A Groundbreaking Mathematical Result

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Thanks Leibniz, very cool! Without complex numbers or the notion of negative arguments in roots, Leibniz shook the mathematical community back in the days by denesting radicals! We are going to explore his result for sqrt(1+sqrt(-3))+sqrt(1-sqrt(-3)) giving us sqrt(6). As a corrolary using the conjugate and difference of two squares, we also find out the value for sqrt(1+sqrt(-3))-sqrt(1-sqrt(-3)) in the process. Enjoy! =)
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[@PapaFlammy69]

*_Thanks for watching, I hope the video was to your liking

[@anissaada8660]

Can you do a video of math of second secondary

[@oni8337]

yes

[@guest_informant]

More generally, is this right SQRT(a+(SQRTb)i) + SQRT(a-(SQRTb)i) = SQRT(a^2+2a+b) In the particular case in the video a = 1 b= 3 so this formulation gives SQRT(1+2+3) = SQRT(6)

[@Manoj_b]

I think I'll be a interesting vedio for fluid dynamics of collision of vortex Paradox........!.... But I have been watching the vedios from end of 2019 and I never got bored , Because I love mathematics and physics.😁

[@Manoj_b]

@@guest_informant even 1×2×3=1+2+3. But may be ......!🤔

[@tiagolevicardoso7674]

Hill: question mark is a valid variable. Proof: Let ? be a variable Q.E.D

[@RedVio972]

Permission denied

[@luggepytt]

I would like to address the question that Jens was kind of begging the audience for an answer: When is it valid to say that √(z1) × √(z2) = √(z1 × z2) if z1 and z2 are complex numbers? First, we need to clarify what we really mean when we write √(z). We want the expression √(z) to have a single, well-defined value, but the equation w² = z has two solutions (unless z = 0). So which one shall we choose? The common practice is to choose the one with a positive real part. Or, when the real part is zero, the one with a positive imaginary part. This is called the principal square root of z. (And the definition works when z is real too.) Some examples, following this definition: √(4) = 2 (and not -2) √(2i) = 1+i (and not -1-i) √(-2i) = 1-i (and not -1+i) √(-4) = 2i (and not -2i) An alternative, but equivalent, definition of the principal square root is this: If you write z as r × e^(i𝜋ϑ), where r > 0 and -𝜋 < ϑ ≤ 𝜋 then √(z) = √(r) × e^(i𝜋ϑ/2) If we follow this definition, then it is easy to see that the equality √(z1) × √(z2) = √(z1 × z2) holds under the following circumstances: Suppose z1 = r1 × e^(i𝜋ϑ1) and z2 = r2 × e^(i𝜋ϑ2) Then everything is fine as long as -𝜋 < (ϑ1 + ϑ2) ≤ 𝜋 In the special case with two complex conjugates with non-zero imaginary parts, as we had in this video, ϑ1 + ϑ2 = 0, so the equality holds. So Jens' calculations were flawless. Hooray!

[@wisdomokoro8898]

You're just Amazing 😊😍

[@Freakschwimmer]

Damn :D I can understand what you are saying but I'm totally mesmerised you were able to put it down so nicely :)

[@grapix1184]

Thx!

[@luggepytt]

Well, this is embarrassing… I just happened to show this post to one of the frogs in my garden pond, and she immediately spotted a horrendous blunder that I had made. In several places, I have written things like “e^(i𝜋ϑ)” when I really meant “e^(iϑ)”, without the “𝜋”. I guess I was so pleased with the pretty 𝜋 symbol that I had found under “Insert special character” that my subconscious mind wanted to use it as much as possible. But if you disregard all those extraneous 𝜋s, the argument is still valid, my frog friend reassured me. So thanks, Blanche, for pointing this out! What would I do without you and all your other amphibian friends?

[@anshumanagrawal346]

@@luggepytt lol

[@YourPhysicsSimulator]

"Mathematicians are serious, boring and quiet people" What mathematicians trully are doing: 0:09

[@nikhilnagaria2672]

More like 0:30

[@YourPhysicsSimulator]

@@nikhilnagaria2672 hahaha that one too

[@herbie_the_hillbillie_goat]

Was that a sneeze, a cough, or wheeze?

