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09:05
$\phi$ $\emptyset$ $\varnothing$
I like $\varnothing$ more than $\emptyset$ despite the LaTeX command name
@LeakyNun How about $Ø$?
@TobiasKildetoft latex :'(
@KaustabhaRay empty set has no element
$\varnothing \in \varnothing$ is false
wait I'm missing the context
I didn't know he meant the empty set
oh
coz I saw him asking about phi
right, he confused phi and empty set
yeah, the var one looks more 'set'-y
thats....the weirdest adjective i've used in a while
09:16
set-like
that works
Going by that, $ {\varnothing} \in {{\varnothing}}$ would be true I guess
@KaustabhaRay But you were just told that it is false
Wait I meant to write { \varnothing }
Parenthesis in latex isn't showing up on my device
If you want { } to show up, put backslashes in front.
\{ \varnothing \} renders as $\{ \varnothing \} $
09:24
Oh sorry for that
I meant $\{ \varnothing \} \in \{\{\varnothing\}\}$ should be true I guess
looks good
@LeakyNun Hello ! :D
@KasmirKhaan hi
How are you? :)
09:39
Nice nice =P
Am doing some homomorphism and isomorphism examples atm
but what is the important thing about those?
I mean like structually ?
@KasmirKhaan The importance is precise that they preserve the structure
@TobiasKildetoft Hmm thanks :D
homomorphism is one criteria right?
f(xy) =f(x)f(y)
For groups, yes
but isomorphism is that criteria plus bijectition
09:43
Yes am doing groups =p
what is monomorphism and epimorphism ?
my teacher mention them without defining them properly
mono means injective and epi means surjective
so they are like hmm defintions that stand on it own?
(though I still think epimorphisms should be referred to as "epic", since monomorphisms are referred to as "monic")
not sure what you mean by standing on their own
haha that would be cool name for it ._.
09:45
like we need those "names" , as in those names arent other names for something we allready know
That reminds me of one of my first abstract algebra classes, where we learned about all that.
My supervisor taught it, and was like
"we will shorten isomorphisms to 'iso's', monomorphisms to 'mono's', epimorphisms to "epi's", and homomorphisms to... 'morphisms'. "
@KasmirKhaan For one thing, those names have very general meanings that work beyond cases where injective and surjective make sense
if we use surjective and injective, we dont need to use mono and epi right?
also, the terms become shorter which can be nice
@TobiasKildetoft that clears it thanks alot Tobias :D
one last Q =p
09:47
Unfortunately, there are also cases where injective and surjective make sense and these don't agree with mono and epi
For an inverse function to be defined, the function need not be onto right?
if we have the homomorphism criteria but we have only one of the criteras either injective or surjective
@KaustabhaRay For a right inverse, you need surjective. For a left inverse you need injective
does that make up something ?
@KasmirKhaan You mean without specifying which of the two we have?
09:48
@TobiasKildetoft all cases =p
if we have say hom + injective without surjective
or
hom + surjective without injecticve
does those things exist? and do we work on them ?
@KasmirKhaan We usually don't distinguish epis or monos that are not isos
all righty :D
since that is usually not necessary
Now I just have to keep doing more examples on hom and isos to get this into my head =p
At least not in the case of groups
09:50
yes we keep this on groups only sicnce am doing abstract algebra
As far as I recall, they do play important roles in the study of Artin algebras
hmm never heard of this =p
@KasmirKhaan do you want elementary group theory exercises?
