For what it's worth, I strongly agree with the value of understanding multivalued functions. Unfortunately, many mathematicians dismiss multivalued functions as "relations" because they insist that functions can only return one value, rather than a set. To accomodate those, the convention has developed that the radical sign √ denotes the principal square root, normally taken to be the square root whose complex argument lies in the semiclosed interval ( -π/2, +π/2 ], which corresponds within real numbers to taking the positive square root. It remains true that all numbers (except 0) have two square roots, each the negative of the other, but the radical sign only denotes one of them. One result of that acomodation is that √w * √z ≠ √wz unless arg(w) + arg(z) < π.
Excellent explanation. All that needs to be added is that the value of the principal root √ depends on your convention for the principal argument, Arg(z). Though this is usually taken as Arg(z) in ( -π,π ], which has the advantage that Re(√z) is non-negative, it is possible and often useful (when doing complex integration, for example) to use the convention Arg(z) in [0, 2π).
No (competent) mathematicians say functions have to return a single value (in the sense of a single number) rather than a set. Maybe some math teachers do, because they believe they shouldn't confuse kids, since they study functions from R to R (meaning they get as an input ONE real number and they output ONE real number). But anyone who has studied math in a university has seen functions from all sorts of different domains returning all sorts of things, including sets. Now, there is a precise sense in which functions return only one value. Functions are defined such that if a=b, then f(a)=f(b). This tells you that a function can't give you a different output with the same input. However this doesn't say the output can't be a set. It just means that f(1) can't capriciously decide to be 1 on some occasions, -1 on some other occasions. However, nothing stops it from being the set {-1,1}. I'm sure you know all that, it's just a bit odd that you said many mathematicians dismiss such functions, which is not the case. What mathematician has never encountered a function that returns something other than a single number?
I use a joke as an analogy: - White chocolate isn't actual chocolate; American cheese isn't actual cheese; multivalued functions aren't actual functions. Of course this is a matter of how you define those words. In the case of the multivalued functions, one may consider the codomain as subsets containing all the possible answers, thus making it return a single object. Or maybe redefine the classical notion of function or even define another concept, a multimap.
@@jimskea224 Thank you for the kind words. I feel that defining the argument principal square root as being on the interval ( -π/2, +π/2 ] should imply that we considering the number itself as being on the interval ( -π, +π ] but I think you're quite right to explicitly note that. I think it is much more common to use the alternative convention that the principal value of a complex number has its argument in the interval [ 0, 2π ) (making the branch cut below the positive real axis) when we're dealing with higher order roots. I believe the reason is the ( -π, +π ] convention (making the branch cut below the negative real axis) is useful for square roots is that it if it is used when the number is real and positive, it returns the same result as we might expect when we take the principal square root of a real number. It doesn't seem possible to get the same fortuitous result with higher order roots (the differing real and principal complex cube roots of -1, for example). So it's probably easier when dealing with nth-roots of a complex number to start with a number whose argument lies in [ 0, 2π ) giving a principal nth-root with its argument in the interval [ 0, 2π/n ).
this is also related to one of my favorite concepts in math, monodromy. since complex nth roots are defined in terms of the logarithm, they inherit some strange behavior around 0 and geometrically, if you start at a point like z=1 you can choose a taylor series expansion for sqrt(z), which is well-defined for points near z=1. if you wanted to define sqrt(-1), you can then extend the domain by taking taylor series expansions for nearby points with the restriction that they agree with the original series and repeat with a slightly larger domain until you get to z=-1. everything is fine until you realize that extending the domain gives you different series at z=-1 depending on whether you extended in a clockwise or counterclockwise direction around 0. its not so bad in the case of sqrt because it is only a difference of sign (and differs by an nth root of unity for nth roots) and we can describe the multivaluedness in terms of its monodromy group (which is jus isomorphic to Z/2Z for sqrt) but can get pretty crazy with othera functions, like my favorite, the generalized hypergeometric functions which all have singularities at 0, 1 and infinity
Roots do not have to be defined in terms of the complex logarithm. In fact, typically the logarithm is defined in terms of the exponential function, which in turn is defined as a power series, which requires that natural number exponents are already defined. So for instance, we need x³ = x·x·x first, and similarly for xⁿ for every other natural number n. So the simplest way to define roots is also the original way: the inverse of the power. The function mapping (real or complex) x to xⁿ has some inverse, and each element in the inverse image of y = xⁿ is called an nth root of y. I think this perspective makes a lot more sense, and it makes zʷ = exp(w log z) a theorem for rational w rather than a definition. (Equality here means that the sets of values for each side of the equality are identical.)
@ sure roots of perfect nth powers can be defined without the complex logarithm but my point is that extending the domain to actually make sense for negative real values usually requires defining it with the logarithm so that the usual array or complex analytic tools can be applied
@@Emma-jx1cd I guess it's the same either way. Whichever way you define it, you need to prove the relevant theorem. For instance, if I define ⁿ√z = exp(1/n log z), I still have to prove that [ⁿ√z]ⁿ = z (for every branch). Otherwise, it's not much of a "root" is it? Similarly, I can define the real logarithm as the integral of 1/t from 1 to x, but if I do, I still need to prove that exp(log x) = x for all real x. So it doesn't really matter which one I call the definition.
@@EebstertheGreat the definition in terms of the logarithm does matter! the point of my comment is that applying tools of analytic continuation to extend the domain requires working with complex differentiable functions to begin with. the concept of monodromy is not unique to nth roots and the underlying cause is the singularity at z=0, so i chose to use a definition for nth roots where it is easier to see the singularity!
Fellow hypergeometric enjoyer I see xD are you sure about the singularities of the generalised hypergeometric functions? I seem to recall that pFq is entire for p
As with many “viral” problems, the question is ill-posed: Is _√a_ to designate the “principal value”, or is it asking which _x_ fulfill _x²_ = _a_ ? Mixing both meanings just feeds viral memes. In the equation sense, there are two solutions, written as _x_ = _±√a_ but even then _the symbol_ “√” means “principal value”. So here, if the question is about “√” as principal value, then √-4 √-9 = 2 _i_ × 3 _i_ = -6 is the only correct answer. But if the question is the implied equation √-4 √-9 = _x_ then that eq. nominally has four answers, which degenerate to 2 unique one, just as shown.
Isn’t it a language or terminology problem? I would say “multivalued functions” aren’t functions, and there are two inverse functions for z^2. You need to specify which is indicated.
I was going to say the same thing about the "definition" of √a: Is it only the principal value (which I was taught)? Or is it equivalent to a^(1/2)? Not well defined.
The problem here is that you don't have a way of specifying which value of i is the positive one. There is no prinicipal value of the square root of (-1), because one can use a left-handed or right-handed complex plane.
If you follow the "new math" formalism the square root symbol denotes the "principal value" which is exclusively positive, so by that reasoning, √-4 just doesn't exist.
It's sad that in such a long video Michael Penn doesn't make time to define what he means by √z. In my view he should have explained that there are two (valid) conventions on the meaning of √z, the multi-valued version and the principal value version, and stated that he is going the multi-valued route. That would be more honest than pretending the multi-valued definition of √z is the only game in town.
If sqrt is considered a function (not multivalued), eg: sqrt(9)=3 and not -3 (we take the principal value, the positive value), then the sqrt(xy)=sqrt(x)sqrt(y) property doesn't hold when you go outside of positive numbers. Thus, sqrt(-4)sqrt(-9) can only be calculated separately. sqrt(-4)=2i; sqrt(-9)=3i and thus, the product is -6.
But the properties of square roots still apply, e.g. √(a)√(b) = √(ab). Thus, √(-4)√(-9) = √[(-4)(-9)] = √36 = 6. This is what Dr. Penn showed, that either answer is equally correct, because it's a problem with a multivalued answer. This becomes obvious when you put it in terms of a complex number represented as an exponential, re^(iθ), which always equals re^[i(θ+2nπ)] for all n∈ℤ.
The rule √a√b = √(ab) always has restrictions, no matter how you define the square root function. I believe Professor Penn is wrong to use the rule and ignore the restrictions. If a and b are real then the restriction is that they can't be both negative. But even if a and b are arbitrary complex numbers then the rule is the sum of arguments must be in the principal branch, usually either [0, 2π) or (-π, π]. If the sum is not in the principal branch then √(ab) will have the wrong sign. For complex numbers on the negative real axis, the argument is π. So if a=-4 and b=-9, then arg(a) + arg(b) = 2π and that's not in the principal branch. Even if you define the square root function on a Riemann manifold you still have the same problem. There, when you multiply two values you have to take care when the values are on different sheets, and if they are on the same sheet, you have to check that √(ab) is still on the same sheet. There is simply no way you can have √a√b = √(ab) hold without restriction.
If you want √a to satisfy (√a)^2 = a for positive and negative values of a, then clearly √a√b = √(ab) cannot hold for a = b = -1 because it would mean -1 = 1.
You certainly do not have to check if ab is on the same sheet if a and b are on the same sheet. As long as they are on the same sheet you will always get the same answer. It doesn't even matter on which sheet they are. If they are on different sheets you are going to get something different, and indeed the convention is to take them both on the same sheet. To see why it's the same, let a and b be on the same sheet. Then adding 2π to their phases takes them both to the other sheet. The product of them both on the new sheet has a phase 4π relative to their product when both are on the same sheet. Half that and you get 2π, so the square root has the same phase as what you started with. If they were on different sheets (which is somewhat unconventional but you can do it) then you would get the other answer, the one with minus. While it is conventional to take them on the same sheet when nothing else is specified, it's not really necessary.
In my experience, √ means principal square root, whereas an exponent of ½ would be multi-valued. We take √4 to be 2, not -2, and take √-1 as being i not -i (in fact, due to the symmetry of the complex numbers, we could even take this choice of principal value of √-1 as being the *definition* of i). With that understanding, √xy = √x√y should only be assumed to hold if x or y is a nonnegative real number. The principal value of √-4 would be 2i,, and the principal value of √-9 would be 3i, so the correct answer would be -6, not 6. √-4 x √-6 is simply not equal to √(-4 x -6). If you allow that, you also get nonsense like 1 = √1 = √-1 √-1 = i² = -1.
I'm not sure you can define i as the principal value of √-1, since the concept of principal value requires that the complex numbers already be set up as background - and that obviously includes i itself.
4:20 I honestly was blown away when you labelled the axes in the complex plane as R and iR ... I had always thought (= been taught to think) about them als Re and Im, but R and iR makes *so much more sense*! Multiplication by i is a 90° rotation ... of course this should also apply to the axes! Thank you, I really gained insight here.
The last case near the end shows why we don't need to look at any values for m and n other than 0 and 1. Adding or subtracting any integer value to m or n just adds or subtracts that many copies of iπ to the exponent of e, and we just bounce around between the positive and negative values we get from just looking at 0 and 1.
In fact, you don't need to look at any value of m other than 0, since we are interested solely in the value of e^(nπ + mπ) which takes on its two possible values { 1, -1 } when n ∈ { 0, 1 }, and any integer value of m simply repeats those in exactly the same way, making m redundant.
@@BrianDominy You should both reconsider because the good professor got this one wrong. I admire and respect Professor Penn greatly, and he is a more accomplished mathematician than I am, but we all make mistakes. The "multi-valued function" debate is a red herring. The rule √a√b = √(ab) always has restrictions, no matter how you define the square root function. If a and b are real then the restriction is that they can't be both negative. But even if a and b are arbitrary complex numbers then the rule is the sum of arguments must be in the principal branch, usually either [0, 2π) or (-π, π]. If the sum of the arguments is not in the principal branch then √(ab) will have the wrong sign. For complex numbers on the negative real axis, the argument is π. So if a=-4 and b=-9, then arg(a) + arg(b) = 2π and that's not in the principal branch. Even if you define the square root function on a Riemann manifold you still have the same problem. Recall the reason we have a multi-valued square root function in the first place is because e.g. both (-2)² and 2² are 4. The square root function is multi-valued but the square function is definitely *not*. So if the square root function splits 4 into -2 and +2 then the square function *merges* them back into one. That merger is the whole reason we needed multi-valued functions in the first place. When you say √4 is ±2 you are saying in one branch it is 2 and in another it is -2. But 2² and (-2)² are both single valued and they're both 4 (sorry for repeating myself, but that's why we needed the multi-valued square root in the first place.) So √-4√-4 DOES NOT equal ±4. The square root function is multi-valued but the square function isn't. (√4)² is just 4, and (√-4)² = (2i)² is just -4 NOT ±4. For the same reason √-4√-9 DOES NOT equal ±6, it is just -6.
No matter how you define the square root function, whether it be multi-valued, or defined on a Riemann manifold, the rule √a√b = √(ab) will always have restrictions. If a and b are real then the restriction is that they can't be both negative. But even if a and b are arbitrary complex numbers then the rule is the sum of arguments must be in the principal branch, usually either [0, 2π) or (-π, π]. If the sum is not in the principal branch then √(ab) will have the wrong sign. For complex numbers on the negative real axis, the argument is π. So if a=-4 and b=-9, then arg(a) + arg(b) = 2π and that's not in the principal branch.