[@stevenxw]

we've gone from "good morning fellow mathematicians, welcome back to another video" to "GOOODIUQWEGHFawuihkefhuio"

[@nikhilnagaria2672]

@@stevenxw the video about the sines (third most recent): check it out ;)

[@redhotdogs3193]

you can also do this by writing 1+sqrt(-3) as 2exp(pi*i/3) and 1-sqrt(-3) as 2exp(-pi*i/3) this then simplifies into sqrt(2)(exp(pi*i/6)+exp(-pi*i/6)) which equals sqrt(2)(2cos(pi/6)) which is sqrt(6)

[@oni8337]

genus

[@aneeshsrinivas892]

I did the same

[@angelmendez-rivera351]

This assumes that you already have complex analysis at your disposal, which Euler obviously did not have when he worked with this problem.

[@anshumanagrawal346]

@@angelmendez-rivera351 exactly my point

[@renanrodz7221]

Someone: "Do you have plans for this weekend?" Me: "It's complex..."

[@_mishi]

complex*.. Its complex

[@Vaaaaadim]

Sneezed for his intro, what a time to be alive

[@pardeepgarg2640]

Alt title : A Math breaking Ground Result

[@parasb21]

You know what i guess the catch here is changing that question mark to x . I solved i similar question 2 years back was stuck for like two days until i realised "X" is the saviour !!

[@sttlok]

You can still use the question mark as a variable lol

[@parasb21]

@@sttlok Haha true but that wont be an intuitive thought to use ? as a variable lol

[@sttlok]

@@parasb21 lol, integrate d^2 in respect to d.

[@hardikjoshi8557]

Loved your DJ flammy version in the last video.. Today I adjusted my neck at the angle of 45 degrees so to see my gf showing her nested radicals. Thanks sir you did a great help solving the problem. Now I can enjoy my remaining day and will share this story with Leibnitz😂 ❤️ 🇮🇳💟🇩🇪 Your boi always:)

[@ricardoparada5375]

This reminds me of a similar process when u use the rational roots theorem to find out if a number is rational or not. It’s really nice

[@Brien831]

these were roots of the polynomial I was given in my abstract algebra class. We were asked to determine the galois group and all the indermediate field extensions. In general are polynomials of degree 4 with these kinds of roots very interesting! Everyone likes the D4 group after all .

[@PapaFlammy69]

yas! :)

[@realkabirc303]

This Problem was really easy, I didn't go into the algebra, I just converted in the Polar form and then simplified and got root 6.

[@PapaFlammy69]

nice :)

[@alberteinstein3612]

Hello fellow math lover :D

[@realkabirc303]

@@alberteinstein3612 Hi👋👋

[@MrRyanroberson1]

oh man you got so close to something even cooler. now that you have x and y, you can take x+y and x-y to find algebraic simplifications for sqrt(1±sqrt(-3)) to be 2(sqrt(6)±sqrt(-2))

[@PapaFlammy69]

oh, nice!

[@alberteinstein3612]

Papa Flammy always cracks the perfect amount of adult jokes and it’s absolutely hilarious

[@brandonklein1]

I mean if we want to be very careful, we can take the square roots as complex numbers and assure that the complex parts cancel. I would guess that if we did this and went through, we would find 2 more solutions due to the +- on √z for complex z.

[@angelmendez-rivera351]

Not necessarily. If the square root function on the complex numbers is defined using the atan2 function and the exponential function, then the result is unique, and it will be the case that sqrt(z*) = sqrt(z)*.

[@brandonklein1]

@@angelmendez-rivera351 that's fair.

[@CauchyIntegralFormula]

You end up with either x^2 = 6 (as seen in the video) or x^2 = -2

[@brandonklein1]

@@CauchyIntegralFormula Then indeed there are 4 solutions.

[@XAE-yc9rr]

> 5:00 AM >slides into view like a boss >"Gyee...*coughs*! " This is already a great start of the day.

[@vigneswarans1636]

Hey flammy, theres also an approach via a std method called square root of complex no where u assume each term as x+ or -iy , simply sqare, find x and y ,then basic addition.

