I wonder if its possible to graph the functions and the complex solutions together, and see if they cross each other (similar to graphing them with real numbers)
It is conditionally true. It is true if x can be shown to be equal to or larger than 0 (in geometric context, for instance), or if it can be shown that the sign for x is meaningless, like in the calculation for standard deviation where all you're interested in is the size and not the direction. In plain algebra, it is false. Because the root is the inverse operation to the exponent. √x = x¹/²
Regarding equation 2. If we assume that x is a real number, there are no solutions because √ always means "principle/positive square root", so it can't take negative values. If we assume that x is complex, then we have to choose from two possible meanings of √. If we understand √ as "principle square root", then there are, again, no solutions, because the principal (complex) square root of any real number is a real positive number, or a product of a real positive number and i. If, however, we don't assume that √ is a principal square root, then there is a solution: x=1, because (-1) * ( -1) = 1, and thus -1 is one of the (complex) square roots of 1. In general, n-th root of x is a n-valued function for complex values of x, and none of those n different roots can be considered any "better" than others, so in general we can't discard any of them when talking about n-th roots of complex numbers. Sometimes, though, it is convenient to talk about roots as single-valued functions, so the notion of principal roots is still useful for complex numbers. It is very important to know exactly what the mathematical notation that we're using at any given point actually means. As a sidenote, more often than not √ DOES mean "principal root", so the answer in the video is still correct, most of the time.
That's not a sqrt(x)=-1, that's x²=1 Similarly, not root_n(x)=-1 but xⁿ=1 Finding root of a complex polynomial is not the same as taking a root of a complex number, because taking a root is a function, 1 output, while solving an equation can have many solutions.
@@_rd_5043 A complex n-th root of z is a multi-valued function, with n distinct branches (values) for all complex z except 0. It is most definitely not 1 value, unlike real root functions. The principle complex n-th root, however, IS a 1-value function, taking a value of a certain ("principle") branch of the multi-valued root. If we only consider the principle root to be the "correct" root, we lose certain properties of exponents that we're used to from real analysis (like √(xy) = √x * √y for example). See also: roots of unity.
What's important is the wording or symbols used. As you've mentioned, there are always n, n-th roots of any number, this is true. There are roughly 3 forms this question can be asked in, how you interpret them is ambiguous and largely dependent on what you were taught. The nth root of a number can be mathematically expressed as: nroot(a), a^(1/n) and b^n=a Doing these one at a time b^n=a would always give n solutions, where b is the nth root of a, as this question does not require one distinct solution, there is nothing much to it. nroot(a) takes the principle root, defined as the root with the smallest argument in the complex plane out of the solutions, when multiple have the same, it takes the positive one. Lastly, a^(1/n) is much more interesting, some people would tell you "Fractional indices always give a positive", this means the solution for a would always be the, if any, positive root of a, however, this is a bit of a misunderstanding. Fractional indices also simply take the principle root, just as nroot(a), these are identical. Now going back to the video, you say that if sqrt(x) is not assumed to take the principle root, then 1 is a solution because -1 is a complex second root of 1, there's 2 things to clear up here. Firstly, -1 is not a complex root, it's just a negative one, no matter. Secondly, you are right that if such an assumption is made then 1 could be considered and option, however, as I stated previously, sqrt simply is not defined that way. the n-th root of z has n solutions, however, if a question is stated simply as "nroot(z)" then the only "proper" answer is the principle root, even in the complex plane. Now this was a whole lot of talking for a pretty boring conclusion, you are mostly right, if a question is asked "Find the nth root of z" then there are n answers, however nroot(z) or z^(1/n) would always give just the principle root, per definition.
The principle square root is not used in the first question, so it is odd that it is required in the second question. For consistency, wouldn't it make sense to say, "If question 1 is accepted to have an answer, question 2 must be accepted to have an answer for the same reason (the reason being that the root does not have to be positive)."?
It's worth noting that |x| is more correctly defined as √(x*conj(x)), which is only equal to √(x²) when x is real. In a video specifically about imaginary numbers, this distinction is actually pretty important.
On √x = -1, wouldn't x=i⁴, i⁸, i¹², ... work since, √i⁴ = i² = -1? Or is that disallowed because all those solutions reduce to 1 and you must reduce first, therefore √i⁴ = √1 = 1 ≠ -1?
√(i⁴) would be some number on a angle beetween 1 and i, cause in i² • i² the second i² would rotate the first i² to (-1)² by removing the i from itself so (-1)² • (1)² = 1, and √(1)=1, the √(x^y)=x^(y/2) doesn't works for imaginary/complex numbers as far as i know
I'm afraid not. The argument of the principal square root is half the argument of the supplied value but this has to be in a certain range. The argument of i is π/2 so we can legitimately say the argument of i⁴ is 2π. Which, if allowed would make our principal root -1 but the argument of the supplied value has to be in the range (-π,π]. 2π is not in this range so we have to take the equivalent argument of 0 which results in our root being 1 not -1.
@@CalebAnderson-gn3ne It's not pretending, it's asserting a valid definition of a square root that is different than yours. Maths is about definitions, and the definition of the root symbol is ambiguous. Claiming that √(x) = |x¹/²| is the only correct definition is misguided and unhelpful.
As a math professor and putting this kind of video for everyone to see, I think that you should at least talk about principal (and non principal) square root when saying "there is no solution to problem 2". Many might not understand otherwise...
@@Itoyokofansqrt have no well defind in C, if we make a definition of it, it turn out need to me multivalue function or it will break some of previous rule in R. Cannot have both in the same time.
What if we're in a hyperbolic space where we can define a negative norm? I know mathematicians hate hyperbolic space for some reason، but it's really important in physics
Where in physics are hyperbolic spaces important? The only example I can think of is in General Relativity, when describing a possible geometry for the universe. But since we know from measurements that our universe probably is flat, not hyperbolic, that doesn't seem so important. And what has the space being hyperbolic to do with having a negative norm?! As far as I know, spaces with negative norms are called "Pseudo-Euclidean" - not "hyperbolic".
@bjornfeuerbacher5514 It is important in relativity because (not space) but spacetime is hyperbolic. Edit: you define a norm in a "Minkowski" space with 3+1 dimensions as sqrt( -(ct^2) +(x^2) +(y^2) +(z^2)) and yeah I've heard it being described as pseudo-norm
@@Gomamon64 I've never heard that called "hyperbolic space". As far as I know, that's called "pseudo-euclidean space". A hyperbolic space would be a space with a constant negative curvature. But Minkowski space(time) has zero curvature.
@@bjornfeuerbacher5514 (Only need to fact check the negative norm part "it's the norm squared what's negative") but yeah it is hyperbolic since when you try to keep a constant norm squared you get hyperboloids, in 1+1 dimensions you get standard hyperbolas (an (x^2) - (y^2) = constant) where y is the time x c Edit: in special relativity we didn't really focus on squaring a tensor in product with itself, another way to think of it is that the metric is the identity matrix with one of its ii entries negative 1
@@Gomamon64 I know that you get a hyperboloid in pseudo-euclidean spaces when the norm (squared) is constant. But since when does that justify calling the space itself "hyperbolic"? Again, I've never seen it called that way.
Yes. Even if x is complex, |x| will always return a non-negative real number. Because the absolute value of a complex number is its distance from the origin on the complex plane, which is always real and non-negative
@@Ninja20704 That got me thinking, if we expand to non-metric spaces, e.g. Minkowski space then we could end up with potential negative "absolute" value if we define it as "Minkowksi distance to origin". Special Relativity regularly deals with "negative" distances (often called time-like separation). So |x|=-1 could have an infinite number of solutions if you think beyond Euclidean metric :).
