Mathematics - 2017-11-25 (page 3 of 4)

I used the fact that the characteristic of $F$ isn't $2$ while showing that the extension is separable and I don't think I needed it while showing that $E$ is normal

But aren't finite degree extensions of finite fields always separable?

MatheinBoulomenos

it is not needed to show normality, that's correct

Kasmir Khaan

So I is an ideal in B, its preimage is an ideal of A that we know

I need to show that ker f in contained in the preimage of I

hmm

I dont know how to start that argument

the intersection of all the preimages of all the ideals of B

should be the kernel of f

if i would belive the statement that i want to prove

so as sets , if we draw circles , they all intersect at the kernel of f

MatheinBoulomenos

What does it mean that ker f is contained in the preimage of I under f?

you're thinking way too complicated

@AlessandroCodenotti finite characteristic need not mean finite

Kasmir Khaan

well it means if I pick an element in kernel of f, it should be mapped to I

Alessandro Codenotti

4:09 PM

Oh, of course

Kasmir Khaan

OHHH

I see :D

all elements of ker f will be mapped to 0

and since I is in ideal of B

it must contain 0

Alessandro Codenotti

Let $F=\Bbb F_2(t)$, then $F[X]/(X^2-t)$ is the counterexample, that was the next exercise

Kasmir Khaan

Did I get it right mathein ? :D

Alessandro Codenotti

Because $X^2-t$ is irreducible but $\sqrt{t}=-\sqrt{t}$ so it doesn't have distinct roots

MatheinBoulomenos

@Kasmir yes, now you know why the preimage contains ker(f)

Maneesh Narayanan

4:19 PM

@LeakyNun Now I really understood , why did you ask that question :). It is really a school level question. I should be careful about reading all assumptions. Thank you.

Kasmir Khaan

@MatheinBoulomenos Yes :D

@MatheinBoulomenos evrry ideal contains 0, and the preimage is the fiber over 0

so those elements in ker (f) are automaticlly there :D

MatheinBoulomenos

yes

Kasmir Khaan

I notice that many of the things I dont get is because of lack of understanding the defintions

But Ill try to get better at that :D

thanks again mathein :D

and thanks leaky :D

Leaky Nun

@KasmirKhaan so you’re using pro terms now :p

Kasmir Khaan

@LeakyNun haha =p

if you mean fiber , that took me a while to get why they called it fiber not just a preimage of 1 element ><

Leaky Nun

4:32 PM

right

gian

4:45 PM

If I remove the pinch point of a pinched torus I just get a cylinder right?

Balarka Sen

yup

gian

just making sure :)

Balarka Sen

you get a twice punctured sphere, which is homeomorphic to the open cylinder

MatheinBoulomenos

Hi @Balarka

Balarka Sen

Hey @MatheinBoulomenos

MatheinBoulomenos

4:47 PM

Apparently Tate cohomology does have a topological interpretation in terms of $BG$, but I know basically no homotopy theory, so I don't understand it

Leaky Nun

@BalarkaSen o people with no shapes

Liad

is it true that $|\sum_{n=1}^N \dfrac{z \ ^ n - z \ ^ {n+1}}{n+k}| \le |\sum_{n=1}^N z \ ^ n - z \ ^ {n+1}| $ ? $z\in B(0,1)$

Balarka Sen

@MatheinBoulomenos That's pretty cool

anakhronizein

Anyone know much about Okounkov-Vershik approach to rep theory of the symmetric group?

Liad

5:08 PM

i need to prove that $\sum z \ ^ n / (n+k)$ converge for all $k $ and $z\in B(0,1) - \{1\}$ , someone can help? the hint is to multiply the partial sums by $1-z$

MatheinBoulomenos

@Liad en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem

that takes care of the interior

not sure about the boundary

Liad

maybe you can try to see how the hint can be used? we get $\sum_{n=1}^N \dfrac{z \ ^ n - z \ ^ {n+1}}{n+1}$ this seems like a telescoping sum

MatheinBoulomenos

it doesn't really telescope due to the denominators

Liad

yea

hence my other question :P

if it can be bounded by the telescopic sum

MatheinBoulomenos

I just wouldn't use a telescoping sum. Ratio test, root test, Cauchy-Hadamard all give you the convergence radius easily

Liad

5:22 PM

yea they are, but there is a problem with $|z| = 1$

Daminark

Hey @Balarka, @Liad, and @Mathein!