[@michapiechota4260]

Playing around with the problem by myself Ive got some messy answear with roots, and its approximately 2.88523 which is cleary not root of 6 Im pretty sure that the logic behind solution is correct. Is there more than one solution to this problem? Im quite new to imaginary numbers so sorry if thats a dumb question

[@Manoj_b]

It's amazing that it's solved by Leibniz before 300 years before Euler ..🎉. He must be happy I think.

[@PapaFlammy69]

:)

[@AliAhmed-zb4nj]

Please I want solution to this ODE: Y'=sqrt(x²+y²) Can you help me

[@akselai]

>algeBRUH Very groundBRUHking indeed, Papa.

[@thebeerwaisnetwork8024]

Very nice

[@chessematics]

Me when I saw the thumbnail: YOU WANT TO FOOL ME THE COUSIN OF CARDANO?

[@saadmoquim8624]

ive been on this channel long enough to know that an alternate cover would be the euler fire emoji

[@PapaFlammy69]

ayyyy :p

[@zildijannorbs5889]

fuck this im doing a compilation of all the good morning fellow mathematicians

[@tambuwalmathsclass]

I didn't try this, but I believe if I were to simplify it, I'll end up with a complex Solution

[@tambuwalmathsclass]

Incredible I got √6 after u, v and y substitution. Permission to record the video

[@md.iftakhar7478]

Hi. Here is a problem from my exam. I was running out of time and couldn't come up with a solution. The problem is, Integration of (arctan(tanx/a))dx. Even now I am struggling with this problem. It will be great If you make a video on how to solve this problem.

[@mkjaiswal11]

Since I am in Class 9 and I have radicals in course, there would be no doubt if this question come in my exams.

[@nikhilnagaria2672]

Your name says that you are in class 7 (?), and also this won't come in your exams until you're giving that je- jee.

[@neilgerace355]

0:05 Well this puts a different slant on things.

[@Mystery_Biscuits]

Soooo, is Flammy's Wood rebranded Trivial? If so, that's great and all but what happened to the trivial videos (namely, the outtakes of your more complicated videos) and if not, what happened to Trivial?

[@PapaFlammy69]

they are still there, just unlisted :)

[@Mystery_Biscuits]

@@PapaFlammy69 oh sure sure. Nice 👍

[@johnny_eth]

The numbers inside the radicals are complex, so you need to work with both roots. If you work with the polar form, you reach two possible values ±sqrt(6) and ±sqrt(2).i. By squaring everything in the start, you are basically merging both roots into its squared value and hence get one answer.

[@mr.unknowngamer2109]

😮wow

[@stvp68]

Like how at some angles, the shirt looks like it says lgb 😁😁

[@Mike-fr9hp]

if x equals to sqrt(1+sqrt(-3))+ sqrt(1+sqrt(-3)) but at the same time x equals to sqrt(6), sqrt(1+sqrt(-3))+ sqrt(1+sqrt(-3))=sqrt(6)

[@elysiummaybe8574]

Mathematicians stereotype: Nerds Mathematicians: 1:47

[@tszhanglau5747]

Recording videos at 5am do be a bruh moment

[@PapaFlammy69]

ayyyy lmao pog sus

[@utkarshsharma9563]

Amogus moment

[@MathAdam]

1:08 "Rested Nadicals" -- This is what happens when you record your videos at 5 am. :D

[@PapaFlammy69]

:D

[@goodplacetostop2973]

12:21

[@bennetthudson9537]

this timestamp is trivial

[@goodplacetostop2973]

@@bennetthudson9537 This timestamp is trivial and is left as an exercise for the reader.

[@herbie_the_hillbillie_goat]

This timestamp is beyond the scope of this lecture and can safely be ignored.

[@anshumanagrawal346]

That's a Good Place To Stop

[@abhigyan7987]

Nice information ℹ️ℹ️ . Watching your videos from india ❤️ sir

[@laurentreouven]

-i sqrt(2) as i sqrt(2) works for y, how to choose wich us the answer ? Not sure its possible

[@johnny_eth]

You use both.

[@laurentreouven]

@@johnny_eth a number that have two value is not a number

[@nikhilnagaria2672]

@@laurentreouven both are equivalent

[@laurentreouven]

@@nikhilnagaria2672 how do you know ?

