@venkovic No, OP is correct, multiplicative property of roots holds only for real numbers (in general, cube roots or higher are included), and you can't take a square root of a negative number in reals, so in reals √-2 is already wrong. "But we talk about complex numbers". Ok, if you insist, then saying that it holds for complex numbers "as long as at least one number is non-negative" is also wrong, because for any complex number z (and that includes 2) something like z≥0 (non-negative that is) doesn't make sense, since complex numbers are an unordered field. And also unless you specifically say that √ in your √-2 stands for the principal root you can't even say that √-2=i√2 (it should have ± in the front), it will be wrong. But for a principal root multiplicative property also doesn't hold just like it does not hold for an algebraic root in complex numbers. So all of this is completely wrong, this is not how you deal with complex or real numbers, for the love of God stop trying to mix the two. Basically, you here try to create your own function from real to complex numbers, use for it a symbol (√) that is already stands for like up to three to four functions and pass it as standard mathematics. This is not how any of it works
There's a lot of confusion in the comments stemming from something that, unfortunately, they don't teach you until complex analysis class, which is that the √x operation is not the same as (x)^(1/2) operation, but rather, the former is a specific case of the latter specified on something called a "principal branch". Basically, if you take the equation y = x^(1/n), then there are n possible solutions of x on the complex plane, which makes x^(1/n) not strictly a function. If you want to define a function and call it the "nth root of x", you have to basically define it as "that solution to y = x^(1/n) that happens to lie in THIS region in the complex plane". The square root happens to be one of those functions, where it's basically "the square root of x is the solution to y = x^(1/2) that falls on the _right half_ of the complex plane." The reason this means that the first method doesn't work and the second does is because you can show that, if you specify a branch, then a product of square roots is not generally equal to a square root of products. This can be proven using complex analysis, but here's an intuitive explanation. The principal branch square root requires an output in the right half of the complex plane. So, for a product of square roots, each factor must lie in the right half of the complex plane, but their product doesn't have to. However, by definition of principal square root, a square root of a product _must_ lie in the right half of the complex plane.
@@sandromaspindzelashvili5767 A plane (C) consisting of the set of all complex numbers z of the form z = x + iy, where is the real part ( your every day real numbers ), y is the "imaginary" part, and i is the imaginary unit ( along the y-axis ) with the property i^2 = - 1, or i = sqrtc( -1), where sqrtc is the COMPLEX square root function. You can think of it as an expansion of your normal real numbers line into a plane to both sides of it - i.e. numbers that lie outside it ( or on it ). These complex numbers and complex functions have a lot of interesting and very useful properties - for instance in calculating otherwise difficult integrals and sums of various series, solving differential equations, AC power distribution, spectral analysis of analog and digital signals ( Fourier series & Fourier transforms, Laplace Transforms, analog and digital filters, control systems ), acoustics, physics ( oscillations ) and quantum mechanics etc. They also combine the properties of numbers with the properties of vectors (additions, subtraction, "length" ( magnitude ) - multiplication & division both involving a scaling of the magnitude and a rotation of the complex number ("vector") in the complex plane. All our normal numbers and operations on them are really just special cases of these complex numbers and these extended operations on them - it's like some sort of higher dimensional shadow world, so to speak 😉 - and they do actually make mathematical sense - satisfying certain conditions - , even though they may seem very weird at a first glance 😬 ). There are several really cool videos about these topics on TH-cam, so I suggest that you watch those, as they explain these fascinating concepts far better that I am able to do here - also graphically of course.
I'm 32 now, haven't ever needed to use this information once in my life outside of high school but being given this crash course about negative square roots makes me remember that I used to LOVE this kind of stuff. It felt like puzzle solving, unlike a lot of other parts of math you learn in hs.
Imaginary numbers are a tool used a lot in physics and esp in engineering to solve real world problems which would otherwise require rigorous calculations, high school sets u up with the basics, college is where u apply them.
When extending the square root function to the complex plane (which cannot be avoided here), you _have_ to specify the branch that you're using, and also mention whether the extension to the negative reals was made through the upper or the lower half-plane, which would indicate whether sqrt(-1) is "i" or "-i". In fact, in the example from the video, the two roots don't necessarily have to be from the same branch. If you take different branches for each root, you get +4, and -4 otherwise. P.S. no, the imaginary unit is _not_ defined as sqrt(-1) (because of this whole thing with multivalued functions that would make a lot of mess), rather, "i^2 = -1" is its property resulting from the multiplication rule in the complex plane (or, equivalently, it's defined as one of the roots of x^2 +1 = 0, doesn't matter which one, as long as it's fixed throughout the whole theory). P.S. #2 Don't you find it strange that sqrt(-2)×sqrt(-8) cannot, according to the author, be written as sqrt((-2)×(-8)), but sqrt(-2) = sqrt((-1)×2) = sqrt(-1)×sqrt(2) no problem? -1 seems like a privileged individual that you can take in and out of the root, but not any other number 😅 (again, it all comes down to the specific branch and where you insert the branch cut)
So applying the "privileged individual" status of -1 that I learned from the video, I can prove that sqrt(16)=-4. sqrt(16)=sqrt(-(-16))=i*sqrt(-16)=i*i*sqrt(16)=i^2*4=-4
√16 is +4 ... Solving for n where n^2 = 16 is +-√16 so +- 4.. it is true to say when not zero a number has two square roots. but the √ symbol means principal square root ie positive so the video is correct. √-1 is i So √-2 is (√2)i so √2*√8*i^2 = √16 *-1 =-4.
@@gonzalobarragan8076 This is incorrect because the property √(ab) = √(a) × √(b) can only be applied when both a and b are non-negative. √[(-2) × (-8)] ≠ √(-2) × √(-8)
@@gonzalobarragan8076 the property '√a•√b = √(a•b)' you used is only applicable for if a and b ≥ 0. Since in the given question, a and b are negative which means a and b < 0. Therefore the property cannot be applied in this case You should have paid attention in these small things also.
I changed the sqrt(-8) to sqrt(4) and sqrt(-2) then sqrt(-2) and sqrt(-2) just multiply to become -2 and then that’s multiplied by sqrt(4) which is 2. Final answer is -4
More needs to be said on this matter. One should say something about "multi-value" functions of root and real and complex principal root. Without precisely determining the definition on what the "sqare root" is (since there are more different functions called the same name) the discussion on this topic can easily open the door of philosophy.
sqrt(a) as defined in high-school maths is ill-defined when 'a' is allowed to be negative. In high school we define sqrt(a) as the nonnegative root of the polynomial x^2 - a. E.g. x^2 - 4 → x ∈ {2, -2} so we define sqrt(4) = 2. However the roots of x^2 + 4 are {2i, -2i}. Neither of them are positive; they fall outside the real number line! Ok, then define sqrt(-a) = sqrt(a)i (for positive real 'a'). You can do this, but the property sqrt(x)sqrt(y) = sqrt(xy) no longer holds! This is due to the fact that unlike the positive real numbers, the upper imaginary line is not closed under multiplication!
