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08:00
to be fair the original question I was going off of that gave me that knowledge of periodic fulfilling certain qualities was saying "rationally periodic". I just didn't realize what they meant.
@euclid idk,sorry :(
However, right now I want to work with zero term algebras a bit cause I have been procrastinating for 4 days playing minecraft when I should be organising those scribbles in the other room, thus I have to quickly get back on track.
technically this is my leisure.
I have to get back on... waiting for the semester to start monday
I guess I could work on making the airship in my mod work right.
i technically have a half finished model, lol.
My leisure has a time limit most of the time (I worked with zero ter algebras purely out of curiosit), due to all those annoying mundane daily task that need to be done
oh well yes of course
08:03
I plan to get those scribbles organise before start preparing for something I have been procrastinated for a year: Study for driving license
how old are you?
touche
i have a permit
gotta get dem hours in
If not that goddamn promise in 2015 to my friends, I don't need to be so annoyed by it every night in dreams
what promise?
that you'd get a license?
@Secret wanna see something cool?
08:06
the promise of getting that godammn license by the first half of this year
sure
there's a reason everything else is at a weird angle.
why is mario a ball?
cause I cannot rotate models properly
:p
(it's filler)
you do know what he's standing on, right?
not really sure, looks like the floor of the starting area in SM64 judged by the hills surrounding it
Nope. He's walking on a ball.
and the background is the starting level of gamejolt.com/games/neen-ten-doo-nightmare/185160
I'm modding in planets
as easily as I can. :p
atm it's just one planet at a time. Cannot jump to other planets
no enemies
@Secret you ok?
08:16
yup
just working in the background
ok
im going to bed
goodnight
night
@Secret other than model files that mod is actually computer code entered in game into a cheat box
so I can easily share it whenever if you're ever interested
not right now, modelling is predicted to come at least 5 years later, I am getting the ground level stuff done first for my plan
um what...?
im talking about the game
08:20
yeah I know, the model files will not be of interest to me currently until I go back to computer graphic stuff and animations, which will be some time in th future once I have my physics and maths foundation solidified
@TheGreatDuck are making game?
modding game
in the weirdest way possible
Wow, unlimited everything?
@Secret I'm saying other than sending one the model files letting someone play the mod is as easy as sending them the code. :p
@Ramanujan what do you mean by that?
Ah I see
08:21
@Ramanujan I'm modding this game gamejolt.com/games/neen-ten-doo-nightmare/185160
@Secret I know. It's late and we're all talking weird and not understanding. :P
Can you mod 8 ball pool,so one can purchase Galaxy cue
wat...
I never claimed I could mod arbitrary games
Dude, I cannot mod arbitrary games.
@Ramanujan in that game... modding is a 'feature' imgur.com/FEq1o2T
08:25
there's no restriction on what code can go in the cheat box
well barring game maker's syntax
but that goes without saying
:p
You mod games by cheat engine or something else? :p
no
that box is built into the game
it's literally a box the guy who made the game put into it
And you are using that box for modding?
and because he made an object (for cutscenes) that executes an arbitrary string of code every frame, I can literally instantiate... whatever I want.
@Ramanujan yup
only thing restricted is graphics
cause graphics in game maker are... slow and overly bloated.
well and obviously efficiency is an issue like any other game.
So every game maker built that kind of boxes?
08:28
no...
it's just a function one can use in their code
if they choose to
you don't know code do you?
like as in programming?
Are you computer scientist?
that is my major
and math
(im an undergrad in college)
supposedly I'm stronger in the math (barely)
I didn't learned coding,i even don't know how to use Java :P
08:30
well you're only 16, right?
no surprise you're not in college.
well unless you just like programming there would be no reason for you to learn it right now
Do you know DHMO? He is of 17 and very talented :D
Ok,good night to you
08:32
yes I do know DHMO
he's only 17?!?
wow he has a mouth
@Ramanujan well basically there is a function that makes a box appear to enter text in (like if you want the player to enter their name and are too lazy to make one that looks nice)
and then there's a command that takes in an arbitrary string of code and executes it on the spot.
a string is basically a line of text
the guy who made the game put that in I suppose as a cheat box for well.... cheating
they also have a moonjump button
but anyway
Well,how you discovered there is box ?
the cheat box basically lets me execute whatever code is necessary to set up whatever mod I wish to do.
