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03:00 - 19:0019:00 - 00:00

19:00
It's probably not.
It's just sending a message as the constant term in a polynomial, and you distribute coordinate pairs to different people, and then if ever they need to reconstruct the polynomial, you need a certain number of them, from which you can interpolate.
I'm incapable of explaining anything, but in any event, it's just Lagrange interpolation over a finite field, and I know the only way you can have a field is with a prime modulo, but I don't understand why we must have inverse. I know it's helpful, but it doesn't seem necessary. But I don't think I was very clear. I'mma just surf google.
I've got a quick question on some notation I've not seen before, and how I'd plot the set of points on the $x-y$ plane.
$ \{ (x,y) : x \in [ 1, 2 ], y \in [ 1, 2 ] \}$
Would this just be a square that I'd need to shade in, in the $x-y$ plane?
@Anthony If $F$ is a field, $F[x]$ is a PID. $\Bbb Z[x]$ is not a PID.
Maybe that's why?
PID?
Principal Ideal Domain.
19:04
Oh god algebra.
@Anthony Find me the proof that your algorithm works. I assume there's a proof somewhere. Once you see it, we can work backwards to see why we can't do it over $\Bbb Z$.
@Khallil Draw the coordinate axis and the desired area is the square from $x=1$ to $x = 2$ and $y = 1$ to $y = 2$.
Nonono!
@BalarkaSen I thought so! Thanks!
I don't care about $\mathbb Z$ I think they do that because we're working with computers, and we want things to be uniformly random. I just was wondering why we can't do it modulo a composite number.
19:05
\mathbb Z
Oh. Okay then.
@BalarkaSen Oh hold on. If say, our inequalities for $x$ and $y$ were strict as opposed to not strict, would I need to draw a dotted line on the vertices of the square, to show that $x \neq 1,2$ and $y \neq 1,2$, instead?
I know it isn't nice, but I feel like it should be possible.
@Anthony Maybe because it's not a field?
Yeah, but you can still define a polynomial over it, can't you?
@Anthony It's not that you don't have inverses. It's that you have zero divisors.
19:07
@Khallil Yeah, you can do that.
$2*3 = 0$ mod 6.
That's bad.
@Anthony Zero divisors.
Probably.
@Mike beat me to it.
19:08
Having zero divisors will make polynomial multiplication a fair hell.
But you can still define a polynomial over a non-prime modulo field, can't you?
Yes, you can.
no such thing as a non-prime modulo "field" :)
You can define a polynomial ring over any ring
Ugh
I just realized that
Can't it have powers of a prime, or something?
Oops. Yeah, you can have fields with prime powers.
19:11
No, @Balarka
There are fields with $p^n$ elements for any prime $p$, any integer $n>0$
But I think @Anthony means whether $\Bbb Z/p^n\Bbb Z$ is a field, which it's not
Obviously I need a cup of coffee.
@MikeMiller Ah. Champion.
I confused with elements and orders.
Coffee is good.
@Anthony $\Bbb Z_4$ still has zero divisors. $2 \cdot 2 = 0$.
There is no field with zero divisors.
19:12
A wise man once told me "A good cup of coffee should practically taste like charcoal."
@MikeMiller ffee isn't.
@Anthony Dude, you live in the bay area. You need to go to Philz.
Philz?
The best coffee I've ever had.
I see.
I don't love coffee, but I'll try it.
For you, Mike.
19:14
They opened one in Santa Monica, and that makes me happy.
Hello
Any one here?
We're all ghosts here
I happened to see a video about [Imagining 10th Dimension][1]. In that (17:03m)
he say that Zero contains every possible values by referring to an author [GEVIN GIORBRAN][2] . It goes like this.

- 1+(-1)=0
- 2+(-2)=0
- 3+(-3)=0
- **∞+(-∞) = 0**

Considering Zero as the set containing -∞ and +∞, it can be the largest possible value. Any other values is simply the absence of the same negative value.