@@vadimbich4602 but even in non-euclidean metrics, the distance metric you use must be non-negative for it to be considered a distance metric because it is one of the conditions required. A function d(a,b) must satisfy the 3 conditions to be a valid distance metric: 1) d(a,b) >=0 for all a and b, and is equal to 0 if and only if a=b 2) d(a,b) = d(b,a) i.e. symmetry 3) d(a,b)
@@Ninja20704 Minkowski space is considered pseudo-metric specifically because Minkowski metric (as in distance, not the property of space) violates normal metric rules. It is still technically a distance in that space, by definition. Minkowski space is a 4d space where distance is defined as Minkowski metric where two distinct points can be at 0 distance (light-like and violarion of rule 1) and can be at negative distance (time-like, at least in a popular definition of Minkiwski metric signature), violating rule 3.
The usual convention is that the radical sign √ has, in fact, two different interpretations depending on whether its argument is a non-negative real number or an arbitrary complex number. That is, as seemingly inconvenient as at is, the expression √x̅ is defined ambiguously. For it to have a definitive meaning, the context should clearly specify the restrictions on the value of x. If x is a non-negative real number, then √x̅ is defined as the non-negative number y such that y² = x. This is the usual definition we all learned in schools and it is very convenient, because this approach makes the square root into a well-defined function on the set of non-negative real numbers. If x is a complex number, then √x̅ is defined to have *two* values {y₁, y₂}, which are the solutions of the equation y² = x. The only exception is x = 0, for which √x̅ is the *single* value 0. In order to distinguish this definition of √ from the usual one studied in school, we sometimes call them "the complex square root" and "the real square root", respectively. They are closely related, but they are not the same thing. When we need the complex square root to be a well-defined (non-multivalued) function, we may split it into the so-called "branches", which are usually denoted as (√z̅)₀ and (√z̅)₁. The former usually corresponds to what was referred to in this thread as the "principal value of the square root" and the latter is just -(√z̅)₀. This is a rather ad hoc notation, thought, as the choice of the branches may be different depending on the circumstances. Namely, it depends on the region of the complex plane in which we want the branches to be continuous. So, since values like √-̅4̅ can not be interpreted using the real square root, we are forced to use the complex square root instead. Thus, by definition, √-̅4̅ = ±2𝒊. Or, in the more technical terms, √-̅4̅ is the two-element set {2𝒊, -2𝒊}. Likewise, √-̅9̅ = {3𝒊, -3𝒊}. This implies that √-̅4̅ • √-̅9̅ = {+6, -6}. Or, more shortly, √-̅4̅ • √-̅9̅ = ±6. It should be noted, however, that those kinds of arithmetic operations on sets are somewhat tricky, and so we would generally simply avoid them altogether in favor of operating on the elements of the sets instead. And note also that even though we may write something like z = √w̅ for complex numbers z and w, this notation is somewhat informal. Technically, it should be (and it usually is) written as z ∈ √w̅, meaning that z is an element of the set of numbers that are the square roots of w. I hope this makes things a little more clear. by @fimmind
sqrt(1) being +1 and not -1 is just a convention from the pre-imaginary numbers times, isn't it? If we go to the complex plane, the operation of sqrt() = ^(1/2) means simply cutting a phase angle by ratio of 2 and number 1 (having phase of 2n*Pi) nicely fits the solution, because phase of -1 on the plane is (2n+1)*Pi, right?
The absolute value has the following four fundamental properties: • |a| ≥ 0 - *non-negativity* • |a| = 0 a = 0 - positive-definiteness • |a⋅b| = |a|⋅|b| - multiplicativity • |a + b| ≤ |a| + |b| - the triangle inequality
@@cyberagua These four properties are always satisfied only when you're working with subsets of complex numbers, so other set of numbers may not satisfy all four properties
because if theres a square root in an expression, it always uses the principal root, and the prinicipal root has its outcome as always positive, the absolute value of a nornal root
If you square -3 you get 9 and the square root of that is 3. So it returns the absolute value. Technically you can't cancel a square and an outer square root if the number that's squared is negative.
Any tips for solving word problems? My brain gets caught around everything thats written down, and I have trouble coming up with answers. Im on an honors algebra level, 8th grade
@@bprpmathbasics Two towns, Town A and Town B, are 300 miles apart. A train leaves from Town A at a constant speed of 60 miles per hour, but it is delayed and departs 2 hours after a car leaves Town B. The car is traveling at 40 miles per hour. 1. Let f(t) represent the position of the train (in miles) from Town A after t hours. Similarly, let g(t) represent the position of the car from Town B after t hours. Write functions f(t) and g(t) to represent their positions, considering the delay of the train. 2. The vehicles meet when the distance between them is zero. Use an absolute value inequality to express the condition when the distance between them is less than or equal to 10 miles, representing the “meeting zone.” Solve for the time t when this occurs. 3. How far has each vehicle traveled when they enter this meeting zone? Bonus: If the speed of the car decreases by 5 miles per hour after every hour of travel, and the train increases its speed by 10 miles per hour after the first hour of travel, how would this affect the meeting time and the distance each has traveled?
@@bprpmathbasics Two towns, Town A and Town B, are 300 miles apart. A train leaves from Town A at a constant speed of 60 miles per hour, but it departs 2 hours after a car leaves from Town B. The car, initially traveling at 40 miles per hour, begins slowing down at a constant rate of 5 miles per hour per hour, starting 1 hour after its departure. Let f(t) represent the position of the train from Town A after t hours since its departure. Let g(t) represent the position of the car from Town B after t hours since the car’s departure. a. Write a function f(t) to represent the position of the train after t hours, considering the 2-hour delay. b. Write a piecewise function g(t) for the car’s position. The first part should represent its motion in the first hour (constant speed), and the second part should account for the car slowing down after the first hour. The two vehicles meet when the distance between them is zero. However, define the “meeting zone” as the condition when the distance between them is less than or equal to 10 miles. Write an absolute value inequality to express the condition when the distance between them is within this meeting zone, and solve for the time t when they enter this zone. |f(t) - g(t)| \leq 10 Solve for the exact time t when they will meet (i.e., when the distance between them is exactly 0 miles). How far has each vehicle traveled by this time? Use your functions from Part 1 to find the distances. Suppose after 3 hours of travel, the train increases its speed by 15 miles per hour for each subsequent hour, while the car’s speed decreases by 10 miles per hour per hour after 2 hours of travel. Adjust your functions for f(t) and g(t) accordingly, and recalculate the time and place where they will meet.
A logistics company operates a delivery service between two cities, City A and City B, which are located exactly 18 miles apart along a straight highway. The company’s central warehouse is situated between the two cities, 5 miles west of City A and 13 miles east of City B. The company uses a complex fee structure that depends on the distance from the warehouse and whether the delivery occurs within certain specified zones. 1. Zone 1: For any delivery within a 10-mile radius of the warehouse (excluding City A and City B), the delivery fee is calculated using the function f(x) = 7|x - w| + 25 + p , where w represents the location of the warehouse on the highway, x is the distance (in miles) from the warehouse to the delivery point, and p is an additional surcharge based on the road conditions, which is a linear function of distance: p(x) = 0.5|x - 3| . The surcharge only applies when the delivery occurs within the first 3 miles from the warehouse.
in complex system sqrt(x)=-1,by Euler's equation we have x=e^(4n+2).pie.i, so x=1 for all integer n we check it sqrt(1)=sqrt(i^4)=i^2=-1 Do have any wrong in process??
