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10:34
hmm....
Still have no idea what motivates us to have the powers axiom...
I know subsets are important, but why are the collection of all possible subsets so important that we want to ensure they are all sets...?
@Perturbative Hey
Heya @BalarkaSen
How'd your exam go? :)
Pretty good, it was really easy
Awesome
Hello!!

Let $a_n=\sin (\sin (\sin (\ldots \sin 1 ))), n\in \mathbb{N}$.

We can write the sequence in the recursive form $a_n=\sin (a_{n-1})$ with $a_0=1$, right?
I want to show that the sequence is monotone and bounded.

It holds that $|\sin (x)|\leq |x|$. So, we have that $|a_n|=|\sin (a_{n-1})|\leq |a_{n-1}|=|\sin (a_{n-2})|\leq |a_{n-2}|\leq ...\leq |a_1|=|\sin (a_0)|\leq |a_0|=1\Rightarrow |a_n|\leq 1$, so the sequence is bounded. Is this correct?

How can we show that the sequence is monotone?
10:42
So what kind of maths will you study now? @Balarka
Yo! @AlessandroCodenotti
@MaryStar If $\sin(x)\le x$ and $a_n=\sin(a_{n-1})$, what can you say of $a_n$ and $a_{n-1}$ ?
Good question
Hm
Hi @Perturbative
@Hippalectryon Ah we have that $a_n=\sin(a_{n-1})\leq a_{n-1}$ and so the sequence is decresing, right?
10:45
@AlessandroCodenotti Have uni's closed yet over in Italy?
@MaryStar Yep
Great! Is the part about boundness correct? @Hippalectryon
@Perturbative we have lectures until the 20th of December
@MaryStar Yep! Actually, you had actually shown the sequence was decreasing in that part, since you wrote $a_n\le a_{n-1}\le\dots\le 1$, you just didn't need the abs values (for showing it was decreasing. They're useful for showing it's bounded)
Ahh I see @AlessandroCodenotti
10:51
So I guess I am going to gear back with ODE's, definitely keep learning h principles, and start doing this covering space-Galois theory-Riemann surface mishmash
Ah Ok! Thank you! @Hippalectryon
@Perturbative So you were asking for textbooks yesterday, right?
Yep, just let me know which ones you want to use
So, since $a_n$ is monoton and bounded, and so it is convergent. Let $a$ be the limit. How can we find the limit? We have that $a=\sin (a)$, how can we continue?
I think I asked you this before but how much alg. top. do you know?
10:56
Nothing much past the basics, like fundamental group, and seifart van kampen theorem
But I'm studying more at the moment
@Hippalectryon Do you have an idea about the limit?
@MaryStar Do you have any theorems/... about the limit of recurrent sequences ?
@Perturbative Ah, excellent. So we should pick up Hatcher, chapter 1.
1.3 is covering spaces, so that serves our purpose well. I will revisit 1.3. myself because I have forgotten a chunk of that
For Galois theory, I think Morandi is a good book, which I will be using. Forster for Riemann, hands down
Finally, to sort it all out, Szamuely, "Fundamental groups and Galois groups"
That has the whole theory of covering spaces/Galois theory/Riemann surfaces summarized and interconnected togather
@BalarkaSen Hmmm, I wasn't to much of a fan of Hatcher's book, some of the stuff I felt wasn't done rigorously (like how he introduced CW complexes), I guess I wasn't just used to that at the time, but I don't have any problem with using it, I'll just look elsewhere if I need more rigorous definitions
@Hippalectryon Which theorem could we use here?
11:02
@Perturbative That's a standard complaint. My standpoint is that you should get used to that. The natural language of topology is pictorial, not symbolical :)
@MaryStar For a sequence $u_{n+1}=f(u_n)$, if a limit exists, it has to be a point such that $f(x)=x$
One should strive for pictorial rigor, rather than symbolical rigor, exactly what Hatcher does
Ahhh, I was actually getting worried because there seemed to be a discrepancy between what most topologists talk of Hatcher and the high praise it gets and someone learning AT for the first time like me feel when reading it
Like I when I switched from Munkres to Hatcher, it felt alien to me
But then when I'd go somewhere else, find a more rigorous proof/definition of something in Hatcher I'd always come back and be like "You know what I like Hatcher's explanation/exposition more"
I'll take a look at the other books now
Yes, so I said in my previous comment we have that if $a$ is the limit, then $a=\sin (a)$. Is the only solution of that equation $a=0$ ? Do we justify it with the uniqueness of the limit? @Hippalectryon
@Perturbative Oh, you have read Munkres? Good, that's where I started learning AT
It also took me some time to get accustomed to Hatcher, especially chapter 0
11:06
@MaryStar It is the only possible solution because it's the only solution of $\sin(x)=x$. All that's left is to show that the limit exists. Which you have done (bounded monotone)
@BalarkaSen I only read the first half on General Topology, but I've read a (tiny) bit of the AT stuff
I actually really like Munkres's exposition
of algebraic topology too
@Hippalectryon Ah ok! Thank you! :-)
@BalarkaSen I'll use it alongside Hatcher if I need to look up anything
Yo @Daminark
That'd be cool
11:11
When are you thinking of starting going through the material? @BalarkaSen
How about today? :D
That's cool with me
Thought you might want to sleep after your exams :p
We have the sequence $a_{n+1}=\frac{1}{2+a_n}$ with $a_1=0$. I want to show that $(a_n)$ is a Cauchy sequence.