So 1 is the absence of (-1) from the whole set and so on. When put together will come to the conclusion that 1>2>3... which make no scene.
I usually consider 0 as the largest element in the multiplicative sense. "Containing" integers is just stupid.
what's wrong with the concept is that, as far as I can tell, it's complete nonsense
that "imagining 10th dimension" video is not even wrong.
19:21
The only way the maximality of 0 makes sense is putting it in the peak of your divisibility poset. One can identify it with $\prod p$
@BalarkaSen be careful about using notation like that to mean something that's not "the product over all primes"
Yes. I agree.. But if we consider set a+b=c and we can say that set c contain a and b right?
since what you really mean is the unique element that all primes divide
@IndrajithIndraprastham In that sense, n contains everything below n.
It's a nonsensical concept.
Yes. Later in the video this theory was mixed with matter and anti mater. Do you suggest asking this questiion in physics.sx ? Or is this just bullshit?
19:25
Well, in fact, if you are in $\Bbb Z$, not only $0$ but $1$ is also the largest element. $1 = -1 + 2 = -2 + 3 = -3 + 4 = \cdots$ so $1$ "contains" all elements too.
It's just bullshit.
I forgot about my no-swear words campaign.
One can't even debunk it because it is like an hour-long video adding more and more things that just make no sense
seek to 17minut...
I've seen it before, it's not worth your time (or mine, or Balarka's, or physics.SE's, or...)
19:27
@BalarkaSen According to that consept .. 1 is only created when -1 from the whole set is taken out. Which makes it lessthan 0
and so on
I am some what adjusted to that concept. can some one logically prove that wrong.. please
I need my logical state back.. LOL
@IndrajithIndraprastham That doesn't make any sense.
@MikeMiller So a positive point is that we have an analogy for galois extensions : the characteristic subgroups.
@BalarkaSen N contain everything below it can be true?
according to him it was like this .. If set a+b= c . we can say that c contain a and b. similarly inf+(-inf)=0 implys zero contain -inf and +inf. so making it the largest. ..
19:42
@IndrajithIndraprastham Not only everything below it but all the integers too. $n = (n - k) + k$.
@BalarkaSen If you extend your analogy more I'd like to see it.
@MikeMiller Is there anything more to do other than mimicking the fundamental theorem?
I'd like to have your suggestions.
Well, there's certainly more than a little bit of Galois theory. But right now I'm not sure what your comparison is. Analogues of Galois theory that I know say "this thing is like groups; this subthing is like subgroups; this special subthing is like normal subgroups"
Say, covering space theory, or Galois theory itself.
If you want to mimick the fundamental theorem you've gotta find what characteristic subgroups correspond to.
Last time you were messing with this you were looking at extending that exact sequence, which is still something you can try to do, but that might involve learning more category theory/homological algebra than you want right now.
I don't get it. I am not a mathematician or a physicist. I am just a programmer. This is what I understood. I am not saying that what I say is right. I am just asking you to prove this wrong so that I can go sleep well..
@MikeMiller You're right. I just mimicked the main theorem about 1 --> Gal(K/F) --> Gal(K/E) --> Gal(E/F) --> 1, not the correspondence.
@MikeMiller wait i was under the impression that we had a short exact sequence.
@IndrajithIndraprastham what do you want me to disprove?
19:50
@BalarkaSen Nope, it's not exact on the end, remember?
You've got a left-exact sequence.
Ah
1 --> Aut_H(G) --> Aut_N(G) --> Aut_N(H)
So maybe that's what you want to do. Hunt down some nice correspondence, and extend your exact sequence (I'd start with the first, maybe. Dunno.)
rather than saying this is bullshit (which i agree) just prove that it is not possible.. that zero can never be the largest number and that logic which he said is wrong.
@IndrajithIndraprastham i have already told you that according to your definition of 'largeness', not only 0 but every integer is 'largest'.
$n = (n - k) + k$ for all $k \in \Bbb Z$.
@MikeMiller you really think that there is more than what meets the eye? blue is totally skeptic about the idea.
and i admit i too am, to some extent.
