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04:00
@Null your "field" must be an "ordered field"
Also, 0 is addition identity in any field
ah ok, then not, because ordering is certainly not required for my statement to prove
@Null I don't understand.
Your statement contains ordering and it does not have to be ordered?
I just thought of a way to prove it and played around a little
so what is your doubt now?
mmh, we don't need ordering for the statement appearantly
but ordering would make things easy haha
04:04
@Null: In any ordered field, if $0<x<1$, then $x^2<x$.
all I'm saying^^
@TedShifrin just multiply x<1 and x=x ?
Using $x>0$, of course.
When you write $<$ or $>$, you're using ordering!
@Null because that uses norm
Hey guys, I am trying to define a cayley table abstractly, suppose I have a magma of $n$ elements $(M,\circ)$ such that any pair of products will be equal to one of the $n$ elements in the underlying set $M$. Suppose now I impose an ordering onto $M$ and call this new ordered set $S$, can the following set notation give the cayley table of $(M,\circ)$?

$$T=\{\ x\lvert \forall (a,b) \in S \times S, a\circ b=x\}$$

or I will just end up with an ordered set (probably in dictionary order) that has entries of the form $(a,b) a\circ b$ instead of the desired form $a\circ b$?
04:05
No, norm doesn't give inequalities.
@TedShifrin I'm looking at the axioms, so let me prove it lol
Good practice for you, @DHMO.
@TedShifrin you mean with norm, norm in a division ring?
Using (1) on $x < 1$ we have $0 < (1-x)$
Also, we have $0 < x$
Then by (2) we have $0<(1-x)x$
So $0 < x - x^2$
And then by (1) we have $x^2 < x$.
04:07
you and everything on proofwiki
I didn't mention norm. I'm just saying that a notion of inequality is unrelated to a notion of norm.
Yup, that's it, @DHMO.
@TedShifrin it was nice lol
BTW, how do you prove that $-(-x)=x$ in a ring?
Or, better, yet, $(-1)(-1)=1$?
@TedShifrin Additive inverse of additive inverse is itself?
@TedShifrin I proved that in my sandbox in my proofwiki account lol
does it make much sense to talk about absolute convergence without respect to an ordered field?
04:10
For the second, one needs to understand why multiplying by $-1$ gives the additive inverse. That's not trivial.
i mean, |z| in C, means some real number, but what would |apple| mean?
@DHMO: Try to make that proof in about 3 or 4 lines, not 20. :)
Let c = -x.
By definition, x+c = 0 = c+x
Then by definition, -c = x
@TedShifrin you would need zero-absorption theorem which is not one of the axioms
my aim was to use only axioms
Yeah, by uniqueness of additive inverse (which needs to be proved, but it's easy).
I want only axioms, as well, @DHMO.
If two random variables are independent, their covariance is 0. isn't it?
04:11
@TedShifrin well I aim to use only one axiom in one step
Sure, sure, @DHMO.
You didn't do the most straightforward argument.
hi @TedShifrin can you explain to me the difference between uniform convergence and pointwise convergence?
Is it possible in 4 lines? @TedShifrin
@DHMO my name is Fawad Mirza
@Ramanujan I thought your name is Ramanujan lol
04:13
@Jacksoja: The issue with uniform convergence is that you get convergence at roughly the same rate everywhere. If you draw an epsilon band around the graph of the limit function, for large enough $n$, all the graphs will be inside it.
Fawad is a Pakinstani name :o
@TedShifrin from what i understadn uniform is stronger condition than pointwise is that correct?
Yes, as I just explained.
@DHMO it's a Muslim name :)
@TedShifrin: did you get a lot of rain?
04:14
@Ramanujan oh, so you're a Muslim Indian
Yup, it's rained most of the day, @robjohn.
@DHMO yes :)
@TedShifrin we got about 7/8"
what is pointwise convergence then ?
@DHMO: I guess I first want to prove $0\cdot a = 0$ for all $a$. Then this result follows easily.
04:15
@TedShifrin rained most of the day until about 3:00
Pointwise convergence gives you the limit function. For each $x$, you find $\lim\limits_{n\to\infty} f_n(x) = f(x)$.
@DHMO iam in Hyderabad
can you give me a good lecture or reading material about this please?
@Ramanujan but there's not much Muslims there lol
04:16
my book dont explain this very well , it was 2 lines "proof"
I always recommend Spivak's Calculus, @Jacksoja.
@TedShifrin and then if you chain the steps into one whole paragraph you would get something similar to my proof
in some sense my proof is one-line.
No, yours is still too long-winded.
@TedShifrin what would yours look like?
"proving a result first and then prove this" counts as two steps
So $0=0\cdot a= (1+(-1))\cdot a = a + (-1)\cdot a$. Adding $-a$ gives $-a = (-1)a$.
On the other hand, $0\cdot a = (0+0)\cdot a = 0\cdot a + 0\cdot a \implies 0\cdot a = 0$.
That's a lot less than 15 lines.
04:23
we used different styles
you can't compare apples and oranges
Excuse me?
@TedShifrin thank you sir
ill try to get that book
It's great, @Jacksoja.
@TedShifrin Look at my proof; I chained everything together in one long line
xxx=xxxx=xxxx=xx=xxx=xxx=xxx=xxx=xx=xx=xxx...
DHMO dont argue with Ted ull never win :)
2
04:25
@DHMO: But yours is far more complicated than it needs to be. I agree that mine needs a few more lines to justify "definition of 0" and "distribute law", "definition of additive inverse," but yours is unnecessarily complicated.