We have that $|a_m-a_n|=\left |\frac{1}{2+a_{m-1}}-\frac{1}{2+a_{n-1}}\right |=\frac{|a_{n-1}-a_{m-1}|}{|(2+a_{m-1})(2+a_{n-1})|}$, right? How could we continue? Could you give me a hint?
@MaryStar What can you say on the sign of $a_n$ ?
11:27
Can we say something although we don't know if the sequence is decreasing or increasing? @Hippalectryon
@MaryStar yes
We have that $a_1=0$, then $a_2=\frac{1}{2}$, then $a_3=\frac{1}{2+\frac{1}{2}}$
So, will $a_n$ be positiv? @Hippalectryon
Alright. Now, what does that tell you on $(2+a_{m-1})(2+a_{n-1})$ ?
About the sign? It will also be positive. @Hippalectryon
About the value, knowing that $a_k\ge0$
11:33
Does someone know wyzant?
@Hippalectryon We have that $(2+a_{m-1})(2+a_{n-1})=4+2a_{n-1}+2a_{m-1}+a_{m-1}a_{n-1}\geq 4$ right?
Yep. Therefore what can we say about $|a_m-a_n|$ ?
@Hippalectryon bonjour <3
@Lagranian Salut
@Hippalectryon ff j ai problemé avec la math
11:37
So, we have that $|a_m-a_n|=\frac{|a_{n-1}-a_{m-1}|}{|(2+a_{m-1})(2+a_{n-1})|}\leq \frac{|a_{n-1}-a_{m-1}|}{4}\leq \frac{|a_{n-2}-a_{m-2}|}{4^2}\leq \ldots \leq \frac{|a_{1}-a_{m-(n-1)}|}{4^{n-1}}=\frac{a_{m-(n-1)}}{4^{n-1}}$, right? @Hippalectryon
@MaryStar That looks right. Now you should be able to conclude.
@Lagranian lequel ?
@Hippalectryon Algebra
et calculus
@Hippalectryon Do we have to show that $a_{m-(n-1)}\leq 1$ ?
@MaryStar We want to show it's a cauchy sequence, i.e. that for any $\epsilon>0$ there exists $N$ such that $m,n\ge N\Rightarrow|a_m-a_n|\le\epsilon$
@Lagranian plus précisément ?
@Hippalectryon uno what durability means?
11:43
@Lagranian Not in maths
It's in what?
probs physics yh
In physics, when studying materials ? I guess
yh right
but my problem is to find what topic it is in
I'm looking as durability
it seems wrong
Materials physics, I'm fairly sure
have u heard something like
$$\alpha = \dfrac {Surface }{Volume}$$
11:45
Well to be honest, "durability" isn't well defined in physics. It's used in different contexts with different meanings
what do u think about it
or what I need to search?
lets give an example
Why do you need that for exactly ? What's the underlying question ?
imagine a ball that is with $r$.
and a ball that is with $2r$ too.
I'm gonna find the durability between them.
The first ball has the durability $\frac {1}{a}$
The second ball has the durability $\frac {1}{2a}$
Where is that problem from ?
thats from my textbook
and its called by ''durability''
11:48
What texbook is it ?
thats not english
as I mean,
this textbook calculated the durability between them
and it says that the durability is
$$\alpha = \dfrac {Surface }{Volume}$$
I couldn't find anything in physics while searching
@Hippalectryon I got stuck right now... We want to show that $\frac{a_{m-(n-1)}}{4^{n-1}}\le\epsilon$, right? But we still have a term of the seqeunce $a_{m-(n-1)}$
@BalarkaSen Let me know what time you want to start today, and if you want to create a room or anything
Well, I've never heard of durability being defined that way @Lagranian :(
@Hippalectryon What do you know about this?
or does it seem wrong?
11:50
@Perturbative There's already a room we created a while ago
@MaryStar We don't want to show that $\frac{a_{m-(n-1)}}{4^{n-1}}\le\epsilon$ for all $m,n$, but that it holds for $m,n\ge N$
The surface-area-to-volume ratio, also called the surface-to-volume ratio and variously denoted sa/vol or SA:V, is the amount of surface area per unit volume of an object or collection of objects. In chemical reactions involving a solid material, the surface area to volume ratio is an important factor for the reactivity, that is, the rate at which the chemical reaction will proceed. For a given volume, the object with the smallest surface area (and therefore with the smallest SA:V) is the sphere, a consequence of the isoperimetric inequality in 3 dimensions. By contrast, objects with tiny spikes...
but why...?
@Lagranian I don't know anything about this :( I've done a fair share of physics but I've never used that definition
@Secret What is it?
That ratio basically talks about how much surface area is exposed for some interaction given some fixed volume, but without knowing more details on why they ask you to calculate that, I have no idea what that problem is trying to tell you to do
11:54
Imagine a pencil with $7$ lenght
and $3$ lenght.
however, since the two objects in your questions are balls, I am suspecting they want you to calculate $\frac{4\pi r^2}{\frac{4}{3}\pi r^3}$
Which pen does make you slog on?
$7$ or $3$?
I mean which is harder to break?
this is called by like that
as seen at the top
$\frac{4\pi r^2}{\frac{4}{3}\pi r^3}$
I think this is on my textbook
let me check it out
and yes, it seems true.
Can you explain it? @Secret
Hi, in an MO answer (mathoverflow.net/a/286895) the following comment is made (ST is set theory, emphasis added):
@Lagranian I am not really sure, but one consequence for large balls having smaller of that ratio is the reason why we use small pellets instead of one large ball of the same volume in chemical reactors because of the massive surface area available for reactions
> all the constructions in homotopy type theory are stable under isomorphism, so if you prove a property of (let's say) a group, then this property is true for any isomorphic group. This is not true in ST and this problem is usually swept under the carpet.
Does somebody know what is meant here? I have never heard of such a problem, and am super interested in reading about it (eg an example or something ;)
12:01
@Secret You are right.
I'm gonna share a question from my textbook