It's sort of due to definitions. "Square root" refers to the "second roots of x", per defintion, you always have n solutions to the nth root of a number, in other words, there are 2 second roots of x. Now that means that 1 and -1 are indeed both "square roots of 1", but sqrt(1) as an operation takes what is known as the principle root, as would nroot(x). This means there is always only 1 outcome from the operation, in the case of square roots, that would always be the positive root, in this case 1. I'm not sure if that answered your question?
At 1:28 "don't forget the plus or minus ...." There he makes a mistake. The √ symbol has been defined to indicate the principal root of a value. And that means that √ (x^2) is defined as just one specific value, even when x is complex. So no "plus or minus" because now in the video √ (x^2) has two different values. A value x can have two square roots but only one of them may be written as √ x
Two points: • when solving equations like x² = a with a > 0, it is not correct to write √x̅² = ±√a̅, since the _arithmetic_ real-valued square root is defined to always be non-negative; a more correct way is just writing x² = a => x = ±√a̅ (when a is positive) • when solving irrational equations over the field of complex numbers ℂ, the _algebraic_ complex nth root of a non-negative complex number is normally defined to have n distinct complex values, so now √−̅1̅ = ±𝒊 and √1̅ = ±1 These are just two different conventions: • the _arithmetic_ real-valued and single-valued square root √4̅ = +2 • the _algebraic_ complex multivalued square root √−̅4̅ = ±2𝒊 Some algebra books I've seen distinguish between the arithmetic and algebraic notations, by adding a subscript or an underscript plus sign: • √₊̅4̅ = +2 - arithmetic • √−̅4̅ = ±2 - algebraic
abs(x) = -1 has no solution Meanwhile four-vectors that can have scalar product of himself and himself (technically norm of that vector in power of two) negative.
We could just define a new Nummer j where sqrt(j) = -1. (Like we did with the Im. Number). In this case we can actually calculate a result. (But it is only valid when j is valid.) So there is no real or complex solution but a "j" solution for Q2 and Q3. This just came to my Mind and maybe there is a better way to descibe this but this should work.
Nah. The thing is when you're talking about square root of 1 it's actually +/- 1. But when you write the √ sign, it's not actually the square root, but positive square root (at least in US). So it just can't be negative by definition, so there just is no j, for which √j = -1 in any number field.
@@F1r1at Oh. That is interresting. Oh now I remember somthing from the Notes a good Friend gave me from Univetite. But idk what it is calles in the US. In German the concept is Called "Abbildungen" and there are like these strikt rules. One of them was f: R -> R. Which means that a real Value that is put into a function should only result in one real Result. So f: x -> sqrt(x) has only the positiv Result because of this Rule. And that is why sqrt(x) cant be negativ. Right?
@@c42xe It's just because of the definition, that √a is a positive result for square root of a. Cause like result of actual square root (I don't recall what it's official name is) is the number that if squared will give you the number under the root. So like 2^2 = 4 and (-2)^2 = 4, so square root of 4 is 2 and -2. But since it's hard to use functions that have more then one output people decided that function sqrt (or √) will return just the positive answer. At least that's how people explained that thing to me. So basically what √ does - is finds all the roots and leaves only the positive one. That's why it can't be negative.
@@F1r1at you're mistaking taking a root with finding a root. One is a function and the other are solutions to a polynomial. Not the same thing. Odd roots have negative outputs, even roots don't, that's just the byproduct of being a function.
The first equation works because you can imagine that it works, but the other two equations don’t work because you cannot imagine that they work. SQRT(x) = -1 if you square both sides you get X = 1 but if you retake the Square root you get back +/- 1 so it’s only partially indeterminate, the answer is null, 1. I think this would be the equivalent of a super position of states, the answer is both in determinant and real at the same time.
I don't think the notion that square root symbol is a principal root automatically applies to complex numbers. When I was taught about square root on C, we started with multivalued functions and transitioned to Riemann surfaces and 4d space to represent square root on C. I think unless explicitly stated that we're talking about principal root sqrt(x)=-1 definitely has solutions on C.
A question: Why simplify the first equation to i = root(-1)? Why not simply use the i²=-1 definition? Especially since you didn't give the definition of your root sign, and many people only know the principal square root, and using that one your substitution is simply wrong, while i²=-1 holds true still. Hell, your second example immediately starts off with a reminder that root(x)² is not equal to x, which makes it all the more confusing why this choice was made. Also in videos like these it's always weird to arbitrarily choose the vector space, couldn't I simply choose a vector space where I define these functions in a way that these equations can be solved? After all, the meaning of even |x| and root(x) are bound by axioms and thus maleable as you want. Hell, I could make it easy and not use a vector space at all. The solution is closer to "none the vector space R, one in C, and we could craft xyz so that [...]"
In general, it's the difference in notation between saying x^2 = 25 and saying sqrt(25) = x. The first is a quadratic equation and it has 2 solutions, even assuming that we are only using the principal root. The second is not a quadratic, and it only has 1 solution assuming the principal root. With x^2 = 25, when assume the principal square root we still get 2 solutions, not because sqrt(25) is +/- 5, but because sqrt(x^2) = |x|. And |x| is where the +/- really comes from. So when we take the root of both sides, instead of getting x = +/- 5 what we really get |x| = |5|, or just |x| = 5. Which means that x is either 5 or -5. The same applies to the video, we get |x| = i, therefore x is either i or -i. (there is a slight hiccup here, depending on your definition of absolute value, as the distance from origin definition would mean that only positive real numbers can be solutions and mean that |x| cannot be i, but if we don't rely explicitly on absolute value for choosing the principal root, then i and -i work fine) For sqrt(25) = x, the principal root simply gives us 5 = x. So -5 isn't a solution because it's not the principal root of 25. In regards to the video, sqrt(x) = -1 has no solutions because -1 is not the principal square root of any number, even in the complex world. All of his solutions are consistent with the principal root.
There's no need to overcomplicate it. sqrt(x) as an operation, just as nroot(x), always takes the principle root, in the case of square roots this is always positive. It is undeniably true that sqrt(-1) is i, and i only because of this rule. For the first one, the definition sqrt(-1) = i is (probably) used because in every case, the 2 second roots of a number are equal in magnitude, but in opposite directions, that is, they are equal to each other times -1. This means he can use one solution to easily find the other, solving sqrt(-1) as i and saying +/-.
sqrt(x) in the complex world pretty much just halves the rotation and square rooted magnitude, although to get -1, you have to get 180 degree rotation which means your initial rotation is 360, so positive real numbers... and you already see how it doesn't work because 360 degrees is just 0 degrees and you can't divide it so basically square root can output any complex number except negative real numbers then i think absolute value for complex numbers just gives its magnitude and magnitude is always only positive
sqrt(x) would always give the positive root of x, because the principle second root of any number will always be positive. nroot(z) is defined as taking the principle nth root of z, and does not always have to be positive, it must simply have the smallest argument in the complex direction, in other words, it is the one where k=0 in the form sqrt(a^2+b^2) x e^(Theta/n + 2k x pi/n). Where a and b are the magnitudes of direction in the real and complex direction. In other words, it is not entirely true that the square root operation can output any complex number, it would only output the principle square root, even when taken of a complex number.