MatheinBoulomenos

Hi @Daminark

Balarka Sen

@Daminark Heyyy

(That's pretty gooood)

Liad

@Daminark hi

Daminark

Wait I don't get it Balarka. Also how's it going for all of you?

Balarka Sen

5:37 PM

It's the classic old Idubbbz meme

Daminark

Oh ffs idubbbz... Kek

gian

Can someone give me an example of how the long exact sequence of homology groups of a relative pair $(X, A)$ where $A \subset X$ helps me compute the relative homology groups?

Balarka Sen

Consider, eg, the pair $(D^n, S^{n-1})$

You could take a look at the piece $H_i(S^{n-1}) \to H_i(D^n) \to H_i(D^n, S^{n-1}) \to H_{i-1}(S^{n-1}) \to H_{i-1}(D^n)$

MatheinBoulomenos

@BalarkaSen one of my profs wrote a survey "Higher dimensional class field theory
(from a topological point of view)" and some things in the introduction seem very similar to what we've been doing in our Galois theory <-> fundamental group analogy thing

Balarka Sen

The homology of the disk vanishes (we're above degree 0)

So you get a short exact piece $0 \to H_i(D^n, S^{n-1}) \to H_{i-1}(S^{n-1}) \to 0$

That means $H_i(D^n, S^{n-1})$ is isomorphic to $H_{i-1}(S^{n-1})$

Run an induction argument to show these groups are zero if $i \neq n$ and $\Bbb Z$ if $i = n$.

That's useful.

@MatheinBoulomenos Wow pretty cool. Link it to me!

MatheinBoulomenos

5:50 PM

@BalarkaSen mathi.uni-heidelberg.de/~schmidt/papers/rims.pdf

gian

awesome example. but what if the connecting homomorphism isn't so nice?

Balarka Sen

@Mathei Thanks!

@gian The connecting homomorphism has a pretty explicit description, actually

You know how elements of $H_n(X, A)$ are represented as relative $n$-cycles $\xi$ in $X$?

gian

Yes

Balarka Sen

Namely, $\xi$ is a chain so that $\partial \xi$ is inside $A \subset X$

The connecting homomorphism $H_n(X, A) \to H_{n-1}(A)$ is nothing but the map $[\xi] \mapsto [\partial \xi]$

gian

Oh I see

Balarka Sen

5:54 PM

It's not obvious that this is what it is from the algebraic proof by Snake lemma

(algebra sucks, yes)

MatheinBoulomenos

@BalarkaSen :O

Balarka Sen

are you T R I G G E R E D?

skullpatrol

>8(

Daminark

@Balarka A N G E R Y R E A C C

skullpatrol

-_-

Kasmir Khaan

5:59 PM

@MatheinBoulomenos can you explain to me the evaluation homomorphism ?

MatheinBoulomenos

Snek lemma is love, Snek lemma is life

Kasmir Khaan

R[x] --> R

like how does it work

gian

@BalarkaSen, unfortunately that trick doesn't work on my problem :(

MatheinBoulomenos

@KasmirKhaan you fix one element in $a \in \Bbb R$, then you just plug that into every polynomial

i.e. $P(x) \mapsto P(a)$

Kasmir Khaan

hmm

gian

6:01 PM

I'm trying to compute the local homology at the pinch point of a pinched torus

Kasmir Khaan

is that important?

do I need to spend time on it for the exam or just move on?

MatheinBoulomenos

evaluation homomorphism are very useful in dealing with polynomial rings

Balarka Sen

@gian Oh yes you need a different trick for that

Kasmir Khaan

oh okay :D

Balarka Sen

You want to apply excision theorem

Kasmir Khaan

6:02 PM

we might have a question on that , only 3 Days left for me Before exam, want to use them as productive as I can on rings

gian

Oh okay, I'll be back then :D

Balarka Sen

You can do it geometrically too

gian

Geometrically?

MatheinBoulomenos

Just draw a picture and pretend it's obvious

Balarka Sen

Yeah relative chains in $C_n(X, X - x)$ are singular simplices with image containing $x$

@MatheinBoulomenos STAHP

Daminark

6:04 PM

Geometrically? Okay bye I'm out of here

Balarka Sen

@Daminark #ragequit

skullpatrol

cya

Daminark

Really I'm staying around but I just wanted to do that real quick

Balarka Sen

You wanted to do that for a long time, didn't you?

