Mathematics - 2017-08-22 (page 3 of 4)

DaneJoe

17:12

Is there a way to directly equate any single operator of a product variable to a factorial?

like if I did \prod sin(k) is there a way to turn that into sin(k)! or anything like that?

oh btw the log thing has 6 stars now

Balarka Sen

17:28

@SteamyRoot Underrated comic there. Really underrated.

Semiclassical

@DaneJoe definitely not.

the case that Balarka did amounted to the observation that 6=1*2*3=1+2+3, i.e. 6 is a perfect number.

in general there's no relation like that, and certainly not for anything like the sine function

DaneJoe

@Semiclassical Well there's already special cases where it's true, I'm just wondering if there's a general case. I wouldn't be so sure about the "definitely" part. There's definitely for sure relationships that use the product of the sine function.

Semiclassical

Product of two sines, yes.

You're asking for products of arbitrarily many sines.

DaneJoe

mm no I'm thinking more of the gamma function

but anyway, as I said I'm not talking specifically only prod sin(k) = sin(k)!

that statement obviously isn't true

Semiclassical

You're looking for relations of the form $f(\Gamma(x))=\Gamma(f(x))$?

DaneJoe

17:32

I'm just wondering about cases with a form like that

like sin(k!) or (-1)^kk!

Semiclassical

actually, what I just wrote isn't very sensible

DaneJoe

so like any formula that comes to mind that "looks" like that

but not limited to that

Semiclassical

something more like $\prod_{k=1}^n f(k)=f(n!)$.

DaneJoe

but no more like prod f(x) = f(k!)

that's the ideal case

but obviously the real case would be more complicated than that

Semiclassical

well, suppose you take logs of both sides

DaneJoe

17:34

right then you have the sum

Semiclassical

then that becomes $\sum_{k=1}^n \log f(k)=\log f(n!)$

DaneJoe

big whoop

id like to keep it as a product however

or a factorial

Semiclassical

so if you let $g(x)=\log f(x)$ then that's equivalently $\sum_{k=1}^n g(k) = g(n!)$.

the main advantage being that the left-hand side is additive rather than multiplicative, and that $g(k)=k$ is now a solution.

DaneJoe

yeah I was thinking about that but there's something that doesn't quite work

but it could be the indeces im using

ill try it again

oh right

Semiclassical

here's the problem I see. for $n=1$, that becomes $g(1)=g(1)$ which is always true. for $n=2$, it's $g(1)+g(2)=g(2!)=g(2)$. so $g(1)=0$.

DaneJoe

17:36

you didn't transform the boundaries of summation

at all

Semiclassical

right.

DaneJoe

but neither did I

Semiclassical

and they shouldn't.

DaneJoe

seems like they should

Semiclassical

no. $\log(f_1f_2\cdots f_n)=\log f_1+\log f_2+\cdots +\log f_n$.

taking the log changes the product to a sum but does nothing to the relevant indices.

DaneJoe

17:37

if you do a sum from k=1 to n, but k turns into log(k), then it seems like you'd have log(k)=1 to log(n) or something along those lines

Semiclassical

again, no.

that's entirely not true.

DaneJoe

there's a lot of implications if you really actually say it's "no"

in fact you'd be the greatest mathematician of the century

Semiclassical

no, I'd be anyone who understands that the log of an n-fold product is the n-fold sum of the logs.

DaneJoe

you still don't see the general implication of that generalized case, so I'll leave and come back later and see if anything changes

Will Hunting

Listen to Semi, he knows stuff.

DaneJoe

17:39

i wish he did but i already tried the method he's suggesting, and it didn't work

and it was an ideal assumption

Semiclassical

$\log \left(\prod_{k=1}^n f_k\right) = \sum_{k=1}^n \log f_k$, full stop.

(assuming all the $f$'s are positive etc.)

Will Hunting

The log of the product is the sum of the logs, what's so hard about that? And yes, period, on the other side of the Atlantic.

Akiva Weinberger

Semi's American

Semiclassical

lol

Will Hunting

Yes, I know, and what I said still holds, full stop.

Also, lol, period.

Akiva Weinberger

17:42

I do like the fact that all American sentences end in menstruation

Semiclassical

Am I accidentally using the British expression instead of the American one? That'd be kinda funny.

Akiva Weinberger

(Well, excluding questions and exclamations)

Wietlol

hya

Will Hunting

British say full stop and Americans say period.