@@mxcoyl Both are solutions, since for z = exp(𝒊φ) we have: wₖ = √z̅ = z¹ᐟ² = exp(𝒊(φ+2πk)/2), k ∈ ℤ, so for φ = 0 we get two root values: • w₀ = exp(𝒊(0+2π·0)/2) = exp(0) = 1 • w₁ = exp(𝒊(0+2π·1)/2) = exp(𝒊π) = -1
So I had to talk about equation 2 but since many people talked about this, I won't tell you. However, I have something to talk about the third equation because that inspired me to rewrite algebra.
The third equation most definitely has no solutions (neither real, nor complex), while the second equation most definitely has a solution x = 1 with the definition of the multivalued complex square root, so that √1̅= ±1 = {1, -1} ∋ - 1.
Sqrt(1) is only 1, not - 1, not +/- 1. It is not the same as solving x^2 = 1. Have you ever heard anyone say sqrt(2) = +/- 1.4142… sqrt(3) = +/-1.732… or etc?
@@Ninja20704, yeah, I know that a root usually is understand as positive only value. But in common way its negative value can't be ignored otherwise we would be loss some solutions. (But sometimes we get extra.) It is known that root of N power is an N-arity function. So square root has 2 values, + and -. If you image 1 and -1 on complex plane, you can see that the 1+0i possible to turn on π angle twice to get 1+0i again. Two sequential equal turns is same that squaring thus exp(iπ) aka -1+0i is possible solution of √1
@@Qraizer we say one solution is sqrt(1) and the other is -sqrt(1). We write the negative sign infront of the sqrt to get the negative answer. Just think about how we solve x^2 = 2. We say one answer is sqrt(2) and the other is -sqrt(2). If you are talking about complex numbers, then the sqrt will only return the principal answer, which is when we write the inside using the principal argument, the argument that lies in (-pi,pi]. Again, no one ever says sqrt(-1) = +/- i
@@Ninja20704 , yes, we use only the principal values of roots. Because it is convenient for us when function has just one value. But this don't change the fact that root of N power is N-arity function. When we consider ℝ instead of ℂ, for odd N we get only 1 real value due to rest are complex. But for even N we can get 2 real values. If we want. I think that strict solution of √x = -1 is not question of our desires but is question of exact properties of square root as a function.
In the first example, you indicate that the sqrt is (+/-)... but in the second example you say -1 is an extraneous solution, but if you indicate the sqrt as (+/-) as in the first example, then 1 and -1 are both valid solutions. Why do we have (+/-) i as two valid answers in the first case, but not (+/-) 1 as two similarly valid answers in the second case?
@@_rd_5043 The real issue is that sqrt(x) = -1 for real numbers has no solutions because sqrt(x) simply IS the principal (positive) square root. sqrt(z) = -1 as a single-valued complex function only has a solution if you pick the appropriate branch (and it is 1 in that case). You can't always rely on exponential rules to work in complex numbers when one of the exponents has a fractional part, but I believe you can choose a branch of the function z^(fractional part) so that it does work in whatever specific case you're dealing with. I don't know if it works with irrational fractional parts, but I'm 99% sure it does with rational ones. Could be wrong though, I was never a complex analysis guy.
I disagree on question 2, as 1 is only an extraneous solution if you consider the square-root function to be single-valued, which I don't think is a fair assumption to make in complex land, beyond computation, when it provides a different result than considering multivaluedness. Considering sqrt(x) as the inverse to x², such that (sqrt(x))² = x, we want sqrt(x) = -1. Square both sides, x = 1, and this is fine, as sqrt(1) = ±1 through consideration of the other branches of the square-root function.
Maybe I'm unpopular with this, but in my opinion the idea that √x = |x¹/²| needs explicit assertion, it is not implicitly true. It is sometimes true, like in geometric context where negative lengths can't exist directly it can be taken as true, and it is used in the definition of standard deviation, but it's not generally true and therefore needs an assertion. Maths should not depend on pedantry, and taking a root is the inverse operation of exponentiation, and should be treated as such.
taking the square root of a number should return 2 possible values, right? A positive and a negative? So sqrt(x)=-1 would absolutely have the solution of x=1 because sqrt(1)=±1, and one of those sides is -1. It works???
"taking the square root of a number should return 2 possible values, right? A positive and a negative? " No. The square root of a (non-negative) number is _defined_ as the non-negative number whose square is the original number. "sqrt(1)=±1" No. sqrt(1) = 1 only. sqrt(1) = -1 is wrong. Look into a math textbooks how the "square root" of a number is defined, if you don't believe me.
@@bjornfeuerbacher5514 V1=±1, because (±1)²=1, but 1(=V1) is the main, arithmetic, principal value. E.g., 2+V9=2±3=5 or -1; V1-V(16)=1-4=-3. Remember that.
solve sqrt(x)=-1 has no silution at all? I'm stuck somehow: Simplifying: sqrt(x) =-1 -1 =e^(i(pi+2npi)) Prinzipal value n=0 -1 = sqrt[(e^i(pi+2npi))²] = sqrt[e^(i(2pi+2×2npi))] For sqrt(e^(i0pi)) it doesn't work, since its =1 but e^(i2pi), e^(i6pi) and so on?
When you are new to the complex world and realise that (-i)² is a triple negation.
2 месяца назад+1
X^2 = -1 had no solution before a formular to solve cubic equation which cause the founding of complex number So I believe sqrt(x) = -1 and abs(x) = -1 will be solvable oneday.
The so-called "principal" roots are defined arbitrarily and are generally useless in complex algebra, since, for example, the cube root ³√−̅8̅ has three distinct values: ³√−̅8̅ = {−2; 1+𝒊√3̅; 1-𝒊√3̅}, and it's hard to decide which of them is more "principal" than the others 🤷♂️ And the same goes for the complex square root as well.
@@cyberagua Nope. √−8 = −2 is the only correct solution, and it isn't arbitrary at all. However, x³ = −8 has three solutions: x₁ = −2 ∨ x₂ = 1 − i√3 ∨ x₃ = 1 + i√3, but that's a completely different question.
@@cyberagua WolframAlpha gives the solution which lies in the first quadrant of the complex plane, i.e. non-negative values on both real and imaginary axis. Other calculators return -2 for ³√-8. What do you mean, what grade am I in?
fun fact: on the first equation he wrote, and proved, that the solution for X² is ±i but, for the second equation ( √x=-1) he claimed that √1=1 instead of √1=±1. as a psychological trick that some people may not notice, he gave 3 equations, one that only has solutions under certain conditions, one that has a similar solution but claimed it doesn't has any and an absurd equation that, clearly, has no solution. this last equation seems to validate that the second also doesn't has a solution despite having one that is coherent with the way he solved the first.
