user147690
It's $\theta_j$ for each $j$ in the indexing set, so it is all $j's$
user147690
and each $p\in \Bbb N$
user147690
So all right ideals being all ideals given by arbitrarily long left tensors onto $\bf f$
user147690
Or am I missing what you mean still
10:41
@AlexClark No, it is all of the chosen generators, but only one at a time
so sure, if we just relabel the same generators it will not change. But if we pick a completely different set of generators it is not clear to me that it will not
user147690
Do you read a right ideal as, say for $\theta_1^4$, $\quad\quad\theta_1\otimes \theta_1\otimes \theta_1\otimes \theta_1 \otimes U(n_+)$?
user147690
Because I think that is what was intended
@DavidZ - I simply wish to understand why is that function linear
how do I interpret equality in generalized triangle inequality? For complex numbers $z_1,\cdots,z_n$ the generalized triangle inequality is $|z_1+\cdots+z_n|\le |z_1|+\cdots+|z_n|$. Is this statement correct: $|z_1+\cdots+z_n|=\sum |z_i|$ if and only if at most one number is non zero or $\mbox{arg}(z_i)\equiv \mbox{arg}(z_j)\mod 2\pi$ for all $i,j$ such that $z_i,z_j$ are non zero.
@AlexClark Right
@Shahab Why would you have at most one of them be non-zero?
@Shahab Ohh, never mind, I read it as an "and"
10:54
I am trying to formulate the correct statement. My doubt is whether the above statement is correct or not. (I know the proof if all are non zero)
@Shahab It is easy to reduce to the case where they are all non-zero, as the zero ones do not contribute to either side
user147690
@TobiasKildetoft But then there doesn't seem to be an relevance in the order of $\theta_1,\cdots,\theta_j$?
@AlexClark Right, but what if you replace $\theta_1$ by $\theta_1+\theta_2$?
user147690
Oh
@TobiasKildetoft That is fine. But my question is whether the above statement is exactly correct or not.
10:58
@Shahab If it is correct when they are all non-zero, then this is the correct generalized version (and it certainly looks correct to me).
@Shahab Note that the part about at most one non-zero number is contained in the second part (i.e. you don't need to mention it)
So it is suffices to say: $|z_1+\cdots+z_n|=\sum |z_i|$ if and only if $\mbox{arg}(z_i)\equiv \mbox{arg}(z_j)\mod 2\pi$ for all $i,j$ such that $z_i,z_j$ are non zero?
@Shahab yeah
@TobiasKildetoft: Thanks.
user147690
@TobiasKildetoft Good call, the first paper didn't have a constraint, Peters has them as chevalley generators
user147690
That should make it fine then I imagine
11:14
@AlexClark Certainly puts a lot more restrictions on them (no idea if it suffices, but I have not seen this before)
user147690
@TobiasKildetoft Thanks for that though, that definitely changed the way I would phrase my write up :D
user147690
In my mind I could just write the super diagonal entries as $\theta_1,\theta_2$ so on, but I hadn't even written that
11:49
ok i misread the biginning
:29005300 i havent read the question context, that is linearity definition of course
since y is a convolution between two signals but it lacks som additional context
12:11
A homotopy of a regular function is a function field with cardinality zero. You can see this by applying the tensor product to the Grassmannian. But of course it lacks rigid diffeonesis between the cohomologies, especially in a Minkowski space.
@anon It will be interesting to see if tylerl-uxai produces anything better. I wonder why he says that ChatJax is a "stupid system"?
user174558
@robjohn Mean Square meets Green Square
i think he wanted this change implemented @robjohn
@skillpatrol All that does is provide a box into which one has to type MathJax to see it on that page.
he didn't like the click and drag method
12:19
@skillpatrol the click and drag is just for installation. His "solution" requires another page and copy and pasting of the LaTeX on that page.
Each piece of LaTeX
hi @robjohn .. sorry to bug you but in case you had a moment
@Anush what's up?
12:36
@robjohn
Hello
12:50
Hola qué pasa
user174558
Hallo.
Ha!
user174558
Buongiorno, Bonjour, Guten Tag, Buenos dias!
בוקר טוב (boker tov)!
Goeiemiddag!
12:55
שלום (shalom)
user174558
Is there any good reason to learn Hebrew if you are not going to Israel?
Better to read the Bible in the original, I guess
user174558
I think Biblical Hebrew is different from modern Hebrew though.
It is, but not massively.
user174558
The same goes for Biblical Greek and modern Greek.
user174558
12:58
And Koranic Arabic and modern Arabic.
Maybe you want to watch the TV show Srugim without subtitles?
Israel's not a bad place to visit, though. I'm going next week.
user174558
I know Israel has many beautiful women.
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