Gaussian Primes Visually

Interact with the primes at thegraycuber.g...

[@matthijshebly]

I wonder how the Game of Life would evolve on that grid

[@daboffey]

I was wondering the same thing:)

[@TheGrayCuber]

people.math.harvard.edu/~knill/primes/life.html

[@matthijshebly]

@@TheGrayCuber Awesome!

[@PRIMARYATIAS]

@@TheGrayCuberNice, It ends ultimately in a stable cyclic state.

[@matthijshebly]

@@PRIMARYATIAS *for the area shown

[@stanleydodds9]

It's worth noting that your definition of "primes" is actually the definition of irreducibles. They happen to be equivalent for sufficiently structured rings, like unique factorisation domains, but in that case it's a bit circular to say that this general definition of a prime gives rise to unique factorisation. Really, it's the true definition of a prime that gives rise to unique factorisation, that being: p is prime, by definition, if p|ab => (p|a or p|b). In words, a prime is a number such that whenever it divides any product, it must divide at least one of the factors. Also a formal statement of unique factorisation would allow for an additional arbitrary unit as a factor - this makes it clear that both 1 and -1 (and any other units) are an empty product of primes. Furthermore, only non-zero elements have unique factorisation, since zero is often the special case when it comes to multiplicative properties of a ring.

[@TheGrayCuber]

Yes, this is a great point, and it is one that I plan to clear up in an upcoming video where I introduce some rings that aren't UFD. I figured this definition would be more palettable as an introduction since it's more similar to the definition many people are familiar with, and it is true if only using Z or Z[i]

[@santerisatama5409]

Yes, irreducible = unique = holistic whole. That's the deep insight of Euclid's intuitive foundation of mathematics, which object oriented axiomatics tend to cover up and forget. We can also construct an operator language in which are the unique combinatorics for what has the numerical interpretation 1/1 when is defined as the denominator element and < and > which are not part of the denominator element are defined as the mark and antimark elements of numerator. How would lateral irreducibility be defined in this case? Mutual mirror symmetries of the operator strings could be said to be mutually reducible monogamic entanglements, when interpreting the operators as "pure Dirac" notation... But are chiral concatenation pairings and >< of strings < and > genuinely monogamic in this case? Or coprimes relative to each other? To complexify the question even further, strings and >< are simple one digit bit rotations of each other to both directions, the two possible perspectives of the same loop. On the other hand, the chiral concatenations and >

[@tomkerruish2982]

Subscribed! Cool tidbit: If you mod the complex plane by the Gaussian integers, you get a torus. This isn't too surprising if you're old enough to have played the arcade game Asteroids. However, the same thing happens if you use the Eisenstein integers, meaning that gluing together opposite sides of a hexagon also produces a torus. (An octagon produces a two-holed torus, but then you need to introduce a hyperbolic metric so the corners will fit together, and it just gets more complicated from there.)

[@user-pr6ed3ri2k]

wait where did you find information on how to compute x mod y on the complex numbers?

[@tomkerruish2982]

@user-pr6ed3ri2k You identify z with z+m+ni for all complex z and integers m,n. Essentially, you cut out a 1×1 square, and whenever you move off one side, you return on the opposite side. It's a topology thing. Alternatively, you map z to frac(re(z))+i frac(im(z)). Edit: This is specifically concerned with how to mod out by the Gaussian integers. Sometimes, I write far too briefly and sacrifice clarity.

[@drdca8263]

Why do you need to introduce any kind of metric? (:P)

[@user-pr6ed3ri2k]

@@tomkerruish2982 How would this definition be related to standard modulo? like, let's say I'm trying to use real mod 3 on the entire complex plane. it's clear that 3*n+[0, 3) are still unique residues like in the real world, but then there could be 3*n + x*i + [0, 3) for arbitrary x since there doesn't seem to be a way to reduce them (complex numbers) down to [0,3)... Is it allowed to multiply by i? this whole thing seems to get a lot messier with complex numbers as the base of the mod....