The rule z^(m/n) = [z^m]^(1/n) does not hold if m/n is not the simplest form of the fraction and z is non-real (or real but negative). So (i^4)^(1/2) ≠ i^(4/2)
For equation 3, you should also mention, for the equation abs(x)=sqrt(x^2), that this works as long as x is a real number. Otherwise, you would get the following for x=i, abs(i) = sqrt(i^2) = sqrt(-1), but sqrt(-1) = i therefore, abs(i) = i, which is not true at all since abs(i) = 1
Because sqrt(x^2) = abs(x), not x, as he explained near the end of the video. It is often treated as if sqrt(x^2) = x because it works when x >= 0, but if x < 0 or is imaginary it is important that the abs(x) is used
The rule of exponent z^(m/n)=[z^m]^(1/n) does not apply if m/n is not the simplest form of the fraction and z is non-real (or even negative real numbers). So [i^4]^(1/2) ≠ i^(4/2)
Because x^(1/2) =/= sqrt(x) Really, the fundamental operation as described with real values, as you see here, is poorly defined in complex values. It results in multiple 'branches' of possibilities - as 1=i^4=i^8 etc so the 'general square root of 1' (vaguely defined) can equally be 1, i^2, i^4 etc. The square root for complex values is defined to be the *principal* result of all such equivalences. That is, effectively, the value being square rooted is first reduced to 'simplest' form before the operation proceeds, therefore giving a consistent result for equivalent inputs. The language I use here is deliberately vague as a lot of the precision makes more sense using polar form, where i=e^(pi/2) [or, technically, i=e^(pi/2+2*n*pi) where n is an integer]
The same reason why 1=sqrt(1)=sqrt(-1 * -1)=sqrt(-1)sqrt(-1)=i*i=-1 is not valid. Many of the exponential laws we're so attached to simply don't work anymore with complex numbers
Why does sqrt(1) not equal one? Look, If we write 1 as e^(2i*pi) and then take the sqare root of it which is by definition: e^(i * pi), because you have to devide the exponent by 2, you get -1 as an answer! So where is my mistake?
You are right and Steve is wrong here. Over the complexes, the square root of a nonzero complex number is normally defined to have two distinct values: √−̅4̅ = ±2𝒊 = {2𝒊; − 2𝒊} and the cube root of a nonzero complex number has three distinct values: ³√−̅8̅ = {−2; 1+𝒊√3̅; 1−𝒊√3̅} While over the non-negative reals we can define a so-called _principal_ square root, it makes very little sense (or rather no sense at all) over the set of complex numbers ℂ.
The real number -1 is a completely valid square root of 1, I have no idea why there's so much racism between principal and non-principal roots when -1 and 1 have the same square. Unless the square root is considered a function and not an operator, but it's still misleading. As for |x| = -1, I don't even know why that needed to be explained with anything other than "The absolute value of any number is strictly non-negative", because that is (practically) the whole truth. Obviously there are complicated reasons as to why it's true, but that's a whole other topic.
Sqrt(x) is range-restricted. You used the exact principle that means sqrt(x)=-1 has a solution in your video. Sqrt(-1) is + OR - i. You said that, for reference, and a rule like that about an operation always applies, or never applies. Therefore, sqrt(1) is +/-1, or x=-i is not a valid solution for x^2=-1. The reason sqrt(x) the function is range-restricted is because otherwise it wouldn’t be a function. However, if you make it a relation instead, and use its maximal range, it is just a sideways parabola. I assume you knew all the info I just said, I just wanted to show my reasoning for why sqrt(x)=-1 has a solution that isn’t just pedantic. (Or at least, no more pedantic than mathematics must be)
Why is x≠-1 in equation 2? After all if x = 1, -1^2=1, but so does 1^2… To me, in equation 2, x=±1, But, as some others pointed out, [e^2i(pi]=1 On top of this, if you take a square root of a number, you are basically doing x^1/2. Back to that problem, if we take the square root [e^2i(pi)] it’s equal to -1.which is the same solution, just in a different form, which would reinforce the solution of x=-1. On top of this, if we were to consider that the square root of i is i/±1. Neither of these are equal to -1, because it’s either -i or just i. However, we know that i^2=-1 and -i^2=-1. That doesn’t matter though, even if it does restate the equation.
Because, unless specified otherwise, sqrt(x) means the principal square root, and that is defined to be the positive root over the real numbers. There are lots of good reasons for that mainly related to making the square root function single-valued and making sqrt(1) = 1 true.
Hold on. For the second case, since you are taking a square root of 1, you have two possible answers, 1 or -1, don't you. I thought square root give two possible answers. One answer may work or two answers may work. So, I am confused.
but that doesnt make any sense, the absolute value of something is always positive, if |A| = -1 then it would literally defy the entire purpose of the "absolute value"
Is sqrt(x^2)=abs(x) true for everything?
ruclips.net/video/IFWyGPYTCPI/видео.html
real numbers only right?
I wonder if its possible to graph the functions and the complex solutions together, and see if they cross each other (similar to graphing them with real numbers)
@@YarinGD Consider watching "Imaginary Numbers Are Real" by Welch Labs
@@YarinGDда.
It is conditionally true. It is true if x can be shown to be equal to or larger than 0 (in geometric context, for instance), or if it can be shown that the sign for x is meaningless, like in the calculation for standard deviation where all you're interested in is the size and not the direction.
In plain algebra, it is false. Because the root is the inverse operation to the exponent. √x = x¹/²
last time i was this early Archimedes was still finding the area of a circle
It's 2πr
@ZDTF bro that's the perimeter
Oh my bad G
Nah you are fine
^^ Bro forgot to take his meds 💀
Regarding equation 2.
If we assume that x is a real number, there are no solutions because √ always means "principle/positive square root", so it can't take negative values.
If we assume that x is complex, then we have to choose from two possible meanings of √.
If we understand √ as "principle square root", then there are, again, no solutions, because the principal (complex) square root of any real number is a real positive number, or a product of a real positive number and i.
If, however, we don't assume that √ is a principal square root, then there is a solution: x=1, because (-1) * ( -1) = 1, and thus -1 is one of the (complex) square roots of 1.
In general, n-th root of x is a n-valued function for complex values of x, and none of those n different roots can be considered any "better" than others, so in general we can't discard any of them when talking about n-th roots of complex numbers.
Sometimes, though, it is convenient to talk about roots as single-valued functions, so the notion of principal roots is still useful for complex numbers.
It is very important to know exactly what the mathematical notation that we're using at any given point actually means.
As a sidenote, more often than not √ DOES mean "principal root", so the answer in the video is still correct, most of the time.
That's not a sqrt(x)=-1, that's x²=1
Similarly, not root_n(x)=-1 but xⁿ=1
Finding root of a complex polynomial is not the same as taking a root of a complex number, because taking a root is a function, 1 output, while solving an equation can have many solutions.
@@_rd_5043 A complex n-th root of z is a multi-valued function, with n distinct branches (values) for all complex z except 0. It is most definitely not 1 value, unlike real root functions.
The principle complex n-th root, however, IS a 1-value function, taking a value of a certain ("principle") branch of the multi-valued root. If we only consider the principle root to be the "correct" root, we lose certain properties of exponents that we're used to from real analysis (like √(xy) = √x * √y for example).
See also: roots of unity.
What's important is the wording or symbols used.
As you've mentioned, there are always n, n-th roots of any number, this is true.
There are roughly 3 forms this question can be asked in, how you interpret them is ambiguous and largely dependent on what you were taught.
The nth root of a number can be mathematically expressed as:
nroot(a), a^(1/n) and b^n=a
Doing these one at a time
b^n=a would always give n solutions, where b is the nth root of a, as this question does not require one distinct solution, there is nothing much to it.
nroot(a) takes the principle root, defined as the root with the smallest argument in the complex plane out of the solutions, when multiple have the same, it takes the positive one.
Lastly, a^(1/n) is much more interesting, some people would tell you "Fractional indices always give a positive", this means the solution for a would always be the, if any, positive root of a, however, this is a bit of a misunderstanding. Fractional indices also simply take the principle root, just as nroot(a), these are identical.
Now going back to the video, you say that if sqrt(x) is not assumed to take the principle root, then 1 is a solution because -1 is a complex second root of 1, there's 2 things to clear up here.