When given the problem 'What is the value of _/-1 x _/-1?', I proceeded to answer it by writing _/-1 x _/-1 = _/((-1) x (-1)) = _/((-1)^2) = _/1 = 1, only to be told afterwards that you can't simply apply the laws of multiplying roots for real numbers to imaginary numbers. Ive always found this to be a non-answer, and had I been taught that complex numbers are fundamentally periodic, as expressed elegantly by the Argand plane and the polar form, I might not have gotten bogged down trying to find 'THE right answer' to this problem. Thank you for your clarification, Professor Penn.
In my opinion, the √ notation should be avoided entirely once you're dealing with numbers outside of the non-negative reals. Like 99.999 % of mathematicians agree that √9 should only return 3. Otherwise in, say, the quadratic formula there would be no need to write +-√..., since the fact that there is possibly more than 1 solution would already be baked into the √ notation. So even though there are two different numbers (3 and -3) that when squared yield 9 it's common agreement that the √ notation only stands for one of those. I'd argue that it's not reasonable to all of a sudden lift that "restriction" when extending the domain to the complex plane since the whole "we-pick-the-non-negative-solution business" is somewhat arbitrary in the first place. My proposed solution would be to only write √a when a is real and non-negative and keep the convention of meaning the principal root and introduce similar but new notation for all other a whose precise meaning is to be determined. Or we could agree that √a always only stands for the solution to x^2 = a with the smallest non-negative argument (angle w.r.t. the real axis) or something similar which is of course quite arbitrary once again, but which could be easily extended to nth roots.
I'm not sure where you get the percentage on mathematicians. Sounds like pulled from ass. Mathematicians TREAT roots as multivalued, always, without exception. Engineers? Of course. Sqrt(9)=3.
Alternative perspective: instead of considering multivalued functions, define the function on its domain, i.e. select a branch. Then you realize that sqrt(x)*sqrt(y) is not in general equal to sqrt(x*y) for complex values of x and y For instance: domain complex plain but the negative imaginary axis Sqrt : z= r exp(i t) -> sqrt(r) exp( i t/2), where t lives in (-Pi/2, 3Pi/2) with this choice. Then sqrt(-9)= sqrt(9)* exp(i Pi/2) And sqrt(-4)=sqrt(4)* exp(i Pi/2) Then their product is 6 * exp(i Pi) = -6 On the other hand Sqrt((-4)*(-9))= Sqrt(36)=6. So you conclude sqrt(x)*sqrt(y) not equal to sqrt(x*y)
In addition, suppose the different evaluation of the function Sqrt(i) = 1 * exp(i Pi/4) Sqrt(-1+i)= sqrt(sqrt(2)) * exp(i 3Pi/4 * 1/2) Hence their product is 2^(1/4)*exp(i 5 Pi/8) in the meanwhile sqrt((-1+i)* i) =sqrt(-1-i)=2^(1/4)*exp(i 5Pi/4*1/2)= 2^(1/4)*exp(i 5Pi/8) So in this example sqrt(x*y)=sqrt(x)*sqrt(y) seems to be valid, but this is the case just because the sum of their phases is not exceeding the branching cut Pi/2 + 3Pi/4 < 3Pi/2
Just to be boring as a last example one could consider the function sqrt' on the complex plain but the positive real axis so that Sqrt' : z= r * exp(i t) -> sqrt(r) * exp(i t/2) for t in (0, 2Pi) Now It is clear that sqrt(-4)= 2 exp(i Pi/2) and sqrt(-9) = 3 exp(i Pi/2) Therefore their product is -6. On the other hand (-4)*(-9) is real and positive, hence our function sqrt' is not even defined on their product
By convention (to make it a function) we say that the square root of 36 is 6. But we also know that the -6 X -6 is also 36. So the square root of 36 is also -6, except we reject this answer as the square root of 6. Why exactly? Why do we allow multiple answers for the square root of negative numbers but not positive numbers? Do we do this in 'regular' math to not 'accidentally' get into imaginary numbers when we do not want to?
There is an argument to be made that √(1) being 1 has a stronger motivation than √(-1) being i: There's two candidate solutions for √(1), those being 1 and -1. Those two numbers are easy to tell apart, in particular, 1 is idempotent under multiplication (1×1=1), but -1 is not idempotent under multiplication ((-1)×(-1)≠-1). If we then look at √(-1), we once again find two candidate solutions, let's call them i1 and i2. It's easy to see that i1=-i2 and i2=-i1. But beyond that, we will find that the two numbers can't be told apart without relying on some mechanism that canonizes one of the two values as the "true" solution (such a mechanism would be e.g. extracting the imaginary part, or extracting the complex argument) - but that canonization is exactly what we're trying to motivate in the first place! So it turns out that the two numbers don't exhibit any differences between their respective properties, meaning there's an ambiguity here that we can only resolve arbitrarily, rather than being able to rely on such different properties of the two candidates.
We don't "reject this answer". The √ sign is not a problem that needs to be solved by giving an answer. It is simply a symbol that has a commonly accepted precise definition at least on the non-negative real numbers. √36 = 6. Why do we not have a special symbol for the other root? Because it is easy enough to add a minus sign: -√36 = -6.
there is a convention to only consider the positive root even though we understand that y = x^2 has two solutions. i was taught to always add a +/- after evaluating a square root.
sqrt(a)*sqrt(b) = sqrt(ab) is a rule for positive a and b. It also happens to hold if only one of a or b are negative. But it does not hold if a and b are negative. sqrt(z) is, by convention, not defined to be multivalued under "standard" rules. You can of course define it that way in some other context, but said context would need to be explicitly stated.
Thanks. You have validated my answers. There are videos saying that you can't have two negative numbers under the sqrt, like sqrt of (-a * -b). I was very confused by that. The videos I have seen are forgetting phase shifts to the numbers. Thanks for clearing my mind. Now, I can sleep better knowing that I was correct all along.
The thing is, you CAN'T have two negative numbers under the radical. The rule √a√b = √(ab) always has restrictions, no matter how you define the square root function. The phase shift of a number in this context is just it's argument, and there are restrictions on the argument of √(ab). If a and b are real then the restriction is that they can't be both negative. But if a and b are arbitrary complex numbers then the rule is the sum of arguments must be in the principal branch, usually either [0, 2π) or (-π, π]. If the sum is not in the principal branch then √(ab) will have the wrong sign. For complex numbers on the negative real axis, the argument is π. So if a=-4 and b=-9, then arg(a) + arg(b) = 2π and that's not in the principal branch. Even if you define the square root function on a Riemann manifold you still have the same problem. There, when you multiply two values you have to take care when the values are on different sheets, and if they are on the same sheet, you have to check that √(ab) is still on the same sheet. There is simply no way you can have √a√b = √(ab) hold without restriction.
Multivalued functions also occur in real analysis with more than two variables. I constructed one which is analytic from R^2 to R^2, but holomorphic only in the unit circle.
3:57 I very briefly thought "wait, which Euler's formula" before I realized that there's only 1 that could be relevant here. As the old joke goes, most math theorems are named after the second person who discovered them, because they can't ALL be Euler's theorem.
The main reason should be that rules that always hold on the real numbers can't always be generalized over the complex numbers. (a^p)^q = a^(p•q) is another one you can't just generalise, or you'll find any non-zero complex z is actually a positive real number. Put it in polar form, z = r e^(a i), with r = |z| and 0
You don't really need to bring the problem to the complex domain! Just observe that sqrt(4) can be both 2 or -2 - since 2*2=4 and (-2)*(-2)=4. This is sufficient to show that square root is multi-valued.
im just a college math level but i feel that the convention of' taking positive root' of -4 , cannot be simply combined under the typical root a x root b = root ab (trivially) i think this is a failure of the convention without first being resolved to i (root -1) x root 4 and -6 should be the correct resolution in view of the 'taking positive root' convention if not, a full complex analysis of the arguments
I'm a big fan of Mr Penn and his channel but I cannot agree with this. If we use the notation sqrt(z) we must make a convention which root is meant. Otherwise we destroy the transitivity of equality which is part of the base of mathematics. If we write sqrt(-4)*sqrt(-9)=6 and sqrt(4)*sqrt(9)=-6 it follows 6=-6. The whole Zermelo-Fraenckel-system doesn't work if we abolish the transitivity of equality. One possible convention is to take the half of the argument between -pi exclusively and pi inclusively. So the notation becomes clear. When making such a convention we must consider that the equality sqrt(xy)=sqrt(x)*sqrt(y) does not always hold. So the paradox disappears and we get -6 as the only correct answer instead of 6. Another way to avoid the paradox is not to use the notation sqrt(a). Instead of saying that we calculate sqrt(a) we could say that we solve x²=a. My English is not perfect. So I hope I was able to make clear my point of view. Greatings from Germany!
The way I was taught (in Germany) is that you can never take a square root (or even a cubic root, etc.) of any number that isn't a nonnegative real number. So the notation i = √ -1 that seems to be common in the English speaking world would indeed be invalid. The definition of i that we were taught was i² = -1.
I am of the conviction that the square root symbol is undefined on numbers other than non-zero reals. There is only one place I have come across where using the square root symbol on general complex numbers is useful, and that's the quadratic formula. And in that case you only need to mix it with extremely limited arithmetic, so you avoid any shenaniganigans. And if you're more of a purist than me, you can see it as not really being a square root but rather a mnemonic for "insert each of the two square roots in here, and calculate the result for each one". Other than that you will not need the square root symbol √ on general complex numbers, and your life will be easier if you simply excise it from your life. _Especially_ if you're teaching or taking an introductory class.
Everything he said after 2:30 might be true but I just explained it by saying the square root of 36 is ± 6 not 6. And the square root of - 1 is ± i, not i. So, either way you end up with the same answer, ± 6.
No, x^2 = 36 has two solutions +/-6, but √36 is simply notation for the positive root 6. The other root can be written down by prefixing a minus sign: -√36 = -6.
I raised this up in a Reddit thread about half a year ago, and got ripped to shreds in the comments. The basis of the problem being a function by definition can only have one output, and f(x)=sqrt(x) is only positive by convention. Otherwise, you can do this: 1 + sqrt(1) = 0.
I think you didn't really address the philosophical question of why we should allow square root to be a multi-valued function among the complex numbers, but not among the reals? Presumably there are problems in complex analysis if this is not allowed, but then what of the reals? Do problems occur in real analysis if we were to allow the square root of 4 to be ±2, instead of insisting by convention on the positive +2 value only? (And presumably it would be acceptable to say the square root of 4 + 0i is ±2 ?)
to only define square root of positive real numbers the positive root is genuinely just a convention. It's a convention that makes some formulas work. It's done that way because it makes some school math easier, lets people just compute stuff without a need of thinking (for example any discussion about which root to take is moot, since all of this is predefined). It's usually not a problem, but it occurs! The easiest thing is the formula for roots of a quadratic polynomial. The +- is an artifact of our definition. If the roots are complex, then suddenly the +- doesn't play any role, since square root is multi-valued. The way out of this mess is to admit that square root was multi-valued to begin with, but we bastardized it for pedagogical reasons.
@@ib9rt I take it you were guessing, and you did guess correctly. Complex root does not spare any number, for any complex input you get two outputs. In complex analysis there are possible trajectories that take you from one branch of the root to another, so it is important to keep track! When you ensure your trajectory does not change branch, you safely choose one to work with. In real analysis, this problem is nonexistent, so algebraic root is rational to use.
@@Czeckie √ is just notation. For positive x, √x is the positive root. For negative x, you could define √x = i(√-x). In this case √-4√-9 = -6. If instead you define √x = { i(√-x), i(√x) } for negative x, then that is fine with me. However, I will note that √-4√-9 now has no meaning, unless you come up with a definition of multiplying two sets of two numbers.
I thought it was the case that there were two answers to just sqrt(36). You can multiple 6*6 or (-6)*(-6) and get 36, so they're both roots. In that case it doesn't matter if you take the 'i's out and get -1 first, that's then -1 * (+/- 6) which is still +/- 6
I guess what should be observed is that in the first case they implicitly assume real numbers and a definition of sqrt where only the positive number x with x^2=y is considered to be the solution. But if we just consider x^2 = y it could be reasonably (and without much mathematical rigor) argued that sqrt(36) can be -6 and 6 and both should be solutions. In the second example complex numbers are assumed... what then changes a lot of things, as you have shown.
x^2 = 36 is indeed an equation with two solutions. But √36 is not an equation. It is just a notation for the positive solution, i.e. √36 = 6. If you want to consider the notation √-4 and √-9 and wonder what is √-4√-9, then you have to decide what the notation √-x means for positive x. If you decide that it means xi, then √-4 = 2i, √-9 = 3i, and therefore √-4√-9 = 2i * 3i = -6. If you decide that √-x = {-xi, xi}, then √-4√-9 is meaningless, until you define what it means to multiple two sets of numbers.
The property sqrt(a) * sqrt(b) = sqrt(a * b) is not valid when a or b is negative. It only holds true for real, non-negative numbers. In the field of complex numbers you might have sqrt(-1) * sqrt(-1), which equals i * i, or -1. But sqrt(-1 * -1) = sqrt(1) = 1. Hence, sqrt(-1) * sqrt(-1) is not equal to sqrt(-1 * -1).