@Blank Space no of course it doesn't change anything.. but it's just a definition to avoid confusion. You can do your own maths with your own definitions, but I'm gonna stick with the ones I learned :)
@Blank Space "Although the question mentioned at the beginning has two solutions with different signs for even-numbered root exponents and positive radicands, the notation with the root sign always stands for the positive solution." This is from wikipedia, but I do understand your thought process, and it really doesn't matter, run with whatever you like. This will be my last reply, have a good one.
@@masteroogway2853 ye i know what you are reffering to, i may have just missunderstood this, but i never learnt that a square root is equal to +- of said number
One of the rules in mathematics says that you can multiple root of a negative number. You have to first change it into a imaginary number by taking out the negative.
@@overdose8329 the square root function is defined in such a way that it only gives positive outputs. A common misconception is that it gives two answers, but that‘s simply not the case. After all, there is a reason we maticilously write out +-sqrt in quadratics and Not just +sqrt.
@@YouHaveToBeTheChange, nope. The square root is a single valued function, defined as the principle branch of the inverse of the square. You've clearly learned that you need to remember the +- when solving a second degree polynomial equation, which is correct and important but doesn't apply to square roots.
You should consider in this case that sqrt(x²) is + or - x, both for the fist case and the second, so the answer is + or - 4, and that's because with negatives taking the square root means solving the equation x²=whatever number you want to take the square root of, and that is + or - x Source: my father, university professor of maths (or at least that's how he explained to me, I'm reading many other explanations here which might be things he didn't tell me) PS sorry for formatting, I'm writing from mobile
Isn't that because if you have for example √9=3 as a result from √x²=y, then x is either +3 or -3 since sqaring either results in 9, and y is thus ±x as a result of two values for x. It doesn't mean that y is negative. √x² = |x| according to what I was taught and I just checked that wolfram alpha affirms that
@Retired Bore as a software engineer, I don't even know what you mean by "mathematics for programming". The square root function is 'defined' to yield the positive solution
@Retired Bore there are multiple square root implementations for computers and so there's no "the square root function". And no, I'm not mixing them up. Feel free to look up the definition of square root, or functions in general and see they don't have multiple answers.
@Retired Bore here you go "Every nonnegative real number x has a unique nonnegative square root, called the principal square root, which is denoted by √x" I pulled it from wikipedia "square root" so you can go and argue with the sources there instead of here.
first method is wrong because the square root of a negative number is not defined in the field of real numbers, instead it is well defined in the field of complex numbers, I think that this should be said …
He did not define that we are dealing with complex numbers. Therefore I was free to define that we are dealing in real numbers and then both solutions are wrong. Tataaa!
In the world of real numbers, sqrt(x)*sqrt(y) = sqrt(x*y) it's only valid if x and y are positive real numbers or at least x or y are 0. So, sqrt(-2)*sqrt(-8) = sqrt(-2*-8) is not valid, and so, having no solution at all. In the complex world, well... you do what was done in the video.
He forgets to mention that this is (of course) the complex square root function, which "happens" to be "identical" to the normal real square root function on the positive part of the real axis & in the point x = 0 ( + 0i ). Yes, you are right - in complex analysis it's allowed to work with socalled multivalued (!) functions - like a group of different plausible branches that satisfy the given conditions. So in this case he will need to look at all combinations of +/- i sqrt(2) & +/- i sqrt(8) ( where sqrt refers to the real square root function), which produces these 4 results, which turns out to be just two: 1) i sqrt(2) x i sqrt(8) = -1 x sqrt(16) = -4 2 i sqrt(2) x ( -i (sqrt(8) ) = +1 x sqrt(16) = 4 3 -i sqrt(2) x i sqrt(8) = +1 x sqrt(16) = 4 4 -i sqrt(2) x ( -i sqrt(8)) = -1 x sqrt(16) = -4 So +4 is actually a solution (contrary to his claim! ), and so is -4.
@@isjosh8064 No, both + 4 & - 4 are actually "solutions" to this expression - you need to check all the 4 ( = 2 x 2 ) possible combinations of the two (primary) branches of the complex square root function.
@@Bjowolf2 When you say “solutions" I imagine you’re thinking of an equation like: x^2 = 16 where the solutions are +/- but those are what x can be to satisfy the equation. But /x = -5 has no solutions even though -5 squared is 25. The definition of domain of /x only allows a positive input and returns a positive input.
@@isjosh8064 Yes, for ordinary (real) square root this is true, but for COMPLEX square roots, which he is clearly working with here ( with -4 & -9 under the square root signs ), different rules come into play, since they have two possible (primary) branches. ( These socalled multi valued functions are permitted in complex analysis ). So sqrtc(-4) = +/- i x sqrt(4) = +/- 2i And likewise for sqrtc(-9).
@@georgesas7090 no the square root is a function meaning every input has one output. You are thinking of solutions to x^2 = 16. Which is different than sqrt(16) as sqrt(16) and -sqrt(16) Are both solutions to that equation
@@BlackHoleSpain Yikes, I can’t think of anyone, literally no one who’s successful that doesn’t have an amazing credit score. Paying cash is fine if you want to stay poor.
Everyone makes jokes like this, but in practice, there are second order differential equations that appear in the study of finance and economics. Their solutions sometimes require finding the roots of a polynomial which are often complex numbers.
@@cdula26 Credit score literally doesn't matter once you're making consistent reliable income. If you have cash, people have goods and services they want to offer you.
@@acex222 Even my real estate mentor who brings in a net income of $800,000 a month has a great credit score and doesn't pay cash for his 500k cars or 5 million dollar yacht. It's better to finance things and use all the cash you would have spent on buying more assets and investing the rest.
Under that logic... 4=✓16=✓i^2(-16)=i✓-16=i^2✓16=-4 You can't avoid the fact that square root has 2 values. Sure, you can talk about the principal branch but it's unavoidable with complex numbers.
What??? The whole point is that you can NOT use the square root factoring rule unless the factors are positive real numbers. His method of evaluating square roots of negative numbers is perfectly valid.
@@Tulanir1 I did exactly as the video. For him ✓-2=i✓2 and that uses the same rule I'm using. If you can't truly use that rule unless it's a positive integer, then he can't even answer the problem.
@@kevinmartincossiolozano8245 surely when you go from sqrt(i²*-16) to i*sqrt(-16) youre skipping an intermediate stage where you get sqrt(i²)*sqrt(-16) which means youve factored with a negative number
@@soroushhaidary7934 square root is a function which means you have only one output (positive 4 in this case) you would be right if it was an equation like x²= 16 , only then x=4 v x=-4
@@soroushhaidary7934 You can also see how there are contradictions when you say sqrt(16) = +-4 Let’s assume sqrt(16) = +-4. Then because 4 = sqrt(16) and sqrt(16) = -4, our conclusion is that 4 = -4, which is false.