[@tomkerruish2982]

@user-pr6ed3ri2k Instead of thinking of two numbers being equivalent modulo 'whatever' as meaning they have the same remainder under (integer) division by w/e, think of it meaning that their difference is a multiple of w/e. For example, if we say that x and y are equivalent modulo 6, we mean that x-y is in the set {..., -12, -6, 0, 6, 12, ...}. Saying that two complex numbers w and z are equivalent modulo the Gaussian integers means that w-z is in the set of Gaussian integers.

[@graf_paper]

Seems like primes that are 3 (mod 4) are gaussian amd primes that are 1 (mod 4 are not) for the reason that primes 3 room 4 cannot be expressed as the sum of 2 squares.

[@hobbified]

Right, 4n+1 is equal to (2n+i)(2n-i), and 2n±i clearly has a smaller modulus, so 4n+1 can't possibly be (Gaussian) prime.

[@lesoleil2465]

​@@hobbifiedwhy 4n+1=(2n+i)(2n-i)

[@ericbischoff9444]

no, it's not true, 4n² + 1 = (2n + i) (2n - i) but it was a nice try, there must be some truth behind that error

[@LeonardoSaobya]

@@ericbischoff9444 make n root

[@MrDarkPage]

The integer primes that are the sum of two squares are not gaussian prime because: a²+b² = (a+ib)(a-ib), both those are the same distance from origin, so they are either both gaussian prime or both divisible by smaller gaussian primes. I don't know if that encompasses all integer prime that are not gaussian primes though

[@dcterr1]

Excellent video! I have a PhD in algebraic number theory from UC Berkeley, so I'm well-versed in everything you discuss in this video, Gaussian primes in particular, but I must say that you give a very good, intuitive description of them! I look forward to your next video on Eisenstein primes and I hope you make even more! (You might want to go into general number fields, UFDs, and perhaps even prime ideals and class numbers, but I don't know how advanced you want to get!) Meanwhile I'll have to check out your app!

[@sonicwaveinfinitymiddwelle8555]

okay skibidi toilet

[@PatrickStaight]

I think I'll check out the app as well

[@dcterr1]

@@PatrickStaight I think you'll enjoy it, and perhaps you'll even be able to figure out how to initiate Level 1, which I still can't do, go figure!

[@StupidusMaximusTheFirst]

"That's beautiful man..." - The Dude.

[@StefanReich]

New information has come to light, man

[@artis.magnae]

Nice video, I like the reasonable pace compared to most other math videos nowadays.

[@benjfr5723]

Great work. I love how it goes back the basics of arithmetics to give definitions.

[@themathhatter5290]

It's also worth noting; if a+bi is a Gaussian Prime, so is a-bi. You can combine this with the rules for units multiplication to see you only need to check Gaussian integers of the form 0

[@themathhatter5290]

EYPHKA! The insight that allows me to show a+bi is in the same associative group as a-bi has worked again! First, why can we do this? It is because in the complex plane, it is ambiguous if i or -i is the "correct" solution to x^2=-1. Moving over, and treating i as one solution, we can factor it as x^2+1=(x-i)*(x+i)=0. Neither solution is more correct. So in addition to multiplying by roots of unity, or the units [1,i,-1,-i], i and -i are interchangeable, allowing us this narrowed, 1/8 search of the plane, because we can apply these, essentially rotations and reflections to get the whole plane. But w is also the root of an equation with ambiguous "correct" definition in this same way! x^3=1 has a solution in the real numbers, x=1. But we can divide that factor out as it's not interesting, though it will return as a unit. This gives the equation x^2+x+1=0. One of the solutions is w, but the other is w^2; let us go through. If w^2+w+1=0, then w^3=1 by factorization. Then, substituting w^2 in for w, w^4+w^2+1=w*w^3+w^2+1=w*1+w^2+1=w^2+w+1=0. They are both solutions, and interchangeable. But this entirely defines the Eisenstein integers! Because it may seem like we need a+bw+cw^2, but we can rearrange the defining equation for w; w^2=-1-w. So there is no need for w^2 in the written form. All Eisenstein integers can be written in the form a+bw, and plotting them, we see a hexagonal pattern, much like how the Gaussian integers make a square pattern. Now the final trick. There are six units in the Eisenstein integers, [1,1+w,w,-1,-1-w,-w]. So we only need to search a sixth of them to find all primes and their associates. But surely we can do better, like we did with the Gaussian integers; there we found a plane of reflection that had a simple rule. In essence we can send (a,b) to any of their positive of negative versions, and swap a and b. How do we do this in the Eisenstein integers? Instead of swapping i for -i, we swap w and w^2, or w for -1-w. So if we have an Eisenstein integer (a,b), we get its reflection as (a-b, -b). The other multiples of roots of unity act as rotations, so choosing the computationally easiest 1/12 slice, we only need to check Eisenstein integers of the form a-bw, where a>=b>=0. This region and its associates covers the field of Eisenstein integers. Q.E.D.

