*_Thanks for watching, I hope the video was to your liking
@anissaada86603 жыл бұрын
Can you do a video of math of second secondary
@oni83373 жыл бұрын
yes
@guest_informant3 жыл бұрын
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_b3 жыл бұрын
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_b3 жыл бұрын
@@guest_informant even 1×2×3=1+2+3. But may be ......!🤔
@luggepytt3 жыл бұрын
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!
@wisdomokoro88983 жыл бұрын
You're just Amazing 😊😍
@Freakschwimmer3 жыл бұрын
Damn :D I can understand what you are saying but I'm totally mesmerised you were able to put it down so nicely :)
@grapix11843 жыл бұрын
Thx!
@luggepytt3 жыл бұрын
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?
@anshumanagrawal3463 жыл бұрын
@@luggepytt lol
@tiagolevicardoso76743 жыл бұрын
Hill: question mark is a valid variable. Proof: Let ? be a variable Q.E.D
@RedVio9723 жыл бұрын
Permission denied
@redhotdogs31933 жыл бұрын
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)
@oni83373 жыл бұрын
genus
@aneeshsrinivas8923 жыл бұрын
I did the same
@angelmendez-rivera3513 жыл бұрын
This assumes that you already have complex analysis at your disposal, which Euler obviously did not have when he worked with this problem.
@anshumanagrawal3463 жыл бұрын
@@angelmendez-rivera351 exactly my point
@YourPhysicsSimulator3 жыл бұрын
"Mathematicians are serious, boring and quiet people" What mathematicians trully are doing: 0:09
@nikhilnagaria26723 жыл бұрын
More like 0:30
@YourPhysicsSimulator3 жыл бұрын
@@nikhilnagaria2672 hahaha that one too
@herbie_the_hillbillie_goat3 жыл бұрын
Was that a sneeze, a cough, or wheeze?
@stevenxw3 жыл бұрын
we've gone from "good morning fellow mathematicians, welcome back to another video" to "GOOODIUQWEGHFawuihkefhuio"
@nikhilnagaria26723 жыл бұрын
@@stevenxw the video about the sines (third most recent): check it out ;)
@renanrodz72213 жыл бұрын
Someone: "Do you have plans for this weekend?" Me: "It's complex..."
@_mishi3 жыл бұрын
complex*.. Its complex
@realkabirc3033 жыл бұрын
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.
@PapaFlammy693 жыл бұрын
nice :)
@alberteinstein36123 жыл бұрын
Hello fellow math lover :D
@realkabirc3033 жыл бұрын
@@alberteinstein3612 Hi👋👋
@parasb213 жыл бұрын
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 !!
@sttlok3 жыл бұрын
You can still use the question mark as a variable lol
@parasb213 жыл бұрын
@@sttlok Haha true but that wont be an intuitive thought to use ? as a variable lol
@sttlok3 жыл бұрын
@@parasb21 lol, integrate d^2 in respect to d.
@Vaaaaadim3 жыл бұрын
Sneezed for his intro, what a time to be alive
@Brien8313 жыл бұрын
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 .
@PapaFlammy693 жыл бұрын
yas! :)
@pardeepgarg26403 жыл бұрын
Alt title : A Math breaking Ground Result
@MrRyanroberson13 жыл бұрын
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))
@PapaFlammy693 жыл бұрын
oh, nice!
@hardikjoshi85573 жыл бұрын
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:)
@XAE-yc9rr3 жыл бұрын
> 5:00 AM >slides into view like a boss >"Gyee...*coughs*! " This is already a great start of the day.
@ricardoparada53753 жыл бұрын
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
@brandonklein13 жыл бұрын
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-rivera3513 жыл бұрын
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)*.
@brandonklein13 жыл бұрын
@@angelmendez-rivera351 that's fair.
@CauchyIntegralFormula3 жыл бұрын
You end up with either x^2 = 6 (as seen in the video) or x^2 = -2
@brandonklein13 жыл бұрын
@@CauchyIntegralFormula Then indeed there are 4 solutions.