@navybrandt I see. Well I'm sure you are aware of Euler identity, another foundation. Right? So, let's see where it leads. e^iπ = -1 => raise to 2 => e^(2iπ) = (-1)^2 = 1 = e^0 => 2iπ=0 => i=0. Right?
@@pelasgeuspelasgeus4634 e^(2iπ) = 1 is indeed true, but it doesn't imply that 2iπ = 0. Why? Because e^z is periodic in the complex domain with period 2πi. This means e^(z+2πi) = e^z. So, e^0 = e^(0+2πi) = e^(0+4πi) = e^(0+6πi) = ... = 1, which simply reflects the periodicity of the exponential function in the imaginary direction. It does not mean 2πi = 0, because exponentials are periodic, not injective, in the complex plane.
This is nothing more than a muddying of the water. The quantity I. is a very special thing And everything we do with it Must go back to the Definition that I times I Equals -1. Proper handling of this special quantity gives us only 1 answer To the question posed.
Actually, when it comes to so-called indeterminate forms, all the common operations are multi-valued functions. When their outputs are an indeterminate form, their real value is the entirety of R or C, depending on what your codomain is.
Is it just me? I prefer to hold onto “=“ to be an “equivalence relation”, meaning it has the property, (amongst others), that a=b and a=c implies b=c. With “multi-valued functions”, this is lost which seems a pretty big sacrifice.
Also, allowing both branches for the square root “function” would mean we have wasted so much ink over so many years writing: x=(-b ”+/-“ sqrt(b^2-4ac))/(2a).
This is ignoring the fact that the radical symbol and raising something to a fractional power *are not the same thing* and do not mean the same thing mathematically. By standard convention, using the radical symbol _only returns the principal root._ It is *not* a multi-valued function. That is why, even in cases which don't involve complex numbers, if it matters, "plus or minus" must be added to the radical to indicate both (real) roots are desired. Because of this, the statement "√{x} ⋅ √(y) = √(x⋅y)" *is only true if "x" and "y" are not both negative.* (This is also a fairly well-established caveat for this rule among most mathematicians, I believe). Therefore, you cannot use that rule in this case, and the answer "6" for that problem is actually objectively wrong. (And most of your explanations are done wrong, because you are writing/saying, for example, "√z" when what you actually _mean_ is "z^(1/2)" instead (which are _not mathematically the same thing_ ).)
I've seen enough popular youtubers explaining this example incorrectly. It is a relief I'm on the same page with math professor, yet I'm baffled by the amount of confusion caused by the incompetence of many "teachers"
I agree. I have found another video explaining it correctly before this video. You might want to check out channel " Science'n'me" titled "Why is the square root of Negative one equal to i". This video explains what people are missing in learning numbers. Every number has a phase shift comes with it. That's what missing in our early learning of math. Check the video out. I think you will like it. It has taken me over 60 years to find out about number system we have been learning is half truth.
I like your explanation about multivalued functions, but typicall the √ symbol is defined as giving the principal square root. So under that standard assumption there actually is a single answer for this problem, namely -6. In particular √x √y = √xy just isn't true unless x,y >= 0.
I'm not a mathematician rather an engineer. When I was in high school in the 80s, this caveat was never taught. When UA-camrs like bprp went over this, I was struggling to understand, why the caveat? After watching this video, my Spidey sense was right. I just didn't have a word for it - the square root is multi-valued for complex numbers.
Second comment - this is not a trivial thing - Michael's topic in video I mean. Consider drawing a circle. Is it 360° 720° or .... I claim it is any line drawn with a fixed radius of 360° or more around a fixed central point will describe a circle. Mathematicians are a bit lazy at times and when they say something like a circle has 360° they mean many important things Maybe a geodesic circle will be one that has infinite collection of points all equidistant from a fixed central point and covers 360° But that might mean one single infinitesimally small point is counted twice. so in order to be well-behaved the starting point of circle is closed (or open) and ending point is open (or closed). They cannot be both closed and they cannot both be open (why?) So if you draw a circle and run your pencil back and forward a couple of times just so that start and end points "meet" nicely. don't worry! It will more than likely be a perfectly drawn circle even if the compass spanned more than 360° and squiggled back and forward a bit to make the circle look better. Ideal shapes and ideals are well , ... sort of idealistic ideals. And in math that is also important 🙂
x^2 = 1 x1 = 1, since 1*1=1 x2 = -1, since -1*-1=1 Why over-complicate it, with complex numbers, Euler formulas, polar form, periodic trigonometry, etc.?! Because we can! Because the answer is not so straightforward with the cube root. Because we must to gain a deeper understanding what's happening.
does sqrt(x)*sqrt(y) = sqrt(x*y) hold for all x,y in the complex world? i was told that this can lead to a contradiction i.e., 1 = sqrt(1) = sqrt((-1)*(-1)) = sqrt(-1) * sqrt(-1) = i^2 = -1 (at least this is what i was told in an undergrad complex variables course, maybe there's something wrong with the argument)
If you allow the sqrt 'function' to be multivalued, the apparent paradox goes away. {1,-1} = sqrt(1) = sqrt(-1 x -1) = sqrt(-1) x sqrt(-1) = {i,-i} x {i,-i} = {ixi,ix-i,-ixi,-ix-i} = {-1,1,1,-1} = {-1,1}. You have to take into account all possible combinations of course.
@@ronald3836 Maybe there should be! My point is simply that the need to arbitrarily pick a single square root of -1 as the reference of √-1 is what leads to the trouble. -1 = exp((2n+1)pi.i) for all integer n, so √-1 = exp((2n+1)pi.i/2) = exp(n.pi.i + pi.i/2) = exp(n.pi.i) x exp(pi.i/2) = ±1 x i = ±i. Certainly by convention √-1 is just i, but that is because we have no way of distinguishing i from -i before labelling one (which one? we have no means of knowing or even saying...) 'i'. It is so to speak a random dubbing of a square root of -1 which makes i i and not -i.
@@russellsharpe288 To extend √a from being defined for positive a to also being defined for negative a, you indeed need to pick one of the two solutions to x^2 = -1, let's call them i and -i. But once you have made this choice, you will see that √-4√-9 = -6. Either √-4√-6 = 2i * 3i = -6. Or √-4√-6 = -2i * -3i = -6. Nothing useful is gained by letting √ be notation that allows (√-4)^2 = 4.
@@pelasgeuspelasgeus4634 In a universe where we define √-a = i√a for a positive, where i is an element in the complex numbers satisfying i^2 = -1. (Or where i=X in the ring R[X]/(X^2+1). Indeed, X^2 = -1 module X^2+1.)
@ronald3836 Those tricks are for idiots. It doesn't have any actual meaning because imaginary numbers aren't countable and therefore useless. Another fancy arbitrary invention.
There's something wrong with the square root function. Why do they accept only a positive solution? (2)^2 = 4. (-2)^2 = 4. Then SQRT(4) = +/- 2. The real plan is a limitation of our thinking, a mere projection of the complex plan.
I se the question and I go... we cant use classic rules, bcs we are dealing with complex numbers we need more robust method... turns out its not that we cant, but we need to know when and why its OK and if we took care of all possibilities. Bcs in the end both were valid solutions but neither were the whole solution.
I love how he complexified (!) the problem just to torture the viewer interested in the problem. I would have said, simply, every square root has two values: +/-. sqrt(36)=+6 and -6. sqrt(-4) has two values: +2i and -2i sqrt(-9) has two values: +3i and -3i Then we get to the end of his video. You can multiply those together in four ways: 2i x 3i = -6 -2i x -3i = -6 2i x -3i = +6 -2i x 3i = +6 That's all. sqrt(36) = +/-6. It's wrong to think sqrt(36) is ONLY +6.
@@pelasgeuspelasgeus4634 LOL, is it countable? No, but pi is not countable. sqrt(2) is not countable. 0.10110111011110... is not countable. What's that have to do with it? Who said square roots have to be countable?
@greg55666 Irrationals aren't countable in the sense they have infinite digits after comma. Imaginary numbers aren't countable because they aren't representing anything real. Do you get the difference? Lol?
@@pelasgeuspelasgeus4634 It would probably be better if you explained it to me. Irrational numbers are not countable, according to you. Imaginary numbers are also not countable, according to you. You started by saying "countable" was the issue, now it turns out that according to you there are different kinds of "not countable," and your rule has nothing to do with "countable" at all, you mean something completely different. Two messages, two totally different rules. Maybe you should explain what you REALLY mean? (Also, the original problem is about sqrt(-4): not a real number. So this whole conversation is completely irrelevant to the question at hand, since the question itself is clearly not limited to "countable" of any kind.)
@greg55666 I see. Well, have you ever bought sqrt(-4) apples? And don't say that similarly you can't buy π apples, because there's an easy answer to it.
Isn't this just pq formula that any square root can gives 2 answers. -6*-6=36 as much as 6*6, without graph you just don't think of it that way. Going via i just gives you the proof that it actually is that way without having to draw it out
Which only goes to show how many problems could be easily solved if they'd just change that damn rule about multiplyiing negative numbers by negative numbers. But no, that's top easy. Instead you create "imaginary" numbers, keeping hundreds, if not thousands of internet mathematicians employed deciphering equations that include the square roots of negative numbers. My own opinion is that once one sees a square root of a negative number in any equation, one knows the answer will be +/- some number. Forget about the +/- for the time being, find the solution, and add on the +/- at the end. In the real world, most solutions will be either negative or positive, not both, so you can use whichever works.
Thanks for this. I get tired of being shot down by the high school and Jr. college kids when I suggest that sqrt() is a multi valued function. Where did this "principle root" BS come from anyway? It was never a part of my educational trip.
@@flowingafterglow629 Well, I learned what he showed here. If you ever do electrical engineering or physics the multi-valued aspect of teh square root must be taken into account or bad things could happen.
@@uni-byte Yes, that is why you use +/- sqrt(value) If as I said sqrt(x^2) = abs(x) then you use +/- sqrt(x) to get both values. But in both cases, sqrt(x) is a positive value.
There are several problems with this video, some of them have already been pointed out, as for example the distinction between the principal root and fractional powers. What I find more terrifying about this video is that the argument laid out here can be used to explain that sqrt(4) * sqrt(9) is 6 and -6. What Michael and others also seem to not to notice is that this "viral problem" is a typical problem in College Algebra (a first algebra course at the college level in the United States) which just got outside the book. In a college algebra course the square root of negative numbers is defined, so sqrt(-4) = 2i and sqrt(-9) = 3i under that definition (in general, if $a < 0$ sqrt(a) = i*sqrt(|a|)). Therefore sqrt(-4)*sqrt(-9) = -6. The reason why this is given as a problem in College Algebra is to make students realize that sqrt(x)*sqrt(y) is not always sqrt(xy). The bottom line: this is not a problem in complex analysis, this is a problem in College Algebra with a very restrictive definition of square root.
Agreed. Most commenters (probably high school students?) seem to believe that √a is an "equation" that needs to be "solved". Of course that is not true. √a is NOTATION for the positive solution to the equation x^2 = a (if a is positive). The other, negative solution can be written simply as -√a.
At 1:01 I was taught to say sqrt(n)=|m| In this case I would have said sqrt(36)=|6| and then explained that |6| can be +/-6. The explanation with the famous Euler’s formula is very very beautiful. Thanks
When you mint a lovely new square root of a negative number, you can use bombelli-i or anti-bombelli-i. Both are equally valid. bombelli-i x bombelli-i = -1 anti-bombelli-i x anti-bombelli-i =-1 But!!!! anti-bombelli-i x bombelli-i = 1 So let Sqrt((-4)x(-9) = 2bombelli-i x 3anti-bombelli-i = +6 Sort of the same as the professor, but weirder and wronger
Am I crazy, or is √36=±6, not SOLELY 6? We see that kind of interaction routinely with multiple equations: ±6 satisfies the first equation, but ONLY -6 satisfies the other, so -6 is the sole possible answer. Kinda weird that the positive answer is the invalid one though; usually it's more like "A triangle cannot be 3cm X 4cm X -5cm, so only positive answers are valid." Oh, well, No Myth is still a pretty good song even if I can recall no others now. ;)
√36 is not an equation. It is notation. x^2 = 36 is an equation. It has two solutions x = 6 and x = -6. You can write the positive solution as √36. You don't to need to have the same or a separate symbol for the negative solution, because you can just write -√36.
So is √2 + √3 + √5 + √7 ONE value or 16 different values? Is it just one value but √-2 + √-3 + √-5+ √-7 is 16 different values? What it only half of them are negative? Then is it only 4 values? What if I don't know? Does the number of possible values of an expression depend on the values of some of the variables so that √a + √b + √c + √d can have either 1, 2, 4, 8, or 16 values, depending on how many of a, b, c, and d are positive reals? This is insanity.
The answer to all those questions depends on how you define the square root, which really depends on the question you are asking. There are two numbers which square to any particular number, which one do you want to use as your square root in that context?