Akiva Weinberger

(and also 99% of my posts here)

Wietlol

17:43

anyone can clarify the "0.999... = 1" statement?

Akiva Weinberger

If they were different there'd be something between them.

Semiclassical

sure. multiply both sides by 10. what do you get?

Will Hunting

Not another recurring decimal question, lol.

Wietlol

@Semiclassical well... more the part, when is this statement true and when is this statement false

Semiclassical

It's always true.

Wietlol

17:44

for example

the question was "get a random number"

where "a random number" is literally ANY number

Will Hunting

The thing is, 0.999... is 1, and one will have to invoke the decimal expansion construction of the real numbers for that.

Wietlol

what is the chance to get the number 5?

well, that is 1 / inf

Akiva Weinberger

Yeah so when there are infinitely many possibilities, "probability 0" does not mean "impossible"

Semiclassical

I wouldn't say there's even a meaningful probability in that case.

Wietlol

which is written down as 0.000... or 0.000...1

Will Hunting

17:45

There is the dedekind cut construction, the cauchy sequence construction, the decimal expansion construction, and a few others.

Akiva Weinberger

The probability is 0.

Semiclassical

0.000....1 is not the same as 1-0.9....

Wietlol

so the probability is not 0% right?

Alessandro Codenotti

this question is ill-posed, without specifying a probability distribution you're choosing the numbers with

Akiva Weinberger

It is. @Wietlol

Will Hunting

17:46

Check out different books for the different constructions then.

Semiclassical

0.000...001 only makes sense if there's finitely many 0s there.

Wietlol

but 0% is never occurring

that is the definition of 0%

Akiva Weinberger

When there are infinitely many possibilities, "probability 0" does not mean "impossible" @Wietlol

Semiclassical

there is no such then as 0.0000...0001 with an infinite number of zeros in between.

Wietlol

on the other hand, we can revert the maths, 0 * inf = ... uhm... 0?

Semiclassical

17:47

no.

0*inf is not determinate.

Wietlol

0*anything = 0

Will Hunting

@AkivaWeinberger That depends on the definition of impossible, lol.

Wietlol

basic math rules

Semiclassical

again, no. not if you're doing infinite

there is no determinate meaning to 0*infty

Will Hunting

What is a point? A point doesn't really exist, because no matter how small we draw a dot, it can be made even smaller.

Wietlol

17:48

but infinitly many zeroes summed up is still 0

GFauxPas

I'm late to the game, are we talking limits or probability?

Akiva Weinberger

That's a good point. There are other reasons for leaving 0*infinity undefined.

Semiclassical

Probability.

Akiva Weinberger

(Set theory supports the idea that it's zero, in that the Cartesian product of the empty set with an infinite set is the empty set.)

skullpatrol

What number is in between 1.000... and 0.999...? @Wietlol

Semiclassical

17:49

the way I'd say it: If 0.000...001 means that there's a finite number of zeros between the decimal and 1, then this is nonzero and positive.

Wietlol

on another side, for any non-neutral number 'n' "a / n * n = a" is correct right?

Will Hunting

There is no such thing as 0.000...0001 actually.

Mr. Xcoder

@skullpatrol $0.(9)$ (i.e $0.99999...$) is $1$

Wietlol

@skullpatrol 0.(0)1

Semiclassical

but if one intends it as an infinite number of zeros, then to the extent that that is meaningful then it's just 0.

Will Hunting

17:50

Because you cannot have infinite number of zeroes followed by one.

skullpatrol

^

Semiclassical

Here's a quick argument. Suppose 0.9999... < 1.

Akiva Weinberger

@Wietlol What's 0.000...1 divided by 10

Semiclassical

any inequality is preserved under multiplication by a positive constant. so let's multiply both sides by 10:

9.999.... < 10.

Akiva Weinberger

What's 0.000...1 divided by 2? Or 0.000...1 multiplied by 5? @Wietlol

Semiclassical

17:52

...ah, dang. this approach doesn't work.

Wietlol

@Semiclassical it does work

but i dont agree with the next step

Akiva Weinberger

@Semiclassical Subtract 0.9999...<1 from both sides?

Mr. Xcoder

@Semiclassical What are you trying to prove?

Akiva Weinberger

Worthiness.

Wietlol

ow wait, it doesnt work :D

skullpatrol

17:52

Recall "..." means "and so on". It never ends.