@LeakyNun Yes sir! much appriciated :D
I could show you my solutions and you can help me with em ( if thats ok with you ofc ) :D
you really need to improve your rigor lol
09:53
Working on that ><
I send those exercices that you helped me with to my teacher btw =p
I know they were not 100% as u wanted them but let see what he sais ><
Let $G$ be a group. Prove that $Z(G) = \{z \in G \mid \forall g \in G [gz=zg]\}$ is a normal subgroup of $G$.
(Center of group, Z for German Zentrum)
gonna copy this and solve it later because we did not do normal subgrous yet =p
then ignore the word "normal" and do it
@LeakyNun okay =p do you have examples on homomorphism and isomorphisms? :D
because next lecture is on those
I need to get a feeling so i can follow the lecture tomoorow with eas
examples?
09:58
@KasmirKhaan Let $g\in G$ and consider the map $G\to G$ given by $x\mapsto gxg^{-1}$
Show that this is an isomorphism.
woo, conjugation
@LeakyNun Yes examples =p
@TobiasKildetoft I was about to type that, but he asked for examples not exercises
@TobiasKildetoft thanks working on this now =P
i meant exercices sorry
><
@LeakyNun
@LeakyNun Well, it is an example given as an exercise
09:59
[Philosophy of maths] To be expanded: How to deal with maths that is fundamentally beyond human understanding due to limitation in our brains wiring. What other reliable tools or thinking we have besides maths that help us to investigate reality?
Also, if you want to know why group theorists don't see a difference between addition and multiplication, try proving that $(\mathbb{R}_{>0}, \cdot) \cong (\mathbb{R},+)$
f(AB)= MABM'
f(A) F(B) = MAM' MBM' = MA(M'M)BM' = MABM'
So we have homomorphim, now need to show it is 1-1 and onto.
@TobiasKildetoft
@TobiasKildetoft I feel like many assumptions are missing, how do we know that M' exists?
@TobiasKildetoft should this be done in SL(n,R) ?
@KasmirKhaan what the hell is M?
and no, you should stop looking for actual defined groups such as SL(n,R).
You're studying abstract algebra.
$G$ is any group. Period.
Oh i thought it was over matrices because of gxg' >< conjugation
never mind then :D
let me think again @LeakyNun
Kasmir.
Abstract algebra.
Groups.
Isomorphism.
10:06
Yes yes :D
let x,y belong to G.
f(xy) = gxyg'
f(x) f(y) = gxg' gyg' = gx(g'g)yg' = gxeyg' = gxyg'
thus we have homomorphim.
@LeakyNun
[Random] To be expanded: N-[5,\infty)={0,1,2,3,4}, N-2N=2N+1
go on @KasmirKhaan
Assume f(x)=f(y)
gxg' =gyg'
g'gxg'g = g'gygg' ( multiplication by g on the right and g' on the left )
(g'g)x(g'g) = (g'g)y(gg')
x=y so the map is 1-1
@LeakyNun now I need to show onto criteria and am done =P
@KasmirKhaan go on
hmm ontoness is trivial ( if I got this right )
let z be an element of G , gzg' also belongs to G
f ( ) = gzg' , for all z in G.
we can write any z in G as gzg' since we have a group
I mean what else can I say ? =P
@LeakyNun
10:16
you only proved that f is a function.
you haven't proved that it is onto.
i'm onto you
to prove ontonees , we pick any element z in G, it can be written as gzg'
I mean hmm, I dont know how to say it in symbols
@LeakyNun Hold on I think I got it =p
@LeakyNun let z be an element of G.
z= gg'zgg'
z =g (g'zg) g'
let w =(g'zg)
so f(w) =f(g'zg) = g g'zg g'
you forgot to prove that w is an element of G
ready for the next level?
Yes :D
and g'zg is in G because G is closed =p
Let $G$ be a group. Let $\operatorname{Aut}(G)$ be the set of all isomorphisms from $G$ to $G$. Show that $\operatorname{Aut}(G)$ is a group under function composition.
10:29
@LeakyNun all righty, i might need some time for this one =p
11:27
0
A: Understanding $\sqrt[i]{z}$ for $z\in\mathbb{C}, z\ne0$