@BalarkaSen No, not really. But it seems like it's something you want to keep pursuing, and it could be a fun project. Trying to do some math instead of just learning it.
Rather: I don't know if there's more than meets the eye, and I don't plan to find out, I've got tests to study for :P
20:04
haha, that's fine. i'll do some digging about this thing then.
20:39
Can anybody help me out with sketching this set of points on the $x-y$ plane?
$\{ (x, x+y) : x \in \mathbb{R}, y \in \mathbb{Z} \}$
I think that set of points describes the entire $x-y$ plane. Can anybody verify that?
it doesn't
I think I'm misinterpreting what the set is trying to describe. Would I be right in saying that in words, it'd read:
"The set of all points of the form $(x, x+y)$ such that $x$ is real and $y$ is an integer"?
Oh hold on, I think I've got it.
Would it be a set of lines of the form $y=x+c$, where $c$ is an integer?
@MikeMiller
20:51
Cool, thanks!
I've got another question on some notation I've not seen before.
$\{ (x,y) \in \mathbb{R}^2 : (y-x)(y+x) = 0 \}$
What does the first part mean and how is it read?
Also, I've managed to find that I ought to sketch the lines $y = \pm x$.
"the set of (x,y) in R^2 such that..."
$\mathbb{R}^2$ being two dimensional space i.e. the $x-y$ plane?
Boo @Mike @Khallil
Hey, @Ted!
How are you?
21:03
Getting psyched for classes tomorrow morning ...
LOL
Ours are in a month and a half
Well, we finish beginning of May.
That's pretty early!
We start early ...
@MikeMiller May I ask why it was necessary for the author to stipulate that the set of all points of the form $(x,y)$ be in $\mathbb{R}^2$?
Fair point. I guess it can only be a good thing if the exams don't coincide with the scorching Summer heat.
21:06
Suppose you were in $\Bbb R^8$ with the same equation?
@TedShifrin I can't even begin to fathom what $\mathbb{R}^8$ is. =S
That's not my problem :) $\Bbb R^3$, then.
I think he means "why does (x,y) not automatically mean elements of the plane, since there are two 'coordinates'?"
Oh, as opposed to $(x,y,z)\in\Bbb R^3$? Somehow we need to specify the domain.
You just need to know $x$ and $y$ are the only coordinates. Knowing real/complex, etc., might be helpful or necessary ...
@MikeMiller That's exactly what I mean!
So it's just for the purpose of being explicit enough to avoid obscurity?
21:12
You need to know what domain we're in ....
@TedShifrin I think I see what you mean. Lets say we were working with a set of coordinates in the complex plane. How would you then denote the set of all points of the form $(x,iy)$? So just like for the Cartesian plane, you'd say $(x,y) \in \mathbb{R}^2$, what would you write down for the complex plane?
$(x,y)\in\Bbb C^2$ ...
Sorry for the late reply, but why would it be $\mathbb{C}^2$ as opposed to $\mathbb{C}$?
I thought $\mathbb{C}$ was the set of all complex numbers of the form $x+yi$, where $x,y \in \mathbb{R}$ i.e. I thought that $\mathbb{C}$ could be represented on an Argand diagram which is analogous to the Cartesian plane.
@Ted - How would you describe $\mathbb{C}^2$?
21:31
Ordered pairs of complex numbers. It looks like. $\Bbb R^4$.
Oh, ok. I think I understand it a bit more now. I need to do some more reading on complex numbers!
22:14
2+2/2*2-2=? anyone halp
Is that how it's written or have you forgotten parentheses etc.?
If it's the former, note that BIDMAS is the order in which you must do the operations.
(Brackets, indices, division, multiplication, addition and finally subtraction.)
@Khallil sry man i know only word BDSM
@crazypotato Well, BIDMAS is what you should be going by.
2
22:36
hi
1. Do all multiplications and divisions in order from left to right.

2. Do all all additions and subtraction in order from left to right
2 Answer
hi
Note @crazypotato "in order from left to right" is important :-)
22:59
Hey, @skullpatrol.
Could you answer a quick question on the Cartesian product on sets?
I need to find $B^3$ i.e. $B \times B \times B$ given that $B = \{ a, c \}$.
I've only ever found the Cartesian product of two sets before, so I'm unsure of how to begin.
23:24
@Khallil: Ordered triples instead of ordered pairs ...
@TedShifrin Is there an intuitive way of finding all the ordered triples like the rectangular array for finding ordered pairs?
23:44
@Khallil: Hence you go from $\Bbb R^2$ to $\Bbb R^3$. You need a 3-dimensional box.
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