Daminark

Nah I just Mathein's thing and wanted to continue the train along a different direction

Idle

6:10 PM

conjectures are false until mathematically proven

Kasmir Khaan

@MatheinBoulomenos when we have two polynomials , and we want to find the ideal generated by them , how does it work?

MatheinBoulomenos

it's generated by the gcd

Kasmir Khaan

same as integers?

hmm

MatheinBoulomenos

you follow Euclid's algorithm

Kasmir Khaan

if we have I = x^2+x and J = x^2+2x

for example

MatheinBoulomenos

6:14 PM

these are easily factored

Kasmir Khaan

Let F be a field and I , J are elements in F[x]

I don't know how to work that out

can you show me? ( I did not spend alot of times on rings , just want to get some feel now Before exam, am a bit too late but ill give it a try ) =p

MatheinBoulomenos

I=x(x+1) and J=x(x+2)

so gcd(I,J)=x

Kasmir Khaan

and if i wanted to find a generator for I+J

that would be (x) ?

MatheinBoulomenos

yes

Kasmir Khaan

this is not obvious to me, what is the anology with integers?

or there any?

Leaky Nun

6:17 PM

@AlessandroCodenotti "uscire" "esco" "esce" @_@

(cc @MatheinBoulomenos)

MatheinBoulomenos

but you should really pay attention to the difference between an element and the ideal generated by it

Leaky Nun

@MatheinBoulomenos deja vu

MatheinBoulomenos

What is the ideal (4)+(6) in the integers? @Kasmir

Kasmir Khaan

by (x) i mean , kx such that k in our field

well that would be (2)

Leaky Nun

@KasmirKhaan that isn't how you generate an ideal

Daminark

6:18 PM

@Mathein "an ideal"

Leaky Nun

you have to have x^2 as well

Kasmir Khaan

leaky are you sure?

Leaky Nun

@KasmirKhaan @_@

MatheinBoulomenos

(x) is the set of all x*f, where f is any polynomial

Kasmir Khaan

so hmm

Leaky Nun

6:19 PM

precisely

Kasmir Khaan

from I and J

we can get anything then

exept constant

am thinking right?

Leaky Nun

@KasmirKhaan an ideal is a ring (why?)

Kasmir Khaan

ideal is a subring also

so is a ring by defintion

Leaky Nun

why is it a subring

Kasmir Khaan

because of closure under addtion

and multiplication

Leaky Nun

6:20 PM

why multiplication

Kasmir Khaan

well (I,+) is a subgroup

Kevin Driscoll

@Idle Do you actually think that the truth value of open questions is 'false' and then suddenly they become true when a proof is produced?

Kasmir Khaan

and a in I and r in I , ar in I

Leaky Nun

@KasmirKhaan where is this from?

Kasmir Khaan

but r can also be in R and still the Product in I

definition of ideal

Leaky Nun

6:21 PM

right

Kasmir Khaan

leaky

Idle

@KevinDriscoll I was refering to the collatz conjecture, still not proven.

Leaky Nun

good

Kasmir Khaan

let me focus on my polys

:D

Leaky Nun

then x in I, x in R, x^2 in I

Idle

6:22 PM

and computational verifications are just partially authentic.

Kevin Driscoll

@idle Ok sure, so in that specific case do you think that the collatz conjecture is false, and then if someone produces a proof it suddenly becomes true?

Daminark

@Idle the point is that there's no "false until proven true", you have "known true", "known false", and "unknown"

Idle

I said this before wait ...

Kasmir Khaan

so once we have x

Idle

Oct 16 at 1:11, by Idle001

i wonder how far the credibility of such proofs

Kasmir Khaan

6:22 PM

we have all x^n

Leaky Nun

@KasmirKhaan precisely

Kasmir Khaan

what I dont get is that what does F[x] mean

when F is not specified

Idle

@KevinDriscoll yes, with a dustless mathematical approach.

Kasmir Khaan

any field?

Leaky Nun

@KasmirKhaan F usually denotes fields

Idle

6:23 PM

not just checking the first billion cases with some PL.

@Daminark the "unknown" isn't even a stable situation.

unlike "known false and true"

Kevin Driscoll

@idle Why do you think the truth value of the underlying claim has anything to d with whether a proof has been produced? I mean its a claim about the natural numbers. So the true or falsehood of the claim turns on the properties of the natural numbers. But a proof or counter-example to the collatz conjecture doesn't change anything about the properties of the natural numbers. It merely makes them apparent. So how can a proof change the truth value of the underlying claim?