@definitelynotofficial7350 But we have conventions for this, and this all is why we have those conventions. In any ordinary context where it isn't clearly specified otherwise, the √ symbol means principal square root or n-th root if n is written. Yes, the principal n+th root is defined even for complex numbers. There are two common conventions regarding *which* of the n values are the "principle" root (or which is the principle sheet), but under none of the conventions is √a more than a single value. Even if you're talking about multi-valued functions and reimann surfaces, you still need a way to write √a that is distinct from -√a and isn't ±√a either. I think Prof. Penn made a mistake in his videos. When he choses n=0, m=0 or n=1, m=1 the values are from the same sheet and he gets the right answer, but when he choses n=0, m=1 or n=1, m=0, he's multiplying values from different sheets when requires a phase adjustment (in this case a sign change). Without the adjustment, he gets an erroneous answer. √-9×√-4 = -6 only, not ±6.
@@definitelynotofficial7350 Do you really want to define the "+" sign as it is used in analysis as something that operates on sets of numbers? This is not a useful concept (nor an existing one). (In combinatorics there is probably a use for such a "+" sign.)
It all comes down to what information the person who wrote the problem meant to convey. Was he using the √ to represent the principal root or all roots? I say the words that sounds like “the red book” which also could be written, “the read book”. What does that phrase mean? You could argue that it means the color of the book. Or you could argue that it means the book has been used by being read. You could argue it means both. So which is it. The correct answer is that it means what I meant it to mean when I said it. If I don’t give the context of the statement, then you can’t give a correct answer. Whether you say it is the color, or the use of the book, or both, you are not correct, you are just listing possibilities. With √ for real numbers, convention is that it means the positive root, otherwise there is no way to differentiate between the two roots of x^2=4. (Note that he used this in the video to change √ 36 to 6, not ± 6.) In the complex world, it is ambiguous unless it is put into context as to what the author meant to convey as the meaning. When we use the quadratic formula and end up with a negative discriminate, we pull out an i, not a ±i. So in that context it is just the principal root. So what I have seen is based on the radicand. If the radicand is a non-negative real number, then √ means a non-negative value. If the radicand is complex (a + bi) and b is not zero, then it is multi-valued. If the radicand is a negative real number then it means a postive real number times i, AND √ a • √ b = √ ab is NOT VALID. This identity is only valid with real numbers. If this is a step in the solution of a problem, then that context is going to determine what is meant. Just like if a told you to hand me the “red” book. While you could interpret that both ways, you would know I meant the color based on the context.
Great explanation, but then, why can't real functions be multivalued? The square root function over the reals is a perfect example. Isn't the definition of a function as something that returns a single value unnecessarily constrained? At least it's inconsistent over complex numbers.
No, because that is what "function" means. A "multivalued function" is not a function (and hardly a useful concept, e.g. multiplying two sets of outcomes of a "multivalued function" is just silly).
@@ronald3836 But that begs the question. In fact, that "multivalued" property is exactly what a "function" means in the complex plane. My question is, Why do you restrict multivalued results from functions operating in the real numbers but not in the complex numbers?
I'm sorry, but I don't get one thing: don't you learn as soon as you learn square roots, that the result is +/- the root, because the negative solution is also correct? So the first line is already wrong, because in school we would immediately write sqrt(36)=+/-6, which makes the second line not contradict the first one, since the second line would read -1 * (+/-6), which comes out to the same. No deep dive needed, basic math. Why is this puzzling people? Why is this a "viral" problem?
No, what you learn is that when you solve x^2 = a (let's assume a is positive now), then one solution is x=√a (the positive-valued solution) and the other solution is x=-√a.
Your work through is very clear but I feel like the conclusion could have been made more precise. Saying "both are correct" is a bit misleading. I think it would have been better to state that the flaw in the original argument is that an assumption has been made about the square root function that is false, namely it is not multiplicative in general, particularly on the negative reals. One must first define "a" square root function. There are two possible definitions. Given one definition you get one answer, given the other you get the other answer. However, both statements are certainly not simultaneously true.
Overkill? SqRt (4) =x is solved with x=2 but also -2. With this open approach SqRt (4) . SqRt (9) can take the values 6 and -6. No need for complex numbers.
@@TheEternalVortex42 Negative surface area doesn't mean much either. The true pressure to accept complex numbers was that they were necessary for solving cubic equations via standard formulae.
@ it was affluence of existential dread rather than anger, but nice to see that you take everything the wrong way. Let me know next time you find a comment online that you have nothing good to say, but decide to say something anyway. I would love to hear your storyline on such
the rule you used only works for positive numbers the answer is only -6 "multivalue results" only exist as sulotions for unknown variables but this expression doesn't have such variables and all the functions are one to one therefore there is ONLY one answer
@@RuleofThehyperbolic "square roots are one to one unless explicitly said otherwise" -- Only by mathematical convention. The entire reason why this problem is viral is because the general populace won't unanimously accept the problem as obeying this convention. So you cannot assume this problem obeys this convention.
I just watched him demonstrate that the rule also works for negative numbers as long as you are aware of and take account of the fact that any non zero number has two distinct square roots
I am confident enough to say that BOTH initial series of transitions are wrong. Neither 6 nor -6 are the answer to this question. It's either n/a if we use algebraic square root which is explicitly undefined for negative values, or ±6 if we use complex root of power 2 as √(-1) is not i, it's ±i, and multiplying ±i by independent ±i results in ±1. The only correct root-based definition for i is a solution for x² + 1 = 0, or such number that i² = -1
Why is -6 not a potentially valid answer? If, for a>0, we define √-a to be i√a, then √-4√-9 = 2i * 3i = -6. (And if instead we take √-a = -i√a, we still get -6.) The problem with the video is that it seems to define √-a as a set { -i√a, i√a }, and that it seems to define A * B for sets A and B to be the set { a * b: a in A, b in B }. This is possible (but unconventional and not useful in analysis) and should have been stated.
In some sense a mulitvalued function is the negation of the very notion of function, as the the defintion of a function is something that, applied to a given argument, gives an unique result. Of course you can say that a multivalued complex function on complex numbers is a function the value of which is a SET of complex numbers, but this still has some problems with notation. When you write a square root in any formula, you are merging apples and oranges by mixing notations designing complex numbers and notations designing sets of complex numbers. You can solve this by assuming that in such fomulas single complex numbers are assumed to mean the set containing just this number (a singleton), but then you have to give a definition of what are the meaning of usual operators (+, -, exp, ...) when applied to sets. Th usual convention is to consider that the extension of these operators to sets is that each operator applied to sets gives as result the set of all result obtainable by applying it to any of the members of its arguments. But the with this convention sqrt(z²) = {-z, z} = - {-z, z} = - sqrt(z²). It would be easy (but false) to conclude that sqrt(z²) = - sqrt(z²) implies that sqrt(z²) = 0 for any z. This shows that you can't use all usual notations with multivalued "functions", because these are implicitely functional notations involving univalued functions.
I agree that the problem presented is one where clarity requires being extra careful about notation. I also recognise that makers of YT videos have to eat and pay the rent, but... The thumbnail of this video did *not* say sqrt(-4)sqrt(-9)={6,-6} which would have alerted me to an interesting discussion about what multiplication and sqrt might mean. (And it was an interesting discussion.) What it asserted as "both true" was that the expression equaled 6 and that it equaled -6. Equality is transitive, any arbitrary pair of numbers are not in general equal and vaccination is safe and effective. Claims to the contrary are clickbait.
I don't necessarily agree. In order to understand lower levels of math, it is not necessary to think of functions as possibly being multi-valued. This concept is often introduced in complex analysis classes. Also, viral math problems also spread outside of the US, so if there is a problem, it is not specific to the US. I don't know of any country where complex analysis is a required part of schooling for everyone. In short, I think it's completely reasonable for someone, even someone from a place with a functional mathematics education system, could be unsure about how to solve this problem. Other viral math problems may be simple enough that someone should be able to solve it with basic mathematics education. (I've seen several problems about the order of operations that fit in this category.) However, I would not generalize and say that viral math problems expose dysfunctions in math education. instead, I would say that they show that people care about math and are trying to use their problem-solving skills along with the rules they learned to justify an answer they think is right. Even if they get the answer wrong, the fact that they are engaging in the process and willing to learn the right answer in the end is more important to a "functional" mathematics education system, in my opinion.
you're violating principle roots, which in other videos you advocate for... do you just change the rules randomly as it suits you? that's some impressive rigor you got there, bud.
I can't believe I'm even saying this but I think Michael Penn is wrong. In grade school I was taught that √a * √b ≠ √(ab) when both a and b are negative. But even in the complex domain, √a * √b ≠ √(ab) when arg(a) + arg(b) ∉ (-π, π]. (or if arg(a) + arg(b) ∉ [0, 2π), depending on convention). But under either convention if a=-9 and b=-4 then arg(√a) = arg(√b) = π, so arg(√a) + arg(√b) = 2π which is not an element of ( -π,π ] OR [0, 2π), so either way √-9 * √-4 ≠ √36. In general identities like √a * √b = √(ab) are not valid if the operation will cause the value to "jump" out of the chosen branch cut.
You're not writing the square root of the product about correctly. Put grouping symbols around it as (you use your radical symbol) sqrt(ab). What you are writing is equivalent to [sqrt(a)]*b.
@@bowlineobama if you're talking about phase shifts, then the values have to be from same sheet. If you multiply two values from different sheets, then you have to phase shift the result accordingly.
For what it's worth, I strongly agree with the value of understanding multivalued functions. Unfortunately, many mathematicians dismiss multivalued functions as "relations" because they insist that functions can only return one value, rather than a set. To accomodate those, the convention has developed that the radical sign √ denotes the principal square root, normally taken to be the square root whose complex argument lies in the semiclosed interval ( -π/2, +π/2 ], which corresponds within real numbers to taking the positive square root. It remains true that all numbers (except 0) have two square roots, each the negative of the other, but the radical sign only denotes one of them. One result of that acomodation is that √w * √z ≠ √wz unless arg(w) + arg(z) < π.
Excellent explanation. All that needs to be added is that the value of the principal root √ depends on your convention for the principal argument, Arg(z). Though this is usually taken as Arg(z) in ( -π,π ], which has the advantage that Re(√z) is non-negative, it is possible and often useful (when doing complex integration, for example) to use the convention Arg(z) in [0, 2π).
No (competent) mathematicians say functions have to return a single value (in the sense of a single number) rather than a set. Maybe some math teachers do, because they believe they shouldn't confuse kids, since they study functions from R to R (meaning they get as an input ONE real number and they output ONE real number). But anyone who has studied math in a university has seen functions from all sorts of different domains returning all sorts of things, including sets.
Now, there is a precise sense in which functions return only one value. Functions are defined such that if a=b, then f(a)=f(b). This tells you that a function can't give you a different output with the same input. However this doesn't say the output can't be a set. It just means that f(1) can't capriciously decide to be 1 on some occasions, -1 on some other occasions. However, nothing stops it from being the set {-1,1}.
I'm sure you know all that, it's just a bit odd that you said many mathematicians dismiss such functions, which is not the case. What mathematician has never encountered a function that returns something other than a single number?
This guy is Micheal Penn fan #1
I use a joke as an analogy:
- White chocolate isn't actual chocolate; American cheese isn't actual cheese; multivalued functions aren't actual functions.
Of course this is a matter of how you define those words. In the case of the multivalued functions, one may consider the codomain as subsets containing all the possible answers, thus making it return a single object. Or maybe redefine the classical notion of function or even define another concept, a multimap.
@@jimskea224 Thank you for the kind words. I feel that defining the argument principal square root as being on the interval ( -π/2, +π/2 ] should imply that we considering the number itself as being on the interval ( -π, +π ] but I think you're quite right to explicitly note that.
I think it is much more common to use the alternative convention that the principal value of a complex number has its argument in the interval [ 0, 2π ) (making the branch cut below the positive real axis) when we're dealing with higher order roots.
I believe the reason is the ( -π, +π ] convention (making the branch cut below the negative real axis) is useful for square roots is that it if it is used when the number is real and positive, it returns the same result as we might expect when we take the principal square root of a real number.
It doesn't seem possible to get the same fortuitous result with higher order roots (the differing real and principal complex cube roots of -1, for example). So it's probably easier when dealing with nth-roots of a complex number to start with a number whose argument lies in [ 0, 2π ) giving a principal nth-root with its argument in the interval [ 0, 2π/n ).
this is also related to one of my favorite concepts in math, monodromy. since complex nth roots are defined in terms of the logarithm, they inherit some strange behavior around 0 and geometrically, if you start at a point like z=1 you can choose a taylor series expansion for sqrt(z), which is well-defined for points near z=1. if you wanted to define sqrt(-1), you can then extend the domain by taking taylor series expansions for nearby points with the restriction that they agree with the original series and repeat with a slightly larger domain until you get to z=-1. everything is fine until you realize that extending the domain gives you different series at z=-1 depending on whether you extended in a clockwise or counterclockwise direction around 0. its not so bad in the case of sqrt because it is only a difference of sign (and differs by an nth root of unity for nth roots) and we can describe the multivaluedness in terms of its monodromy group (which is jus isomorphic to Z/2Z for sqrt) but can get pretty crazy with othera functions, like my favorite, the generalized hypergeometric functions which all have singularities at 0, 1 and infinity
Roots do not have to be defined in terms of the complex logarithm. In fact, typically the logarithm is defined in terms of the exponential function, which in turn is defined as a power series, which requires that natural number exponents are already defined. So for instance, we need x³ = x·x·x first, and similarly for xⁿ for every other natural number n. So the simplest way to define roots is also the original way: the inverse of the power. The function mapping (real or complex) x to xⁿ has some inverse, and each element in the inverse image of y = xⁿ is called an nth root of y. I think this perspective makes a lot more sense, and it makes zʷ = exp(w log z) a theorem for rational w rather than a definition. (Equality here means that the sets of values for each side of the equality are identical.)