Firstly, -1 is not a complex root, it's just a negative one, no matter.
Secondly, you are right that if such an assumption is made then 1 could be considered and option, however, as I stated previously, sqrt simply is not defined that way.
the n-th root of z has n solutions, however, if a question is stated simply as "nroot(z)" then the only "proper" answer is the principle root, even in the complex plane.
Now this was a whole lot of talking for a pretty boring conclusion, you are mostly right, if a question is asked "Find the nth root of z" then there are n answers, however nroot(z) or z^(1/n) would always give just the principle root, per definition.
@@westy9447 To properly appreciate this way of thinking, one has to go all the way to Rienman surface. But that is really neat.
The principle square root is not used in the first question, so it is odd that it is required in the second question. For consistency, wouldn't it make sense to say, "If question 1 is accepted to have an answer, question 2 must be accepted to have an answer for the same reason (the reason being that the root does not have to be positive)."?
It's worth noting that |x| is more correctly defined as √(x*conj(x)), which is only equal to √(x²) when x is real. In a video specifically about imaginary numbers, this distinction is actually pretty important.
|x| = √(x²) is only generally true for real values of x.
sqrt(x * conj(x)) works though.
@@chaosredefined3834 OK, I should have said only generally true. I'll edit my post.
it is only true if you take the prinipal root which is only defind for real x
@@CalebAnderson-gn3ne Principal root is defined for complex numbers as well. The radical symbol (√) refers explicitly to the principal root.
I don't get it: How is the absolute value of x equal to the square root of x²?
I wish they had just made another symbol for principal root and ±√
On √x = -1, wouldn't x=i⁴, i⁸, i¹², ... work since, √i⁴ = i² = -1? Or is that disallowed because all those solutions reduce to 1 and you must reduce first, therefore √i⁴ = √1 = 1 ≠ -1?
√(i⁴) would be some number on a angle beetween 1 and i, cause in i² • i² the second i² would rotate the first i² to (-1)² by removing the i from itself so (-1)² • (1)² = 1, and √(1)=1, the √(x^y)=x^(y/2) doesn't works for imaginary/complex numbers as far as i know
I'm afraid not. The argument of the principal square root is half the argument of the supplied value but this has to be in a certain range. The argument of i is π/2 so we can legitimately say the argument of i⁴ is 2π. Which, if allowed would make our principal root -1 but the argument of the supplied value has to be in the range (-π,π]. 2π is not in this range so we have to take the equivalent argument of 0 which results in our root being 1 not -1.
@@reizinhodojogo3956x^(y/2) = ±√(x,y) for all values of x and y where x≠0. This holds true for complex values as well.
The later statement is correct.
sqrt(i^4) IS NOT i^2 but |i^2| which is 1
I like how square rooting a negative number is allowed in a significant part of higher maths but squaring a negative number is not
it is not that squaring a negative is not allowed, it really is pretending that a principal root is the inverse of squaring that is not allowed
@@CalebAnderson-gn3ne It's not pretending, it's asserting a valid definition of a square root that is different than yours. Maths is about definitions, and the definition of the root symbol is ambiguous. Claiming that √(x) = |x¹/²| is the only correct definition is misguided and unhelpful.
As a math professor and putting this kind of video for everyone to see, I think that you should at least talk about principal (and non principal) square root when saying "there is no solution to problem 2". Many might not understand otherwise...
What about e^x=-1? What would that be? Probably no solution, but I would like to see if there is anything interesting involved
The answer is iπ (and in fact, i*odd*π). You can see my video here ruclips.net/video/Jqt67tJCiwQ/видео.htmlsi=GJNrBy4D02FC5UjD
so... is sqrt(e^2ipi) = -1
then if x = e^2ipi
sqrt(x) =-1
@@Itoyokofan "sqrt(e^2ipi) = -1"
No. e^2ipi is equal to 1, so sqrt(e^2ipi) is 1, not -1.
@@Itoyokofansqrt have no well defind in C, if we make a definition of it, it turn out need to me multivalue function or it will break some of previous rule in R. Cannot have both in the same time.
X is "πi" Or not?
Solving x^4=1
ruclips.net/video/8qhGzsCyklQ/видео.html
±1, ±i
What if we're in a hyperbolic space where we can define a negative norm?
I know mathematicians hate hyperbolic space for some reason، but it's really important in physics
Where in physics are hyperbolic spaces important? The only example I can think of is in General Relativity, when describing a possible geometry for the universe. But since we know from measurements that our universe probably is flat, not hyperbolic, that doesn't seem so important.
And what has the space being hyperbolic to do with having a negative norm?! As far as I know, spaces with negative norms are called "Pseudo-Euclidean" - not "hyperbolic".
@bjornfeuerbacher5514
It is important in relativity because (not space) but spacetime is hyperbolic.
Edit: you define a norm in a "Minkowski" space with 3+1 dimensions as sqrt( -(ct^2) +(x^2) +(y^2) +(z^2)) and yeah I've heard it being described as pseudo-norm
@@Gomamon64 I've never heard that called "hyperbolic space". As far as I know, that's called "pseudo-euclidean space".
A hyperbolic space would be a space with a constant negative curvature. But Minkowski space(time) has zero curvature.
@@bjornfeuerbacher5514
(Only need to fact check the negative norm part "it's the norm squared what's negative") but yeah it is hyperbolic since when you try to keep a constant norm squared you get hyperboloids, in 1+1 dimensions you get standard hyperbolas (an (x^2) - (y^2) = constant) where y is the time x c
Edit: in special relativity we didn't really focus on squaring a tensor in product with itself, another way to think of it is that the metric is the identity matrix with one of its ii entries negative 1
@@Gomamon64 I know that you get a hyperboloid in pseudo-euclidean spaces when the norm (squared) is constant. But since when does that justify calling the space itself "hyperbolic"? Again, I've never seen it called that way.
1) ±i; 2) Ø in R, 1 in C; 3) Ø in elem. alg., -1 in higher alg.
I could tell |x| = -1 didn't make sense at a glance, because there's no number where its absolute value is negative, right?
Yes. Even if x is complex, |x| will always return a non-negative real number. Because the absolute value of a complex number is its distance from the origin on the complex plane, which is always real and non-negative
@@Ninja20704 That got me thinking, if we expand to non-metric spaces, e.g. Minkowski space then we could end up with potential negative "absolute" value if we define it as "Minkowksi distance to origin". Special Relativity regularly deals with "negative" distances (often called time-like separation). So |x|=-1 could have an infinite number of solutions if you think beyond Euclidean metric :).
@@vadimbich4602 but even in non-euclidean metrics, the distance metric you use must be non-negative for it to be considered a distance metric because it is one of the conditions required.
A function d(a,b) must satisfy the 3 conditions to be a valid distance metric:
1) d(a,b) >=0 for all a and b, and is equal to 0 if and only if a=b
2) d(a,b) = d(b,a) i.e. symmetry
3) d(a,b)
@@Ninja20704 Minkowski space is considered pseudo-metric specifically because Minkowski metric (as in distance, not the property of space) violates normal metric rules. It is still technically a distance in that space, by definition. Minkowski space is a 4d space where distance is defined as Minkowski metric where two distinct points can be at 0 distance (light-like and violarion of rule 1) and can be at negative distance (time-like, at least in a popular definition of Minkiwski metric signature), violating rule 3.