@@Dra3oonBoth are correct but only because we included complex numbers in the beginning. If you only have sqrt(16) then is 4, but if you have Sqrt(-16) × sqrt(-1), then its both -4 and 4.
Normally, it depends on how you _define_ the √□ operation. And as long as you explain the full chain of reasoning and it starts with a reasonable definition, I'd say the reasoning is valid. Of course, if you have a test-like question where you have to produce the same text as what's written on the sheet labeled "correct answers"... Then it's a game of either "memorize this particular textbook and not another textbook" or "guess the answer". The answer here can be any of "4", "-4", "±4", "expression is ill-defined".
Square root is a function, meaning it can only return one result, the positive one (for real numbers) You need to take both values when you have x²=4 x=±2 but it's because x²=4 x²-4=0 (x+2)(x-2)=0 x=±2
@@gabrieleymat6332 A "√□" is a symbol. Nothing internally inconsistent would happen if you were to define it as "all numbers that produce □ when multiplied by themselves". For example, in a residue field modulo 7, 3²=4²=2. In this case, trying to find a "principal" value of √2 would be very much futile. And even if you insist to take a positive value for real numbers (why? because a particular textbook said so? how you would arrive to this conclusion if you were inventing all math from scratch?), you still have the same problem with complex numbers: i and -i don't differ in any reasonable way. In some cases, it may be important to take the value in "the same direction" for both roots: you can write √-1x√-1 = ixi *or* √-1x√-1 = (-i)x(-i) but not √-1x√-1 = ix(-i). Like when solving cubic equations, you get an expression ³√a+³√b and then you have to sum the "right" pairs of cubic roots of a and b.
@@sytherplayz yes but I think "solutions" isn't the right word in this case because we aren't solving a polynomial equation but nonetheless we both are saying the same thing that the √-4 doesnt belong in the sets of real number system.
Square root always has two solutions: positive and negative; unless we calculate square root of 0, which has only one solution. Also, multiplying numbers under the same exponent is allowed. Extracting i is allowed, but unnecessary. Both solutions should yield the same results: Square roots of 16 are 4 and -4. By using the other method we arrive at the same results: square roots of 16 multiplied by -1 gives -4 and 4, which is the same thing.
It's specifically √x not x^2=y , the answer in second one is x=±√y Where √y is the principal solution x^2=y , This is done because √x is defined to be one-one function, for each x you put you can get only one answer, the negative ans is discarded, this principle soln is put in x=±√y.
When we ask for all the square roots of 16, it's ±4. But when we write it with a square root symbol such as √16, it's only the principal square root, which is +4. Hence, the reason when we simplify expressions such as √4+√9, we simplify it to +5, not ±5 or ±1.
@@BlackSakura33 maybe you forger that square root of as positive no. is positive For eg. √4=2 and not -2 But if you put as x²=4. Then x has values as 2 and -2 And please stop hating and stereotyping other countries, coz we are also being hated and stereotyped at international level Spread peace
@@kartavyasharma7266thats only true when x is a real number. If you inclide the complex plain, both negative and positive solutions are real solutions. And because we included the complex plain at the start, we cant ignore it. Look at the start of the problem, sqrt(-2)=i×sqrt(2) AND -i×sqrt(2) Same for the 8, and then we can see that theres actually 2 solutions, + and - 4
I would argue that both solutions are right and wrong at the same time. Since √16 = ±4 and -√16 = ±4, the solution should be ±4 in both cases. Choosing one arbitrarily is just that, an arbitrary choice. Basically, the rules for arithmetic with roots hold. You can't just disregard the second solution and say that these rules don't hold for some complex numbers.
@@IoDavide1 square root by definition refers to the positive root of x²=16. square root as an operator itself doesn't give you both the roots. Because if that was the case then we couldn't raise both sides of an equation to fractional powers. Suppose sqrt(1) =+-1 1=1, square rooting both sides (raising powers in both side to 1/2) We would get 1=-1 as a solution. Hence by convention 16 to the power 1/2 (ie. Sqrt(16)) gives only 4 (the absolute value)
@@ComposedBySam this second comment make less sense then the first. The square root has not determined sign, so you have always two results: + and - You seems still confused with primary school definitions
@@IoDavide1 then do a favor. Try harder to understand what I wrote. And if you cannot google it. the solution of x² = 16 and square root of 16 are different things
Incorrect, √(-2) × √(-8) ≠ √(-2 × -8) because the expression √(ab) = √(a) × (√(b) only applies when both a and b are non-negative. Since both a and b are negative in this situation, we have to use complex numbers to solve the expression. √(-a) × √(-b) = i√(a) × i√(b) = i^2 × √(a × b) Since i^2 = -1, and √ only returns the positive root, the answer to √(-a) × √(-b) only yields negative solutions. The end result is only -4, +4 is not a solution.
I concur that you take "i' out of the square roots first, but root(16) = +/-4....multiply +/-4 by -1 and you still get +/-4 (which is also found by the 1st method, but that 1st method is still the wrong method). I think this simple problem is just a bad example of why the indicated method is the correct procedure.
@@BlackCat-fx9kb you are right that sq root should be positive but here you see ✓16 = ✓(+4)^2 and it can also be ✓16= ✓(-4)^2 therefore they both are positive only. So sq root of 16 will give +4 or -4.
Consider: sqrt(16) = +/- 4 Just accept the square root as naturally and inherently multivalued instead of arbitrarily defining it as positive. Gets at the heart of the matter and resolves all discrepancies without any overcomplicated notions of branch cuts etc. Now both methods are equally correct: sqrt(-8)*sqrt(-2) = sqrt(16) = +/- 4 sqrt(-8)*sqrt(-2) = i*i*sqrt(16) = -1*+/-4 = +/-4
Nobody denies the square roots of 16 are 4 and -4, but √ does NOT yield the square roots - it yields the POSITIVE square root. If you want both, you need to write this x² = 16 -- > x = +/ - √16 because √16 = 4 There is no discussion on that - it's simply using the defintion of the function/symbol √ If you use a different definition, you're solving a different problem. If you solve a different problem, you get a different answer.
you get the same answer either way if you just dont forget the +- sqrt rather than just +sqrt this is actually the exact reason you think you're getting 2 different answers
Not correct. The square root of 16 is 4. -4 is not the square root of 16. -4 is one of the solutions for the quadratic equation x^2=16. Square root of 16 is a number and a number cannot assume multiple values while the equation will have two solutions
@@ovidiucroitoru2290that property only works for square roots of positive numbers Also for the OP : no, √16 = x and x^2 = 16 is not the same thing, x for the 1st is 4, while x for the 2nd can either be -4 or 4. The guy in video talks about the 1st
@@ovidiucroitoru2290You would get the square root of -16 with that result which is undefined in real numbers. Hence why one needs to express it as a complex number like he does in the video
this is how government calculate your pension
funny 🤣🤣🤣
Fuuuuck
Makes sense. It's complex and perpetually negates your progress. Our governmental system only makes sense in our imagination.