@Ramanujan They actually tell you in game at one point.
"press this key and you can enter cheat codes"
Like int that photo?
no
that's the actual box
There's a random sign that says "press '.' to enter cheat codes"
08:36
and well... experiment enough and you find things.
I might show you the mod sometime. I don't work on it much mind you.
I am currently trying to work on the final level
which is gonna be a combination of piloting bowser's airship and galaxy planets
cause... I can and so I shall.
and if somebody is willing and capable of making it through whatever randomness I cook up they deserve something awesome in the final level. :p
anyway
cya
2
Q: Do there exist "weak ring homomorphisms" that aren't (genuine) ring homomorphisms?

goblin(All my rings have $1$, and all my ring homomorphisms preserve $1$.) Suppose $R$ is a ring. Philosophically, it may be the case that $R$ has too many zero divisors for a particular method "$\mu$" of studying it to work. One way to rectify this is to try to find a bigger ring $S$ together with a ...

breaking homomorphism on multiplication...?
08:54
@Secret second property can be deduced from the first, can it not ?
yeah, by plugging a-a for example since we are dealing with a ring
I was more thinking of $f(0+0) = f(0)+f(0)$
well theoretically, 2nd can stood all by itself, but then we will no longer be dealing with a ring as an additive identity demands the equation 0+0=0 is true
Didn't quite get that
By the way, where does your passion for weakened algebraic-structures come from ? I see you discussing those a lot :p
hi chat
09:00
The major motivation is because they are really the only ones that can allow division by zero (i.e. a multiplication of zero, not just an involution or reciprocal or pseudoinverse like meadows and wheels).

The minor, but more general reason is due to my learning style of going from general to specific, as reflected by how I read about semigroups before groups
weak structures allow a lot more freedom to experiment with how axioms interact with each other, with category theory handling the formalisms
I am a chemist at the core, after all, I like to do experiments
This is why mathematicians found my attitude to maths kinda weird. In fact ,my logic can sometimes go backwards, which is quite a pain when it comes to proving things
typo: multiplicative inverse
And you've always wanted to divide by zero, is that it ?
I see, you like to fiddle with "simplicity"
I always want to divide by zero, or prove that the universe does not allow it by having the forbiddne of division by zero be ingrained in the logical structure of mathematics
am i allowed to ask something here?
Another way of allowing that is to consider more complex structures than $\Bbb R$
@Amila Of course
See the guidelines on the top right of your screen
modules are nice because they are like vector spaces with torsion
09:06
I never got into modules that much, though I hope I will in the future
its a linear algebra question.
how to say that span(v)=span(basis of V)?
It could be said that I am kinda obsessed with the mathematical properties of zero objects in general, because everytiem when people talked about them, they said weird things happen. I wnat to understand down to the core why they are weird, how they are weird and what we can and cannot do with them at the level of universal algebra or even category theory
Do you mean how one proves it ? @Amila
In particular for the case of division by zero, I want to either prove the people wrong that you cannot divide by zero in any means, or prove them right to the level of a universal law so that nobody will want to touch the topic anymore
09:08
I think it probably comes from the fact we define multiplication as distributive over addition, thus having a neutral element for addition messes things up
That seems ambitious :p
I am a perfectionist, I only want perfect answers to everything, even if the answer is undecidable, because undecidability is still a property hence an answer
@Amila What have you tried ?
Aren't we all ..?
if something is impossible it is either 100% impossible, or impossible under a condition. Perhaps the better way to say is:...