[@themathhatter5290]

It is worth pointing out; a prime and its conjugate are not in the same associative group. The product of m*z, where m and z are both complex, is the conjugate of conj(m)*conj(z), but not equivalent to either conj(m)*z or m*conj(z). All four will however lie on the samr circle centered about the origin.

[@stevemcwin]

This is a really interesting topic and the video is well made. I hope to see more of your content!

[@NexusCubed2950]

Horizontically

[@broyojo]

i appreciate no music

[@Rikri]

Since each non-gaussian prime is a product of a gaussian prime and its conjugate, we know they're of the form (a+bi)(a-bi) = a^2+b^2; so are the primes that aren't gaussian primes a sum of two squares?

[@NoNameAtAll2]

can you rephrase the question?

[@coppertones7093]

yes

[@Ardient_]

yep they always are! its quite cool imo

[@wyattstevens8574]

Exactly! The only ones that already are Gaussian primes are the ones 1 less than a multiple of 4.

[@yoyoyogames9527]

yup did a uni project about it, a presentation about how many ways natural numbers are the sum of two squares and used exactly this to turn it into a problem of counting Gaussian factorizations. btw when Gaussian primes are in the conversation the normal integers are usually called rational integers and so "non-Gaussian primes" are called rational primes.

[@johnchessant3012]

The norm map N : Z[i] -> Z given by N(a+bi) = (a+bi)(a-bi) = a^2+b^2 is multiplicative, so N(x) being prime in Z guarantees x is irreducible in Z[i]. The converse is not true since N(x) could be composite but none of its proper divisors are the norm of any element of Z[i]. Since the sum of two squares can never be 3 (mod 4), any 3 (mod 4) prime p is irreducible in Z[i].

[@graf_paper]

Great color scheme for this video. Chefs kiss!

[@FunkyThousandMiles]

This what i need after a day at the office, thank you!!!

[@user-xn9um6ne2q]

Can't wait for your next video!

[@santerisatama5409]

Word "unit" is a mistranslation of Euclid's term 'monas' (from which Leibnitz 'monadology' originates from). The word actually means 'unique', meaning the distinctive mutual property between coprimes and their lesser cousins the "solitary primes". The fundamental elements of Euclid's number theory are coprime fractions, from which also the lemma of "solitary" primes and their unique gcd multiplications are derived from. Thanks, this presentation was really helpful for me, giving also some basic intuition of what Gaussian rationals could mean, when we replace the mistranslation "unit" with each of the unique coprimes as the uniques, instead of limiting our perspective only to "natural numbers/integers" and filtering out mathematics with that sieve. I'd love to hear your perspective also on Gaussian rationals.

[@bozhidarmihaylov]

Excellent palette :) Well done

[@jennycotan7080]

The Gaussian prime plot looks... aesthetically amusing. Gonna see if someone make it a tile pattern...