In applied mechanics, bending (also known as flexure) characterizes the behavior of a slender structural element subjected to an external load applied perpendicularly to a longitudinal axis of the element. The structural element is assumed to be such that at least one of its dimensions is a small fraction, typically 1/10 or less, of the other two. When the length is considerably longer than the width and the thickness, the element is called a beam. For example, a closet rod sagging under the weight of clothes on clothes hangers is an example of a beam experiencing bending. On the other hand, a...
given me being a science major and not an engineering major, all of this is extremely complicated
@Secret It's important you to get what I mean.
Are you sure?
Imagine that there are two ball.
@Arrow I am not sure whether there is non-negligible probability that somebody notices it there, but I have reposted your question also in Geometry & Topology chatroom.
The first ball has $r$ and $h$.
The second ball has $2r$ and $h$.
From the equation,
$\frac {\pi r^2}{\pi r^2h}= \frac {1}{h}$ for first ball.
$\frac {\pi 2r^2}{\pi 2r^2h}= \frac {1}{h}$ for the second ball
so we see that the both ball has same durability.
what is the physical meaning of $h$?
12:12
lenght
It sounds to me that these are not balls, but cylinders (balls are specified only by radii)
Yeah, that was my mistake.
They are cylinders.
Let's give an example too
also, isn't the surface area of cylinders $2\pi r^2 + 2\pi r h$?
I probs wrote the equation wrong
That was cross-sectional area of cylinders
indeed this is math
not physics
However, I couldn't find the correct definition
so it seem they are asking the cross sectional area vs volume ratio of cylinders.
12:16
exactly.
What do you think now?
Then the equation is $\frac{\pi r^2}{\pi r^2h}=\frac{1}{h}$ thus that quantity depends only on the length of the cylinder. However I have no idea what that quantity means and it does not seemed to relate to how bendable the cylinder is...
And
Textbook says that '' If the lenght increases, then the durability decreases''.
So like
Imagine two pen
with $7$ and $4$ lenght.
Which is harder to break?
So its about the lenght
What is this "durability" in the original language. Maybe I should check which of the many quantities about bending strength it is referring to?
Why do you think it's bending?
two pen, one is longer and one is shorter. If they have the same cross sectional area, then the long one is going to break easier than the short one
12:23
@Secret You got what I mean now!
and before anything breaks, it bends slightly before fracture kicks in
Exactly.
I mean that
and looking for the its definition
or wait
would it be Strenght Of Materials?
Have a look
The specific strength is a material's strength (force per unit area at failure) divided by its density. It is also known as the strength-to-weight ratio or strength/weight ratio. In fiber or textile applications, tenacity is the usual measure of specific strength. The SI unit for specific strength is Pa m3/kg, or N·m/kg, which is dimensionally equivalent to m2/s2, though the latter form is rarely used. Specific strength has the same units as specific energy, and is related to the maximum specific energy of rotation that an object can have without flying apart due to centrifugal force. Another way...
force per unit area of fracture over density, sounds about right. So for a rod with uniform density (which is basically a cylinder), its specific strength is proportional to its cross sectional area to volume ratio
I think that's not it.
oh wait
that's it!
its specific strength is proportional to its cross sectional area to volume ratio
However, that's not what I'm looking for.
12:39
Even though that quantity is related to specific strength, I have no idea whether cross sectional area to volume ratio has a special name to refer to it
If "durability" is really the same as its english meaning in that it is alluring to the specific strength, then I guess that is the best we can desctibe this quantity, but if "durability" means something else in english, then I have no idea.
@Lagranian can you give me the term in the original language where you get "durability" from since you said your book is not english?
Could someone give a hint on this (just to let me going): Spaces $S^{2] \times \Bbb{RP}^\infty$ and $$\Bbb{RP}^{2}$ have same homotopy groups.
@Secret Hold on.
Let me create a room with you
13:06
We cannot define an uncountable set to be one where there is an element that is unreachable by any countable sequence, because that's topology not set theory
13:16
Anyone here wants to give a more complete or an alternative answer to this question: stats.stackexchange.com/q/256271/82135 ?
13:40
hmm... let's try this:
Let $S$ be a set. Define the model function $f$ (which is basically an axiom) on any set such that $f(S)$ is a set defined as follows:
@Secret set theory has supermum as well
you can treat $\omega_1$ as an unreachable ordinal in some sense
and in other senses $\omega$ itself is not reachable
Axiom of inaccessibility (replaces the powerset axiom): For any poset $S$ of cardinality $\kappa$, the inaccessible set $\mathcal{I}(S)$ is a set that contains $S$ and an element $x$ such that it cannot be expressed as any union of the form:
$$\bigcup_{\alpha < \kappa} y_{\alpha}$$
where $y_{\alpha}$ are subsets or elements of $S$
This ensures $\mathcal{I}(S)$ can never inject into $S$, while $S$ can inject into it, thus fullfilling the minimal requirement for higher cardinality
And therefore, using this axiom, we thus obtain $\omega,\omega_1,\omega_2$ etc.
3
Q: Why is $\omega_1$ an accumulation point here, but has no sequence converging to it?