Semiclassical

$a<b$ and $c<d$ doesn't imply anything about $a-c$ or $b-d$.

Wietlol

yet again, i know all those proofs, but i dont think they are always valid

Semiclassical

they are.

Wietlol

@AkivaWeinberger NaN

NaN / 2 = NaN, NaN * 5 = NaN

Akiva Weinberger

I mean 0.000...1 is "Not a Number" as well

Wietlol

17:53

just like inf is not a number

its comparable to 1/3

Semiclassical

it's not a number, and yet you're trying to do arithmetic like it is a number

Akiva Weinberger

1/3 is a number. confused

Wietlol

1/3 = 0.(3) but 0.(3) != 1/3

Akiva Weinberger

0.(3) does equal 1/3

In any case, there are two relevant facts. (a) The real number system has no infinitesimals. (b) There do exist systems with infinitesimals, but these usually will have more than one infinitesimal.

So if you want 0.000...1 to be a thing you'd want to specify which infinitesimal it is.

Wietlol

@Semiclassical well... the arithmetics werent there at all

Semiclassical

17:54

you wrote NaN/2= NaN. that's arithmetic.

moreover, if it's not a number, you can't say it's greater than or less than another number.

Wietlol

just someone trying to say
> 100% / inf = 0.(0)
> 0.(0)% = 0%
> 0% * inf = 0%
> 100% = 0%

Akiva Weinberger

I mean if you asked a computer to computer NaN/2 that's probably what it'd return

@Wietlol Yeah so the usual resolution to this is to say that most of those operations aren't defined

Wietlol

a / n * n = a (given n is not 0)

Semiclassical

and given a is a number.

Wietlol

yes

but that is completely irrelevant

in most examples, a would be 1 or 100

Akiva Weinberger

17:57

If everything were defined, math would be so much simpler. As in, more boring. I think a lot of the structure in math comes from the fact that certain things we'd want to work don't.

Semiclassical

here's a fairly concrete example. pick a random real number between 0 and 1 (not including the endpoints)

what's the probability that you pick 1/2 ?

Akiva Weinberger

0.1234127346172384617324691823647632798467...

Wietlol

1/inf

skullpatrol

Infinity is not a real number.

Semiclassical

^

Wietlol

17:58

@skullpatrol so, what is the answer then?

Semiclassical

Zero.

Akiva Weinberger

0

Balarka Sen

0

skullpatrol

^

Balarka Sen

sniped come on

Akiva Weinberger

17:58

Daminarked

Wietlol

by what calculation?

Balarka Sen

sniped, @Akiva

Akiva Weinberger

@BalarkaSen That's what I said

Balarka Sen

I said sniped before you said daminarked

so you are sniped slash daminarked

Wietlol

its 1 / totalNumberOfOptions

Akiva Weinberger

17:59

@Wietlol The probability it's between 0.4999 and 0.5001 is 0.0002, right?

Semiclassical

That's applicable if there are finitely many options.

Akiva Weinberger

So it's less than 0.0002

Balarka Sen

That's not how probability is defined for continuous distributions.

Semiclassical

^

Wietlol

yes... sort of

i think

Akiva Weinberger

18:00

Similarly it's clearly less than any positive number. So it's zero

Wietlol

probably have to include one of the endpoints to make it correct

Akiva Weinberger

TL;DR Probability doesn't work right for infinite options. So we had to step in and fix it

and it works kinda weirdly but it works

but you get weird results sometimes

Semiclassical

What one needs to distinguish in probability are: "X occurs with probability zero" and "X never occurs."

Wietlol

but this fights 2 logic rules now

Semiclassical

If there's finitely many options, the two are identical.

Balarka Sen

18:01

In general instead of a probability mass function you have a probability density function $f$ corresponding to the random variable $X$ you are working with and $\Bbb{P}(a \leq X \leq b) = \int_a^b f(x) dx$ by definition.

Wietlol

one, 0% chance should never occur

Semiclassical

If there are infinitely many, they are not.

Wietlol

that is why it is 0%

Semiclassical

Again, no.

"X occurs with probability zero" = "X almost never occurs"

2

skullpatrol

^

Wietlol

18:02

who made this completely illogical statement?

i want to kill someone

i only deal with logic

and logically 0 = 0

Semiclassical

Well, there's two scenarios.