The usual convention is that the radical sign √ has, in fact, two different interpretations depending on whether its argument is a non-negative real number or an arbitrary complex number. That is, as seemingly inconvenient as at is, the expression √x̅ is defined ambiguously. For it to have a definitive meaning, the context should clearly specify the restrictions on the value of x.
If x is a non-negative real number, then √x̅ is defined as the non-negative number y such that y² = x. This is the usual definition we all learned in schools and it is very convenient, because this approach makes the square root into a well-defined function on the set of non-negative real numbers.
If x is a complex number, then √x̅ is defined to have *two* values {y₁, y₂}, which are the solutions of the equation y² = x. The only exception is x = 0, for which √x̅ is the *single* value 0. In order to distinguish this definition of √ from the usual one studied in school, we sometimes call them "the complex square root" and "the real square root", respectively. They are closely related, but they are not the same thing.
When we need the complex square root to be a well-defined (non-multivalued) function, we may split it into the so-called "branches", which are usually denoted as (√z̅)₀ and (√z̅)₁. The former usually corresponds to what was referred to in this thread as the "principal value of the square root" and the latter is just -(√z̅)₀. This is a rather ad hoc notation, thought, as the choice of the branches may be different depending on the circumstances. Namely, it depends on the region of the complex plane in which we want the branches to be continuous.
So, since values like √-̅4̅ can not be interpreted using the real square root, we are forced to use the complex square root instead. Thus, by definition, √-̅4̅ = ±2𝒊. Or, in the more technical terms, √-̅4̅ is the two-element set {2𝒊, -2𝒊}. Likewise, √-̅9̅ = {3𝒊, -3𝒊}. This implies that √-̅4̅ • √-̅9̅ = {+6, -6}. Or, more shortly, √-̅4̅ • √-̅9̅ = ±6.
It should be noted, however, that those kinds of arithmetic operations on sets are somewhat tricky, and so we would generally simply avoid them altogether in favor of operating on the elements of the sets instead.
And note also that even though we may write something like z = √w̅ for complex numbers z and w, this notation is somewhat informal. Technically, it should be (and it usually is) written as z ∈ √w̅, meaning that z is an element of the set of numbers that are the square roots of w. I hope this makes things a little more clear.
by @fimmind
sqrt(1) being +1 and not -1 is just a convention from the pre-imaginary numbers times, isn't it?
If we go to the complex plane, the operation of sqrt() = ^(1/2) means simply cutting a phase angle by ratio of 2 and number 1 (having phase of 2n*Pi) nicely fits the solution, because phase of -1 on the plane is (2n+1)*Pi, right?
Sure, that's perfectly correct! 👍
Во втором x = 1 , т.к. √1 = ±1, а не 1 (см. формулу Муавра) (если мы работаем с комплексными числами)
Совершенно верно 👍
The absolute value of a split-complex number can be negative
But then it's not an absolute value (in the usual sense). But nevertheless, can you please provide an example?
@cyberagua
The hyperbolic unit j has an absolute value of -1
@@JJean64 Can you please provide a source?
The absolute value has the following four fundamental properties:
• |a| ≥ 0 - *non-negativity*
• |a| = 0 a = 0 - positive-definiteness
• |a⋅b| = |a|⋅|b| - multiplicativity
• |a + b| ≤ |a| + |b| - the triangle inequality
@@cyberagua
These four properties are always satisfied only when you're working with subsets of complex numbers, so other set of numbers may not satisfy all four properties
4:44 I don't get it - the square root of x² cancels the square, leaving x
How is that the absolute value of x?
because if theres a square root in an expression, it always uses the principal root, and the prinicipal root has its outcome as always positive, the absolute value of a nornal root
If you square -3 you get 9 and the square root of that is 3. So it returns the absolute value. Technically you can't cancel a square and an outer square root if the number that's squared is negative.
Any tips for solving word problems? My brain gets caught around everything thats written down, and I have trouble coming up with answers. Im on an honors algebra level, 8th grade
Do you have a problem that you are working on?
@@bprpmathbasics Two towns, Town A and Town B, are 300 miles apart. A train leaves from Town A at a constant speed of 60 miles per hour, but it is delayed and departs 2 hours after a car leaves Town B. The car is traveling at 40 miles per hour.
1. Let f(t) represent the position of the train (in miles) from Town A after t hours. Similarly, let g(t) represent the position of the car from Town B after t hours. Write functions f(t) and g(t) to represent their positions, considering the delay of the train.
2. The vehicles meet when the distance between them is zero. Use an absolute value inequality to express the condition when the distance between them is less than or equal to 10 miles, representing the “meeting zone.” Solve for the time t when this occurs.
3. How far has each vehicle traveled when they enter this meeting zone?
Bonus:
If the speed of the car decreases by 5 miles per hour after every hour of travel, and the train increases its speed by 10 miles per hour after the first hour of travel, how would this affect the meeting time and the distance each has traveled?
@@bprpmathbasics Two towns, Town A and Town B, are 300 miles apart. A train leaves from Town A at a constant speed of 60 miles per hour, but it departs 2 hours after a car leaves from Town B. The car, initially traveling at 40 miles per hour, begins slowing down at a constant rate of 5 miles per hour per hour, starting 1 hour after its departure.
Let f(t) represent the position of the train from Town A after t hours since its departure. Let g(t) represent the position of the car from Town B after t hours since the car’s departure.
a. Write a function f(t) to represent the position of the train after t hours, considering the 2-hour delay.
b. Write a piecewise function g(t) for the car’s position. The first part should represent its motion in the first hour (constant speed), and the second part should account for the car slowing down after the first hour.
The two vehicles meet when the distance between them is zero. However, define the “meeting zone” as the condition when the distance between them is less than or equal to 10 miles. Write an absolute value inequality to express the condition when the distance between them is within this meeting zone, and solve for the time t when they enter this zone.
|f(t) - g(t)| \leq 10
Solve for the exact time t when they will meet (i.e., when the distance between them is exactly 0 miles). How far has each vehicle traveled by this time? Use your functions from Part 1 to find the distances.
Suppose after 3 hours of travel, the train increases its speed by 15 miles per hour for each subsequent hour, while the car’s speed decreases by 10 miles per hour per hour after 2 hours of travel. Adjust your functions for f(t) and g(t) accordingly, and recalculate the time and place where they will meet.
A logistics company operates a delivery service between two cities, City A and City B, which are located exactly 18 miles apart along a straight highway. The company’s central warehouse is situated between the two cities, 5 miles west of City A and 13 miles east of City B. The company uses a complex fee structure that depends on the distance from the warehouse and whether the delivery occurs within certain specified zones.
1. Zone 1:
For any delivery within a 10-mile radius of the warehouse (excluding City A and City B), the delivery fee is calculated using the function f(x) = 7|x - w| + 25 + p , where w represents the location of the warehouse on the highway, x is the distance (in miles) from the warehouse to the delivery point, and p is an additional surcharge based on the road conditions, which is a linear function of distance: p(x) = 0.5|x - 3| . The surcharge only applies when the delivery occurs within the first 3 miles from the warehouse.
Show us a 3d plot of (1) showing it crossing 0 at im = +/- 1?
You need a 4d plot because both the input and output are complex which are each 2d. And unfortunately we are unable to see or visualise 4d.
@@Ninja20704 not true, we pick one and time shift it
in complex system
sqrt(x)=-1,by Euler's equation we have x=e^(4n+2).pie.i, so x=1 for all integer n
we check it
sqrt(1)=sqrt(i^4)=i^2=-1
Do have any wrong in process??