Dr. ZeusI'm trying to understand the following. I'll denote by $X$ the set of all ordinals that are at most $\omega_1$, the first uncountable ordinal. To avoid confusion, they can be equal to $\omega_1$ as well. Equip $X$ with the order topology. Then how come $\omega_1$ is an accumulation point of $X\...

I guess this question really helps on higlighting the gist of an uncountable well ordering, it is inaccessible from below via unions
Set unions are also independent of the topology being used, thus I think this should help make uncountablity more constructive
So we now have:
System 3: Extensionality, Regularity, Specification, Pairing, Union, Replacement, Infinity, Inaccessibility, Existence of an inaccessible cardinal (optional), dependent choice (optional)
Axiom of infinity is still necessary because of the following demonstration:
Take $\varnothing$. Apply inaccessibility, obtain $\{\varnothing,a\}$
hmmm....
something's amiss...
...For any poset $(S,<)$ of cardinality $\kappa$, there exists a set $\mathcal{I}(S)$ such that $S \subsetneq \mathcal{I}(S)$ and for all $T \subseteq S$ and $x \in T$ there exists an element $y$ such that $\bigcup_{T} x < y$...
typo: there exists a supremum $y$
14:30
Revised:
Axiom of inaccessibility : For any poset $(S,<)$ of cardinality $\kappa$, there exists a set $\mathcal{I}(S)$ such that $S \subsetneq \mathcal{I}(S)$ and maximal elements $y$ corresponding to every possible chains formed by all subsets $T$ of cardinality $\alpha$ of $S$ where all $x$ are comparable, such that:
uh, still typo
Axiom of inaccessibility : For any poset $(S,<)$ of cardinality $\kappa$, there exists a set $\mathcal{I}(S)$ such that $S \subsetneq \mathcal{I}(S)$ and for every possible chains formed by all subsets $T$ of cardinality $\alpha$ of $S$ where all $x$ are comparable, there exists a maximum $y$ such that:
14:50
2
Q: Marginalization of conditional probability