Kenny Lau$$\begin{array}{rcl} z^{-i} &=& \exp(-i\ln z) \\ &=& \exp(-i\ln(re^{i\varphi})) \\ &=& \exp(-i[\ln_\Bbb R (r) + i\varphi + 2ni\pi]) \\ &=& \exp(-i\ln_\Bbb R (r) + \varphi - 2n\pi) \\ &=& \exp(\varphi - 2n\pi) [\cos(\ln r)-i\sin(\ln r)] \\ \end{array}$$ For an understanding, we must view number...

Why does the argument of $z^i$ depend on the magnitude of $z$ and the magnitude of $z^i$ on the argument of $z$?
8
Q: Can science work without mathematical formulations?

Remi.bErnst Mayr in his last book titled "What Makes Biology Unique?" argues that many of the theories in biology do not need any mathematical support. He says that much of biology is only conceptual and cannot be describe by mathematical formulations. In the meantime, he also argues that these propert...

Hmm... qualitative reasoning...
but isn't that logic and hence part of mathematics?
1
A: How is this 'explanation' of the unreasonable effectiveness of mathematics even an explanation?

jobermarkWhat we as humans can understand is automatically limited by our processing apparatus. If part of the world were inherently contradictory, we would find some way to pretend that it wasn't, simply because inherent contradiction is beyond the bounds of what we can reasonably abide. We might give ...

> If mathematics is so deeply a part of human understanding that it filters everything we perceive or interpret in this way, how is it mysterious that everything the filter lets through happens to agree very well with mathematics?
So...
What tools do we actually have in investigating reality besides mathematical models?
@Secret Do you like philosophy?
I do, but mostly metaphysics and epistemology
Ok
11:42
For me, philosophy is the study of concepts, and it has a lot of intersection with mathematics because maths is never lack of concepts
@Secret is philosophy taught in schools(in your country)?
I can do, limited amount of the more humanities side of philosophy such as ethics, but the arguments are not formal because I never had a chance to have formal philosophy training
@Abcd Most unis have philosophy courses
@Secret yes, but not schools
That I am not sure, I have not heard of philosophy being taught as a course in high school
@Secret okay. Sometimes I feel tempted to read philosophical stuff related to science, abstractness
11:47
Having some knowledge on the philosophy of science is a good thing, it helps you understand how we use science
yes...
12:08
My recent interest in the philosophy of mathematics and science is wondering about mathematics that is fundamentally beyond human understanding, and what other thinking methods or tools one can use to do science
For the former, it seems we are getting close to such state as evidenced by how we start to have trouble understood some proofs done by AIs
12:20
If we can base insight to God on binary logic, we have got it made.
It is not even clear what non-logic means, given we even have logical systems that allow genuine contradictions
But is binary logic enough to capture all insights of God?
@Secret I am halfly joking. It is Chris Langan that has said this quote originally in one of his videos on Youtube.
I see
key board fail, I am eating
Usually, when I think about theology questions, the first thing is not whether God exists, but rather, if He/She/It/whatever exists, what happens when we interact in different scenarios
12:27
theology assumes that God exists.
In other news, UK visas for Hong Kong people are being delayed for a long time
@LeakyNun I assume you exist because I exist.
The outer circle should be deleted.
With nothing surrounding everything.
0
Q: Complex algebra using cube root of unity

Rajesh DachirajuIn complex algebra we use square roots of unity. Is there any algebra where instead we use cube roots of unity? For example, properties like, as in usual complex algebra $z = x+iy$, we say $|z| = \sqrt{x^2+y^2}$, where in the new algebra $z = x+\omega y$, we say $|z| = (x^3+y^3)^{1/3}$, where $\o...

Everything is nothing, and nothing is something, while something arises from nothing
@Secret everything is not nothing.
nothing is some "thing" though
12:37
@Secret no it's not
Nothing is nothing, but the concept nothingness is a concept
Nothingness is maths is easy to define though:
$\emptyset$
$\phi, \phi$
Now back to some more closer to earth stuff:
$\Bbb{N}$
$|\Bbb{N}|=\aleph_0$
$|2\Bbb{N}|=\aleph_0$
$|\Bbb{N}-2\Bbb{N}|=|2\Bbb{N}+1|=\aleph_0$
$|\Bbb{N}-\{6,7,8,...\}|=5$?
I wonder if there's a maths concept that capture this
The concept here being:
Let $A,B$ be sets, $B \subset A$ and $|A|,|B|$ be their cardinalities
Suppose $|A|=\aleph_0,|B|=\aleph_0$, then the following holds:
$|A|-|B|=c \in \Bbb{N}\cup\{\aleph_0\}$
where the value of $c$ depends on the details of A and B
12:52
Please help me understand the bold part of this question: Find the value of a for which $(ax^2+3x-4)/(a+3x-4x^2)$ takes all values for all real values of x. I don't want the solution.
@Abcd think about it yourself.
Let's take $y =(ax^2+3x-4)/(a+3x-4x^2)$ (Leaky's solution)
@Abcd don't copy my solution.
Leaky argues, for all y there's a root for x, therefore discriminant is greater than zero.
I understood neither the bold part of the question nor Leaky's argument.
think about it yourself.
13:08
O wait, $|A|-|B|$ makes no sense if A,B are infinite, because infinite cardinals have no additive inverses
> On the other hand, subtracting two concrete sets is perfectly doable and legitimate mathematically. But we cannot conclude anything on the cardinality of the result, unless we know what are the specific sets (or at least more than just "two sets", which is the most general notion).
7
A: Subtracting two infinities