[@GameJam230]

The pattern for which primes are gaussian primes is dependent on their remainder when dividing by 4, as gaussian primes need to be 3 (mod 4). Further, given Dirichlet's Theorem of Arithmetic Progressions, it would follow that the density of integer primes that ARE gaussian primes is equal to the density of integer primes that are NOT gaussian primes, and so about half of all primes should be gaussian. If you actually want to see a REALLY cool extension of Dirichlet's Theorem on the primes, see what happens when you have TWO arithmetic progressions, (ax+b+c) and (ax+b-c), and you want to know when BOTH are simultaneously primes. The amount of values x for a given a and c will be the same for any value of b you plug in, implying Dirichlet's Theorem extends to pairs of primes with a specified difference, 2c. This was a paper I started working on as a hobby project and got decently far with it, but I have no credentials so I can't get anyone qualified to look into it further than I know how to, and the paper only got as far as the last one or two steps of recreating the original proof (but modified for my case), where it can't move forwards without a proof of Polignac's Conjecture. But, if Polignac's Conjecture WERE shown true, this extension would say that not JUST are all even numbers the gap of two primes infinitely many times, but ALSO that they occur infinitely many times in each pair of my progressions for a given a, b, and c.

[@pandorairq]

The correct definition of a prime is the following: "A number is called prime if and only if it has exactly two divisors". Its the simples and easiest to understand definition and also explains why 1 is not a prime (it has only 1 divisor)

[@chemistrycuber]

This is a very entertaining video. I love the way it starts with the basics and you work your way up to the Gaussian primes from there. I hope that you reveal the pattern of which radii are Gaussian primes and which are the square roots of real primes in the next video because I doubt I will be able to reach the answer myself - but I will certainly use your website to think about it more! And the Argand diagram on the grid produces some oddly hypnotising symmetry lol. Is this the kind of stuff that you study in a maths degree at university? Or is that maths more applied?

[@TheGrayCuber]

Depends on what type of maths program, there are degrees for both pure and applied maths. This could be covered in a number theory or abstract algebra course - both pure maths. Though an applied student could take those if they'd like

[@NobodyYouKnow01]

TIL I learned that 1's role as the multiplicative identity prohibits its inclusion as prime or composite. Similarly to how 0 can be neither positive not negative because of its role as the additive identity.

[@simonstrandgaard5503]

beautiful

[@alipourzand6499]

Great video! Hope that sometimes you will make another about Eucliden algorithm for complex gcd ☺

[@ld7496]

awesome vid, sad its new and your next one isnt out yet :D

[@TheGrayCuber]

It'll be out within a week!

[@pedropiata648]

Hope u get to 5k subs soon ☺️

[@guigazalu]

To define complex integers primes (ie. primes on the gaussian integers), I usually start by defining a total order over the universe of work: 1. Order by the radius 2. Order by the angle: 1, i, -1, -i. Then, after the sieve, I end up with (i) as a prime, then (1 + i), then 2 isn't and so on. Ex.: 2 = (i)³(1 + i)²

[@ankitjain5701]

it is very obvious that pattern is mirrored about both axis; the challenge is to find pattern in a single quadrant.

[@dlfon99]

I know that defining i as the square root of -1 is a common shorthand, but I think it's more accurate to say that i²=-1, since this cam be extended to quayernions and further hypercomplex number systems.

[@daboffey]

What would be a nice addition to the web tool would be a control that allows one to zoom in or out.

[@TheGrayCuber]

I just added a setting here for a zoomed out view: thegraycuber.github.io/quadratic.html

[@daboffey]

@@TheGrayCuber Looks nice.

[@egohicsum]

great video ❤️

[@thomascrouson6085]

oh yay, I've actually been very interested in primes lately, so seeing this video was an absolute treat :) Do you have any idea if primes are a thing in quaternions? I've been meaning to look into that, but I've been putting it off cause quaternions are so hard to wrap my head around lol.

[@TheGrayCuber]

There are kind of prime quaternions, but they dont work exactly the same without commutativity

[@m9l0m6nmelkior7]

How can you be sure that this algorithme works, and that no two composite numbers factor out to a prime ? Well I guess the norm is a good proof : the norm of the product is the product of the norms, so even in gaussian integers, you can proove (a version of) the fundamental theorem of arythmetics !