I WANT TO QUANTIFY IMPOSSIBILITY AT THE LEVEL OF RIGOR OF MATHEMATICS
ssory caps
Good luck then :)
@Astyx Well for that at least for associative structure, I now have some idea on how the messing up happens. One of my calculation showed that if 0 is an additive identity and you allow a0 to be nonzero, then a0 will also become an identity. For finite assoiciative ring like structures (which is called associative division by zero algebra in the other room), division by zero ensure all zero terms are unique, hence by that a0 rule up there, you end up having the + structure completely messed up into a left or right null semigroup
09:13
I know span(basis of V)=V.
V contains basis of V
What do you know about $span\circ span$
or put in another way, if you allow 0+0=0, then division by zero in finite structures guareentee all elements became one side identities
intuitively i know span span = span
Make it rigorous :)
What is you definition of span ?
all the linear combinations of the given set of vectors
09:16
Right, this shouldn't be too hard
hey stop i think i got it.
@Secret I'll give this some thinking
deleted because it does not makes sense to me for some reason
Saying it's "messed up" seems a bit weird to me. It's only that we don't define it since it cannot be consistent with our other axioms for the structure
Anyways I gotta go now
See you later
ok bye
footnote: I think it kinda justify the adjective "messed up". They are not compatible with ring axioms, pseudorings axioms, group axioms and field axioms. And there are many past attempt by many people trying to introduce it, but getting a contradiction in the end. It seems if an interesting division by zero algebra exists, it must be one that is very hard to get it right in the construction process. Therefore it becomes more interesting why it is incomplatible with so many algebraic structures
point is, unlike other algebraic construct, as far I know besides the russel paradox (which is later fixed by using proper classes) division by zero has the largest number of incompatibilities with other mathematics stuff in general, this number of incompatibility raised the curiosity on why is it managed to pull this off, as if there is some underlying pattern that allow it to be incompatible with so many axiomatic systems waiting to be understood
The meadow guys came close to that by defining a pseudoinverse of zero, with resulting axioms not very different from the field axioms and richer behaviour on the zero terms than wheels
09:29
Have you heard of surrreals ?
I do, they introduce a system of infitesimals and infinite numbers via transfinite induction, and it contains every nice number system we knew of. Anything else more than that I have not read up yet
The hyperreals are a subset of surreals, and they are used in nonstandard analysis
I like infinite and infintesimal numbers, but in terms of the potential number of incompatibilities, they are rather tame
It seems to me it defines something which gets close to division by 0 (even though it's not). However I know very little on the topic so I might be completely wrong
Can we every time say that $x=\sqrt (x^2)$ ?
They don't have a zero inverse, but they do have the loose notion of divison by zero in that you can divide by infitesimals by any nunber
@Ramanujan No : $|x| = \sqrt{x^2}$
09:34
In fact, as far I know from the literature, there are no known algebraic systems that has a zero inverse and an additive identity
For reals, for complex numbers this does not hold
For sec∅ ?
@Secret But then they'd have to give up on associativity or something like that, right ?
@Ramanujan $1+\cot ^2 \theta = \cosec ^2 \theta $ Right?
sec${\emptyset}$ ?
09:37
@AmilaPasan :P I was so concentration on fast typing
That's what I am suspecting. One of the things I am trying to proof in my project is whether all interesting division by zero algebra is nonassociative. So far, I have confirmed this to be true for finite structures, and this is where I am now in infinite structures

Proofing a property to be dependent on associativity is sometimes very hard because you basically need to pick the correct associative law to demonstrate that. This is why a side effect of doing this project is investigating associativity as an abstract structure
And can we write x=√x^2?
@Ramanujan I answered that did I not ?
@Secret And you want division by zero to be allowed with every element right ?
@Astyx No, I only need one zero inverse, even if one sided, is enough. These structures are already restrictive enough as my experience shown, it might be impossibel to introduce another zero inverse
Oh right
09:41
@Ramanujan $\sqrt{x^2} = |x|$
Can we write cosecant x = √ (cosecant x)^2 ?
