@SayanChattopadhyay yeah, I was being dumb. thanks!

Plotting such a function on the real number line would yield a function made entirely of holes, logically speaking.

@BalarkaSen you haven't defined binary operation yet for that group lol

@AMDG You will immediately ran into a contradiction here unless it is not associative

What not being associatve?

since x(/0/0)=x/0=/=(x/0)/0

x/(0/0) would reduce to x/0 in this arithmetic

I know about modular arithmetic but I have forgot it

Oh wait sorry I didn't see that there

I haven't been procrastinating for 2 months after reading like 90 page

(x/0)/0 is not equal to x/(0/0)

(x/0)/0 = x . x/(0/0) = x/0

then for all mirrored reals, I would say (though I wouldn't be able to prove) that each of them maps to a function (or relation) of y.

@AMDG division not associative 0_0?

Bruh, floating point arithmetic isn't even associative for even basic arithmetic operations on the reals...

Deal with it :P

wouldn't 0/0 be undefined ?

I don't know about floating point arithmetic

No because the mirror of 0 would be 0.

cs stuff or numerical analysis stuff?

No, just some stuff that I was recalling from high school :)

:P I don't know about it lol

I was thinking about this sort of stuff in high school.

thank God I was thinking about it was taught u like that in high school lol

"Hey, yeah, you all remember when we were taught about division by zero arithmetic back in 9th grade?"

lol

I see

(x/0)/0 = x reminds me of one of the meadow axioms

you may be dealing with something similar there

If you break associativity, most of the time the system will survive, but do check it more closely as they are very hard to make

Currently the most successful approaches in the literature are Wheel axioms

along with their wheel distributive law

You know, what I heard about in the "borderline" section of your discord screenshot there, I kind of like the idea of "Synthetic Maths" if I understood it correctly... that is to say, I prefer to contemplate abstract things and make them practical as well as find their relationship to the natural order.

Intuition can be good, but don't let it blind you completely, that's why academic rigor is also needed

But yeah, it implies that if you keep dividing by zero, you just go back and forth between the two sets.

The magnitude and sign do not change as a result of the operation.

that's standard for an involution

both Meadows and Wheels have that

Cool, so then I suppose we can define the lambert W function on the reals finally with division by zero :P

Well more like an inverse of x^x or y^y

That's what I meant

It is not that straightforward

zero terms (things like 0x, x/0, etc.) tend to fall outside the usual sets

"(x^x)/0 = ln(x)/W(ln(x))"

also it is currently unclear if such systems can be extended to exponentiation

Well 0x still has the usual behavior of 0x= 0, regardless. (x/0)*0 = 0.

well that explains why you must break associativity

otherwise 00=0 => 0=1

Yes, but only for operations involving division by zero and nothing else. :)

@Astyx Thanks! Now it makes even more sense! I understand now that $\overline{F}$ has same elements as $G$ means F's elements were relabeled as element of $G$. Wished they defined g as elements of $G$ earlier so it would not create that much confusion.

x/0 is just a conversion operation, so I could just as well clarify (x/0)*0 by saying mirror(x) = x/0, x*0 = mirror(x)*0.

0 is the only intersection between the two things which is also why the conversion would even be possible in the first place assuming this were an actually accepted formal system.

now I makes clear after reading it once more time

Thanks to everyone evolved in conversation. Wish you guys have Great day! success!

I was involved. I interrupted the string of comments in the conversation several times. Haha!

Hm, with regards to exponentiation, the question is what 0^0 is in this context and in all contexts with regards to the new number system extending the reals that is not complex (but can be extended by complex numbers). It would still be perfectly fine and safe to say that even with division by zero defined, 0^0 can still be left undefined.

for n not equal to 0, 0^n = 0, so (x^n) / (0^n) = (x/0)^n, right?

Even if you set n=0, and you could say that 0^0 = 1, then it would resolve to (x^0) / (0^0) = (1) / (1) = 1

Or wait, was 0^0 already defined... I can't remember...

0^0=1 is a natural choice, but I think more care is needed here as your structure is nonassociative

you cannot just expect the index laws will hold

I don't know what the index laws are.

Oh I see now what those are

0^0/0^1 = 0^(-1) = 1/(0) = 1/0; 0^0/0^1 = (1) / (0). As an example.

and just for the other part: 0^1/0^0 = 0^(1) = 0; 0^1 / 0^0 = (0) / (1) = 0

I think something like 0^(x/0) is where things will get interesting as the x root of 0^0.

@Stupidquestioninc He has. "Multiplication is defined by adding component wise" with mod 2 arithmetic

@SayanChattopadhyay it would be nice to say binary operation is ...

anyway I am feeling too sleepy since I just sleep for 2 to 3 hour so good night

Ok

@Secret Fun fact: when I was thinking about this division by zero stuff in high school, I recall I had to walk around one of the athletic fields as a sort of detention for something, I can't remember what, and I was just thinking about this stuff as I walked around, and in particular I recall thinking about the plots of such functions as each point being a hole on the Euclidean plane. Odd thing to remember honestly.

$$0^{\frac{x}{0}}$$?

Yes

Supposedly that would just be the x root of 1 for all values of x.

But yeah, I spent my time in high school as a contemplative thinking about things that are actually meaningful unlike the garbage they "taught" us, or rather the attempt thereof.

My grades were abysmal, but of course, I didn't care about that. I cared about actually learning.