[@nikhilnagaria2672]

@@laurentreouvenok I ask you a question: what is i?

[@adityaekbote8498]

Nice

[@gragasapmidlane6761]

"dont ask me why im doing dis at 5am" also him 30minutes before "Raids with americans"

[@Lolwutdesu9000]

I bet a lot of people will do a lot of "woodworking" once they see your OF.

[@veralgupta8182]

Papa flammy at 5am so fresh so cool Me when I wake up at 8am : 😑😴😑

[@PapaFlammy69]

:D

[@CauchyIntegralFormula]

Fundamentally, square root is multi-valued except at 0. For positive real x there's a canonical choice (called the "principal" square root) but across the entire complex plane I don't think there's a compelling reason to pick one over the other (and doing so leads to annoying continuity problems) Edit: The square root rule can be more generally handled as sqrt(ab) = +-sqrt(a)sqrt(b), with the choice of plus or minus determined by how you pared sqrt down to be single-valued

[@calambuhayjr.josevirgiliog2094]

is there a word raw written in your long sleeves? p.s.: do you have a compiled documents where I can study your trivial math problems? You helped me broaden my perspective in number theory... my idol forever.

[@WriteRightMathNation]

What makes it "groundbreaking"?

[@siranguru]

Lets take x = sqrt(3/2) and y= sqrt(-1/2) The first term becomes sqrt((x+y)^2) The second term becomes sqrt((x-y)^2) And you get both 2*sqrt(x) and 2*sqrt(y) as the solution

[@siranguru]

x and y are symmetry i.e. exchangeable

[@anonymanonym6641]

And I thought I had to rewrite 1 as 2²- sqrt(-3)²...

[@angelmendez-rivera351]

The complex square root function can be carefully defined in terms of the real square root function, and this is analogously true for any complex nth root function as well. In this case, we have that sq : R+ -> R+, sq(x) = x·x, where R+ denotes the nonnegative real numbers. It can be proven that sq is invertible, and this inverse we call root, so root : R+ -> R+, root(x)·root(x) = root(x·x) = x for every nonnegative real x. To define the complex square root function, first, you need to know about the complex absolute value. It can actually be proven that for every real number z, if z* denotes the complex conjugate of z, Im(z·z*) = 0, which is to say that we can use root[Re(z·z*)] as the norm of z, and it can be proven that if Im(z) = 0, then root[Re(z·z*)] = |z|, so this makes for the canonical extension of the absolute value to the complex numbers. Now, for complex numbers z, you can define arg(z) := atan2[Im(z), Re(z)], and define the exp function via its Maclaurin series. From here, the most natural definition of the complex square root function is given by sqrt(z) := exp[arg(z)·i/2]·root(|z|). Why is this definition the most natural? Because for nonzero z, this is also equal to exp({ln(|z|) + atan2[Im(z), Re(z)]·i}/2) = exp[log(z)/2], where z would be equal to exp[log(z)]. Given sqrt(z) := exp[arg(z)/2]·root(|z|), it is now trivial to show that sqrt(x·y) = exp[arg(x·y)/2]·root(|x·y|) = exp[arg(x·y)/2]·root(|x|)·root(|y|), while sqrt(x)·sqrt(y) = exp([arg(x) + arg(y)]/2)·root(|x|)·root(|y|). Dividing the second equation by the first gives us that sqrt(x)·sqrt(y) = exp([arg(x) + arg(y) - arg(x·y)]/2)·sqrt(x·y). With this, if x and y are positive real numbers, then exp([arg(x) + arg(y) - arg(x·y)]/2) = 1, but if x is positive and y is negative, then exp([arg(x) + arg(y) - arg(x·y)]/2) = -1, so sqrt(-1)·sqrt(-1) = -sqrt[(-1)·(-1)]. This resolves the contradiction. In your problem, y = x*, so arg(y) = -arg(x), and arg(x·y) = 0, hence exp([arg(x) + arg(y) - arg(x·y)]/2) = 1. Therefore, you can say, safely, that sqrt[1 + sqrt(-3)]·sqrt[1 - sqrt(-3)] = sqrt([1 + sqrt(-3)]·[1 - sqrt(-3)]). In general, this can be done with the complex nth root function. What you do is say rt(n, z) = exp[arg(z)/n]·root(n, |z|), where root is the function with nonnegative real numbers, and rt is its extension. With this, you have that rt(n, x)·rt(n, y) = exp([arg(x) + arg(y) - arg(x·y)]/n)·rt(n, x·y). In fact, this can be made more useful and concise by having h(x, y) := arg(x) + arg(y) - arg(x·y). So rt(n, x)·rt(n, y) = exp[h(x, y)/n]·rt(n, x). The h(x, y) is the important part is this entire formulation: it is what determines what factor you multiply by, for any given index n, when you do the distribution.