🤣 they go straight to imaginary numbers
_Seems fair_ 🍷🗿
-Some Govt.
_The multiplicative property of square roots that √a*√b = √(a*b) only applies when a and b are more than or equal to 0._
Yes many people in comment section do not know this
You are wrong💀💀💀💀if you don't know something then at least fact check yourself first before commenting online bruh embarrassing
@venkovic No, OP is correct, multiplicative property of roots holds only for real numbers (in general, cube roots or higher are included), and you can't take a square root of a negative number in reals, so in reals √-2 is already wrong. "But we talk about complex numbers". Ok, if you insist, then saying that it holds for complex numbers "as long as at least one number is non-negative" is also wrong, because for any complex number z (and that includes 2) something like z≥0 (non-negative that is) doesn't make sense, since complex numbers are an unordered field. And also unless you specifically say that √ in your √-2 stands for the principal root you can't even say that √-2=i√2 (it should have ± in the front), it will be wrong. But for a principal root multiplicative property also doesn't hold just like it does not hold for an algebraic root in complex numbers. So all of this is completely wrong, this is not how you deal with complex or real numbers, for the love of God stop trying to mix the two.
Basically, you here try to create your own function from real to complex numbers, use for it a symbol (√) that is already stands for like up to three to four functions and pass it as standard mathematics. This is not how any of it works
@@kushaljoshi3862 Bruh, he's 100% on point, stop learning maths from some twats on YT, grab a good textbook on complex numbers and read it
No. It applies when ANY of a and b is >=0, because negative number under square root is fully legal notation (and means imaginary result).
Man this guy roasts every regular middle school math student like crazy
Im halfway through a bachelor's in computer engineering and this man is still HUMBLING me. 🤣🤣
i am doing coputer engineering too
first semester math
first week of the first semester
Exactly
Dude how did you get into the college without knowing this?
@@naikubaid you can forget this sort of stuff pretty fast.
@@BRAINSPLATTER16you have never studied a single day of your life, haven't you?
Imaginary numbers are easier to handle than hallucinary numbers
lol
(North + wall) × leaf × wind - [summer + winter] /30people!
Well, I guess you are right
Lmao, should have been the proper term for quaternions
Works best in hungarian 😂
All numbers are imaginary.
There's a lot of confusion in the comments stemming from something that, unfortunately, they don't teach you until complex analysis class, which is that the √x operation is not the same as (x)^(1/2) operation, but rather, the former is a specific case of the latter specified on something called a "principal branch". Basically, if you take the equation y = x^(1/n), then there are n possible solutions of x on the complex plane, which makes x^(1/n) not strictly a function. If you want to define a function and call it the "nth root of x", you have to basically define it as "that solution to y = x^(1/n) that happens to lie in THIS region in the complex plane". The square root happens to be one of those functions, where it's basically "the square root of x is the solution to y = x^(1/2) that falls on the _right half_ of the complex plane."
The reason this means that the first method doesn't work and the second does is because you can show that, if you specify a branch, then a product of square roots is not generally equal to a square root of products. This can be proven using complex analysis, but here's an intuitive explanation. The principal branch square root requires an output in the right half of the complex plane. So, for a product of square roots, each factor must lie in the right half of the complex plane, but their product doesn't have to. However, by definition of principal square root, a square root of a product _must_ lie in the right half of the complex plane.
What is a complex plain. Is it X or Y?
@@sandromaspindzelashvili5767
A plane (C) consisting of the set of all complex numbers z of the form z = x + iy, where is the real part ( your every day real numbers ), y is the "imaginary" part, and i is the imaginary unit ( along the y-axis ) with the property i^2 = - 1, or i = sqrtc( -1), where sqrtc is the COMPLEX square root function.
You can think of it as an expansion of your normal real numbers line into a plane to both sides of it - i.e. numbers that lie outside it ( or on it ).
These complex numbers and complex functions have a lot of interesting and very useful properties - for instance in calculating otherwise difficult integrals and sums of various series, solving differential equations, AC power distribution, spectral analysis of analog and digital signals ( Fourier series & Fourier transforms, Laplace Transforms, analog and digital filters, control systems ), acoustics, physics ( oscillations ) and quantum mechanics etc.
They also combine the properties of numbers with the properties of vectors (additions, subtraction, "length" ( magnitude ) - multiplication & division both involving a scaling of the magnitude and a rotation of the complex number ("vector") in the complex plane.
All our normal numbers and operations on them are really just special cases of these complex numbers and these extended operations on them - it's like some sort of higher dimensional shadow world, so to speak 😉 - and they do actually make mathematical sense - satisfying certain conditions - , even though they may seem very weird at a first glance 😬 ).
There are several really cool videos about these topics on TH-cam, so I suggest that you watch those, as they explain these fascinating concepts far better that I am able to do here - also graphically of course.
hey Schrodinger thank you for the electron wave equation 😊
@@sohanchowdhury1312also thanks to wolf for his bite😢
And this explanation is still complex to us elementary learners but I love the detail and time you put into this
I'm 32 now, haven't ever needed to use this information once in my life outside of high school but being given this crash course about negative square roots makes me remember that I used to LOVE this kind of stuff. It felt like puzzle solving, unlike a lot of other parts of math you learn in hs.
Excuse it's for mathematicians not for normal people
Imaginary numbers are a tool used a lot in physics and esp in engineering to solve real world problems which would otherwise require rigorous calculations, high school sets u up with the basics, college is where u apply them.
Thank you for showing us the true and yet-to-be accepted way to show us how to write absolute value for the negative numbers ❤
I studied this, got a doctorate, forgot all of it 😂
Relatable 😂
me too, at least I thought so... I had a stroke and had to learn everything again. I needed my fingers to add up. But this I remember.....
whats the secret to getting rec letters?
When extending the square root function to the complex plane (which cannot be avoided here), you _have_ to specify the branch that you're using, and also mention whether the extension to the negative reals was made through the upper or the lower half-plane, which would indicate whether sqrt(-1) is "i" or "-i". In fact, in the example from the video, the two roots don't necessarily have to be from the same branch. If you take different branches for each root, you get +4, and -4 otherwise.
P.S. no, the imaginary unit is _not_ defined as sqrt(-1) (because of this whole thing with multivalued functions that would make a lot of mess), rather, "i^2 = -1" is its property resulting from the multiplication rule in the complex plane (or, equivalently, it's defined as one of the roots of x^2 +1 = 0, doesn't matter which one, as long as it's fixed throughout the whole theory).