Wietlol

0% chance is 0 successes out of any number of attempts

Secret

Infinitary combinatorics

In mathematics, infinitary combinatorics, or combinatorial set theory, is an extension of ideas in combinatorics to infinite sets. Some of the things studied include continuous graphs and trees, extensions of Ramsey's theorem, and Martin's axiom. Recent developments concern combinatorics of the continuum and combinatorics on successors of singular cardinals. == Ramsey theory for infinite sets == Write κ, λ for ordinals, m for a cardinal number and n for a natural number. Erdős & Rado (1956) introduced the notation κ → ( λ ) ...

Wietlol

that is logic

Leaky Nun

"logic" is overrated these days

Secret

18:02

Infinity is weird

Wietlol

i dont strike infinity as weird

i programmed it once

Akiva Weinberger

@Wietlol Choose a random number from 0 to 1. A random number in [0, 1]. What are the odds you choose something in [0, 1/2]?

Semiclassical

lol

Leaky Nun

hop in to measure theory to see why {0.5} has measure 0 while [0,1] has measure 1

Akiva Weinberger

That is, something between 0 and 1/2

Will Hunting

18:03

When you have studied enough mathematics, everything will make sense.

Balarka Sen

@Wietlol It's not illogical. Just that probability for continuous distributions is not what you think it is.

Will Hunting

Things only don't make sense when you have studied only a little.

Semiclassical

You programmed something you thought of as infinity into a computer.

Wietlol

thats 50%

Akiva Weinberger

Well, there are infinitely many options and infinitely many success events, so

would you say infty/infty??

Leaky Nun

18:04

3

Wietlol

@Semiclassical and the computer will always say that it is larger than any finite number

Akiva Weinberger

See, "success events divided by possibilities" doesn't work for infinity

Wietlol

and every other rule applied on infinity is also applied correctly

Akiva Weinberger

I mean, yes, the answer is 50%

Wietlol

infty/infty

also correct

but here you have 2 different infinities

Secret

18:05

Is it even possible to use finite amount of computer resource to compute an algebraic element that is larger than any real number?

Wietlol

so, you have to specify their origin

Semiclassical

...

Wietlol

one of them is only half as large as the other

Leaky Nun

@Secret yes

because all of our definable numbers (infinity included) can be defined in a finite number of steps using predicates

via ZFC axiom schema

Semiclassical

The distinction that should be drawn is this. If I pick a random real number from 0 to 1, is 1/2 an allowed possibility? Yes, certainly so.

By comparison, is 2 an allowed possibility? No, certainly not.

Akiva Weinberger

18:06

@Wietlol OK, so if we call the first infinity $j$, the second one will be called $j/2$. Is that what you're saying

Secret

@LeakyNun Interesting, I did not know that infinity can be defined in a finite number of steps

Leaky Nun

@Secret if it can't, you won't be able to talk about it

Wietlol

that would be correct

Semiclassical

So X=1/2 can occur, but X=2 cannot.

Akiva Weinberger

@Wietlol So what kind of number is j?

What's j+1?

Semiclassical

18:07

However, I can guarantee that if you ask a random number generator to produce samples from this sequence, you will never ever ever see exactly 1/2.

Wietlol

iDunno

Leaky Nun

@Wietlol you may want to read about this answer and this answer.

Wietlol

i dont know the value of j

so j+1 = j+1

Balarka Sen

Stop Making Sense - Trailer

►

I don't know why I linked that. Just for the title lol

Akiva Weinberger

Hm I actually don't know how to continue this

Semiclassical

18:08

It is not logically forbidden for X=1/2 to occur, but the probability of it occuring is 0 because there's only one possible way to succeed but infinitely many ways to fail.

Wietlol

@Semiclassical its only 0 because you follow the rule that 0.(0) = 0

otherwise, the probability would have been 0.(0)

Secret

how is $j$ behave like infinity if it cannot absorb finite numbers additively?

Semiclassical

Sure. I see no reason why one wouldn't follow that rule.

Wietlol

@Secret it will still remain a relative number

@Semiclassical so, 1/inf = 0 ONLY IF 0.(0) = 0

Semiclassical

Again, you're reifying infinity into a quantity that one can do meaningful arithmetic on.

Secret

18:11

You need to somehow intoduce a notion that it is larger than all finite numebrs else $j$ is just some algebraic element, not a representation of infinity

Wietlol

then you can choose between following that rule or not

then you have a different situation

@Secret true, j+1 is also infinite

but it is larger than j now

exactly by 1

Secret

that will mean $j$ is some ordinal number

Wietlol

so its definition is j+1

Leaky Nun

@Wietlol are you aware how real numbers are defined?