[@rodomify7681]

I like to define prime as whole numbers that have 2 unique factors. While, the original definition works, this way highlights the important difference with 1 being that it only has 1 factor. For the case of the Gaussian primes you could say the numbers that have 8 unique factors, since that accounts for all units and associates, with units being excluded since they have 4 factors. My guess is that for any number system, Primes are numbers with twice as many factors as there are units.

[@sigmaoctantis5083]

Your conjecture is (nearly) correct: in any integral domain a unit is divisible exactly by all units, while any nonunit x != 0 is divisible by all units as well as by all its associates, the products of x and some unit. Since these products are pairwise distinct (as x is cancellable), there are as many associates of x as units, and hence any such x has at least twice as many divisors as a unit has (of course for a ring like Z[√2] with infinitely many units this "twice as many" does not make much sense). The elements attaining the minimum are exactly the _irreducible_ elements. If the ring has unique factorization, these are the primes, but in general irreducibles are not necessarily prime: the classical example is the ring Z[√-5] with its irreducible elements 2, 3, 1+√-5 and 1-√-5, which are not prime as 2*3=(1+√-5)*(1-√-5), but no factor divides one from the other side of the equation.

[@tcaDNAp]

The graphics look beautiful, and it makes me want to play a dungeon crawler exploring a maze of Gaussian primes Eisenstein primes sound so cool too... Maybe gamers can do some research on the Gaussian moat problem! 🤓

[@MagruderSpoots]

6:55 Did you forget some brackets there?

[@ingiford175]

Yes he did, but other then that it was a nice solid video

[@TheBookDoctor]

Woah. I don't think I've ever randomly stumbled across a KZbin channel from someone I've previously met in real life (at a few PNW comps). I've met IRL a couple of KZbinrs I've already followed, but not the other way around...

[@TheGrayCuber]

No way, welcome to the channel! what's your name?

[@Dent42]

6:30 It’s hard to say, since we can’t see the full grid, but it looks like it’s also diagonally symmetrical, no?

[@petterlarsson7257]

10:15 TheGrayCuber GONE WOKE!? TheGrayCuber USED *THEY/THEM PRONOUNS* FOR NUMBERS!!

[@kales901]

6:57 i think it is importaint to put the complex numbers in parrenthasis, so that nobody can think that a+bi = a-bi

[@davidwright8432]

The way of excusing 1 as a prime is to notice that by definition , a prime is UNIQUELY defined a divisible by itself and 1. But 1 = 1*1 =(1*1)*1 = ... and so on. In other words, 1 is not uniquely factorized by 1 and itself. So fails the uniqueness test, so can't be prime.

[@victorespeto]

Great video! But please, please, use brackets at 7:00

[@JwalinBhatt]

Can the primes be extended to the complex numbers using some other approach? Like find func a function f(n) which maps the integers to the primes and then extended it somehow to take in complex values? We already have the Riemann's prime counting function which gives the number of primes less than a given number but it doesnt gives the nth prime. Maybe that function can be modified somehow.

[@talastra]

Gaussian primes are of the form 4k-1, non-Gaussian of the form 4k+1. Edit: ah, the other comment ...

[@sschmachtel8963]

Aren't Gaussian primes then somehow primes that can not be written as a sum of two square numbers in the form a^2+b^2 = (a+bi)*(a-bi) or how is the connection to pytagorean numbers?

[@pondermatic]

They don’t half remind me of 2D cellular automata. Probably no coincidence. I reckon the prime numbers are actually so-called Gardens of Eden in the cellular automaton of integer multiplication. In which case, to study the primes, it might be helpful to generalise in this way.

[@CM-mo7mv]

the generated voice is nice

[@zoinkerspoinker4326]

i don't reckon it's ml, i checked some other videos and they sound pretty similar

[@TheGrayCuber]

It's not ml, but the voice was generated by my mom over the course of 9 months

[@publicemily1357]

Can confirm

[@calculuslover2078]

What if 1 is a composite number? Seriously.