Math is so hard.
But, honestly, I just don't care. Good night.
@TedShifrin I agree
Math is difficult.
@kayak 무슨 수학 공부하고 있어?
04:26
Statistics, ergodic theory, Partial differential equation.
Do I have to give up? TT
PDE의 예 있어?
examples?
simple math: work hard, become hard ;)
@DHMO your proofs got very quick, nice nice
(i mean those axiom proofs)
@Null which?
@kayak 응
elliptic equations are those.
Parabolic equations are those.
And I study semigroup theory.
04:29
I think it's hard because you're a postgraduate...
@DHMO your last proof a couple minutes ago, came like a bullet :D
I gave three proofs a couple of minutes ago
but I think you're referring to this
23 mins ago, by DHMO
Using (1) on $x < 1$ we have $0 < (1-x)$
Also, we have $0 < x$
Then by (2) we have $0<(1-x)x$
So $0 < x - x^2$
And then by (1) we have $x^2 < x$.
^this
yep it's nice xd
Lol
You two are great mathematicians.
04:32
@kayak i'm just in the final year of high school
you're my senior xd
i am more of a lazy person.
like, damn lazy
Lazy and the level of the year is not metter.
matter
@kayak 한국어로 말하자?
@Null but not with brain
그럴까?ㅋㅋㅋㅋ
04:34
@kayak 응
@Ramanujan can't judge that statement.
I think we have to make a new chat room
for Korean talking room
lol
04:52
That is, will $T$ be the set described in the red box?
@Secret can we make orderings like $0^2<0\cdot 1<0\cdot 2...$?
(note that I don't mean "<" as usual)
What $0$ are we talking about here, is it an element in a field, a group, a semiring, a semigroup or something else?
and how is your < defined?
ordering
i mean, wether you call it now bigger or more right, or more pink, does it really matter?
ok, so its some kind of abstract ordering, rather than an ordering given by some relation between two or more elements,
now what properties do your elements $0,1,2$ etc. have?
sorry
05:03
oh, that $\times$ is the cartesian product
if I recall, a cartesian product of an ordered set with an ordered set gives an ordered set
i meant that, but I now see what your T is
(unrelated) it's amusing how "waystage" (CEFR Level A2) sounds the same with "wastage" (A useless person)
My $M$ is just a generic magma with 10 elements, there's nothing being specified for the elements a to j
I think that the box describes T, but wouldn't any box with all elements aa, ab,...,ji,jj do so?
If I recall, $A \times B=\{(x,y)|x\in A \textrm{ and } x \in B\}$ thus $S \times S$ should have all possible pairings of a and b, not just the diagonal ones plus a couple of cross terms?
hmm, I found something:
In mathematics, given two ordered sets A and B, one can induce a partial ordering on the Cartesian product A × B. Given two pairs (a1,b1) and (a2,b2) in A × B, one sets (a1,b1) ≤ (a2,b2) if and only if a1 ≤ a2 and b1 ≤ b2. This ordering is called the product order, or alternatively the coordinatewise order, or even the componentwise order. Another possible ordering on A × B is the lexicographical order. Unlike the latter, the product order of two totally ordered sets is not total. For example, the pairs (0, 1) and (1, 0) are incomparable in the product order of 0 < 1 with itself. The lexicographic...
so it seems the cartesian product alone does not specify the type of ordering being imposed
05:16
that ordering is not total right
no, it is not total
no, that isn't even an ordering
Because it is not antisymmetric
I think since the product order is defined to be $(a_1,b_1) \leq (a_2,b_2)$ iff $a_1 \leq a_2$ and $b_1 \leq b_2$, it is antisymmetric wrt the a s and b s
but not wrt the ordered pairs themselves as they are not a total order
Well, consider $a_1 < a_2$ and $b_1 > b_2$.
Both statements are false:
$(a_1,b_1) \le (a_2,b_2)$
$(a_2,b_2) \le (a_1,b_1)$
So it is not antisymmetric
The product order need to have the partial ordering of a and b to be facing the same direction, so the example you have above is not a product order (and as you have noticed, not even an order)
Meanwhile trying to look up for the name of the ordering in a matrix
For the first step, this is what I am trying to do in the above definition of $T$, by imposing a product order (or more accurately:
In computing, row-major order and column-major order describe methods for arranging multidimensional arrays in linear storage such as random access memory. In row-major order, consecutive elements of the rows of the array are contiguous in memory; in column-major order, consecutive elements of the columns are contiguous. Array layout is critical for correctly passing arrays between programs written in different languages. It is also important for performance when traversing an array because accessing array elements that are contiguous in memory is usually faster than accessing elements which are...
so that a cayley table behaves like a matrix
05:29
http://math.stackexchange.com/questions/2061088/sequentially-defined-probability-density-function-and-its-umvue-estimator
I proved a question by myself. Anyone who want to check this?
No my answer is wrong.
Will the following set notation give me the desired set that is the rows of the cayley table?
06:06
0
A: Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a(ac)^b(ab)^c}$