Still managed to barely pass high school, though, with a 1.97 GPA. I guess they rounded up and had pity on me or something since at the time, that high school required a 2.0 for graduation.

If it weren't for school, I'd probably have an income as I had a project I was trying to work on which could have given me something to build a little nest of cash with.

Not too social much in here, eh?

that's fine

I am not a very social person and don't like the rules of society either

hmm...

Let 0^(x/0)=y

Then take /0 power both sides

What is /0?

that's the thing that sticks on the right side of the LHS

well by default it should be 1/0, not /0, no?

right because (x/0)(1/0)=(x/0)/0=x

Ok, just making sure

y^(1/0) is \sqrt{0}{y} then. Is that defined?

it is not in reals

with the usual algebraic structure

in fact, lim_x->0 y^(1/x) is undefined there since this limit do not converge

but we are now working in the mirror numbers

so the rules are different

Ok

Well if "in the mirror of x" we'll call it formally has supposedly the same semantics as x in the reals, then such a limit would logically be defined as some mirror of x. (Also, should probably already be obvious, but for rigor: (1/0) = mirror(x). mirror(x)/0 = x. Therefore the mirror of the mirror of x is the real value of x.)

Though we would need disambiguating vocabulary to distinguish "mirrored reals" and "mirror of x".

sound like mirror() is involutive

meaning applying it twice returns what it started with

Ok, then we'll define mirror as an involutive operation since that is what it is.

As I was saying... the latter is context-sensitive, so if x is in the mirrored reals, then "mirror of x" here means mirror(mirror(x)), and vice versa if x is already real: mirror(x).

right so if you have x real, then /x will be mirror of x

Yes

and so //x=x

mirror mirror of x give x

$\Bbb{Q}$ is not a free $\Bbb{Z}$module basically because no two rationals are linearly indepent over $\Bbb{Z}$? More specfically, if $a/b,c/d$ are nonzero and in lowest terms, then $(-bc) \frac{a}{b} + (ad) \frac{c}{d} = -ac + ac = 0$.

Does this sound correct?

correct

@Secret Yeah, so, that means the limit is something along the lines of lim_x->0 y^(1/x) = y^(mirror(1)) for real y and real 1.

So now it is a matter of defining exponentiation across the two sets.

@user193319 yes; of course you also have to exclude that it is free of rank $0$ or $1$

What would rank 0 mean? Rank 1 means it's a cyclic group, right?

(Curiously enough, I have also toyed with the idea of a system where all arithmetic operations of addition, subtraction, multiplication, and division are commutative, resulting in zero never existing except as a digital place holder, and in its place, the real number line is extended with two objects, alpha and omega, representing "the first and the last" real number. Would be approximated by the standard arithmetic with norms or magnitudes of x and y.)

rank $0$ means free on a set of cardinality 0$, i.e. free on the empty set, i.e. the zero module

(In that case, negation becomes involutive between negative and positive reals)

well, rank $1$ means isomorphic to $\mathbb{Z}$

cyclic just means generated by a single element

but free of rank $1$ also requires for that element to be linearly independent

i.e. to not be torsion

In a YT comment about maths: "Day 55 of Quarantine: Am now watching math for entertainment. Send help pls."

"recent studies have shown that intellectual activity and curiosity due to coronavirus has reached an all time high since the renaissance due to increased levels of boredom..."

Is there any particular reason why we don't have operations on the digits of numbers themselves rather than arithmetic quantities? Is there such a formal arithmetic available already?

@AMDG we have language theory if that's what you're asking

Not sure Language theory is the name of the theory to be honest

Formal language theory en.wikipedia.org/wiki/Formal_language

I mean things like actual arithmetic on digits of quantities as opposed to the logical quantities themselves.

Things like bitwise operations.

That's formal language theory I think

Well then I guess I compute 2^x and log_2(x) using formal language theory :)

pow2(x) = 1 << x

log2(u64 x) = 64 - lzcnt(x)

exp(x) = 1 << ((x * 10000)/14426)

cos(x) = haha just kidding I'm still looking for that one

Is the formulaic way to go from a moment generating function to a probability density function a line integral in the complex plane, similar to the Inverse Laplace Transform?

suppose a Borel set $E$, is approximated by a compact set from within, and an open set from without: $K \subset E \subset V$ so that $m(V) - m(K) < \epsilon$. What can be said about the corresponding indicator functions $1_K, 1_V$? $\|1_K - 1_V\| < f(\epsilon)$..? And in what norm?

$\lVert1_k-1_v\rVert_1<\varepsilon$ in the $L^1$-norm

Woops nvm

what makes higher dimensional extensions of p-p plots difficult?

a p-p plot is a plot of two CDF's against each other. so we can parametrise two CDF's $(F(z),G(z))$ with parameter $z.$ Couldn't you construct an extension just by plotting 3 CDF's against each other such as $(F(z),G(z),H(z))?$

when's the last time you saw a 3D graph? @geocalc33

also I made a mistake above. should have two parameters in the last line not just one

actually nvm

Can anyone recommend me some good material(s) about the numerical methods/schemes for the solution in 1, 2 and 3 spatial dimensions wrt Boltzman Transport and Ginzburg-Landau (Real and Complex) equations?

Is there a common notation for the probability distribution denoted by normalizing an arbitrary density $f$? Something like $\tilde{f}$?

Not that I know of

hello

o/