gwgI am working through these examples of computations on Bayesian networks and came across this claim (part of the last sample computation): $$ P(E=e|A=a) = \sum_{c \in C} P(E=e, C=c | A=a) $$ I am newly familiar with marginalization, but I thought that it was: $$ P(A=a) = \sum_{b \in B} P(A=a,B...

Feel free to give more complete and thus useful answers to that question!
15:01
$$ \bigcup_{x \in T} x < y \land \{\bigcup_{x \in T} x\} < y \land \bigcup \{T,y\} = y$$
15:16
Demonstration:
$\varnothing$, $\mathcal{I}(\varnothing) =\{ \bigcup \varnothing < y \land \{\bigcup \varnothing\} < y \land \bigcup \{0,y\} = y \} = \{0,a : 0 < a\}$
typo
$\mathcal{I}(\varnothing) =\{ \bigcup \varnothing < y \land \{\bigcup \varnothing\} < y \land \bigcup \{0,y\} = y \} = \{ \varnothing = 0 < y \land \{\varnothing\} = 1 < y \land \bigcup \{0,y\} = y \} = \{0,1,a : 0 < 1 < a\}$
@MatheinBoulomenos Hello ! :D
Mathein :D
kinna need help with ideal question :D
let phi, A-->B be a surjective ring hom, the map from I --> phi inverse of (I) is a bijection
I mean hmm
doent the preimage of I live in A?
so why does one need this?
15:26
ok this is getting too long, this going to do it in a different way:
the map from I to phi invers of I?
I don't know what you mean
let me call phi for "f"
f :A--> B , A,B are rings
we have that f is a surjective ring homomorphism
the map f :A-->B
I |---> f^-1 ( I)
then my teacher wrote
What is the domain and the codomain of the map $I \mapsto f^{-1}(I)$?
{ideals in B } ---> {ideals in A that contain the kernel of f }
Ah okay
yeah, that's a bijection
15:30
hmm
can you explain to me what it sais?
I mean doesnt f^-1 (I) live in A ?
where does I live here?
In A no ?
it's an ideal of B
we take any ideal of B and get an ideal of A
that happens to contain the kernel of f
15:32
hmm let me Think for a second why that is true ><
and why did he write it that way ?
f:A-->B one would Think that I is in A not B
{ideals in B } ---> {ideals in A that contain the kernel of f }
didn't he wrote that?
he did
but first lines was comfusing
I--> f^-1(I)
anyway now that I got what he meant
ill try to understand why every ideal of B can be viewd as an ideal of A
contaning the kernel of f
The map $I \mapsto f^{-1}(I)$ says you how to do it
hmm i have hard time understaning how to work with such thing
we did not have much practice on that
Maybe an example will help
Consider the projection map $\Bbb Z \to \Bbb Z / 6\Bbb Z$
15:35
okay but does that mean, the map you just wrote , need to be shown its 1-1 and onto ?
okay let me do ur example :D
@Secret I have an open question about models that you might be interested in
Can you figure out the ideals in $\Bbb Z/6\Bbb Z$? Without using that bijection
cc @MatheinBoulomenos
15:36
yes
@LeakyNun I am not sure if my knowledge can reach that level, but go ahead
I know basically nothing on model theory
ideals of Z/6Z are 1Z, 2Z, 3Z, 4Z,5Z and the kernel 6Z
@MatheinBoulomenos your knowledge on fields might help
esp on number theory
okay
@Kasmir that's not right
15:37
hmm
let me think
Also some update: I my consider changing $T$ from being a subset to just elements of $S$, that way I can steer clear of any assumption of powerset axiom
@MatheinBoulomenos when we check for ideals , dont we add and multiply by other elements and see if we leave the ideal or not ?
okay hmm let me check them again
@MatheinBoulomenos grrr 0Z, and 1Z
so zero ideal and whole R
You should right 0+6Z and 1+6Z
15:43
okay
or $(\overline{0})$ and $(\overline{1})$
these two always exist
but there are more
okay hmm
@MatheinBoulomenos can you tell me how to do it?
like how to work it properly
In this case, any ideal will in particular be a subgroup
what do you know about a subgroup of a finite group?
Does it holds that $$([(A\cup B)\setminus (A\cap B)]\cup C)\setminus ([(A\cup B)\setminus (A\cap B)]\cap C)=(A\cup [(B\cup C)\setminus (B\cap C)])\setminus (A\cap [(B\cup C)\setminus (B\cap C)])$$ ? If yes, how can we prove this?
the order must divide 6
so 1,2,3,and 6
i found 1 and 6, so the rest are 2 and 3
15:46
@MaryStar venn diagram
since they are prime we are guranteed to have such subgroup
or turn that into a logic question
@KasmirKhaan yup
that's the two universal approaches to elementary set theory questions
how to see that 1+6Z is all of Z/6Z
because it is a generator?
15:48
it isn't
Under this question, I have doubt How can I check the following statement true or false ' $\sum_{k=1}^{\infty}x_k$ diverges to $\infty$'
When I take $a_n=n$ statement is true. I didn't get any counter example.
3
Q: Properties of $x_k=\frac{a_{k+1}-a_{k}}{a_{k+1}}$ where $\{a_n\}$ is unbounded, strictly increasing sequence of positive reals