@Astyx in fact, one does not really need to worry with zero reciprocals and zero inverses (the losse notion of division by zero), because from the fintie case division by zero demands them to all exist, it would not be suprising that simialr things will hold for infintie cases
so once you can get a zeero inverse in, you can divide by zero for everything
@Ramanujan $\sqrt{\csc^2 x} = |\csc x|$
For the lose notion of division by zero, it turns out introducing them is easier than expected for example:
@DHMO thanks
09:43
this has no zero inverse, but you can meaningfully talk about $\frac{x}{2}$ here
typo: loose
and for an infintie example, you can take the usual integer ring, get all additive inverses to be one sided and then pick an element x and set 0x=x and you are done
(there are probably more ways to do but I have not started studying the infinite case in detail yet)
Point is, it seems introducing zero terms (0/x, 0/0, x/0, 0x) is quite easy for infinite systems as far the propositions suggests, the challenge is to introduce q such that q0=1
@Secret what do you think x will be such that 0x=x?
it will be some element depending on the zero term algebra that is being constructed
for example, in my integer looking example, I have set 02=2
this is not a very good example, however because it is not flexible to extensions
To give a brief rundown here, the key reasons for associative finite division by zero algebras to be incompatible with a lot of field looking axioms is the following:
1. a0+a0=a0 (All zero terms are idempotent)
2. a+a0=a(1+0)=a0 or 0 (zero term domination behaviour to its coefficient a)
3. If q0=1, then all z0 are unque. In finite structure, the action of z on 0 is a bijection hence a permutation
4. Every division by zero algebra has the elements 0,1,q. That's 3 distinct permutation actions and hence they all become additive identitities hence the + structure has a null semigroup of size at
So in the end of the day, the property of the additive identity 0+x=x end up propagated throughout most if not the whole structure by the distributive law and associative law, turning any expression involving them into identities or absorbing like behaviour (at least for finite case)
I'm back
09:59
in Zero term algebra, Jan 4 at 15:18, by Secret
Theorem $\Omega_1$: Division by zero no-go theorem (Associativity): Finite associative division by zero is not interesting
My work is partially done if this is true for the infinite case as well
That's one theorem
Before it follows many other propositions (mislabelled as theorems)
They are not hard to prove but finding the correct associative law is the main challenge
@OskarTegby Let $g(x) = f(x)e^{-x}$. Extend $g$ continuously to $+\infty$ by setting $g(+\infty) = 0$. By assumption, $\int_0^{\infty} g(x) e^{-kx}\,dx = 0$ for $k = 0,1,2,\dotsc$. By Stone-Weierstraß, you can approximate $g$ uniformly on $[0,+\infty]$ with functions $p_m(x) = \sum_{k = 0}^{N_m} a_k e^{-kx}$. We have $\int_0^{\infty} g(x)p_m(x)\,dx = 0$ by assumption. By dominated convergence, we get $\int_0^{\infty} g(x)^2\,dx = 0$, hence $g \equiv 0$ and then $f\equiv 0$.
@TedShifrin Well, we can arrange that :-)
@TheGreatDuck Thanks ! I will try but I am not that well versed in floor function , though I will try
10:15
ah yes,I always forgot about mobiles. I promise I will use imgur links next time to not blow up the chat with my conversations
cool, no more blowing up
Pretty sure negation ring is no different from a division ring, but have to check...
(compare that slide with the ring like slide to get how they are different)
NB: Attempt to demonstrate a reversed ring by modifying the axioms of the usual intergers have so far end up in disaster, collapsing it into a trivial ring
(and no, I am still studying groups, thus I know nothing about e.g. rings except anything that I happened to seemed to understand when other people said it)
11:24
I intend to use the next few weeks (before the semester starts) to revise single-variable calculus from an analysis textbook, in preparation of taking a course in advanced calculus. I was looking various textbooks online and initially I found "Mathematical Analysis, Volumes I and II" by Zorich to be appealing. It seemed to cover the material I needed to review and more than enough to get me through the next term. But I found out that the textbook has no solutions, and that's a bummer. I'm a bit short on time, and I'd ideally like a resource that has solutions. Hence, I have decided to revie
Why don't we care about those last numbers?
because we're being ridiculous.