@ sure roots of perfect nth powers can be defined without the complex logarithm but my point is that extending the domain to actually make sense for negative real values usually requires defining it with the logarithm so that the usual array or complex analytic tools can be applied
@@Emma-jx1cd I guess it's the same either way. Whichever way you define it, you need to prove the relevant theorem. For instance, if I define ⁿ√z = exp(1/n log z), I still have to prove that [ⁿ√z]ⁿ = z (for every branch). Otherwise, it's not much of a "root" is it?
Similarly, I can define the real logarithm as the integral of 1/t from 1 to x, but if I do, I still need to prove that exp(log x) = x for all real x. So it doesn't really matter which one I call the definition.
@@EebstertheGreat the definition in terms of the logarithm does matter! the point of my comment is that applying tools of analytic continuation to extend the domain requires working with complex differentiable functions to begin with. the concept of monodromy is not unique to nth roots and the underlying cause is the singularity at z=0, so i chose to use a definition for nth roots where it is easier to see the singularity!
Fellow hypergeometric enjoyer I see xD are you sure about the singularities of the generalised hypergeometric functions? I seem to recall that pFq is entire for p
As with many “viral” problems, the question is ill-posed: Is _√a_ to designate the “principal value”, or is it asking which _x_ fulfill _x²_ = _a_ ? Mixing both meanings just feeds viral memes.
In the equation sense, there are two solutions, written as _x_ = _±√a_ but even then _the symbol_ “√” means “principal value”.
So here, if the question is about “√” as principal value, then √-4 √-9 = 2 _i_ × 3 _i_ = -6 is the only correct answer.
But if the question is the implied equation √-4 √-9 = _x_ then that eq. nominally has four answers, which degenerate to 2 unique one, just as shown.
Isn’t it a language or terminology problem? I would say “multivalued functions” aren’t functions, and there are two inverse functions for z^2. You need to specify which is indicated.
I was going to say the same thing about the "definition" of √a: Is it only the principal value (which I was taught)? Or is it equivalent to a^(1/2)? Not well defined.
The problem here is that you don't have a way of specifying which value of i is the positive one. There is no prinicipal value of the square root of (-1), because one can use a left-handed or right-handed complex plane.
If you follow the "new math" formalism the square root symbol denotes the "principal value" which is exclusively positive, so by that reasoning, √-4 just doesn't exist.
It's sad that in such a long video Michael Penn doesn't make time to define what he means by √z. In my view he should have explained that there are two (valid) conventions on the meaning of √z, the multi-valued version and the principal value version, and stated that he is going the multi-valued route. That would be more honest than pretending the multi-valued definition of √z is the only game in town.
If sqrt is considered a function (not multivalued), eg: sqrt(9)=3 and not -3 (we take the principal value, the positive value), then the sqrt(xy)=sqrt(x)sqrt(y) property doesn't hold when you go outside of positive numbers. Thus, sqrt(-4)sqrt(-9) can only be calculated separately. sqrt(-4)=2i; sqrt(-9)=3i and thus, the product is -6.
But the properties of square roots still apply, e.g. √(a)√(b) = √(ab). Thus, √(-4)√(-9) = √[(-4)(-9)] = √36 = 6. This is what Dr. Penn showed, that either answer is equally correct, because it's a problem with a multivalued answer. This becomes obvious when you put it in terms of a complex number represented as an exponential, re^(iθ), which always equals re^[i(θ+2nπ)] for all n∈ℤ.
The rule √a√b = √(ab) always has restrictions, no matter how you define the square root function. I believe Professor Penn is wrong to use the rule and ignore the restrictions.
If a and b are real then the restriction is that they can't be both negative. But even if a and b are arbitrary complex numbers then the rule is the sum of arguments must be in the principal branch, usually either [0, 2π) or (-π, π]. If the sum is not in the principal branch then √(ab) will have the wrong sign. For complex numbers on the negative real axis, the argument is π. So if a=-4 and b=-9, then arg(a) + arg(b) = 2π and that's not in the principal branch.
Even if you define the square root function on a Riemann manifold you still have the same problem. There, when you multiply two values you have to take care when the values are on different sheets, and if they are on the same sheet, you have to check that √(ab) is still on the same sheet.
There is simply no way you can have √a√b = √(ab) hold without restriction.
If you want √a to satisfy (√a)^2 = a for positive and negative values of a, then clearly √a√b = √(ab) cannot hold for a = b = -1 because it would mean -1 = 1.
You certainly do not have to check if ab is on the same sheet if a and b are on the same sheet. As long as they are on the same sheet you will always get the same answer. It doesn't even matter on which sheet they are. If they are on different sheets you are going to get something different, and indeed the convention is to take them both on the same sheet.
To see why it's the same, let a and b be on the same sheet. Then adding 2π to their phases takes them both to the other sheet. The product of them both on the new sheet has a phase 4π relative to their product when both are on the same sheet. Half that and you get 2π, so the square root has the same phase as what you started with.
If they were on different sheets (which is somewhat unconventional but you can do it) then you would get the other answer, the one with minus. While it is conventional to take them on the same sheet when nothing else is specified, it's not really necessary.
NEXT: Math professor explains backflip. 👍
Does the Math professor spin around himself during a backflip or does the World spin around him? Both ist true.
@@davidbelgardt3775 One day he's going to flip onto a different sheet of the riemann manifold and disappear. 🫥
In my experience, √ means principal square root, whereas an exponent of ½ would be multi-valued. We take √4 to be 2, not -2, and take √-1 as being i not -i (in fact, due to the symmetry of the complex numbers, we could even take this choice of principal value of √-1 as being the *definition* of i).
With that understanding, √xy = √x√y should only be assumed to hold if x or y is a nonnegative real number. The principal value of √-4 would be 2i,, and the principal value of √-9 would be 3i, so the correct answer would be -6, not 6.
√-4 x √-6 is simply not equal to √(-4 x -6). If you allow that, you also get nonsense like 1 = √1 = √-1 √-1 = i² = -1.
I second this
I third this.
I cannot condone these shenanigans.
I'm not sure you can define i as the principal value of √-1, since the concept of principal value requires that the complex numbers already be set up as background - and that obviously includes i itself.
I wanted to write exactly the same comment, but somebody did it for me.
4:20 I honestly was blown away when you labelled the axes in the complex plane as R and iR ... I had always thought (= been taught to think) about them als Re and Im, but R and iR makes *so much more sense*! Multiplication by i is a 90° rotation ... of course this should also apply to the axes! Thank you, I really gained insight here.
i didn’t even notice…thanks for this comment!
The last case near the end shows why we don't need to look at any values for m and n other than 0 and 1. Adding or subtracting any integer value to m or n just adds or subtracts that many copies of iπ to the exponent of e, and we just bounce around between the positive and negative values we get from just looking at 0 and 1.
In fact, you don't need to look at any value of m other than 0, since we are interested solely in the value of e^(nπ + mπ) which takes on its two possible values { 1, -1 } when n ∈ { 0, 1 }, and any integer value of m simply repeats those in exactly the same way, making m redundant.
@RexxSchneider Indeed. That's why +6 and -6 showed up twice in his analysis of cases.
Great work! This video has been treated with the care and caution it needs!
There are so many junk videos on this topic. I've had to block so many channels for it.
@@BrianDominy You should both reconsider because the good professor got this one wrong. I admire and respect Professor Penn greatly, and he is a more accomplished mathematician than I am, but we all make mistakes.
The "multi-valued function" debate is a red herring. The rule √a√b = √(ab) always has restrictions, no matter how you define the square root function.
If a and b are real then the restriction is that they can't be both negative. But even if a and b are arbitrary complex numbers then the rule is the sum of arguments must be in the principal branch, usually either [0, 2π) or (-π, π]. If the sum of the arguments is not in the principal branch then √(ab) will have the wrong sign. For complex numbers on the negative real axis, the argument is π. So if a=-4 and b=-9, then arg(a) + arg(b) = 2π and that's not in the principal branch.
Even if you define the square root function on a Riemann manifold you still have the same problem.
Recall the reason we have a multi-valued square root function in the first place is because e.g. both (-2)² and 2² are 4. The square root function is multi-valued but the square function is definitely *not*. So if the square root function splits 4 into -2 and +2 then the square function *merges* them back into one. That merger is the whole reason we needed multi-valued functions in the first place.
When you say √4 is ±2 you are saying in one branch it is 2 and in another it is -2. But 2² and (-2)² are both single valued and they're both 4 (sorry for repeating myself, but that's why we needed the multi-valued square root in the first place.)
So √-4√-4 DOES NOT equal ±4. The square root function is multi-valued but the square function isn't. (√4)² is just 4, and (√-4)² = (2i)² is just -4 NOT ±4.
For the same reason √-4√-9 DOES NOT equal ±6, it is just -6.
So basically the convention of only considering the positive root when applying a square root function only applies to positive real numbers.
No matter how you define the square root function, whether it be multi-valued, or defined on a Riemann manifold, the rule √a√b = √(ab) will always have restrictions.
If a and b are real then the restriction is that they can't be both negative. But even if a and b are arbitrary complex numbers then the rule is the sum of arguments must be in the principal branch, usually either [0, 2π) or (-π, π]. If the sum is not in the principal branch then √(ab) will have the wrong sign. For complex numbers on the negative real axis, the argument is π. So if a=-4 and b=-9, then arg(a) + arg(b) = 2π and that's not in the principal branch.
Because only positive real numbers have square roots that are positive real numbers :)
Depends on where exactly the branch cut is put
When given the problem 'What is the value of _/-1 x _/-1?', I proceeded to answer it by writing _/-1 x _/-1 = _/((-1) x (-1)) = _/((-1)^2) = _/1 = 1, only to be told afterwards that you can't simply apply the laws of multiplying roots for real numbers to imaginary numbers. Ive always found this to be a non-answer, and had I been taught that complex numbers are fundamentally periodic, as expressed elegantly by the Argand plane and the polar form, I might not have gotten bogged down trying to find 'THE right answer' to this problem. Thank you for your clarification, Professor Penn.
Would it make sense if the square of the/a square root of x is not x?
2:26 love the pun!
In my opinion, the √ notation should be avoided entirely once you're dealing with numbers outside of the non-negative reals. Like 99.999 % of mathematicians agree that √9 should only return 3. Otherwise in, say, the quadratic formula there would be no need to write +-√..., since the fact that there is possibly more than 1 solution would already be baked into the √ notation. So even though there are two different numbers (3 and -3) that when squared yield 9 it's common agreement that the √ notation only stands for one of those. I'd argue that it's not reasonable to all of a sudden lift that "restriction" when extending the domain to the complex plane since the whole "we-pick-the-non-negative-solution business" is somewhat arbitrary in the first place.
My proposed solution would be to only write √a when a is real and non-negative and keep the convention of meaning the principal root and introduce similar but new notation for all other a whose precise meaning is to be determined. Or we could agree that √a always only stands for the solution to x^2 = a with the smallest non-negative argument (angle w.r.t. the real axis) or something similar which is of course quite arbitrary once again, but which could be easily extended to nth roots.
I'm not sure where you get the percentage on mathematicians. Sounds like pulled from ass.
Mathematicians TREAT roots as multivalued, always, without exception.
Engineers? Of course. Sqrt(9)=3.
I cannot stop watching until it is a good place to stop. I am stuck.
Because this isn't thf place to stop with these ideas it's just start
Alternative perspective: instead of considering multivalued functions, define the function on its domain, i.e. select a branch. Then you realize that sqrt(x)*sqrt(y) is not in general equal to sqrt(x*y) for complex values of x and y
For instance: domain complex plain but the negative imaginary axis
Sqrt : z= r exp(i t) -> sqrt(r) exp( i t/2), where t lives in (-Pi/2, 3Pi/2) with this choice.
Then sqrt(-9)= sqrt(9)* exp(i Pi/2)
And sqrt(-4)=sqrt(4)* exp(i Pi/2)
Then their product is 6 * exp(i Pi) = -6
On the other hand Sqrt((-4)*(-9))= Sqrt(36)=6.
So you conclude sqrt(x)*sqrt(y) not equal to sqrt(x*y)
In addition, suppose the different evaluation of the function
Sqrt(i) = 1 * exp(i Pi/4)
Sqrt(-1+i)= sqrt(sqrt(2)) * exp(i 3Pi/4 * 1/2)
Hence their product is 2^(1/4)*exp(i 5 Pi/8) in the meanwhile sqrt((-1+i)* i) =sqrt(-1-i)=2^(1/4)*exp(i 5Pi/4*1/2)= 2^(1/4)*exp(i 5Pi/8)
So in this example sqrt(x*y)=sqrt(x)*sqrt(y) seems to be valid, but this is the case just because the sum of their phases is not exceeding the branching cut Pi/2 + 3Pi/4 < 3Pi/2
Just to be boring as a last example one could consider the function sqrt' on the complex plain but the positive real axis so that
Sqrt' : z= r * exp(i t) -> sqrt(r) * exp(i t/2) for t in (0, 2Pi)
Now It is clear that sqrt(-4)= 2 exp(i Pi/2) and sqrt(-9) = 3 exp(i Pi/2)
Therefore their product is -6.