Asaf KaragilaSubtraction between infinite cardinals cannot be well-defined. This is a good example why. We know that $|\Bbb N|$ and $|\Bbb Z|$ are both of the same cardinality, but so is $|\Bbb{Z\setminus N}|$. On the other hand, $|\Bbb N\setminus\{k\in\Bbb N\mid k>2\}|=3$ (zero is a natural number here). S...

So, infinite cardinals are strange, in that they are sensitive to details:
Secret.
Subtraction is not well-defined for cardinals.
Period.
Pick the set of naturals $\Bbb{N}$. Now compute the set $A=\Bbb{N}-\{n\}$ for some $n \in \Bbb{N}$. Then $|A|=\aleph_0$. Now compute $B=\Bbb{N}-2\Bbb{N}$, $|B|=\aleph_0$. However, if we compute $C=\Bbb{N}-\{3,4,5,...\}$, then $|C|=2$
An interesting question to ask in general is that: For a countable, linearly ordered set $S$ and some countable subset $X \subset S$, what property $X$ must have such that $S-X < \aleph_0$
@Secret prove that $G/N$ is abelian iff $[G,G] \subseteq N$.
13:32
We wish to regularize the series$$\sum_{k=1}^\infty(-1)^{k+1}\left\lceil\frac k2\right\rceil$$This may be done with zeta regularization:$$f(s)=\sum_{k=1}^\infty(-1)^{k+1}\left\lceil\frac k2\right\rceil k^{-s},\quad\Re(s)\ge2$$The approach requires analytically continuing this, which isn't difficult if you note that
\begin{align}f(s)&=\sum_{k=1}^\infty\frac k{(2k-1)^s}-\frac k{(2k)^s}\\&=\frac12\sum_{k=1}^\infty\frac1{(2k-1)^{s-1}}\\&=\frac12\sum_{k=1}^\in‌​fty\frac{(-1)^{k+1}}{k^{s-1}}+\frac1{k^s}-\frac1{(2k)^s}\\&=\frac12\eta(s-1)+\frac‌​12\left(1-\frac1{2^s}\right)\zeta(s)\\\implies f(0)&=\frac12\eta(-1)=\frac18\end{align}$$\sum_{k=1}^\infty(-1)^{k+1}\left\lceil‌​\frac k2\right\rceil=\frac18$$
@AkivaWeinberger @Semiclassical May be interested ^
@SimplyBeautifulArt you might be interested in this
2
A: A rigorous treatment of the seemingly simple d.e. $x^2 y' - y = 0$

Kenny LauNaive solution for some intuition (with division of zero involved): $$\begin{array}{rcl} x^2 y' - y &=& 0 \\ x^2 y' &=& y \\ \dfrac{y'}{y} &=& \dfrac1{x^2} \\ \displaystyle \int \dfrac{\mathrm dy}{y} &=& \displaystyle \int \dfrac1{x^2} \ \mathrm dx \\ \ln y &=& C-\dfrac1x \\ y &=& Ae^{-1/x} \end...