ÍgjøgnumMeg

@KasmirKhaan What do you mean by "I don't get what F[x] means"? :P You don't understand the notation or what?

Kasmir Khaan

I am really not sure

Leaky Nun

@AlessandroCodenotti hi

Alessandro Codenotti

Hi

Kasmir Khaan

6:27 PM

Like from the first view, i thought we take elements of R

like if we have x

(x) = Rx

Alessandro Codenotti

@LeakyNun ?

Leaky Nun

@AlessandroCodenotti "hey you've changed the vowel" "hey you've changed the consonant"

Kasmir Khaan

I dont know exactly what are the elements generated by a polynomials

for integers i know for example (2) are multiples of 2

Daminark

It's not stable, we hope to move things from unknown to known as much as possible, just that truth or falsehood of a statement is just there, and knowledge is our access to that aspect of the statement

ÍgjøgnumMeg

@KasmirKhaan Surely it's best to first understand the objects your talking about before you start talking about subobjects?

Leaky Nun

6:28 PM

@KasmirKhaan F[X] is a ring for any field F

Kasmir Khaan

hmm help with that?

Idle

@KevinDriscoll en.wikipedia.org/wiki/List_of_disproved_mathematical_ideas

Alessandro Codenotti

It also changed tense and person, I'm not sure what's your point

Semiclassical

'unknown' is, for the most part, not a statement about mathematics but a statement about human knowledge of mathematics

Leaky Nun

@AlessandroCodenotti like they are so different lol

ÍgjøgnumMeg

6:28 PM

@KasmirKhaan You say you don't get what "F[x]" is but you're talking about the ideals generated by elements of F[x]

Semiclassical

and, of course, what humans think they know certainly can also be wrong

Alessandro Codenotti

Yes, but we have plenty of irregular verbs

Leaky Nun

@AlessandroCodenotti also, how can "exire" give "uscire"

Kasmir Khaan

well I get what F[x] is

Leaky Nun

there's like no "u" in the paradigm of "exeo" in Latin

Kasmir Khaan

6:29 PM

its polynomials in the variable x

and coefficient are in the field F

Daminark

So by saying something is true or false before having proven it either way, you may very well be wrong. If you wait until you have a proof and say until then that we don't know, you're all good. Stability is not what we're looking for here by any means @Idle

Kevin Driscoll

@Idle You're confusing the ontological question of whether the statement is true or false with the epistemological statement about whether we know that it is true or false. The existence of previous claims that people thought were true but later were proved false doesn't say anything at all about whether current open questions are true or false.

Leaky Nun

@KasmirKhaan do you understand how they form a ring

ÍgjøgnumMeg

@KasmirKhaan with coefficients in $F$

Kevin Driscoll

It just says that in the past some people believed things that turned out to be wrong

Alessandro Codenotti

6:30 PM

Most occurences of "x" became "sc" in Italian (x isn't a letter in the modern Italian alphabet)

Kasmir Khaan

If we have ´the element 1, for example.

does it mean we have 1 ,x,x^2 ... ?

Leaky Nun

@AlessandroCodenotti I know, but how does "e" become "u"

Kasmir Khaan

do dont have any limit of what the Powers of x are?

Leaky Nun

@KasmirKhaan yes, <1> = R for every ring R

Alessandro Codenotti

Dunno, I'm not a linguist

Leaky Nun

6:31 PM

@KasmirKhaan the sky's the limit

1 min ago, by Leaky Nun

@KasmirKhaan do you understand how they form a ring

Kasmir Khaan

okay 1 was stupid example

Leaky Nun

3 mins ago, by ÍgjøgnumMeg

@KasmirKhaan Surely it's best to first understand the objects your talking about before you start talking about subobjects?

Kasmir Khaan

yes leaky

addition is poly addtion

and multiplication is the usual polys multiplication

Leaky Nun

then surely you can see why x^3 is in <x^2>?

Kasmir Khaan

the units in R[x] are units in R

ÍgjøgnumMeg

6:32 PM

@KasmirKhaan the question is ill-posed, what do you mean "if we have $1$ do we have $1, x, x^2, \dots$? Do you mean "if an ideal of $F[x]$ contains $1$ then do we have $1, x, x^2, \dots$?