On the other hand (-4)*(-9) is real and positive, hence our function sqrt' is not even defined on their product
Or observe that we really want the square of √-1 to be -1. So √-1√-1 = -1 != 1 = √(-1 * -1)
By convention (to make it a function) we say that the square root of 36 is 6. But we also know that the -6 X -6 is also 36. So the square root of 36 is also -6, except we reject this answer as the square root of 6. Why exactly? Why do we allow multiple answers for the square root of negative numbers but not positive numbers? Do we do this in 'regular' math to not 'accidentally' get into imaginary numbers when we do not want to?
There is an argument to be made that √(1) being 1 has a stronger motivation than √(-1) being i: There's two candidate solutions for √(1), those being 1 and -1. Those two numbers are easy to tell apart, in particular, 1 is idempotent under multiplication (1×1=1), but -1 is not idempotent under multiplication ((-1)×(-1)≠-1).
If we then look at √(-1), we once again find two candidate solutions, let's call them i1 and i2. It's easy to see that i1=-i2 and i2=-i1. But beyond that, we will find that the two numbers can't be told apart without relying on some mechanism that canonizes one of the two values as the "true" solution (such a mechanism would be e.g. extracting the imaginary part, or extracting the complex argument) - but that canonization is exactly what we're trying to motivate in the first place! So it turns out that the two numbers don't exhibit any differences between their respective properties, meaning there's an ambiguity here that we can only resolve arbitrarily, rather than being able to rely on such different properties of the two candidates.
We don't "reject this answer". The √ sign is not a problem that needs to be solved by giving an answer. It is simply a symbol that has a commonly accepted precise definition at least on the non-negative real numbers. √36 = 6. Why do we not have a special symbol for the other root? Because it is easy enough to add a minus sign: -√36 = -6.
@@ronald3836 cool!
there is a convention to only consider the positive root even though we understand that y = x^2 has two solutions. i was taught to always add a +/- after evaluating a square root.
sqrt(a)*sqrt(b) = sqrt(ab) is a rule for positive a and b. It also happens to hold if only one of a or b are negative. But it does not hold if a and b are negative.
sqrt(z) is, by convention, not defined to be multivalued under "standard" rules. You can of course define it that way in some other context, but said context would need to be explicitly stated.
There are more complicated rules with complex numbers but sqrt(a)*sqrt(b)=sqrt(ab) still has restrictions even in the complex world.
If √ is redefined as multivalued, √a√b becomes meaningless.
Thanks. You have validated my answers. There are videos saying that you can't have two negative numbers under the sqrt, like sqrt of (-a * -b). I was very confused by that. The videos I have seen are forgetting phase shifts to the numbers. Thanks for clearing my mind. Now, I can sleep better knowing that I was correct all along.
The thing is, you CAN'T have two negative numbers under the radical. The rule √a√b = √(ab) always has restrictions, no matter how you define the square root function. The phase shift of a number in this context is just it's argument, and there are restrictions on the argument of √(ab).
If a and b are real then the restriction is that they can't be both negative. But if a and b are arbitrary complex numbers then the rule is the sum of arguments must be in the principal branch, usually either [0, 2π) or (-π, π]. If the sum is not in the principal branch then √(ab) will have the wrong sign. For complex numbers on the negative real axis, the argument is π. So if a=-4 and b=-9, then arg(a) + arg(b) = 2π and that's not in the principal branch.
Even if you define the square root function on a Riemann manifold you still have the same problem. There, when you multiply two values you have to take care when the values are on different sheets, and if they are on the same sheet, you have to check that √(ab) is still on the same sheet.
There is simply no way you can have √a√b = √(ab) hold without restriction.
Multivalued functions also occur in real analysis with more than two variables. I constructed one which is analytic from R^2 to R^2, but holomorphic only in the unit circle.
well don't leave us hanging!
You forgot to say which one
According to this video, √4√9 = -6.
@@doraemon402 I know, but this is very difficult to explain in the comments.
Another one: e^(iπ)=-1, so we get ln(-1)=iπ
But we know ln(xy) =ln(x)+ln(y)
So 0=ln(1)=ln((-1)(-1))=ln(-1)+ln(-1)=2πi
3:57 I very briefly thought "wait, which Euler's formula" before I realized that there's only 1 that could be relevant here. As the old joke goes, most math theorems are named after the second person who discovered them, because they can't ALL be Euler's theorem.
Thanks!
The main reason should be that rules that always hold on the real numbers can't always be generalized over the complex numbers.
(a^p)^q = a^(p•q) is another one you can't just generalise, or you'll find any non-zero complex z is actually a positive real number.
Put it in polar form, z = r e^(a i), with r = |z| and 0
You don't really need to bring the problem to the complex domain! Just observe that sqrt(4) can be both 2 or -2 - since 2*2=4 and (-2)*(-2)=4. This is sufficient to show that square root is multi-valued.
im just a college math level
but i feel that the convention of' taking positive root' of -4 , cannot be simply combined under the typical root a x root b = root ab (trivially)
i think this is a failure of the convention
without first being resolved to i (root -1) x root 4
and -6 should be the correct resolution in view of the 'taking positive root' convention
if not, a full complex analysis of the arguments
I'm a big fan of Mr Penn and his channel but I cannot agree with this. If we use the notation sqrt(z) we must make a convention which root is meant. Otherwise we destroy the transitivity of equality which is part of the base of mathematics. If we write sqrt(-4)*sqrt(-9)=6 and sqrt(4)*sqrt(9)=-6 it follows 6=-6. The whole Zermelo-Fraenckel-system doesn't work if we abolish the transitivity of equality. One possible convention is to take the half of the argument between -pi exclusively and pi inclusively. So the notation becomes clear. When making such a convention we must consider that the equality sqrt(xy)=sqrt(x)*sqrt(y) does not always hold. So the paradox disappears and we get -6 as the only correct answer instead of 6.
Another way to avoid the paradox is not to use the notation sqrt(a). Instead of saying that we calculate sqrt(a) we could say that we solve x²=a.
My English is not perfect. So I hope I was able to make clear my point of view. Greatings from Germany!
The way I was taught (in Germany) is that you can never take a square root (or even a cubic root, etc.) of any number that isn't a nonnegative real number. So the notation i = √ -1 that seems to be common in the English speaking world would indeed be invalid. The definition of i that we were taught was i² = -1.
Sorry, is it normal in Germany a number raised in 2 to be negative?
I am of the conviction that the square root symbol is undefined on numbers other than non-zero reals. There is only one place I have come across where using the square root symbol on general complex numbers is useful, and that's the quadratic formula. And in that case you only need to mix it with extremely limited arithmetic, so you avoid any shenaniganigans. And if you're more of a purist than me, you can see it as not really being a square root but rather a mnemonic for "insert each of the two square roots in here, and calculate the result for each one".
Other than that you will not need the square root symbol √ on general complex numbers, and your life will be easier if you simply excise it from your life. _Especially_ if you're teaching or taking an introductory class.
My approach to these idiotic viral things:
Don't talk about math with people who don't know anything about math.
That's the way to deal with these viral problems, baffle everyone.
This guys great. Thanks for the video
And to add to that, 6 ≠ -6 because the "equality" here is not always transitive and reflexive when dealing with multivalued functions.
Everything he said after 2:30 might be true but I just explained it by saying the square root of 36 is ± 6 not 6. And the square root of - 1 is ± i, not i. So, either way you end up with the same answer, ± 6.
The sqrt(36) is +/-6? Really now? 😂😂😂
No, x^2 = 36 has two solutions +/-6, but √36 is simply notation for the positive root 6. The other root can be written down by prefixing a minus sign: -√36 = -6.
@@ronald3836You are answering wrong questions. Sqrt(36)=x. What's the value of x?
I raised this up in a Reddit thread about half a year ago, and got ripped to shreds in the comments. The basis of the problem being a function by definition can only have one output, and f(x)=sqrt(x) is only positive by convention. Otherwise, you can do this: 1 + sqrt(1) = 0.
√-4 · √-9 = 2i · 3i = 6i² = -6
I think you didn't really address the philosophical question of why we should allow square root to be a multi-valued function among the complex numbers, but not among the reals? Presumably there are problems in complex analysis if this is not allowed, but then what of the reals? Do problems occur in real analysis if we were to allow the square root of 4 to be ±2, instead of insisting by convention on the positive +2 value only? (And presumably it would be acceptable to say the square root of 4 + 0i is ±2 ?)
to only define square root of positive real numbers the positive root is genuinely just a convention. It's a convention that makes some formulas work. It's done that way because it makes some school math easier, lets people just compute stuff without a need of thinking (for example any discussion about which root to take is moot, since all of this is predefined). It's usually not a problem, but it occurs! The easiest thing is the formula for roots of a quadratic polynomial. The +- is an artifact of our definition. If the roots are complex, then suddenly the +- doesn't play any role, since square root is multi-valued. The way out of this mess is to admit that square root was multi-valued to begin with, but we bastardized it for pedagogical reasons.
@@ib9rt I take it you were guessing, and you did guess correctly. Complex root does not spare any number, for any complex input you get two outputs. In complex analysis there are possible trajectories that take you from one branch of the root to another, so it is important to keep track! When you ensure your trajectory does not change branch, you safely choose one to work with.
In real analysis, this problem is nonexistent, so algebraic root is rational to use.
@@Czeckie √ is just notation. For positive x, √x is the positive root. For negative x, you could define √x = i(√-x). In this case √-4√-9 = -6.
If instead you define √x = { i(√-x), i(√x) } for negative x, then that is fine with me. However, I will note that √-4√-9 now has no meaning, unless you come up with a definition of multiplying two sets of two numbers.
I thought it was the case that there were two answers to just sqrt(36). You can multiple 6*6 or (-6)*(-6) and get 36, so they're both roots. In that case it doesn't matter if you take the 'i's out and get -1 first, that's then -1 * (+/- 6) which is still +/- 6
I guess what should be observed is that in the first case they implicitly assume real numbers and a definition of sqrt where only the positive number x with x^2=y is considered to be the solution. But if we just consider x^2 = y it could be reasonably (and without much mathematical rigor) argued that sqrt(36) can be -6 and 6 and both should be solutions. In the second example complex numbers are assumed... what then changes a lot of things, as you have shown.
x^2 = 36 is indeed an equation with two solutions.
But √36 is not an equation. It is just a notation for the positive solution, i.e. √36 = 6.
If you want to consider the notation √-4 and √-9 and wonder what is √-4√-9, then you have to decide what the notation √-x means for positive x. If you decide that it means xi, then √-4 = 2i, √-9 = 3i, and therefore √-4√-9 = 2i * 3i = -6.
If you decide that √-x = {-xi, xi}, then √-4√-9 is meaningless, until you define what it means to multiple two sets of numbers.
If you take n=1 and m=0, or vice versa, aren't you choosing different branch cuts for each?
I might be missing something in my limited understanding of linear algebra, but I got 6ij.
The property sqrt(a) * sqrt(b) = sqrt(a * b) is not valid when a or b is negative. It only holds true for real, non-negative numbers. In the field of complex numbers you might have sqrt(-1) * sqrt(-1), which equals i * i, or -1. But sqrt(-1 * -1) = sqrt(1) = 1. Hence, sqrt(-1) * sqrt(-1) is not equal to sqrt(-1 * -1).
So the property isn't valid but the expression sqrt(-1) is valid? 😂
@@pelasgeuspelasgeus4634 yes, the square root of -1 equals the imaginary unit i. Thus, i^2 = -1. This is the foundation of complex numbers.
@navybrandt I see. Well I'm sure you are aware of Euler identity, another foundation. Right? So, let's see where it leads.
e^iπ = -1 => raise to 2 => e^(2iπ) = (-1)^2 = 1 = e^0 => 2iπ=0 => i=0. Right?
@@pelasgeuspelasgeus4634 e^(2iπ) = 1 is indeed true, but it doesn't imply that 2iπ = 0. Why? Because e^z is periodic in the complex domain with period 2πi. This means e^(z+2πi) = e^z. So, e^0 = e^(0+2πi) = e^(0+4πi) = e^(0+6πi) = ... = 1, which simply reflects the periodicity of the exponential function in the imaginary direction. It does not mean 2πi = 0, because exponentials are periodic, not injective, in the complex plane.
@navybrandt It seems you make rules on the way. That's ridiculous.
If x^m=x^n then m=n. Simple math even for kindergarten.
Arguing that √-4√-9 can equal 6 is equivalent to arguing that √4√9 can equal -6.
This is nothing more than a muddying of the water. The quantity I. is a very special thing And everything we do with it Must go back to the Definition that I times I Equals -1. Proper handling of this special quantity gives us only 1 answer To the question posed.
Actually, when it comes to so-called indeterminate forms, all the common operations are multi-valued functions. When their outputs are an indeterminate form, their real value is the entirety of R or C, depending on what your codomain is.
Is it just me?