Wietlol

hearing your question, i assume not

Secret

18:12

if it cannot absorb 1 and $j+1 > j$ then the only sensible way to define $j$ is it is some infinite ordinal

Wietlol

even though before you asked it, i would have said yes

Akiva Weinberger

@Secret That, uh,

why

Hyperreals have infinite things that don't absorb stuff for example

Secret

I was almost going to said that

(surreals...)

Leaky Nun

9 mins ago, by Leaky Nun

Akiva Weinberger

@Secret I don't know how helpful this is to be honest

@LeakyNun That's not even the real graph

Fun fact

Leaky Nun

18:14

@AkivaWeinberger heh

the more you know

Akiva Weinberger

I underestimate my understanding of Dunning–Kruger

Wietlol

@LeakyNun that means that... you are either an idiot or an expert

because im quite confident

or is it about myself?

Leaky Nun

@Wietlol how did you draw this conclusion seeing that I did not say I am confident?

Semiclassical

yes. yes, it does.

Secret

Well, suppose $j$ is an infinite hyperreal, what wil that lead us?

Wietlol

18:15

ah, i misread competence

still, then im either an idiot or an expert

Akiva Weinberger

Eh whatever I'm out bye

Leaky Nun

@Wietlol whatever

Secret

1

Q: Hyperreal probability density?

I'm fairly new to the subject of hyperreal numbers and I'm wondering if there exists an infinitesimal number $a$ such that (in some reasonable sense) $$\sum_{n=1}^\infty a=1$$ ? In other words: Is there a uniform (hyperreal valued) probability distribution on the natural numbers? I hope the qu...

4

Q: Probability theory with the hyperreals?

Forgive the undoubted ignorance of this question. I am out of my element both in probability and in nonstandard analysis. A mathematically curious layperson friend recently had a conversation with me where he wanted a notion of "pick a random integer" (meaning uniformly random). I told him that ...

probability-theory reference-request nonstandard-analysis

Wietlol

knowing i passed that first peak, and knowing that i grew in confidence

im closer to being an expert than an idiot

4

:D

Leaky Nun

in Algebraic/Transcendence Theory, Dec 26 '16 at 15:03, by DHMO

The natural numbers are the numbers specified by Peano's axioms, denoted by $\Bbb N$.

in Algebraic/Transcendence Theory, Dec 26 '16 at 15:05, by DHMO

The integers are the numbers in $\Bbb N \times \Bbb N$ under the equivalence relationship $(a,b) = (c,d) \iff a+d = b+c$. A natural number $n$ is represented by the equivalence class $[(n,0)]$. They are denoted as $\Bbb Z$.

in Algebraic/Transcendence Theory, Dec 26 '16 at 15:10, by DHMO

The rational numbers are the numbers in $\Bbb Z \times \Bbb Z^+$ under the equivalence relationship $(a,b) = (c,d) \iff ad = bc$. An integer $z$ is represented by the equivalence class $[(z,1)]$. They are denoted as $\Bbb Q$.

in Algebraic/Transcendence Theory, Dec 26 '16 at 15:12, by DHMO

The real numbers are denoted as $\Bbb R$. They are defined in the following two ways:

in Algebraic/Transcendence Theory, Dec 26 '16 at 15:14, by DHMO

1. Real numbers as Cauchy sequences: Let $(a_n)$ be a sequence of real numbers such that for any given $\epsilon > 0$, there is a natural number $N$ such that for all natural numbers which are greater than $m$ and $n$, $|a_m - a_n| < \epsilon$. Then, this sequence is a Cauchy sequence, and its limit is defined as a real number.

in Algebraic/Transcendence Theory, Dec 26 '16 at 15:16, by DHMO

2. Real numbers as Dedekind cuts: a real number is defined a partition of the rational numbers into two sets, the smaller set containing no biggest element. The real number is the supremum of the smaller set (and the infimum of the larger set).

Will Hunting

18:23

All this can be found in several books.

Read those books for more info. Don't take Math SE as the authority.

Leaky Nun

@WillHunting I concur

Will Hunting

@LeakyNun I just use agree, my vocab is very small, lol.

Leaky Nun

@WillHunting I agree

Will Hunting

Again, I recommend Elliot Mendelson: Number Systems and the Foundations of Analysis for a construction of the real numbers in a few hundred pages.