[@VaradMahashabde]

ohk, so negative times negative gets the negative out front treatment. nice. btw, a visual representation of the rule can be the graph for |x+y|

[@angelmendez-rivera351]

@@VaradMahashabde Yes.

[@Abdul-Akeem_Akinloye]

My fucking head.

[@neilgerace355]

6:04 if you trust the Peano axioms.

[@PapaFlammy69]

:^D

[@sheriff-larue7770]

Who here recognizes when the Redbull or coffee- kicks in? Min 2.48

[@mikejackson19828]

Rested nadicals. 🤣

[@the_nuwarrior]

Cardano ?

[@frozenmoon998]

Flammy's Wood. Flammy's OnlyFans incoming next.

[@pavolgalik499]

The formula Sqrt (a). Sqrt (b) = Sqrt (a.b) holds and is therefore proved only for a> 0, b> 0, (and also a = 0, b = 0). If you want to use Sqrt (Negative number), you must immediately switch to imaginary numbers. Otherwise, you can prove anything, including 1 = -1.

[@nikhilnagaria2672]

It still holds whenever at least one of them is nonnegative

[@lukeerikblue958]

Waiting on that only fans papa

[@arielfuxman8868]

Papa Flammy, how tall are you?

[@tetris450]

Rested nadicals

[@JR13751]

You said x should be positive because it is sum of 2 square roots. But they are square roots of complex numbers, so I don't think they need to be positive.

[@Gameboygenius]

Oh, our boi Leibniz! I recently listened to the last episode of 3blue1brown's podcast with Steven Strogatz. They discussed, let's call it, the mythology of different characters in mathematical history, and Steven though Leibniz was probably the most sympathetic of the bunch living at the time, especially compared to Newton. Overall a really interesting episode.

[@orisphera]

4:25 There are two square roots of each nonzero number. Otherwise, -1=sqrt((-1)²)=sqrt(1)=1

[@angelmendez-rivera351]

No, there are not. The symbol sqrt denotes a function, and a function can have one output. As for your argument that -1 = sqrt[(-1)^2] = 1, you are wrong, because sqrt(x^2) is not equal to x. sqrt(x^2) = |x|. You are thinking of sqrt(x)^2 instead. sqrt(x^2) and sqrt(x)^2 are not equal.

[@orisphera]

@@angelmendez-rivera351 I didn't write there are two outputs of sqrt. I only wrote there are two roots. Read about square root on Wikipedia. It follows from what is written there that square root and sqrt are not the same. By the second part, I meant that if there is only one square root, x is the square root of x² because it is a square root of x² and sqrt(x²) is the square root, too, so they are the same Btw, you can define a type of nonzero number where there are countably many versions of each normal nonzero number and for any such number there is exactly one such number that gives it when squared

[@angelmendez-rivera351]