P.S. #2 Don't you find it strange that sqrt(-2)×sqrt(-8) cannot, according to the author, be written as sqrt((-2)×(-8)), but sqrt(-2) = sqrt((-1)×2) = sqrt(-1)×sqrt(2) no problem? -1 seems like a privileged individual that you can take in and out of the root, but not any other number 😅 (again, it all comes down to the specific branch and where you insert the branch cut)
Wooooooooooooo That's a hell of an effort you've put in there.....i appreciate it.👍
You do not have to, you could also just embrace the square root as multivalued as god intended :3
I came to the comments to make the same complaints, then saw you’d already done a great job!
Lets be real.... i=ln(-1)/pi
;)
So applying the "privileged individual" status of -1 that I learned from the video, I can prove that sqrt(16)=-4.
sqrt(16)=sqrt(-(-16))=i*sqrt(-16)=i*i*sqrt(16)=i^2*4=-4
This is the fun stuff to show middle schoolers who think math can't be quirky and fun.
√16 is +4 ... Solving for n where n^2 = 16 is +-√16 so +- 4.. it is true to say when not zero a number has two square roots. but the √ symbol means principal square root ie positive so the video is correct. √-1 is i So √-2 is (√2)i so √2*√8*i^2 = √16 *-1 =-4.
Thank you~
if the video is correct, then 4=-4 is also true:
4 = √16 = √[(-2) x (-8)] = √(-2) x √(-8) = -4
The real result is both 4 and -4.
@@gonzalobarragan8076 This is incorrect because the property √(ab) = √(a) × √(b) can only be applied when both a and b are non-negative.
√[(-2) × (-8)] ≠ √(-2) × √(-8)
@@gonzalobarragan8076 the property '√a•√b = √(a•b)' you used is only applicable for if a and b ≥ 0. Since in the given question, a and b are negative which means a and b < 0. Therefore the property cannot be applied in this case
You should have paid attention in these small things also.
I haven't learnt really anything about iota, but if it equals -1, shouldn't i^2 therefore equal +1? -1 x -1 = 1. What am I missing?
I changed the sqrt(-8) to sqrt(4) and sqrt(-2) then sqrt(-2) and sqrt(-2) just multiply to become -2 and then that’s multiplied by sqrt(4) which is 2. Final answer is -4
True, I totally forgot that sqrt(-8) = sqrt(4). Very good
@@fashnekwdym √(-8) = √4?
That chalk hitting the board sounds so good
When -1is taken outside the square root it becomes positive because -1 * -1 =+1
@@detac1405bro wtf lmao?
Have you ever studied complex numbers? i is sqrt(-1) not -1
omg your chalk skills are the best asmr
More needs to be said on this matter. One should say something about "multi-value" functions of root and real and complex principal root. Without precisely determining the definition on what the "sqare root" is (since there are more different functions called the same name) the discussion on this topic can easily open the door of philosophy.
Okay, so you have to take out the i before multiplying. Thanks for the free lesson.
don't memorize this, it's wrong
I like these shorts of all these math problems I haven’t seen for decades, so I have forgotten a some of them, these are good reminders.
I'm a 10 th student
But still l watch your videos
Everything have new content which l haven't learnt. So l will learn it. THANKYOU ❤❤❤❤❤
I have complex numbers coming up next year, not yet but this dude probably just talk me how to do them too :o
Basically, take out the negative out before doing the "wrong method" and put it back at the end
it doesnt make sense
a shortcut reminder i would say
@@dekirou320 it’s factorizing, but with the negative symbol only
asian johnny depp
I thought I was done with complex analysis and here it is bitting me in the ass
thanks for the information, i was really used to the first method 😅
Glad to share the knowledge
it makes sense hear me out:
if some imaginary person gives you four apples
you somehow owe 4 apples to someone
😂😂😂😂😂 love this
If you put 4 apple debts in a square and take away 1 line of apple debts, the whole thing will somehow convert into imaginary apples
If you think of the square roots as exponents (1/2), i think the path to the answer is a bit more obvious.
Wow, I haven't done this stuff since high school. Always nice to refresh every now and again! Thanks guy!
You're most welcome. Thanks for the nice comment.
Complex numbers..Haha my favorite 😊😊
This is good, brother❤
Thanks
I wish I had a math teacher like you when I went to school.
Very good explanation. Your voice clarifies it better.
Guy's a good teacher. As I'm getting these right now due to his lessons.
sqrt(a) as defined in high-school maths is ill-defined when 'a' is allowed to be negative.
In high school we define sqrt(a) as the nonnegative root of the polynomial x^2 - a. E.g. x^2 - 4 → x ∈ {2, -2} so we define sqrt(4) = 2.
However the roots of x^2 + 4 are {2i, -2i}. Neither of them are positive; they fall outside the real number line!
Ok, then define sqrt(-a) = sqrt(a)i (for positive real 'a'). You can do this, but the property sqrt(x)sqrt(y) = sqrt(xy) no longer holds! This is due to the fact that unlike the positive real numbers, the upper imaginary line is not closed under multiplication!
Very well explained!
You missed the negative solution of the first method
√16 is only +4. Not ±4.
@@blankspace178 no it's not, sqrt(x) is a function from R+ to R+, hence the result is only positive.
@@blankspace178 I study Maths at uni, maybe this will clarify it for you:
x^2=4 => x=±2
But √4=2
@Blank Space no of course it doesn't change anything.. but it's just a definition to avoid confusion. You can do your own maths with your own definitions, but I'm gonna stick with the ones I learned :)
@Blank Space "Although the question mentioned at the beginning has two solutions with different signs for even-numbered root exponents and positive radicands, the notation with the root sign always stands for the positive solution." This is from wikipedia, but I do understand your thought process, and it really doesn't matter, run with whatever you like. This will be my last reply, have a good one.
I worked so much without imaginary numbers that every time i see something negative inside of a square root my brain goes hey that's illegal!!
After years of doubts, I finally cleared it. Thank you so much sir❤
Great 👍
square root of 16 is +4 or -4. So it doesn't matter if the i^2 exists or not.
no? theres no unknown integer in the equation
I agree, square root of 16 is +/-4, so it's not correct
@@enchantedhamburger8934 what’s 4*4, ok good, now what’s -4*-4, u see it now don’t you
@@masteroogway2853 ye i know what you are reffering to, i may have just missunderstood this, but i never learnt that a square root is equal to +- of said number
But what if it's another square root.
I did, first write -8 as -2 * 4, then you can take out 4 thus giving me 2*root(-2)*root(-2). So ans comes out to be 2 * -2 = -4.