@johnny_eth3 жыл бұрын
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.
@alberteinstein36123 жыл бұрын
Papa Flammy always cracks the perfect amount of adult jokes and it’s absolutely hilarious
@Manoj_b3 жыл бұрын
It's amazing that it's solved by Leibniz before 300 years before Euler ..🎉. He must be happy I think.
@PapaFlammy693 жыл бұрын
:)
@vigneswarans16363 жыл бұрын
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.
@Mike-fr9hp3 жыл бұрын
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)
@akselai3 жыл бұрын
>algeBRUH Very groundBRUHking indeed, Papa.
@tambuwalmathsclass3 жыл бұрын
I didn't try this, but I believe if I were to simplify it, I'll end up with a complex Solution
@tambuwalmathsclass3 жыл бұрын
Incredible I got √6 after u, v and y substitution. Permission to record the video
@siranguru3 жыл бұрын
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
@siranguru3 жыл бұрын
x and y are symmetry i.e. exchangeable
@saadmoquim86243 жыл бұрын
ive been on this channel long enough to know that an alternate cover would be the euler fire emoji
@PapaFlammy693 жыл бұрын
ayyyy :p
@CauchyIntegralFormula3 жыл бұрын
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
@chessematics3 жыл бұрын
Me when I saw the thumbnail: YOU WANT TO FOOL ME THE COUSIN OF CARDANO?
@zildijannorbs58893 жыл бұрын
fuck this im doing a compilation of all the good morning fellow mathematicians
@angelmendez-rivera3513 жыл бұрын
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.
@VaradMahashabde3 жыл бұрын
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-rivera3513 жыл бұрын
@@VaradMahashabde Yes.
@Abdul-Akeem_Akinloye3 жыл бұрын
My fucking head.
@MathAdam3 жыл бұрын
1:08 "Rested Nadicals" -- This is what happens when you record your videos at 5 am. :D
@PapaFlammy693 жыл бұрын
:D
@VaradMahashabde3 жыл бұрын
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-rivera3513 жыл бұрын
The arg function as you define it is onto (-π, +π], not (-π, +π).
@VaradMahashabde3 жыл бұрын
@@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-rivera3513 жыл бұрын
@@VaradMahashabde But they do work well with the definition. I would recommend you check out my post on the main comments section.
@neilgerace3553 жыл бұрын
0:05 Well this puts a different slant on things.
@goodplacetostop29733 жыл бұрын
12:21
@bennetthudson95373 жыл бұрын
this timestamp is trivial
@goodplacetostop29733 жыл бұрын
@@bennetthudson9537 This timestamp is trivial and is left as an exercise for the reader.
@herbie_the_hillbillie_goat3 жыл бұрын
This timestamp is beyond the scope of this lecture and can safely be ignored.
@anshumanagrawal3463 жыл бұрын
That's a Good Place To Stop
@pavolgalik4993 жыл бұрын
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.
@nikhilnagaria26723 жыл бұрын
It still holds whenever at least one of them is nonnegative
@md.iftakhar74783 жыл бұрын
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.
@stvp683 жыл бұрын
Like how at some angles, the shirt looks like it says lgb 😁😁
@mkjaiswal113 жыл бұрын
Since I am in Class 9 and I have radicals in course, there would be no doubt if this question come in my exams.
@nikhilnagaria26723 жыл бұрын
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.
@veralgupta81823 жыл бұрын
Papa flammy at 5am so fresh so cool Me when I wake up at 8am : 😑😴😑
@PapaFlammy693 жыл бұрын
:D
@JR137513 жыл бұрын
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.
@Lolwutdesu90003 жыл бұрын
I bet a lot of people will do a lot of "woodworking" once they see your OF.
@anonymanonym66413 жыл бұрын
And I thought I had to rewrite 1 as 2²- sqrt(-3)²...
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
@gragasapmidlane67613 жыл бұрын
"dont ask me why im doing dis at 5am" also him 30minutes before "Raids with americans"
@calambuhayjr.josevirgiliog20943 жыл бұрын
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.