Jesse P Francis Let $\{a_n\}$ be an unbounded, strictly increasing sequence of positive real numbers and let $x_k=\frac{a_{k+1}-a_{k}}{a_{k+1}}$. Which of the following statements is/are correct? (CSIR NET December 2014) For all $n\geq m, \sum\limits^{n}_{k=m}x_k>1-\bf{\frac{a_m}{a_n}}$ There exist ...

@ManeeshNarayanan I don't mean any offense, but since this is not the first time you asked questions, I have to doubt your credentials
i spent all of the time on Group theory to prepare for exam
did not find time for ring part
grrrr
the ideal generated by 1+6Z is all of Z/6Z
@LeakyNun May I Know, What do you mean?
15:50
At the left side we have that if $x\in ([(A\cup B)\setminus (A\cap B)]\cup C)\setminus ([(A\cup B)\setminus (A\cap B)]\cap C)$ it means that $x$ is in $A$ or in $B$ but in their intersection, or in C but not in the intersection with $A$ and $B$. So it must be in exactly one of $A$, $B$ or $C$. Is this correct? @LeakyNun
hmm but if we take something like
3+4 =7 = 1 mod6
@ManeeshNarayanan in a less formal manner, "do you really have MSc in maths?"
why is 3 an ideal
3 isn't an ideal
(3) = {0,3)
15:51
@LeakyNun
Yes :)
You should distinguish between elements and the ideals generated by them
then why are you asking those questions? @ManeeshNarayanan
@MaryStar I don't know. I don't want to think about that.
I mean the ideal generated by 3
(3) = (0,3)
@KasmirKhaan then you've answered your own question
"why is the ideal generated by 3 an ideal"
I am preparing for NET Exam @LeakyNun
15:52
I meant for the case
@ManeeshNarayanan I mean, those questions you ask are obviously below MSc level
for what case
[3] + [4] = [1]
so in particualar it is not in [3]
yes?
you're asking why [3]+[4]=[1]?
During my Graduation, I was not sincere in studies. Now, I am. @LeakyNun
and we know if I is an ideal, then a in I , x in R, ==> ax , a+x in I
in this case it is not
15:53
@ManeeshNarayanan alright
@KasmirKhaan no, we don't have a+x
and I still don't know what you are asking
oh
it has to be a subgroup of (R,+) and have the property that a in I, r in R, ==> ar in I
a in I, b in I, x in R ==> ax, a+b in I
only that to be an ideal ?
yes, exactly
okay thanks, i mixed things up
so a subgroup of the additive Group (R,+)
and swallows up the multiplication
15:56
yes
grrr ill keep Reading , ill come back shortly :D
@MatheinBoulomenos ehm , still that question of bijection between ideals ><
if you got time ofc
@KasmirKhaan may I help you?
@LeakyNun yes :D
trying to understand this
{ideals in B } ---> {ideals in A that contain the kernel of f }
I only have 15 minutes though
define your variables
that is fine =p
oh i thought you been here for the convo ><

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