RamanujanHint $$ \Huge{x>y}$$ then $$\huge{\frac{1}{y} > \frac{1}{x}}$$

Someone check my answer
06:51
Is this rational? $\displaystyle\sum_{n=0}^\infty \dfrac{2n+1}{(4n+1)^2(4n+3)^2}$
07:25
@DHMO: According to WolframAlpha this becomes $\fracC8$ where $C$ is Catalan's constant. According to wikipedia it is unknown whether it is irrational.
C/8 *
\frac C8
you need some space
ah whoops thanks. but yeah. there's the answer.
thanks
sorry i need to go now
07:29
Bye
cya
@DHMO Hey
잘자~
07:57
@kayak 안 저고 있어
"싶은데"하고 "싶어" 같아?
08:08
No different @DHMO
싶어는 상대방에게 요구하는거고
싶은데는 sometimes, 후회할때 씌이고, sometimes, 상대방에게 요구하는걸로 씌이지
싶은데 can be used for both situations like 1) regretful 2) require or need from others.
싶어 can only be used for 2).
08:26
thanks @kayak
 
1 hour later…
09:55
Can I produce ordered rows this way with that set builder notation shown a few posts above?
10:24
If I know the eigenvalues of two lower triangular matrices $A$ and $B$, can I say something about the eigenvalues of the matrix product $A.B$?
10:36
Hello!! Does it hold that $\{x(1, 0, 0, 0) + y(0, 1, 0, 0) + z(0, 0, 2, 1) + w(0, 0, 0, 1) | x, y, z, w \in \mathbb{R\}} = \mathbb{R}^4$ ?
@MaryStar Hint: Check whether the set form by these 4 vectors are linearly independent
yes @mats, a lot actually. A generic element on the diagonal of $AB$ is of the form $(AB)_{ii}=\sum\limits_{k=1}^nA_{ik}B_{ki}$, now if $A$ and $B$ are lower triangular...
@Secret They are linearly independent.
then by rank nullity theorem, they must span $\mathbb{R}^4$ and hence the statement holds
since they are independent what's the dimension of their span as a subspace of $\mathbb R^4$?
10:39
So, when we have 4 linearly independent vectors, they form a set that is equal to $\mathbb{R}^4 ? It holds because when $A\subseteq B$ and dim(A)=dim(B) then A=B, right?
Ah ok... Thank you!!
@Alessandro I have put the matrices here: math.stackexchange.com/a/2062116/8530
I meant one lower triangular matrix $A$ and one upper triangular matrix $B$.
I don't have the time to read the whole question now sadly
ah, I see, that's a completely different question, I'm not sure about it
10:45
Hello
I would need some help from people who are very good at logical thinking. (if that is how you properly say it in English...)
In this paper on p12 you can see that two images are subdivided in chunks of 4x4 pixels: ti.com/lit/an/bpra065/bpra065.pdf
On page 13 they are applying a formula to calculate $\alpha$
As you can see it contains the following part: $\sum{m,n} (D_{i,j})_{m,n} (R_{k,l})_{m,n}$
My question is:
i,j ranges from 0 to 16 and m,,n from 0 32 what happens when i,j reaches its maximum value (16) and you still have all these blocks m,n?
Isn't there something wrong with this formula?
hi @balarka
Yo
whats new
not much, I'm studying for an abstract algebra exam
Error detecting and error correcting codes plus a couple of isomorphisms theorems, nothing too fancy
10:59
gah programmers. I have a function like exp(float x, unsigned int n) and they can't tell if it is $x^n$ or $n^x$. I'll be damned dead before using variables with more than 1 letter
@Alessandro boo
Ah, we also saw some basic stuff about PIDs, UFDs and Noetherian rings, that was pretty interesting
much better
although my algebra is really bad
Since next year I can choose (almost) freely my courses I think I'll do all of the algebra and geometry offered and stay as far away as I can from analysis :P
analysis is cool, but hard
I like the Fourier theoretic flavor, where all of it - algebra, geometry and analysis - get punched.
11:07
@Sophie those are not true programmers
Trying to find interesting properties of 2017 that might show up in olympiad questions