@@orisphera *I didn't write there are two outputs of sqrt.* You did not write this, but this is an implication of your statement, since you say that -1 = sqrt(1) is true. In fact, it is not. *I only wrote there are two roots.* This is the same thing as saying sqrt(1) has two outputs. *Read about square root on Wikipedia.* I have done, and I have also read many other sources on the subject, and I studied some mathematics in college. I think I know what I am talking about. Wikipedia is not a scholarly source, even though it can serve as a good introductory source for the subject. *It follows from what is written there that square root and sqrt are not the same.* What is written there is also not exactly correct, as the article is an oversimplification of the subject meant for laypeople and beginners, and is not intended to be taken as a source for rigorous mathematics. *By the second part, I meant that if there is only one square root, x is the square root of x^2, because it is a square root of x^2, and sqrt(x^2) is the square root, too, so they are the same.* This is super incorrect. x is not the square root of x^2, and it is not "a square root" of x^2 to begin with, because there is no such a thing as "a square root". You are conflating concepts here. Talking about the square root of 4 is different than talking about the multiset of roots of X^2 - 4. If X^2 - x^2 is considered as a formal polynomial, then the multiset of roots is indeed given by {-x, x}, but what the square root of x^2 is depends on whether x < 0 or x > 0, since the definition of the square root implies that sqrt(x^2) > 0 for nonzero x. This is because the square root of a number is not defined as the multiset of roots of "the associated" quadratic polynomial, but is defined as the output of the square root function, and the radical symbol, which is also substituted for sqrt in text, denotes this square root function being applied to the input. The square root function is defined using the strict lexicographic order < of the algebraic numbers cl(Q), and noticing that sq : {z in cl(Q) : 0 < z} -> cl(Q)\{0} with sq(z) = z·z is a bijection. The inverse of sq, having 0 included in its domain, is the square root function.

[@orisphera]

@@angelmendez-rivera351 I don't understand why you use z for real numbers without having y, but you've convinced me that Wikipedia and serious mathematics have different terminology and in serious mathematics, each number has only one square root and there is always only one root of unity, which is 1 PS I didn't state that -1=sqrt(1) is true. I only stated it's true if there is only one root

[@orisphera]

@@angelmendez-rivera351 “what the square root of x^2 is depends on whether x < 0 or x > 0” - what is fixed? If abs(x) is fixed, it doesn't - the square root is abs(x) in both cases

[@arkonarcaxe1808]

hi papy

[@arkonarcaxe1808]

first

[@jorex6816]

Comment for the algorithm

[@rssl5500]

The answer to your question is sqrt 6

[@juijani4445]

fUgACiTy

[@VaradMahashabde]

Well since the square root itself would not be properly defined, we could a get phase {0, pi} in either, so the total phase could be {0, pi, 2pi} for the product, so a plus minus still remains right? We could define the complex square root as saying that get the part which has a positive real part (corresponding with going from arg(z) to arg(z) / 2), then multiplication works out only when the phase of both the insides added doesn't cross pi or -pi. Othertimes you would have to multiply a negative out front. Since they are conjugates, the phase adds to 0 here, so things are ok for now. BTW assuming arg(z) is onto the set (-pi, pi) EDIT : talking about component values, multiplication always works out when 1. both real components are positive, 2. if both the imaginary components are of a different sign. 3. needs a negative out front if both real component values are negative and the imaginary parts are also of the same sign (so they are in the same quadrant) 3. if the real components are of different signs, the imaginary components are the same sign, and the product of the two keeps the sign of the imaginary components the same. So if (+,+) * (-,+) = (-,+) or (+,-) * (-,-) = (-,-). Otherwise we would have a negative sign out front, for the definition of the square root listed above. Edit : Negative reals are very problematic so kept them out of this.

[@angelmendez-rivera351]

The arg function as you define it is onto (-π, +π], not (-π, +π).

[@VaradMahashabde]

@@angelmendez-rivera351 Well I decided to keep the negative numbers out of this because they don't really work well with either the square root definition or the rule set. I was thinking about it and any inclusion broke things.

[@angelmendez-rivera351]

@@VaradMahashabde But they do work well with the definition. I would recommend you check out my post on the main comments section.

[@AVLZOfficial]

oh interesting video... cheers!!!! ............. #avlzofficial

[@einsteingonzalez4336]

Professor Fehlau, if this result is groundbreaking, I challenge you to another question. Given this series (this is in LaTeX): \sum_{n=0}^{k}(-1)^{\lfloor\frac{n}{2} floor}\binom{k-\lfloor\frac{n+1}{2} floor}{\lfloor\frac{n}{2} floor}x^{k-n} The series converges to 0 when x = 2cos(2π/(2k+1)). Prove it.

[@einsteingonzalez4336]

Can you prove it?