[@1.4142]

The views doubled in the past hour

[@Fine_Mouche]

ah, (+-50, +-25) is the limit of the complex plane, i would have see how is the pattern farer ("more far") :/

[@justinlloyd3]

If one is not prime then you would not be able to construct one as a product of primes thus defeating the definition of primes. It has to be a special more basic level of a prime. Its not that one isnt a prime. Its that it is a super prime.

[@PunmasterSTP]

It's weird to think about generalizing the real integers and then that makes 2 and 5 composite...

[@user-ht1ed3ws2l]

then 1 is the only number that hasn't enough dividers to be prime?

[@rodrigoqteixeira]

Ofc bro would go from 11bld to complex primes. Ofc.

[@kinoseidon]

My only comment is that multiplication/division by 1+i and its associates should be allowed in the game. Maybe there could be some levels that make use of that?

[@mendelovitch]

Do you have the primes spiral mentioned in Rendezvous With Rama?

[@bendono]

"Every number has a unique factorization into primes." That makes one wonder about the the unique factorization into primes for 0 and 1...

[@figboot]

"Every number has a unique factorization into primes" is a non-rigorous way to state the theorem. A more rigorous way to state the proposition of the theorem is that every *integer greater than 1* has a unique factorization into primes.

[@mathgeniuszach]

If it got you wondering, can you create a system in which 0 and 1 have unique factorizations? Major math has often been invented by breaking the rules a little, like allowing for the imaginary numbers. I can't think of one, but maybe you can

[@TheGrayCuber]

The unique factorization for 1 is just the 'empty' factorization, no primes. 0 could not have a unique factorization. If it did, some a*b*c*...*z, then a*b*..*z*2 would also be a factorization of 0 since 0*2=0. I just didn't want to spend too much time within the video explaining edge cases

[@hhhhhh0175]

given the convention that the p-adic valuation of 0 is infinity, 0 = 2^infinity*3^infinity*5^infinity...

[@terdragontra8900]

@@figboot 1 does have a unique factorization

[@DrSegatron]

8:28 Motivation?

[@jannis11]

NaIce

[@user-ro2tm3dp8x]

how did that work it marked 2 and 3 as non primes while they obviously are

[@gustamanpratama3239]

Are there primes too in quaternions, octonions ... and so on, in all the other hypercomplex numbers?

[@TheGrayCuber]

Kind of. There are sets called 'primes' but they don't work quite the same since there is no commutativity

[@nahum8240]

what if you put them play the conway's game of life?

[@ytkerfuffles6429]

it's not obvious to me that you can't have a clash, for example 5 and 3+4i both have radius 5, so which do you choose? edit: or can you prove clashes are not possible edit 2: i think im dumb if you have a clash you can just use both

[@TheGrayCuber]

I'm not sure what exactly you mean by a clash, but those two values do have different factorizations. 5 = (2+i)(2-i) while 3+4i=(2+i)^2

[@ytkerfuffles6429]

@@TheGrayCuber a clash is when a circle of a certain radius has more than one possible gaussian integer so you don't konw which one to choose edit: i know my example isn't prime, but i mean if there was an example like it where they have the same radius and ARE prime, which do you choose when sieving? i guess you would just use both

[@drdca8263]

Neither will be a product of the other multiplied by any Gaussian integer with absolute value greater than 1, so, the order cannot matter, as removing multiples of one and then removing multiples of the other, and removing multiples of the other and then removing multiples of the first one, leave the same set remaining. [strikethrough]However, I think this is kind of moot, as this situation cannot occur[/strikethrough] [I was wrong, it does occur, but only in one type of case]: If they have the same absolute value, then when each is multiplied by its complex conjugate, the result will be the square of the absolute value of the number, and this is the same for the two numbers, as they are on the same circle. So, this produces two factorizations of the same number, which isn’t possible because the Gaussian integers are a Unique Factorization Domain? Ah, but wait, there’s a hole in that argument. What if they are complex conjugates of each-other and not associates of each-other (or, if they are complex conjugates of an associate of the other)? Then, the factorization would just differ in the order (or the order and units). Like, if you have a+b i, a - b i is not in general an associate of it. (a+bi=(-1)(a-bi) iff a=0 , a+bi=(i)(a-bi) iff a=b a+bi=(-i)(a-bi) iff b=-a a+bi=(1)(a-bi) iff b=0 ) But other than that..