Hey @LeakyNun
@SimplyBeautifulArt hi
@LeakyNun Cool I suppose
$\prod_{i,j}[g_i,g_j]\subset N \implies n\prod_{i,j}[g_i,g_j]n^{-1}=\prod_{i,j}[g_i,g_j], \forall i,j \in \Bbb{N}$

$G/N=\{gN,g\in G\}\supset \{g\prod_{i,j}[g_i,g_j],g\in G\} = \{gg^{-1}\prod_{i,j}[g_i,g_j]g,g\in G\} = \{\prod_{i,j}[g_i,g_j]g,g\in G\}$
13:41
$[G,G] = \{[a,b] \mid a,b\in G\}$
I didn't put angle brackets
Guys, any idea what $\mathbb T$ stands for in this context: $\mathbb C^*\cong\mathbb R^*\times\mathbb T\cong\mathbb C/\mathbb Z$
I mean I know the torsion subgroup, but we denote that by $T(G)$, for $G$ a group
what is $*$?
multiplicative group
no
$\Bbb C^*$
should be non-zero right
which means.... multiplicative
13:43
alright
how is it isomorphic to $\Bbb C/\Bbb Z$?
well first I would like to know what $\mathbb T$ means, before doing this exercise :P
Torus.
def?
wiki isn't exactly helpful
$[G,G]=\{\langle[a,b]\rangle |a,b\in G\}$
$[G,G]\subseteq N \implies \{g\langle[a,b]\rangle g^{-1} |a,b\in N,g \in G\}$
$G/N=\{gN,g\in G\}=\{gs\langle [a,b]\rangle,a,b \in N, g,s\in G\} = \{gg^{-1}s\langle [a,b]\rangle g,a,b \in N, g,s\in G\}$
$= \{s\langle [a,b]\rangle g,a,b \in N, g,s\in G\} = \{Ng,g\in G\}$
@ShaVuklia can you include the whole question?
Secret. I said no angle brackets.
@Mr.Xcoder hi
13:47
@LeakyNun Hi
that is the question lol. show those groups are isomorphic :P
Hello I am new here. May I mention my question here to discuss?
@kamranjamil yes
Did you forget to require the $x_i$ to be non-negative integers? — kimchi lover 17 hours ago
My average conversation in the mathematics chat room: Leaky Nun: @Mr.Xcoder Hi – Mr. Xcoder: @LeakyNun Hi – Mr. Xcoder: What do all those signs mean... Quits
13:50
@Mr.Xcoder bye
Oh who said I'm leaving... Not this time ;p
@Secret your second line makes no sense
@Leaky anyhow I asked my classmates, I hope they'll respond soon. but I guess I can first try to show $\mathbb C^*\cong\mathbb C/\mathbb Z$
@ShaVuklia right
$[G,G]\subseteq N \implies \{g\langle[a,b]\rangle g^{-1} |a,b\in N,g \in G\} = \{\langle[a,b]\rangle |a,b\in N\}$
13:52
@ShaVuklia T/F: If $H < G$ and $K$ is a group such that $H \times K \cong G$, then $G/H \cong K$.
since a normal subgroup is invariant under conjugation
@ShaVuklia T/F: If $H<G$ and $K$ is a group such that $G/H \cong K$, then $H \times K \cong G$.
(two questions)
one sec. I can't let go of $\mathbb T$ XD
@ShaVuklia I'm thinking about that
@Semiclassical you might be interested
but is that really strange: Replacing $\langle [a,b]\rangle$ by $N$ will not affect the workings, thus it will mean all quotient groups are abelian?
13:56
it just means that your steps are all wrong
Which question? I just got here
the meaning of $\mathbb T$
at least, that is my interest
1. what does T mean here?
in this context:
15 mins ago, by Sha Vuklia
Guys, any idea what $\mathbb T$ stands for in this context: $\mathbb C^*\cong\mathbb R^*\times\mathbb T\cong\mathbb C/\mathbb Z$
13:57
oh lol
2:
4 mins ago, by Leaky Nun
@ShaVuklia T/F: If $H < G$ and $K$ is a group such that $H \times K \cong G$, then $G/H \cong K$.
3:
2 mins ago, by Leaky Nun
@ShaVuklia T/F: If $H<G$ and $K$ is a group such that $G/H \cong K$, then $H \times K \cong G$.
I think I've seen $\mathbb{T}$ as the set of complex numbers with unit modulus. @ShaVuklia
thattt.......... could be it
that makes sense
$re^{i\theta} \in \Bbb C$
13:58
Semi saves the day
$\Bbb R^*$ would have to be the positive integers though
$\Bbb T$ becomes the circle
Lemme see if I can find a source on that
I mean
we've had the "circle group"
we've had two exercises on that

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