I prefer to hold onto “=“ to be an “equivalence relation”, meaning it has the property, (amongst others), that a=b and a=c implies b=c. With “multi-valued functions”, this is lost which seems a pretty big sacrifice.
Also, allowing both branches for the square root “function” would mean we have wasted so much ink over so many years writing:
x=(-b ”+/-“ sqrt(b^2-4ac))/(2a).
So I don’t like the thumbnail saying “both are true”.
So I have a problem with the thumbnail saying ‘Both are true”.
Not just you.
"Multivalued function" is just a very sloppy concept that can be used to help a student develop the right intuition.
So within the confines of the real numbers, the correct answer would ‘does not exist’?
Well yes, of course. √(-4) doesn't exist in real numbers :)
√x always means the principal value. So √(-4) * √(-9) = 2i * 3i = -6 and no other value.
This is ignoring the fact that the radical symbol and raising something to a fractional power *are not the same thing* and do not mean the same thing mathematically.
By standard convention, using the radical symbol _only returns the principal root._ It is *not* a multi-valued function. That is why, even in cases which don't involve complex numbers, if it matters, "plus or minus" must be added to the radical to indicate both (real) roots are desired.
Because of this, the statement "√{x} ⋅ √(y) = √(x⋅y)" *is only true if "x" and "y" are not both negative.* (This is also a fairly well-established caveat for this rule among most mathematicians, I believe). Therefore, you cannot use that rule in this case, and the answer "6" for that problem is actually objectively wrong.
(And most of your explanations are done wrong, because you are writing/saying, for example, "√z" when what you actually _mean_ is "z^(1/2)" instead (which are _not mathematically the same thing_ ).)
I've seen enough popular youtubers explaining this example incorrectly. It is a relief I'm on the same page with math professor, yet I'm baffled by the amount of confusion caused by the incompetence of many "teachers"
I agree. I have found another video explaining it correctly before this video. You might want to check out channel " Science'n'me" titled "Why is the square root of Negative one equal to i". This video explains what people are missing in learning numbers. Every number has a phase shift comes with it. That's what missing in our early learning of math. Check the video out. I think you will like it. It has taken me over 60 years to find out about number system we have been learning is half truth.
I like your explanation about multivalued functions, but typicall the √ symbol is defined as giving the principal square root. So under that standard assumption there actually is a single answer for this problem, namely -6. In particular √x √y = √xy just isn't true unless x,y >= 0.
I'm not a mathematician rather an engineer. When I was in high school in the 80s, this caveat was never taught. When UA-camrs like bprp went over this, I was struggling to understand, why the caveat? After watching this video, my Spidey sense was right. I just didn't have a word for it - the square root is multi-valued for complex numbers.
The xy- term must have grouping terms around it, such as parentheses, because as it is written, it reads as [sqrt(x)]*y.
And if instead you let it be "the other" root, then you still get -6.
Second comment - this is not a trivial thing - Michael's topic in video I mean.
Consider drawing a circle. Is it 360° 720° or ....
I claim it is any line drawn with a fixed radius of 360° or more around a fixed central point will describe a circle. Mathematicians are a bit lazy at times and when they say something like a circle has 360° they mean many important things
Maybe a geodesic circle will be one that has infinite collection of points all equidistant from a fixed central point and covers 360°
But that might mean one single infinitesimally small point is counted twice. so in order to be well-behaved the starting point of circle is closed (or open) and ending point is open (or closed). They cannot be both closed and they cannot both be open (why?)
So if you draw a circle and run your pencil back and forward a couple of times just so that start and end points "meet" nicely. don't worry!
It will more than likely be a perfectly drawn circle even if the compass spanned more than 360° and squiggled back and forward a bit to make the circle look better.
Ideal shapes and ideals are well , ... sort of idealistic ideals. And in math that is also important 🙂
x^2 = 1
x1 = 1, since 1*1=1
x2 = -1, since -1*-1=1
Why over-complicate it, with complex numbers, Euler formulas, polar form, periodic trigonometry, etc.?!
Because we can! Because the answer is not so straightforward with the cube root.
Because we must to gain a deeper understanding what's happening.
Great vid!
It's also quite obvious when you consider that sqrt( 36 ) = +/- 6. The imaginary calculus just proves the result.
No, x^2 = 36 has two solutions +/-6. But √36 is simply notation for the positive root. The negative root is -√36.
does sqrt(x)*sqrt(y) = sqrt(x*y) hold for all x,y in the complex world? i was told that this can lead to a contradiction
i.e., 1 = sqrt(1) = sqrt((-1)*(-1))
= sqrt(-1) * sqrt(-1) = i^2 = -1
(at least this is what i was told in an undergrad complex variables course, maybe there's something wrong with the argument)
If you allow the sqrt 'function' to be multivalued, the apparent paradox goes away. {1,-1} = sqrt(1) = sqrt(-1 x -1) = sqrt(-1) x sqrt(-1) = {i,-i} x {i,-i} = {ixi,ix-i,-ixi,-ix-i} = {-1,1,1,-1} = {-1,1}. You have to take into account all possible combinations of course.
You are fully correct. Clearly we really want the square of √-1 to be -1, and we don't want -1=1, so √-1√-1 = -1 != 1 = √(-1 * -1)
@@russellsharpe288 There is no serious part of complex analysis which uses such a concept of multiplying sets of values.
@@ronald3836 Maybe there should be! My point is simply that the need to arbitrarily pick a single square root of -1 as the reference of √-1 is what leads to the trouble.
-1 = exp((2n+1)pi.i) for all integer n, so √-1 = exp((2n+1)pi.i/2) = exp(n.pi.i + pi.i/2) = exp(n.pi.i) x exp(pi.i/2) = ±1 x i = ±i. Certainly by convention √-1 is just i, but that is because we have no way of distinguishing i from -i before labelling one (which one? we have no means of knowing or even saying...) 'i'. It is so to speak a random dubbing of a square root of -1 which makes i i and not -i.
@@russellsharpe288 To extend √a from being defined for positive a to also being defined for negative a, you indeed need to pick one of the two solutions to x^2 = -1, let's call them i and -i. But once you have made this choice, you will see that √-4√-9 = -6.
Either √-4√-6 = 2i * 3i = -6.
Or √-4√-6 = -2i * -3i = -6.
Nothing useful is gained by letting √ be notation that allows (√-4)^2 = 4.
Instead of -4 and -9, just use -1. If we accept that negative numbers have square roots, then clearly sqrt(-1) * sqrt(-1) can only be -1.
In which universe a sqrt can be negative?
@@pelasgeuspelasgeus4634 In a universe where we define √-a = i√a for a positive, where i is an element in the complex numbers satisfying i^2 = -1. (Or where i=X in the ring R[X]/(X^2+1). Indeed, X^2 = -1 module X^2+1.)
@ronald3836 Those tricks are for idiots. It doesn't have any actual meaning because imaginary numbers aren't countable and therefore useless. Another fancy arbitrary invention.
There's something wrong with the square root function. Why do they accept only a positive solution? (2)^2 = 4. (-2)^2 = 4. Then SQRT(4) = +/- 2. The real plan is a limitation of our thinking, a mere projection of the complex plan.
You know, math works based on definitions, rules. Violating them isn't a revolution. It's ignorance.
@@pelasgeuspelasgeus4634 You know, maths rules work on reality. Violating reality is stupidity.
@@sorin86ytb Stupidity is to think sqrt can be a negative number...
I se the question and I go... we cant use classic rules, bcs we are dealing with complex numbers we need more robust method...
turns out its not that we cant, but we need to know when and why its OK and if we took care of all possibilities. Bcs in the end both were valid solutions but neither were the whole solution.
I love how he complexified (!) the problem just to torture the viewer interested in the problem. I would have said, simply, every square root has two values: +/-. sqrt(36)=+6 and -6.
sqrt(-4) has two values: +2i and -2i
sqrt(-9) has two values: +3i and -3i
Then we get to the end of his video. You can multiply those together in four ways:
2i x 3i = -6
-2i x -3i = -6
2i x -3i = +6
-2i x 3i = +6
That's all. sqrt(36) = +/-6. It's wrong to think sqrt(36) is ONLY +6.
Is sqrt(-4) a countable value? I mean, did you ever buy sqrt(-4) apples?
@@pelasgeuspelasgeus4634 LOL, is it countable? No, but pi is not countable. sqrt(2) is not countable. 0.10110111011110... is not countable. What's that have to do with it? Who said square roots have to be countable?
@greg55666 Irrationals aren't countable in the sense they have infinite digits after comma. Imaginary numbers aren't countable because they aren't representing anything real. Do you get the difference? Lol?
@@pelasgeuspelasgeus4634 It would probably be better if you explained it to me. Irrational numbers are not countable, according to you. Imaginary numbers are also not countable, according to you. You started by saying "countable" was the issue, now it turns out that according to you there are different kinds of "not countable," and your rule has nothing to do with "countable" at all, you mean something completely different. Two messages, two totally different rules. Maybe you should explain what you REALLY mean?
(Also, the original problem is about sqrt(-4): not a real number. So this whole conversation is completely irrelevant to the question at hand, since the question itself is clearly not limited to "countable" of any kind.)
@greg55666 I see. Well, have you ever bought sqrt(-4) apples? And don't say that similarly you can't buy π apples, because there's an easy answer to it.
Isn't this just pq formula that any square root can gives 2 answers. -6*-6=36 as much as 6*6, without graph you just don't think of it that way. Going via i just gives you the proof that it actually is that way without having to draw it out
Sqrt(36) = + or - 6 That’s pretty much the explanation needed. For convenience people ignore the negative root which is incorrect
No, we just write -√36 for the negative root. Why make life difficult?
Which only goes to show how many problems could be easily solved if they'd just change that damn rule about multiplyiing negative numbers by negative numbers. But no, that's top easy. Instead you create "imaginary" numbers, keeping hundreds, if not thousands of internet mathematicians employed deciphering equations that include the square roots of negative numbers.
My own opinion is that once one sees a square root of a negative number in any equation, one knows the answer will be +/- some number. Forget about the +/- for the time being, find the solution, and add on the +/- at the end. In the real world, most solutions will be either negative or positive, not both, so you can use whichever works.
Thanks for this. I get tired of being shot down by the high school and Jr. college kids when I suggest that sqrt() is a multi valued function. Where did this "principle root" BS come from anyway? It was never a part of my educational trip.
It was mine.
sqrt(x^2) = abs(x)
Learned it in algebra I
It is just notation. √23 is the positive real number that squares to 23. The negative real number that squares to 23 can be written as -√23.
@@flowingafterglow629 Well, I learned what he showed here. If you ever do electrical engineering or physics the multi-valued aspect of teh square root must be taken into account or bad things could happen.
@@uni-byte Yes, that is why you use +/- sqrt(value)
If as I said
sqrt(x^2) = abs(x)
then you use +/- sqrt(x) to get both values. But in both cases, sqrt(x) is a positive value.
There are several problems with this video, some of them have already been pointed out, as for example the distinction between the principal root and fractional powers. What I find more terrifying about this video is that the argument laid out here can be used to explain that sqrt(4) * sqrt(9) is 6 and -6.
What Michael and others also seem to not to notice is that this "viral problem" is a typical problem in College Algebra (a first algebra course at the college level in the United States) which just got outside the book. In a college algebra course the square root of negative numbers is defined, so sqrt(-4) = 2i and sqrt(-9) = 3i under that definition (in general, if $a < 0$ sqrt(a) = i*sqrt(|a|)). Therefore sqrt(-4)*sqrt(-9) = -6. The reason why this is given as a problem in College Algebra is to make students realize that sqrt(x)*sqrt(y) is not always sqrt(xy).
The bottom line: this is not a problem in complex analysis, this is a problem in College Algebra with a very restrictive definition of square root.
Agreed.
Most commenters (probably high school students?) seem to believe that √a is an "equation" that needs to be "solved".
Of course that is not true. √a is NOTATION for the positive solution to the equation x^2 = a (if a is positive).
The other, negative solution can be written simply as -√a.
it is multi valued because the question is in complex domain already 😁😁😁 square root still a normal function with 1 value in real domain
At 1:01 I was taught to say sqrt(n)=|m|
In this case I would have said sqrt(36)=|6| and then explained that |6| can be +/-6.
The explanation with the famous Euler’s formula is very very beautiful. Thanks
That's not really correct, √ is defined as the positive root for real numbers.
When you mint a lovely new square root of a negative number, you can use bombelli-i or anti-bombelli-i. Both are equally valid.
bombelli-i x bombelli-i = -1
anti-bombelli-i x anti-bombelli-i =-1
But!!!!
anti-bombelli-i x bombelli-i = 1
So let Sqrt((-4)x(-9) = 2bombelli-i x 3anti-bombelli-i = +6
Sort of the same as the professor, but weirder and wronger
You can use:
🙂= bombelli-i
🙃= anti-bombelli-i
🙂🙂= -1
🙃🙃= -1
🙂🙃= +1
🙃🙂= +1
Always two solutions to a root. It was a false assumption that the first solution was equal to the second.
-6... I don't need to write it out. Im no longer a slave to showing ny work
Is that like a complex conjugate?
And that's a good place to stop.
infinite series algebra arithmetic, complex number algebra arithmetic, algebra engineering ...
√-4 √-9 = reciprocal of sum of naturals lol
so it's both because we are dealing with complex numbers?