[@TheGrayCuber]

You would use both. 3 + 2i and 3 - 2i are both primes, both have radius sqrt13, but they are not associates. When doing the sieve, I removed the multiples of both at the same time.

[@giuseppepapari7419]

Question: when you remove the multiples of a complex integer n, you remove all numbers that can be expressed in the form of k * n, where k is a complex integer. Am I correct?

[@TheGrayCuber]

Yes that is correct!

[@Bolpat]

The “only 1 and itself” is simply wrong.

[@Peterotica]

which is better, gaussian prime, eisenstein prime, or amazon prime?

[@TheGrayCuber]

Eisenstein > Gaussian >>>>>>>> Amazon. I avoid supporting Amazon - haven't purchased anything from them or their subsidiaries for years

[@_mb_2617]

isnt the figure at 7:04 missing some parentheses?

[@plugplagiate1564]

this proof is pritty odd. i can not say more.

[@eeetube1234]

Why 2+0*i is not prime? And why 0+2*i is not prime?

[@kales901]

how is 2 a prime if 2 is (1+i)(1-i)??

[@Haxihoovis]

Wait, why is 2 not prime here?

[@horacenyc492]

I've never understood the use of the + sign. Is it just a convention? You're not actually adding numbers. Complex numbers are single numbers located in two-dimensional space (whereas the Reals are single numbers located on a one-dimensional number line). Why is it 2 + 4i and not something like (2, 4i)?

[@TheGrayCuber]

It is just a convention. But 2 + 4i does equal the sun of 2 and 4i, so it is a sensible convention. We do the same thing with polynomials

[@0011peace]

not true 2 is special as well as i is the only even prime, Ii beyond is useless as 4i is 1

[@cvabds]

I bet you can't do that in TempleOS haha, I subscribe but please try doing something like that in that coding language

[@Kyoz]

🤍

[@VandalIO]

Is this true ? One is not prime?

[@user-fj9hf4bu9f]

Is there a github for the code?

[@TheGrayCuber]

github.com/thegraycuber/thegraycuber.github.io

[@user-tx4ic1ke6n]

what is oyur 3x3 pb

[@TheGrayCuber]

kzbin.info/www/bejne/qYqxnnV3d96LfKssi=_uH5ptfupAPDTTNS

[@Bolpat]

7:04 PEMDAS?

[@guigazalu]

What is PEMDAS?

[@TheGrayCuber]

Please Evaluate Mathematics Directed by Appropriate Sequence

[@pelasgeuspelasgeus4634]

Can you name a modern device that works thanks to primes???

[@sensorer]

Every single device that uses the internet

[@pelasgeuspelasgeus4634]

@@sensorer that's nonsense. Explain how.

[@TheGrayCuber]

Public-key cryptography is a great example

[@pelasgeuspelasgeus4634]

@@TheGrayCuber Maybe so but is that enough? I mean you presented primes as the corner stone of civilization. Really now?

[@sensorer]

@@pelasgeuspelasgeus4634 your banking apps would not work without that. I mean, any digital banking wouldn't work. ALL OF THE INTERNET wouldn't work. It's all built on the fact that it's easy to multiply primes, but its way harder to factor a big number into primes

[@user-lu9fg7pc9q]

0:31 It's not

[@walterbrownstone8017]

Nope you failed to explain the importance of primes. I don't really understand the relationship between multiplication and addition. Does multiplication emerge from addition and time? Does division emerge from subtraction and time? Is there multiplication without time?