[@nikhilnagaria2672]

@@einsteingonzalez4336 what's the point in this even

[@FreeGroup22]

taking the square root of complex numbers ? you have 2 roots

[@white-ww8lf]

Helli

[@sampson4844]

omg，i am the exact 1000th like

[@giuseppemalaguti435]

Rad2rad(rad10+1)

[@herbie_the_hillbillie_goat]

Ummm Rafael Bombelli predates Leibnitz by about a century.

[@PapaFlammy69]

who? lol

[@giuseppemalaguti435]

Forse....

[@madice5071]

alg comment

[@abidhossain8074]

Early gang

[@mudkip_btw]

Cant find the onlyfans link smh

[@PapaFlammy69]

smdh

[@SuperTommox]

Papa's onlyfans just dropped?

[@MeMyselfForWho]

Yes!

[@gregoired.4660]

yoooo man you really have to stop right negative numbers in sqrt... I mean, you just wrote a contradiction proof of why you cant do that with the -1 stuff. If you juste write i*sqrt(3) instead, now your math are right and moreover, you get ready of any possible misstakes about it. *you seemed to have it right but thats basicly math done wrong gone right. Althought, since sqart(-3) should be written as i*sqrt(3), you are dealing with sqrt of complexe numbers. That makes the argument "x is a sum of you sqrt so it has to be positive" not old anymore. We could go even further and argu that since 1 -/+ i*sqrt(3) is complexe, take the sqrt of that boii is also a probem and we should find their roots without using the sqrt, which is only defined for positive numbers. If I'm not wrong, your problem just dont mathematicly make sens at all and this how I'ld write it : Let x²=1 + i*sqrt(3) and y²=1 - i*sqrt(3), found x+y. Even if it looks similare, you're actually getting read of sqrts and all the problems they implie. That way, i've got x=-/+ sqrt(2)*e^(i*pi/6) and y=-/+ sqrt(2)*e^(-i*pi/6). That gives us 4 solutions : sqrt(6), -sqrt(6), i*sqrt(2) and -i*sqrt(2). Now, for the second problem, since y=-/+ sqrt(2)*e^(-i*pi/6), then -y=+/- sqrt(2)*e^(-i*pi/6), which give the same soluition for x+y than x-y ! You can notice that whose solution are the same as you get, you just dont get every solution !

[@gregoired.4660]

Of course, if i'm not wrong ! btw, sry for bad english, hope that make sens.

[@lukandrate9866]

Finally I found a man who knows that. You could also rewrite square roots as ^(1/2) and everything would be okay because rational power of a complex number has multiple values unlike the square root sign(√) which means that the answer is greater than 0 but you can't compare complex numbers

[@brandonklein1]

This is exactly what I thought would happen under more careful consideration. Thanks for working it out!!

[@gregoired.4660]

@@lukandrate9866 Actually i don't remember had seen a number worte as "(a+ib)^(1/2)" in class, so i dont know. To me, it feels a bit weird to note two roots with a single notation, but you may be right thought

[@angelmendez-rivera351]

@@lukandrate9866 No, that would be doubly wrong. Firstly, you _can_ compare the complex numbers: the well-ordering theorem, which is a consequence of the axiom of choice, tells us that every set has a well-ordering. The complex numbers can be totally ordered using the lexicographic ordering with respect to the real numbers. Given the lexicographic ordering, it is possible to talk about the square root function for complex numbers: for nonzero z, 0 < sqrt(z), where < is the lexicographic strict ordering. Then there are no contradictions, as long as the sqrt function is handled carefully. It can be re-expressed in terms of the exp function and the arg function.

[@Juliasn68]

Do you have a release date for your onlyfans?

[@PapaFlammy69]

69th of the 420th month

[@Juliasn68]

@@PapaFlammy69 Thanks, I'll be noting that.

[@Rafael-oq9vu]

first

[@xa-1243]

Trade offer: You receive: Me telling my friends about your channel I receive: A reply to this comment Deal?

[@diogeneslaertius3365]

Did you dumb down the level of your videos on purpose to get more views? You used to do much more complex stuff before. Now it's at the school level. Sad.

[@PapaFlammy69]

No, didn't do so, I always covered on this channel the things I was working on/interested.

[@cosinev1265]

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