@DHMO you mean because we are bad?
omg, I know nothing about infinite series sums
@Ramanujan the sum does not converge
we're just using some ridiculous methods to assign a number to the sum
11:30
Ridiculous=bad?
yes
Ok
Thanks
@Ramanujan Never, ever, sum infinite (diverging) series like that.
11:49
In reply to the xkcd purity comic: smbc-comics.com/comic/purity
I can totally see myself becoming that dad.
@SteamyRoot do you know who Ramanujan was?
Well, duh
en.m.wikipedia.org/wiki/1_−_2_%2B_3_−_4_%2B_⋯
Hey,i didn't mean this link
In mathematics, 1 − 2 + 3 − 4 + ··· is the infinite series whose terms are the successive positive integers, given alternating signs. Using sigma summation notation the sum of the first m terms of the series can be expressed as ∑ n = 1 m n ( − 1 ) n − 1 . {\displaystyle \sum _{n=1}^{m}n(-1)^{n-1}.} The infinite series...
This
I know the series and that explanation.
But that is not a proof, only a (very dangerous) heuristic.
Ramanujan don't monkey with conditionally convergent series
and more never monkey with divergent series
you can however do whatever you want with absolut convergent series
11:57
Take the same series; but now take it $6$ times
shift the second, third and fourth $2$ positions to the right
and the fifth and sixth $3$ positions.
Haha,but although it's valid in "mathematics"
You'll find that the sum of the $6$ series is now $-1$, so you get $-1/6$.
@Ramanujan No, this "proof" by writing down the series and shifting is never correct. I just proved using the same techniques the sum must be $-1/6$.
@SteamyRoot no,it is 1/2
Let S=1-2+3-4+5-6+7…
6s=1-2+3-4+5+0+0+0+…
6s=9-6
6s=3
S=1/2
Correct?@SteamyRoot
12:12
. math.stackexchange.com/questions/2031295/… is it possible that marked for duplicates link to themselfes in a cyclew of links? at least it feels like that...
Super crazy real projective plane embedding (to be proved)
sorry: the correct term shoudl be immersion (as there is self intersection)
@Ramanujan I don't know how you got to $6s = 1-2+3-4+5+0+0+\cdots$.
Either way, the point is, the technique can give you many different values of $s$, so clearly it's not a good way of assigning a value to $s$.
12:32
Sorry,iam not getting 1/2
It's not wrong
How should we know that value when x is just greater than 0?
@Ramanujan there is no "just greater than 0"
what it means is that cos x < (sinx/x)^3 is true whenever 0 < x < pi/2
for example it is true when x=0.001
12:43
When x is 0.00001?
yes
but it does not mean that it is true when you take the limits of both sides
Is that still true?(inequality?)
yes, the inequality is still true when xd=0.00001
http://imgur.com/a/kup2d
bigger
What is the difference between Venn and Euler diagrams? Wikipedia doesn't explain the terms very well.
An Euler diagram (/ˈɔɪlər/, OY-lər) is a diagrammatic means of representing sets and their relationships. Typically they involve overlapping shapes, and may be scaled, such that the area of the shape is proportional to the number of elements it contains. They are particularly useful for explaining complex hierarchies and overlapping definitions. They are often confused with the Venn diagrams. Unlike Venn diagrams which show all possible relations between different sets, the Euler diagram shows only relevant relationships. The first use of "Eulerian circles" is commonly attributed to Swiss m...