Leaky Nun

@ÍgjøgnumMeg come on, we all know what he meant

Kevin Driscoll

@Semiclassical Did you happen to see that we were wroking on a physics problem here last night at like 2AM?

Leaky Nun

and that "we" is inclusive of him

Semiclassical

lolno

Idle

@KevinDriscoll yes, I don't abrogate this mean of just-in-time check-ups as being a background of next applications, but just making it clear that it's not a very authentic way of tackling a proof, or asserting anything.

Kasmir Khaan

6:33 PM

@LeakyNun no i dont see that

ÍgjøgnumMeg

@LeakyNun I recall you saying a few days ago that one should be precise when talking about mathematics

Leaky Nun

@KasmirKhaan x^2 in I, x in R

@ÍgjøgnumMeg not when he's frustrated about not getting things right

ÍgjøgnumMeg

sigh

Kasmir Khaan

grrrrr

how is x in R?

Leaky Nun

is x a polynomial?

R=F[X] here

Kasmir Khaan

6:34 PM

ahhhhhh

:D

so we take all polynomials

and coefficient in our ring R

Kevin Driscoll

@idle Of ocurse. But thats a veeeeery different claim from the claim that all open problems have a truth value of false and then they become true when a proof is produced.

Leaky Nun

@KasmirKhaan yes, but you're going to confuse yourself by using R as the base ring and as the polynomial ring

Kasmir Khaan

Okay kasmir will keep working on this and come back later

mean while I igot a question for later

Daminark

@Idle if you allow the classification of "unknown", it still fits with what you're saying here

Leaky Nun

I think we all missed one more possible classification

Kevin Driscoll

6:36 PM

@Semiclassical Someone was asking suppose that I'm given an initial position and initial velocity, and also a desired final position. Assume the dynamics are discrete and that at the start of each tiem step you can apply some maximum instantaneous impulse in any direction.

Kasmir Khaan

Suppose R is a not necessarly commutative division algebra, prove that the only two sided ideals of R are zero ideal and R

Semiclassical

oh, that

Leaky Nun

@KasmirKhaan ok

Kevin Driscoll

How do you optimally choose the direction to reach teh desired final position

Daminark

We aim to figure out whether things are true or false, and until we know that, we say it's unknown. If we then use an unknown fact to prove something else, we're still not sure that it's true

Kasmir Khaan

6:37 PM

@LeakyNun dont want solution yet because did not try hard on that, ill keep Reading and come back later

Idle

@kevin Take a proof for a a very practical conjecture like a roadsign before an accident-friendly hotspot or a frequent situation of hazards.

Leaky Nun

@KasmirKhaan ok

Daminark

@Leaky you're gonna talk about independence of the axioms, right?

Leaky Nun

@Daminark from

Daminark

Yeah yeah whatever

Leaky Nun

6:37 PM

:)

Daminark

Anyway what I have in mind is more about ontological truth

Leaky Nun

@AlessandroCodenotti does your idiolect exhibit raddoppiamento fonosintattico?

MatheinBoulomenos

You can't define truth, anyway

Leaky Nun

@Daminark like $\vDash$?

@MatheinBoulomenos $\varphi$ is true in model $M$ if $M \vDash \varphi$

Alessandro Codenotti

@LeakyNun what's that?

Daminark

6:39 PM

Like, from a more philosophical sense, so to speak. It's not formal so much as, we find that our axioms of a set do not give us power to prove that $\aleph_1 = 2^{\aleph_0}$, but in principle that statement still has a truth value

Leaky Nun

@AlessandroCodenotti let's take it as a no

it's where you say "accasa" instead of "a casa"

Kevin Driscoll

I actually dont think you can define truth value independent of a particular model

Alessandro Codenotti

Nah, that's more common in Tuscany and south Italy

Leaky Nun

@AlessandroCodenotti so you're from north Italy?

Alessandro Codenotti

Yep

Leaky Nun

6:41 PM

I see

Daminark

That's why I say it isn't formal, like I'm thinking more... I hesitate to say it in this way but the way Plato talks about "the form of the good" in the Republic is sort of what I have in mind.

user193319

0

Q: Finding a Bounded Measurable Set and Sequence of Simple Functions

Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m(E) < \infty$. Show that for each $\epsilon > 0$, show that there is a measurable set $F$ contained in $E$ and a sequence $\{\varphi_n \}$ of simple functions on $E$ such that $f$ is bounded on $F$ and $m(E-F) < \epsilon$...