other than it's prime :P
most interesting so far oeis.org/A039917
11:23
Is $\displaystyle \sum_{n=0}^\infty \frac{e^n}{n!}$ rational?
it's probably open
but shrug
@BalarkaSen how can it be open?
i don't understand the question
Weird thing I noticed today
actually that's just $e^e$ isn't it?
11:27
@BalarkaSen how come people still cant solve it?
yes it is
well deciding irrationality of numbers is hard.
In all the SO network my profile name renders as ypercubeᵀᴹ.
In my math.se profile, it renders as ypercube(SM).
It probably has to do with the font.
@BalarkaSen but it doesn't look rational at all
Q has L measure 0
that's not a proof
and the lebesgue measure is irrelevant
neither does $e+\pi$ but we don't know whether it's rational either
11:36
$\displaystyle\sum_{m=0}^\infty \sum_{n=0}^\infty \dfrac m {m!n!}$
$e^a$ is transcendental if $a$ is algebraic and nonzero
Lindemann-Weierstrass IIRC
In transcendental number theory, the Lindemann–Weierstrass theorem is a result that is very useful in establishing the transcendence of numbers. It states that if α1, ..., αn are algebraic numbers which are linearly independent over the rational numbers ℚ, then eα1, ..., eαn are algebraically independent over ℚ; in other words the extension field ℚ(eα1, ..., eαn) has transcendence degree n over ℚ. An equivalent formulation (Baker 1975, Chapter 1, Theorem 1.4), is the following: If α1, ..., αn are distinct algebraic numbers, then the exponentials eα1, ..., eαn are linearly independent over t...
$\displaystyle\sum_{m=0}^\infty \sum_{n=0}^\infty \dfrac {m^n} {m!n!}$
@DHMO calm down
11:40
I made a typo in the first time
$$\sum_{p=2}^\infty\sum_{q=2}^\infty\frac{1}{p^q}=1$$
geometric series and then telescope
Schanuel's conjecture of course gives all of this anyway so it's probably true
how is the truth value of this affected by any conjecture at all?
Schanuel's conjecture is almost certainly true, so if this is implied by that it's true too
11:43
@DHMO it's a pretty identity, isn't it
yes it is
hello everyone
just for claryfication
$\alpha = [0,1] \in R$
means -0.0057 is not OK, right?
yes
Is the set of subets of a group with the addition of sets a monoid, or is it more than that ?
damn...
11:47
what is the operation of the new set? @Astyx
$A+B = \{a+b, a\in A, b\in B\}$
messy set
Messy ?
the identity is {0}
I mean, this is quite a messy set
That's not a subgroup in general if neither A nor B is normal unless I'm mistaken?
11:49
A and B are not subgroups, just subsets
And that couldn't possibly be a subgroup because the elements are not elements of G
Ah, right, I misread your message
@Astyx I don't think inverse exists
Definitely not
Except for singletons
because |A+B| >= |A|
so A and B must be singletons for A and B to be inverse
a monoid is a group without inverse right
Yeah
11:52
just associative and identity
so it is a commutative monoid
And if $G$ is infinite, is it regular ?
Anyone can help me on how to correctly formulate a set notation that can pick out all rows of a cayley table. Null and DMHO checked that the set $T$ is indeed the cayley table, but I am not sure whether I can restrict one element of the ordered pair that is in the definition of cartesian product in order to pick out the rows only

I am thinking in terms of arrays like in programming, I am not sure how to translate the programming thinking into formal set notation cause so far people have no idea what I am doing when reading my attempted formulations
@Secret I don't quite get what you want to do
Oh I have to go and eat
Be right back

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