See the example in Overview
The pinching line is now shown
Thanks for the pointer. I'm somewhat surprised that its more or less just a notation difference.
so am i
13:11
There doesn't seem to be a term for the circles/Boxes
As in some called them sets; others sample spaces
Like If I was talking about a single number it wouldn't make too much sense to say the set of 2 is huge nor would it make sense to say the sample space of 2 is huge outside of probability
But for lack of a better word: The sample space/set of any number is freaking enormous. If you made a venn or euler diagram containing every logical statement possible the numbers you wrote would intersect with nearly everything.
13:56
Hi.
If $a+b+c=\pi$
We have : $$\tan a+\tan b+\tan c=\tan a \tan b \tan c$$
What's the geometric interpretation of this ?
Seems to be very symmetric.
when is a point continuous?
14:15
Has anyone here worked with discrete fourier transforms much?
@Owatch Some, not many
Okay. I guess my question would be more or less a technical one about Matlab/Octave. But maybe you have used such software before?
Ah, no, sorry
Aww. :(
Well in case you might know, the algorithm that Octave uses for fast fourier transforms spits out the same answers that mine does for input vectors of length 2^n, but gives something different when the vectors are not of length 2^n. I thought it would just zero-pad the input and return the same thing as I got when I did it, but it doesn't.
Not sure how it works. Thought someone might know.
I need some help
If there were not 0s in matrix then what will be that rectangle stretched to?
14:24
Think I figured it out (Krijn)
They were zero-padding the vector from the left end, not the right end. In any case my output also includes a zero there which they seem to omit in their output. However my inverse transform works okay.
Hm. Theirs still isn't quite the same actually.. never mind.
@Mahmoud Assalamualaikum :D
@DHMO are you here to answer me?
@Ramanujan sorry no idea
Cross cap immersion of the real projective plane, slightly altered to make the double line visible as a circular arc that will otherwise be orthogonal to us in the 4th dimension
@Null You mean, when is a function continuous at a point?
14:30
@AkivaWeinberger yes
@Secret the butt-curve
Informally, when the values of the function near the point are near the value of the function at the point
or, still informally, $f$ is continuous at $x$ if $f(x+\epsilon)\approx f(x)$ for small $\epsilon$
Formally, it's continuous at $x$ if for every $\epsilon$ there exists a $\delta$ such that for all $y\in(x-\delta,x+\delta)$ we have $f(y)\in(f(x)-\epsilon,f(x)+\epsilon)$.
Null: Magnified
(That is, if for all $y$ such that $|x-y|<\delta$ we have $|f(x)-f(y)|<\epsilon$.)
Something like this will happen
Alternatively, if you know some topology, another rigorous formulation is that the inverse image of a neighborhood of $f(x)$ is a neighborhood of $x$. @Null
Wa alaykom salam @Ramanujan
An example is $f(x)=\begin{cases}x,&x\text{ is rational}\\0,&x\text{ is irrational}\end{cases}$
That's continuous only at $0$.
Finally someone to share 3Blue1Brown love with ! :D
For the formal epsilon-delta definition, you can take $\delta=\epsilon$ to show it's continuous at $0$.
@Mahmoud I know right!
14:37
I love that channel, too
I discovered that channel a few days ago in this site only
@AkivaWeinberger Does MathJax not work in the chat? I see all your stuff with $$$ in it...
@HarshaG. Look at the link on the top-right
Thanks @Ramanujan
14:39
Bookmark the "Start ChatJax" link
Did it :)
Welcome @HarshaG..
I am curious about what you people are? I mean are you mathematicians? College students?
@HarshaG. here you can find a dumb school student to a great MIT professor
I am the former I guess @Ramanujan
Preparing for JEE...
14:41
I'm a high school student, + Math enthusiast.
Masters student
@Krijn of maths?
Yeah
Cool! I hope I can do that someday as well
You cannot. Krijn took the last spot.
No more masters students allowed in the world.
14:43
:O
@Owatch really?
Would I ever lie to you?
Can we consider $\max(x,y)=\frac 12 (x+y+|x+y|)$ a multi-variable function ?
And if so, how can I graph it ?
Thanks.

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