One of the rules in mathematics says that you can multiple root of a negative number. You have to first change it into a imaginary number by taking out the negative.
@@mrhtutoring doesn’t the square root of 16 have 2 answers? 4 and -4? So the answer here is -1 * those numbers meaning the answer remains 4 or -4?
@@overdose8329 Square root of 16 is only +16. When you have an equation x²=16, x=±4.
@@overdose8329 the square root function is defined in such a way that it only gives positive outputs. A common misconception is that it gives two answers, but that‘s simply not the case. After all, there is a reason we maticilously write out +-sqrt in quadratics and Not just +sqrt.
Oh yeah, that really cleared things up.
bro writes smoother than how I write on the board
thank you sir i am from India
Thank you!
So elegantly simple yet so easy to get it wrong. Dealing with negative square roots can be tricky.
I love you channel man. It just popped up and it is really helping me out!❤
me who tried to do it before him and managed to get it wrong.
You‘d need to consider that you can have +-i as a prefactor for both of the sqrts!
i = sqrt(-1)
sqrt(-2) = sqrt(-1)×sqrt(2)
Not
sqrt(-2) = +/-sqrt(-1)×sqrt(2)
@@moonchock4390 but y=sqrt(x^2) has two solutions. y=+x and y=-x. For sqrt(-1), this yields +i AND -i.
@@YouHaveToBeTheChange, nope. The square root is a single valued function, defined as the principle branch of the inverse of the square. You've clearly learned that you need to remember the +- when solving a second degree polynomial equation, which is correct and important but doesn't apply to square roots.
@@YouHaveToBeTheChange are you serious lmao
@@TheGlassgubben Sorry to burst your bubble, but the square root of -1 has indeed two values: +/- i.
You should consider in this case that sqrt(x²) is + or - x, both for the fist case and the second, so the answer is + or - 4, and that's because with negatives taking the square root means solving the equation x²=whatever number you want to take the square root of, and that is + or - x
Source: my father, university professor of maths (or at least that's how he explained to me, I'm reading many other explanations here which might be things he didn't tell me)
PS sorry for formatting, I'm writing from mobile
Isn't that because if you have for example √9=3 as a result from √x²=y, then x is either +3 or -3 since sqaring either results in 9, and y is thus ±x as a result of two values for x. It doesn't mean that y is negative.
√x² = |x| according to what I was taught and I just checked that wolfram alpha affirms that
@Retired Bore go ahead and simplify the quadratic equation then, I think you would find it contains an unnecessary ±
@Retired Bore as a software engineer, I don't even know what you mean by "mathematics for programming".
The square root function is 'defined' to yield the positive solution
@Retired Bore there are multiple square root implementations for computers and so there's no "the square root function". And no, I'm not mixing them up. Feel free to look up the definition of square root, or functions in general and see they don't have multiple answers.
@Retired Bore here you go
"Every nonnegative real number x has a unique nonnegative square root, called the principal square root, which is denoted by √x"
I pulled it from wikipedia "square root" so you can go and argue with the sources there instead of here.
I still have a grudge against imaginary numbers
Math fuels me with energy holy i feel like i have power to do any equation now 😂
Thank you for all the great comments.
This is a good refresher.
first method is wrong because the square root of a negative number is not defined in the field of real numbers, instead it is well defined in the field of complex numbers, I think that this should be said …
because in definition, the condition says that number inside root must be positive.
That's true in basic algebra.
However, algebra 2 and on, students are taught imaginary numbers.
√-1=i
And √-9=3i.
댓글 감사합니다.
Kamsabnida
the moment i squared came into play, this just got more complicated
WOW! Great explanation. Thanks for sharing.
My pleasure!
He did not define that we are dealing with complex numbers. Therefore I was free to define that we are dealing in real numbers and then both solutions are wrong. Tataaa!
so true
are you dumb? radical of a negative is a complex number
In the world of real numbers, sqrt(x)*sqrt(y) = sqrt(x*y) it's only valid if x and y are positive real numbers or at least x or y are 0. So, sqrt(-2)*sqrt(-8) = sqrt(-2*-8) is not valid, and so, having no solution at all.
In the complex world, well... you do what was done in the video.
Doesn't dealing with the square root of negative numbers by definition mean we're dealing with complex numbers?
@@spectreone In the world of real numbers sqrt(16) is both 4 and -4. Positive real numbers have 2 square roots that are also real numbers.
"Mommy can we buy real math"
"No, we have math at home"
The math at home:
This is actual maths wdym
@@HatterTobias it's a joke I know
I can't believe I got got by this. It's like being hit by a 'deez nuts' you hadn't prepared for.
Very nice excellent🙏🙏🙏
Thank you! Cheers!
He forgets to mention that this is (of course) the complex square root function, which "happens" to be "identical" to the normal real square root function on the positive part of the real axis & in the point x = 0 ( + 0i ).
Yes, you are right - in complex analysis it's allowed to work with socalled multivalued (!) functions - like a group of different plausible branches that satisfy the given conditions.
So in this case he will need to look at all combinations of +/- i sqrt(2) & +/- i sqrt(8) ( where sqrt refers to the real square root function), which produces these 4 results, which turns out to be just two:
1) i sqrt(2) x i sqrt(8) =
-1 x sqrt(16) = -4
2 i sqrt(2) x ( -i (sqrt(8) ) =
+1 x sqrt(16) = 4
3 -i sqrt(2) x i sqrt(8) =
+1 x sqrt(16) = 4
4 -i sqrt(2) x ( -i sqrt(8)) =
-1 x sqrt(16) = -4
So +4 is actually a solution (contrary to his claim! ), and so is -4.
That symbol means you only hold the positive not the negative. So -4 is the only answer
@@isjosh8064 No, both + 4 & - 4 are actually "solutions" to this expression - you need to check all the
4 ( = 2 x 2 ) possible combinations of the two (primary) branches of the complex square root function.
@@isjosh8064 Which symbol? 😉
@@Bjowolf2 When you say “solutions" I imagine you’re thinking of an equation like: x^2 = 16 where the solutions are +/- but those are what x can be to satisfy the equation. But /x = -5 has no solutions even though -5 squared is 25. The definition of domain of /x only allows a positive input and returns a positive input.
@@isjosh8064 Yes, for ordinary (real) square root this is true, but for COMPLEX square roots, which he is clearly working with here ( with -4 & -9 under the square root signs ), different rules come into play, since they have two possible (primary) branches. ( These socalled multi valued functions are permitted in complex analysis ).
So sqrtc(-4) =
+/- i x sqrt(4) =
+/- 2i
And likewise for sqrtc(-9).
Well... What about the negative solution to the sqrt of 16?