@WriteRightMathNation3 жыл бұрын
What makes it "groundbreaking"?
@Mystery_Biscuits3 жыл бұрын
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?
@PapaFlammy693 жыл бұрын
they are still there, just unlisted :)
@Mystery_Biscuits3 жыл бұрын
@@PapaFlammy69 oh sure sure. Nice 👍
@sheriff-larue77703 жыл бұрын
Who here recognizes when the Redbull or coffee- kicks in? Min 2.48
@Gameboygenius3 жыл бұрын
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.
@the_nuwarrior3 жыл бұрын
Cardano ?
@tszhanglau57473 жыл бұрын
Recording videos at 5am do be a bruh moment
@PapaFlammy693 жыл бұрын
ayyyy lmao pog sus
@utkarshsharma95633 жыл бұрын
Amogus moment
@thebeerwaisnetwork80243 жыл бұрын
Very nice
@einsteingonzalez43363 жыл бұрын
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.
@einsteingonzalez43363 жыл бұрын
Can you prove it?
@nikhilnagaria26723 жыл бұрын
@@einsteingonzalez4336 what's the point in this even
@abhigyan79873 жыл бұрын
Nice information ℹ️ℹ️ . Watching your videos from india ❤️ sir
@ali_ahmed318993 жыл бұрын
Please I want solution to this ODE: Y'=sqrt(x²+y²) Can you help me
@orisphera3 жыл бұрын
4:25 There are two square roots of each nonzero number. Otherwise, -1=sqrt((-1)²)=sqrt(1)=1
@angelmendez-rivera3513 жыл бұрын
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.
@orisphera3 жыл бұрын
@@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-rivera3513 жыл бұрын
@@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.
@orisphera3 жыл бұрын
@@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
@orisphera3 жыл бұрын
@@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
@neilgerace3553 жыл бұрын
6:04 if you trust the Peano axioms.
@PapaFlammy693 жыл бұрын
:^D
@arielfuxman88683 жыл бұрын
Papa Flammy, how tall are you?
@frozenmoon9983 жыл бұрын
Flammy's Wood. Flammy's OnlyFans incoming next.
@tetris4503 жыл бұрын
Rested nadicals
@laurentreouven3 жыл бұрын
-i sqrt(2) as i sqrt(2) works for y, how to choose wich us the answer ? Not sure its possible
@johnny_eth3 жыл бұрын
You use both.
@laurentreouven3 жыл бұрын
@@johnny_eth a number that have two value is not a number
@nikhilnagaria26723 жыл бұрын
@@laurentreouven both are equivalent
@laurentreouven3 жыл бұрын
@@nikhilnagaria2672 how do you know ?
@nikhilnagaria26723 жыл бұрын
@@laurentreouvenok I ask you a question: what is i?
@mr.unknowngamer21093 жыл бұрын
😮wow
@herbie_the_hillbillie_goat3 жыл бұрын
Ummm Rafael Bombelli predates Leibnitz by about a century.
@PapaFlammy693 жыл бұрын
who? lol
@rssl55003 жыл бұрын
The answer to your question is sqrt 6
@adityaekbote84983 жыл бұрын
Nice
@gregoired.46603 жыл бұрын
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.46603 жыл бұрын
Of course, if i'm not wrong ! btw, sry for bad english, hope that make sens.
@lukandrate98663 жыл бұрын
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
@brandonklein13 жыл бұрын
This is exactly what I thought would happen under more careful consideration. Thanks for working it out!!
@gregoired.46603 жыл бұрын
@@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-rivera3513 жыл бұрын
@@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.
@mikejackson198283 жыл бұрын
Rested nadicals. 🤣
@FreeGroup223 жыл бұрын
taking the square root of complex numbers ? you have 2 roots
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@diogeneslaertius33653 жыл бұрын
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.
@PapaFlammy693 жыл бұрын
No, didn't do so, I always covered on this channel the things I was working on/interested.