A month old with no answers :(

Daminark

Like there's some idea we're trying to capture with the axioms, and I'm wondering how much of truth is intrinsic to that idea.

Kevin Driscoll

AH okay. In that case I agree there is some state-of-affairs about the universe and statements about such states-of-affairs are either true or false. But I'm not into philosophy o mathematics to have an intelligent position on whether we should include mathematical proposition in our states-of-affairs of the universe, like Plato would

Daminark

I mean I'm not deep either, it's just that I tend to view model theory in two lights. One as a subject of study in itself because it seems really cool, and the other as being a technical framework of truth

Kasmir Khaan

6:53 PM

handsome folks

the defnition of prime ideal

is that if ab in P, means either a in P or b in P

Daminark

But as far as the discussion we were having above went, whether "morally speaking" we ought to classify statements as true/false/whatever before they're being proven, I find that we're thinking about the latter more, which is why I'd rather push it more in the direction of, is this true or false in a state-of-affairs. On the whole I'm not sure if I buy completely into trying to transplant Plato's theory of the forms into math

Kasmir Khaan

doesnt this hold for all ideals?

Daminark

Nope

Kasmir Khaan

hmm

Leaky Nun

@KasmirKhaan <4> in Z

and generate the witness of the counter-example yourself

Daminark

6:55 PM

Think about $\mathbb{Z}$, prime ideals are meant to represent prime numbers

Kasmir Khaan

so the analogy of prime ideals are the prime numbers?

(4) = 4n

but this is also 2*2

but 2 is not in (4)

hmm

Leaky Nun

great

Daminark

Precisely

Kasmir Khaan

:D

Sometimes kasmir is smart

Okay thanks guys :D

ill keep Reading :)

Leaky Nun

@Daminark what's a prime in the category of a ring with morphisms being divisibility?

Daminark

6:57 PM

Well in rings you think about prime and irreducible as being different things, I believe

So if $R$ is a ring, you say $p$ is prime if $p\mid ab \implies p\mid a \lor p\mid b$

Leaky Nun

@Daminark that's right

Daminark

You say $n$ is irreducible if $n = ab \implies a\in R^* \lor b\in R^*$

Leaky Nun

yes

Silent

"Let $S$ be a set. the law of composition defined on $S$ by $ab=a$ for any $a,b\in S$ is associative. For which sets does this law have identity? " It seems to me that if $S$ is singleton, then this law has identity, otherwise not; because, if $e$ is identity, then $ea=e$ and $ae=e$ for all $a\in S$. Am I right, or missing something? Also, what if $S$ is empty?

hola

Hey guys, can i ask a magnetostatic question here? The people at the physics forum dont wanna help me.

can i?

anakhronizein

7:18 PM

hola hola

"Just ask; don't ask to ask." is the rule here.

If anyone can help, they will do so. If they can't, you get ignored.

Idle

@KevinDriscoll I think the main reason why the reverse tree is disregarded that there exists someway a group of integers that are not blackhole'd to 1. So starting from 1 as a reference just suggests the existence of an arbitrary group of integers forming an arbitrary galaxy.

This can be a nice model of a universe, if this conjecture is proven.

Silent

17 mins ago, by Silent

"Let $S$ be a set. the law of composition defined on $S$ by $ab=a$ for any $a,b\in S$ is associative. For which sets does this law have identity? " It seems to me that if $S$ is singleton, then this law has identity, otherwise not; because, if $e$ is identity, then $ea=e$ and $ae=e$ for all $a\in S$. Am I right, or missing something? Also, what if $S$ is empty?

Plz take a look

Daminark

7:46 PM

Hey @Alessandro!

Also @Silent let's see

So $ab = a$ for all $a,b$ is associative, which is true since $(ab)c = ac = a$ and $a(bc) = ab = a$

Mike Miller

hi @anakhronizein

Daminark

And yeah that's right, since if $a\ne b$, then $ab = a$ and $ba = b$

Usually you exclude $S$ being empty when you talk about these things

Silent

@Daminark thank you!

Kevin Driscoll

Ya saying that the set has identity is saying there exists an element such that..... But if $S$ is empty there do not exist any elements. So there cannot be any such element

anakhronizein

hi @MikeMiller How are you?

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$Mathematics$ Mathematics