Yeah, so the answer is technically ±4. Or maybe we just missed something
They are radicals which have only positive solution
That's waht i understand atleast
@@georgesas7090 no the square root is a function meaning every input has one output. You are thinking of solutions to x^2 = 16. Which is different than sqrt(16) as sqrt(16) and -sqrt(16) Are both solutions to that equation
@@ChessThingsOfficial It's not, the square root funcion is defined assigning only positive values.
@@eldins1813 Oh okay thanks!
30 years old, and every day I wake up I still thank the gods I do not have to deal with this anymore
I forgot all about this rule dealing with 2 negatives in a square root. Thanks for the reminder.😅
“How do I get a credit card and build a great credit score”
School:
Simple. No money? No purchase. Avoid credit at all costs in your whole life!
@@BlackHoleSpain Yikes, I can’t think of anyone, literally no one who’s successful that doesn’t have an amazing credit score. Paying cash is fine if you want to stay poor.
Everyone makes jokes like this, but in practice, there are second order differential equations that appear in the study of finance and economics. Their solutions sometimes require finding the roots of a polynomial which are often complex numbers.
@@cdula26 Credit score literally doesn't matter once you're making consistent reliable income. If you have cash, people have goods and services they want to offer you.
@@acex222 Even my real estate mentor who brings in a net income of $800,000 a month has a great credit score and doesn't pay cash for his 500k cars or 5 million dollar yacht. It's better to finance things and use all the cash you would have spent on buying more assets and investing the rest.
Under that logic...
4=✓16=✓i^2(-16)=i✓-16=i^2✓16=-4
You can't avoid the fact that square root has 2 values. Sure, you can talk about the principal branch but it's unavoidable with complex numbers.
What??? The whole point is that you can NOT use the square root factoring rule unless the factors are positive real numbers. His method of evaluating square roots of negative numbers is perfectly valid.
@@Tulanir1 I did exactly as the video. For him ✓-2=i✓2 and that uses the same rule I'm using. If you can't truly use that rule unless it's a positive integer, then he can't even answer the problem.
@@kevinmartincossiolozano8245 surely when you go from sqrt(i²*-16) to i*sqrt(-16) youre skipping an intermediate stage where you get sqrt(i²)*sqrt(-16) which means youve factored with a negative number
@@PotassiumLover33 That's exactly the same step done in the video. Because ✓-2=✓2✓-1, which means, they have factored with a negative number too!
look into imaginary numbers
Right! You start with negative you have to end with negative to complete the sentence.
√-2.√-8 = ±4 (+4,-4)
no , square root can only give positive values
@@dftsxy5 but -4×-4 is 16. So -4 is on of the answers. The other answer is obviously 4
@@soroushhaidary7934 square root is a function which means you have only one output (positive 4 in this case) you would be right if it was an equation like x²= 16 , only then x=4 v x=-4
@@soroushhaidary7934 You can also see how there are contradictions when you say sqrt(16) = +-4
Let’s assume sqrt(16) = +-4.
Then because 4 = sqrt(16) and sqrt(16) = -4, our conclusion is that 4 = -4, which is false.
TLDW: Square roots are annoying in the complex world, sicne they're even more ambiguous...
i is nothing but root of -1 and i is an imaginary number
never needed this info in the last 55 years of my life
Root of 16 is +4 & -4
Yes but the √ symbol means positive root. Eg quadratic formula always include +-√ not just √
By the graph of the root it always greater than zero
√16 is only +4. Not ±4.
Maturity is when you know the answer is +/-4
It’s not lol. That’s only when you’ve introduced a square root yourself.
More like education system failure, √n is always positive lol.
@@omnipresentcatgod245 he probably meant a square power
@@Dra3oonBoth are correct but only because we included complex numbers in the beginning. If you only have sqrt(16) then is 4, but if you have
Sqrt(-16) × sqrt(-1), then its both -4 and 4.
@@goldbeni thank you
you're helping me a lot
Love this! Dont understand a single word but fo relly doe. LOVE THIS!
I always thought the sqrt of a number is either a positive or negative number...so Sqrt of 9 is 3 or -3....
Normally, it depends on how you _define_ the √□ operation. And as long as you explain the full chain of reasoning and it starts with a reasonable definition, I'd say the reasoning is valid.
Of course, if you have a test-like question where you have to produce the same text as what's written on the sheet labeled "correct answers"... Then it's a game of either "memorize this particular textbook and not another textbook" or "guess the answer". The answer here can be any of "4", "-4", "±4", "expression is ill-defined".
Square root is a function, meaning it can only return one result, the positive one (for real numbers)
You need to take both values when you have x²=4 x=±2 but it's because
x²=4 x²-4=0
(x+2)(x-2)=0 x=±2
@@gabrieleymat6332 A "√□" is a symbol. Nothing internally inconsistent would happen if you were to define it as "all numbers that produce □ when multiplied by themselves". For example, in a residue field modulo 7, 3²=4²=2. In this case, trying to find a "principal" value of √2 would be very much futile.
And even if you insist to take a positive value for real numbers (why? because a particular textbook said so? how you would arrive to this conclusion if you were inventing all math from scratch?), you still have the same problem with complex numbers: i and -i don't differ in any reasonable way. In some cases, it may be important to take the value in "the same direction" for both roots: you can write √-1x√-1 = ixi *or* √-1x√-1 = (-i)x(-i) but not √-1x√-1 = ix(-i). Like when solving cubic equations, you get an expression ³√a+³√b and then you have to sum the "right" pairs of cubic roots of a and b.
What about the set of reals?
Can you give an example of "set of reals?" I need an example to answer your question
-4 is itself a real number. so it is in the set of reals.
@@sytherplayz but the √-4 isn't in the set of real numbers
@@lazygod1854 √(-4) is not in the set of solutions.
@@sytherplayz yes but I think "solutions" isn't the right word in this case because we aren't solving a polynomial equation but nonetheless we both are saying the same thing that the √-4 doesnt belong in the sets of real number system.
ياربي شهل جمال استاذ شهل شرح ❤❤❤❤
I love this lesson it's one of my favorites I have it in final exams next week
This one was tricky
Thank you for sharing
Thank you
Square root always has two solutions: positive and negative; unless we calculate square root of 0, which has only one solution. Also, multiplying numbers under the same exponent is allowed.
Extracting i is allowed, but unnecessary. Both solutions should yield the same results:
Square roots of 16 are 4 and -4.
By using the other method we arrive at the same results: square roots of 16 multiplied by -1 gives -4 and 4, which is the same thing.
It's specifically √x not x^2=y , the answer in second one is x=±√y
Where √y is the principal solution x^2=y ,
This is done because √x is defined to be one-one function, for each x you put you can get only one answer, the negative ans is discarded, this principle soln is put in x=±√y.
When we ask for all the square roots of 16, it's ±4.
But when we write it with a square root symbol such as √16, it's only the principal square root, which is +4.
Hence, the reason when we simplify expressions such as √4+√9, we simplify it to +5, not ±5 or ±1.