Am I crazy, or is √36=±6, not SOLELY 6? We see that kind of interaction routinely with multiple equations: ±6 satisfies the first equation, but ONLY -6 satisfies the other, so -6 is the sole possible answer. Kinda weird that the positive answer is the invalid one though; usually it's more like "A triangle cannot be 3cm X 4cm X -5cm, so only positive answers are valid." Oh, well, No Myth is still a pretty good song even if I can recall no others now. ;)
√36 is not an equation. It is notation.
x^2 = 36 is an equation. It has two solutions x = 6 and x = -6.
You can write the positive solution as √36.
You don't to need to have the same or a separate symbol for the negative solution, because you can just write -√36.
So is √2 + √3 + √5 + √7 ONE value or 16 different values? Is it just one value but √-2 + √-3 + √-5+ √-7 is 16 different values? What it only half of them are negative? Then is it only 4 values? What if I don't know? Does the number of possible values of an expression depend on the values of some of the variables so that √a + √b + √c + √d can have either 1, 2, 4, 8, or 16 values, depending on how many of a, b, c, and d are positive reals?
This is insanity.
The answer to all those questions depends on how you define the square root, which really depends on the question you are asking. There are two numbers which square to any particular number, which one do you want to use as your square root in that context?
@definitelynotofficial7350 But we have conventions for this, and this all is why we have those conventions. In any ordinary context where it isn't clearly specified otherwise, the √ symbol means principal square root or n-th root if n is written.
Yes, the principal n+th root is defined even for complex numbers. There are two common conventions regarding *which* of the n values are the "principle" root (or which is the principle sheet), but under none of the conventions is √a more than a single value. Even if you're talking about multi-valued functions and reimann surfaces, you still need a way to write √a that is distinct from -√a and isn't ±√a either.
I think Prof. Penn made a mistake in his videos. When he choses n=0, m=0 or n=1, m=1 the values are from the same sheet and he gets the right answer, but when he choses n=0, m=1 or n=1, m=0, he's multiplying values from different sheets when requires a phase adjustment (in this case a sign change). Without the adjustment, he gets an erroneous answer.
√-9×√-4 = -6 only, not ±6.
@@definitelynotofficial7350 Do you really want to define the "+" sign as it is used in analysis as something that operates on sets of numbers? This is not a useful concept (nor an existing one).
(In combinatorics there is probably a use for such a "+" sign.)
@@nbooth I agree.
@@ronald3836 Usually you don't, but maybe sometimes you do...
...and that's a good place to stop.
Would it be -6 or 6 in the title
Either this
Sqrt(-9*-4) =6
Or
(Sqrt(4)*i)*(sqrt(9)*i) =-6
But which is it? I guess its -6
It all comes down to what information the person who wrote the problem meant to convey. Was he using the √ to represent the principal root or all roots?
I say the words that sounds like “the red book” which also could be written, “the read book”.
What does that phrase mean? You could argue that it means the color of the book. Or you could argue that it means the book has been used by being read. You could argue it means both. So which is it. The correct answer is that it means what I meant it to mean when I said it. If I don’t give the context of the statement, then you can’t give a correct answer. Whether you say it is the color, or the use of the book, or both, you are not correct, you are just listing possibilities.
With √ for real numbers, convention is that it means the positive root, otherwise there is no way to differentiate between the two roots of x^2=4. (Note that he used this in the video to change √ 36 to 6, not ± 6.)
In the complex world, it is ambiguous unless it is put into context as to what the author meant to convey as the meaning. When we use the quadratic formula and end up with a negative discriminate, we pull out an i, not a ±i. So in that context it is just the principal root.
So what I have seen is based on the radicand. If the radicand is a non-negative real number, then √ means a non-negative value. If the radicand is complex (a + bi) and b is not zero, then it is multi-valued. If the radicand is a negative real number then it means a postive real number times i, AND √ a • √ b = √ ab is NOT VALID. This identity is only valid with real numbers.
If this is a step in the solution of a problem, then that context is going to determine what is meant. Just like if a told you to hand me the “red” book. While you could interpret that both ways, you would know I meant the color based on the context.
So √(-1)×√(-1) = ±1, but i×i = -1?
Pretty sure that we want the square of the square root of x to be x, even for x = -1.
Great explanation, but then, why can't real functions be multivalued? The square root function over the reals is a perfect example. Isn't the definition of a function as something that returns a single value unnecessarily constrained? At least it's inconsistent over complex numbers.
No, because that is what "function" means. A "multivalued function" is not a function (and hardly a useful concept, e.g. multiplying two sets of outcomes of a "multivalued function" is just silly).
@@ronald3836 But that begs the question. In fact, that "multivalued" property is exactly what a "function" means in the complex plane. My question is, Why do you restrict multivalued results from functions operating in the real numbers but not in the complex numbers?
So, √-1 * √-1 = 1 is also correct? And by your argument, √1 = -1 is also right.
Yes you can argue for those answers the same way
And this is why the video is wrong.
Isn't the square root of 36 +/-6?
√36 is just notation for 6.
The solutions to the equation x^2 = 36 are 6=√36 and -6=-√36.
-6
I'm sorry, but I don't get one thing: don't you learn as soon as you learn square roots, that the result is +/- the root, because the negative solution is also correct? So the first line is already wrong, because in school we would immediately write sqrt(36)=+/-6, which makes the second line not contradict the first one, since the second line would read -1 * (+/-6), which comes out to the same. No deep dive needed, basic math. Why is this puzzling people? Why is this a "viral" problem?
When we define the square root as a function we take the principal square root as the definition, so the square function can be inverted.
No, what you learn is that when you solve x^2 = a (let's assume a is positive now), then one solution is x=√a (the positive-valued solution) and the other solution is x=-√a.
Your work through is very clear but I feel like the conclusion could have been made more precise. Saying "both are correct" is a bit misleading. I think it would have been better to state that the flaw in the original argument is that an assumption has been made about the square root function that is false, namely it is not multiplicative in general, particularly on the negative reals. One must first define "a" square root function. There are two possible definitions. Given one definition you get one answer, given the other you get the other answer. However, both statements are certainly not simultaneously true.
Overkill? SqRt (4) =x is solved with x=2 but also -2. With this open approach SqRt (4) . SqRt (9) can take the values 6 and -6.
No need for complex numbers.
The problem asks √(-4) not √4. So you can't really get away from complex numbers.
@@TheEternalVortex42 Negative surface area doesn't mean much either.
The true pressure to accept complex numbers was that they were necessary for solving cubic equations via standard formulae.
@@Apollorion "Pressure to accept complex numbers"? Is this some sort of new denialism?
The answers 6 right
WHERES THE GOOD PLACE TO STOP??? I NEED A GOOD PLACE TO STOP!!!
Stop yelling in all caps, you rude poster.
@ it was affluence of existential dread rather than anger, but nice to see that you take everything the wrong way. Let me know next time you find a comment online that you have nothing good to say, but decide to say something anyway. I would love to hear your storyline on such
there are infinity places to stop!
The line where the finite turns into infinite, that's a good place to stop
Great!
the rule you used only works for positive numbers the answer is only -6
"multivalue results" only exist as sulotions for unknown variables but this expression doesn't have such variables and all the functions are one to one therefore there is ONLY one answer
The thing is square root is NOT one to one when it acts as a complex function instead of a real function
@yuging-q2l square roots are one to one unless explicitly said otherwise
and they work for complex numbers regardless of whethernot
@@RuleofThehyperbolic "square roots are one to one unless explicitly said otherwise" -- Only by mathematical convention. The entire reason why this problem is viral is because the general populace won't unanimously accept the problem as obeying this convention. So you cannot assume this problem obeys this convention.
I just watched him demonstrate that the rule also works for negative numbers as long as you are aware of and take account of the fact that any non zero number has two distinct square roots
To explain it simply, is sqrt( ) with domain C a multifunction?
yes
Depends what you mean by sqrt(z). If you mean (in standard modern notation) z^½ = ±√z then yes. If you mean the principal value, denoted √z, then no.
@@jimskea224 And since the video is about √-4√-9, wiggles seems to be writing sqrt() for √.
I am confident enough to say that BOTH initial series of transitions are wrong. Neither 6 nor -6 are the answer to this question. It's either n/a if we use algebraic square root which is explicitly undefined for negative values, or ±6 if we use complex root of power 2 as √(-1) is not i, it's ±i, and multiplying ±i by independent ±i results in ±1. The only correct root-based definition for i is a solution for x² + 1 = 0, or such number that i² = -1
Why is -6 not a potentially valid answer? If, for a>0, we define √-a to be i√a, then √-4√-9 = 2i * 3i = -6. (And if instead we take √-a = -i√a, we still get -6.)
The problem with the video is that it seems to define √-a as a set { -i√a, i√a }, and that it seems to define A * B for sets A and B to be the set { a * b: a in A, b in B }. This is possible (but unconventional and not useful in analysis) and should have been stated.
@ronald3836 and that would be a wrong definition leading to paradoxes like in video.
In some sense a mulitvalued function is the negation of the very notion of function,
as the the defintion of a function is something that, applied to a given argument,
gives an unique result. Of course you can say that a multivalued complex function
on complex numbers is a function the value of which is a SET of complex numbers,
but this still has some problems with notation. When you write a square root in any
formula, you are merging apples and oranges by mixing notations designing complex
numbers and notations designing sets of complex numbers. You can solve this by
assuming that in such fomulas single complex numbers are assumed to mean the set
containing just this number (a singleton), but then you have to give a definition of
what are the meaning of usual operators (+, -, exp, ...) when applied to sets. Th usual
convention is to consider that the extension of these operators to sets is that each
operator applied to sets gives as result the set of all result obtainable by applying
it to any of the members of its arguments. But the with this convention sqrt(z²) =
{-z, z} = - {-z, z} = - sqrt(z²). It would be easy (but false) to conclude that sqrt(z²) = - sqrt(z²)
implies that sqrt(z²) = 0 for any z. This shows that you can't use all usual notations
with multivalued "functions", because these are implicitely functional notations
involving univalued functions.
I agree that the problem presented is one where clarity requires being extra careful about notation. I also recognise that makers of YT videos have to eat and pay the rent, but... The thumbnail of this video did *not* say sqrt(-4)sqrt(-9)={6,-6} which would have alerted me to an interesting discussion about what multiplication and sqrt might mean. (And it was an interesting discussion.) What it asserted as "both true" was that the expression equaled 6 and that it equaled -6. Equality is transitive, any arbitrary pair of numbers are not in general equal and vaccination is safe and effective. Claims to the contrary are clickbait.
@@peterbrockway5990 I don't think the video was deliberately created as clickbait, so I might just call this video "very misguided'.
I think we can all agree that viral math problems are doing a good job of exposing how dysfunctional math education is in the US.
Not only that but this also proves that the US has a serious problem with ignorant and arrogant self proclaimed know it alls in social media
I don't necessarily agree. In order to understand lower levels of math, it is not necessary to think of functions as possibly being multi-valued. This concept is often introduced in complex analysis classes. Also, viral math problems also spread outside of the US, so if there is a problem, it is not specific to the US. I don't know of any country where complex analysis is a required part of schooling for everyone. In short, I think it's completely reasonable for someone, even someone from a place with a functional mathematics education system, could be unsure about how to solve this problem.
Other viral math problems may be simple enough that someone should be able to solve it with basic mathematics education. (I've seen several problems about the order of operations that fit in this category.) However, I would not generalize and say that viral math problems expose dysfunctions in math education. instead, I would say that they show that people care about math and are trying to use their problem-solving skills along with the rules they learned to justify an answer they think is right. Even if they get the answer wrong, the fact that they are engaging in the process and willing to learn the right answer in the end is more important to a "functional" mathematics education system, in my opinion.
They're not only viral in the US.
No, we cannot all agree.
you're violating principle roots, which in other videos you advocate for... do you just change the rules randomly as it suits you? that's some impressive rigor you got there, bud.
I can't believe I'm even saying this but I think Michael Penn is wrong.
In grade school I was taught that √a * √b ≠ √(ab) when both a and b are negative.
But even in the complex domain, √a * √b ≠ √(ab) when arg(a) + arg(b) ∉ (-π, π]. (or if arg(a) + arg(b) ∉ [0, 2π), depending on convention).
But under either convention if a=-9 and b=-4 then arg(√a) = arg(√b) = π, so arg(√a) + arg(√b) = 2π which is not an element of ( -π,π ] OR [0, 2π), so either way √-9 * √-4 ≠ √36.
In general identities like √a * √b = √(ab) are not valid if the operation will cause the value to "jump" out of the chosen branch cut.
You're not writing the square root of the product about correctly. Put grouping symbols around it as (you use your radical symbol) sqrt(ab). What you are writing is equivalent to
[sqrt(a)]*b.
@@robertveith6383 I edited it for clarity. Fixed some typos too.
I believe a and b can be both negative numbers because you have to include phase shifts to those negative numbers.
@@bowlineobama if you're talking about phase shifts, then the values have to be from same sheet. If you multiply two values from different sheets, then you have to phase shift the result accordingly.
"Two Pi Periodic" is a longer way of saying "Tau Periodic".
Wrong explanation. The only answer is -6