DaneJoe

18:43

Oh yeah I figured out the problem with transcendental numbers

So polynomials of a degree of 5 or more can't have solutions generalized by only algebraic functions...

But transcendental functions are not solutions to polynomials as they are defined

Semiclassical

by algebraic functions, you mean elementary ops + taking radicals?

DaneJoe

So somehow there's a function that can't be described by algebraic operations, and yet somehow it is not transcendental because it is implied it is the solution tova polynomial

Leaky Nun

@DaneJoe a number, not a function.

DaneJoe

I think this is misleading and I think it results from the fact mathematicians assumed all polynomials could be generalized with radicals before the Abel ruffini theorem

Semiclassical

The reason I say this is that the more common definition of "algebraic function" (the one which Wikipedia uses, for instance) is: "In mathematics, an algebraic function is a function that can be defined as the root of a polynomial equation."

Leaky Nun

18:45

I thought you already figured out that not all roots of polynomials can be written with elementary operations

DaneJoe

But remember the discussion

Semiclassical

So under that definition the roots are always an algebraic function of the coefficients.

DaneJoe

There's a difference between an algenraic number and algebraic.operations

That's where the ambiguity comes from

Leaky Nun

@DaneJoe yes, there is

DaneJoe

Right so that's what I'm saying,

Semiclassical

18:46

Under your definitions, yes.

DaneJoe

Under the concensus decision you mean

Leaky Nun

@DaneJoe it's either an equivocation fallacy or an etymological fallacy

Semiclassical

Given that Google pings back the Wikipedia definition of "Algebraic function" if you type that in with quotes...no, I wouldn't call yours the consensus definition by any means.

DaneJoe

Like I said given how old it is, it would seem the terminology is a misnomer stemming from a period from when mathematicians assumed all polynomials could be generalized by algebraic operators

Semiclassical

Only insofar as you insist on "algebraic operators" taking your meaning.

Leaky Nun

18:48

@DaneJoe you can view it so.

DaneJoe

The definition I have is according to multiple sources including wikipedia

So you should be agreeing with me

Leaky Nun

I don't see the point you're trying to make.

Some algebraic numbers cannot be expressed with algebraic operations. So what?

DaneJoe

At least for algebraic operations

Semiclassical

Ah. You're using two phrases which sound similar but are distinct.

"algebraic operation" vs. "algebraic operator"

DaneJoe

It's a misnomer because algebraic operations are simply radicals and basic operations. But, the solutions to all polynomials must include an operation that is not one of those

Leaky Nun

18:50

@DaneJoe ok, it's a misnomer, so?

Semiclassical

The latter isn't used much, but it does have a determinate meaning and that meaning is more in line with algebraic function, algebraic value, etc.

DaneJoe

They should be called "polynomial" operations instead

Semiclassical

Yeah, I'd buy that.

DaneJoe

So I think from now on if I publish anything related to that I'm going to call them polynomial solutions and polynomial numbers

Semiclassical

No. You're conflating things.

Leaky Nun

18:51

@DaneJoe that is, if you publish anything related to that.

Semiclassical

You're correct that the usage of algebraic operations is inconsistent with algebraic number and algebraic function.

But algebraic number is a far more significant concept than algebraic operation.

deostroll

how to argue that this function is not differentiable at $\frac{n\pi}{4}$ desmos.com/calculator/k32dmnlvas

DaneJoe

It's not really conflating, it's updating

Leaky Nun

@deostroll nice programming skills

@deostroll argue that the left- and the right- derivatives differ

DaneJoe

Any argument for "significance" is arbitrary relative to math itself. The terminology comes from an older era, and it's not unheard of to change things. If algebraic numbers are more "significant" the sure people wouldn't mind renaming the group of operators to something like "elementary" or "radical" operators

Leaky Nun

18:55

@DaneJoe which we did.

I call them elementary operations.

or elementary functions

deostroll

@LeakyNun thanks

DaneJoe

No elementary just means +-/*

It doesn't include radicals

Leaky Nun

hmm

Tobias Kildetoft

@DaneJoe Of course algebraic numbers are more important than those operations that are called algebraic here (I am not actually sure I have heard them called that before btw)

Semiclassical

^

DaneJoe

18:57

I think it would be easier to make a new term than to change a current definition

Elementary can stay as it is, but a new term should be for when it includes radicals

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