-4: Baby im not here Im a hallucination
Nice sir🎉🎉
Never argue with an Asian teacher.
Maybe he forgot his mathematics because he stayed in murica for so long. 🤣🤣🤣
√16 = + -4
@@BlackSakura33 maybe you forger that square root of as positive no. is positive
For eg. √4=2 and not -2
But if you put as
x²=4. Then x has values as 2 and -2
And please stop hating and stereotyping other countries, coz we are also being hated and stereotyped at international level
Spread peace
@@kartavyasharma7266thats only true when x is a real number. If you inclide the complex plain, both negative and positive solutions are real solutions. And because we included the complex plain at the start, we cant ignore it.
Look at the start of the problem,
sqrt(-2)=i×sqrt(2) AND -i×sqrt(2)
Same for the 8, and then we can see that theres actually 2 solutions, + and - 4
The real answer is undefined .
idk who u r or how u got on my reccomended but god bless u for save my ass in math today
I am very happy to hear! Seriously.
I need a heavy drink after that episode of ………
I would argue that both solutions are right and wrong at the same time. Since √16 = ±4 and -√16 = ±4, the solution should be ±4 in both cases. Choosing one arbitrarily is just that, an arbitrary choice. Basically, the rules for arithmetic with roots hold. You can't just disregard the second solution and say that these rules don't hold for some complex numbers.
sqrt(16) = +4
Not +-
The solution of x² = 16 is x = +-sqrt(16)
@@ComposedBySam your comment make not sense.
@@IoDavide1 square root by definition refers to the positive root of x²=16.
square root as an operator itself doesn't give you both the roots. Because if that was the case then we couldn't raise both sides of an equation to fractional powers.
Suppose sqrt(1) =+-1
1=1, square rooting both sides (raising powers in both side to 1/2)
We would get 1=-1 as a solution.
Hence by convention 16 to the power 1/2 (ie. Sqrt(16)) gives only 4 (the absolute value)
@@ComposedBySam this second comment make less sense then the first.
The square root has not determined sign, so you have always two results: + and -
You seems still confused with primary school definitions
@@IoDavide1 then do a favor. Try harder to understand what I wrote. And if you cannot google it. the solution of x² = 16 and square root of 16 are different things
Reminder that sqrt of 16 is 4 or -4
Genuinely love this
VERY GOOD, FANTASTIC.❤❤❤❤❤
Actually both 4 and -4 are solutions to the expression. Both ways are correct.
no
Incorrect,
√(-2) × √(-8) ≠ √(-2 × -8) because the expression √(ab) = √(a) × (√(b) only applies when both a and b are non-negative.
Since both a and b are negative in this situation, we have to use complex numbers to solve the expression.
√(-a) × √(-b) = i√(a) × i√(b) = i^2 × √(a × b)
Since i^2 = -1, and √ only returns the positive root, the answer to √(-a) × √(-b) only yields negative solutions.
The end result is only -4, +4 is not a solution.
I concur that you take "i' out of the square roots first, but root(16) = +/-4....multiply +/-4 by -1 and you still get +/-4 (which is also found by the 1st method, but that 1st method is still the wrong method). I think this simple problem is just a bad example of why the indicated method is the correct procedure.
but √(x²) = |x|
4÷√-2 = -2√-2 = -√-8 incorrect
-4÷√-2 = 2√-2 = √-8 correct
So, the answer is -4.
Right...
As we are dealing with complex numbers...
And not the real numbers....
Thanks...
Won't the sqrt(16) be +/-4 and hence the final answer be -/+4?
His conclusion was basically that the square root of 16 is both 4 and -4
Square root of 16 is only 4 which is why this question is tricky
@@Adventurer-te8fl no, sqrt(16) is + or - 4 actually
@@let1742 what ??? Sqrt is always positive, sqrt(16) is equal to 4 only
@@let1742 Completely incorrect a square root cant be negative. Its always positive. So it is 4. Thats why imaginary numbers get used.
@@BlackCat-fx9kb you are right that sq root should be positive but here you see ✓16 = ✓(+4)^2 and it can also be ✓16= ✓(-4)^2 therefore they both are positive only. So sq root of 16 will give +4 or -4.
Consider: sqrt(16) = +/- 4
Just accept the square root as naturally and inherently multivalued instead of arbitrarily defining it as positive. Gets at the heart of the matter and resolves all discrepancies without any overcomplicated notions of branch cuts etc.
Now both methods are equally correct:
sqrt(-8)*sqrt(-2) = sqrt(16) = +/- 4
sqrt(-8)*sqrt(-2) = i*i*sqrt(16) = -1*+/-4 = +/-4
Nobody denies the square roots of 16 are 4 and -4, but √ does NOT yield the square roots - it yields the POSITIVE square root. If you want both, you need to write this
x² = 16 -- > x = +/ - √16
because
√16 = 4
There is no discussion on that - it's simply using the defintion of the function/symbol √
If you use a different definition, you're solving a different problem. If you solve a different problem, you get a different answer.
And you pretend to be a "mathematic fanatic", what a joke !!! And the Earth is flat, that's it ??? Lol
@@extrams0 forgive me for discussing it :P bad definition is bad
I've always liked mathematics. Even when i may not fully understand certain equations. It's literally the language of life. 😏👌🏾
Yes. AI is going to have the whole math system held in perfect order.
No more rounding off numbers.
you get the same answer either way if you just dont forget the +- sqrt rather than just +sqrt
this is actually the exact reason you think you're getting 2 different answers
sqrt of 16 +/- 4, not only +4, so in both ways the final answer should be the same
Not correct. The square root of 16 is 4. -4 is not the square root of 16. -4 is one of the solutions for the quadratic equation x^2=16. Square root of 16 is a number and a number cannot assume multiple values while the equation will have two solutions
-1(√16)=-1(±4)=±4
You only take the positive root when solving squares
Wow that's new information, thank you🙏
I want to be a Mathematician or a statistician someday😊
He explained it well
I wish someone has told me this when I was studying 🙂
But it coole also be +4. -1*sqrt16 has two solutions, since sqrt 16 could either be 4 or -4
Actually he quite messes with exponentiation properties. sqrt(-2)·sqrt(-8)=sqrt((-2)·(-8)) .
I wouldn't trust my kids to this teacher
@@ovidiucroitoru2290that property only works for square roots of positive numbers
Also for the OP : no, √16 = x and x^2 = 16 is not the same thing, x for the 1st is 4, while x for the 2nd can either be -4 or 4. The guy in video talks about the 1st
@@ovidiucroitoru2290 you do realise sqrt of negative numbers are undefined in real numbers which is why complex numbers exist right
@@ovidiucroitoru2290You would get the square root of -16 with that result which is undefined in real numbers. Hence why one needs to express it as a complex number like he does in the video