Where is the real-world application for this?
Why can't square root of 1 be -1? I thought -1 squared is still 1? Or is it because there's no +/- symbol in front of the square root?
It's sort of due to definitions.
"Square root" refers to the "second roots of x", per defintion, you always have n solutions to the nth root of a number, in other words, there are 2 second roots of x.
Now that means that 1 and -1 are indeed both "square roots of 1", but sqrt(1) as an operation takes what is known as the principle root, as would nroot(x).
This means there is always only 1 outcome from the operation, in the case of square roots, that would always be the positive root, in this case 1.
I'm not sure if that answered your question?
At 1:28 "don't forget the plus or minus ...."
There he makes a mistake. The √ symbol has been defined to indicate the principal root of a value. And that means that √ (x^2) is defined as just one specific value, even when x is complex. So no "plus or minus" because now in the video √ (x^2) has two different values.
A value x can have two square roots but only one of them may be written as √ x
For problem no.2 why can't we use i^4?
i^4=1 sqrt(1)=1 not -1
Two points:
• when solving equations like x² = a with a > 0, it is not correct to write √x̅² = ±√a̅, since the _arithmetic_ real-valued square root is defined to always be non-negative; a more correct way is just writing x² = a => x = ±√a̅ (when a is positive)
• when solving irrational equations over the field of complex numbers ℂ, the _algebraic_ complex nth root of a non-negative complex number is normally defined to have n distinct complex values, so now √−̅1̅ = ±𝒊 and √1̅ = ±1
These are just two different conventions:
• the _arithmetic_ real-valued and single-valued square root √4̅ = +2
• the _algebraic_ complex multivalued square root √−̅4̅ = ±2𝒊
Some algebra books I've seen distinguish between the arithmetic and algebraic notations, by adding a subscript or an underscript plus sign:
• √₊̅4̅ = +2 - arithmetic
• √−̅4̅ = ±2 - algebraic
abs(x) = -1 has no solution
Meanwhile four-vectors that can have scalar product of himself and himself (technically norm of that vector in power of two) negative.
Please explain further, we will like to hear
@@Freedom-of-Thought Read up on Pseudo-Euclidean spaces and/or the Minkowski metric.
We could just define a new Nummer j where sqrt(j) = -1. (Like we did with the Im. Number). In this case we can actually calculate a result. (But it is only valid when j is valid.)
So there is no real or complex solution but a "j" solution for Q2 and Q3.
This just came to my Mind and maybe there is a better way to descibe this but this should work.
Nah. The thing is when you're talking about square root of 1 it's actually +/- 1. But when you write the √ sign, it's not actually the square root, but positive square root (at least in US). So it just can't be negative by definition, so there just is no j, for which √j = -1 in any number field.
@@F1r1at Oh. That is interresting.
Oh now I remember somthing from the Notes a good Friend gave me from Univetite. But idk what it is calles in the US. In German the concept is Called "Abbildungen" and there are like these strikt rules. One of them was f: R -> R. Which means that a real Value that is put into a function should only result in one real Result. So f: x -> sqrt(x) has only the positiv Result because of this Rule.
And that is why sqrt(x) cant be negativ. Right?
@@c42xe It's just because of the definition, that √a is a positive result for square root of a.
Cause like result of actual square root (I don't recall what it's official name is) is the number that if squared will give you the number under the root. So like 2^2 = 4 and (-2)^2 = 4, so square root of 4 is 2 and -2.
But since it's hard to use functions that have more then one output people decided that function sqrt (or √) will return just the positive answer. At least that's how people explained that thing to me.
So basically what √ does - is finds all the roots and leaves only the positive one. That's why it can't be negative.
2. totally has a solution, just need to define sqrt in a way where it has multiple return values.
@@F1r1at you're mistaking taking a root with finding a root. One is a function and the other are solutions to a polynomial. Not the same thing. Odd roots have negative outputs, even roots don't, that's just the byproduct of being a function.
You tricked into going in math class in my free time, what is this power?! XD please make more vids I’m loving my time here!
The first equation works because you can imagine that it works, but the other two equations don’t work because you cannot imagine that they work.
SQRT(x) = -1 if you square both sides you get X = 1 but if you retake the Square root you get back +/- 1 so it’s only partially indeterminate, the answer is null, 1.
I think this would be the equivalent of a super position of states, the answer is both in determinant and real at the same time.
I don't think the notion that square root symbol is a principal root automatically applies to complex numbers. When I was taught about square root on C, we started with multivalued functions and transitioned to Riemann surfaces and 4d space to represent square root on C. I think unless explicitly stated that we're talking about principal root sqrt(x)=-1 definitely has solutions on C.
A question:
Why simplify the first equation to i = root(-1)? Why not simply use the i²=-1 definition? Especially since you didn't give the definition of your root sign, and many people only know the principal square root, and using that one your substitution is simply wrong, while i²=-1 holds true still. Hell, your second example immediately starts off with a reminder that root(x)² is not equal to x, which makes it all the more confusing why this choice was made.
Also in videos like these it's always weird to arbitrarily choose the vector space, couldn't I simply choose a vector space where I define these functions in a way that these equations can be solved? After all, the meaning of even |x| and root(x) are bound by axioms and thus maleable as you want. Hell, I could make it easy and not use a vector space at all. The solution is closer to "none the vector space R, one in C, and we could craft xyz so that [...]"
In general, it's the difference in notation between saying x^2 = 25 and saying sqrt(25) = x. The first is a quadratic equation and it has 2 solutions, even assuming that we are only using the principal root. The second is not a quadratic, and it only has 1 solution assuming the principal root.
With x^2 = 25, when assume the principal square root we still get 2 solutions, not because sqrt(25) is +/- 5, but because sqrt(x^2) = |x|. And |x| is where the +/- really comes from. So when we take the root of both sides, instead of getting x = +/- 5 what we really get |x| = |5|, or just |x| = 5. Which means that x is either 5 or -5. The same applies to the video, we get |x| = i, therefore x is either i or -i. (there is a slight hiccup here, depending on your definition of absolute value, as the distance from origin definition would mean that only positive real numbers can be solutions and mean that |x| cannot be i, but if we don't rely explicitly on absolute value for choosing the principal root, then i and -i work fine)
For sqrt(25) = x, the principal root simply gives us 5 = x. So -5 isn't a solution because it's not the principal root of 25. In regards to the video, sqrt(x) = -1 has no solutions because -1 is not the principal square root of any number, even in the complex world.
All of his solutions are consistent with the principal root.
There's no need to overcomplicate it.
sqrt(x) as an operation, just as nroot(x), always takes the principle root, in the case of square roots this is always positive.
It is undeniably true that sqrt(-1) is i, and i only because of this rule.
For the first one, the definition sqrt(-1) = i is (probably) used because in every case, the 2 second roots of a number are equal in magnitude, but in opposite directions, that is, they are equal to each other times -1. This means he can use one solution to easily find the other, solving sqrt(-1) as i and saying +/-.
sqrt(x) in the complex world pretty much just halves the rotation and square rooted magnitude, although to get -1, you have to get 180 degree rotation which means your initial rotation is 360, so positive real numbers... and you already see how it doesn't work because 360 degrees is just 0 degrees and you can't divide it
so basically square root can output any complex number except negative real numbers
then i think absolute value for complex numbers just gives its magnitude and magnitude is always only positive
sqrt(x) would always give the positive root of x, because the principle second root of any number will always be positive.
nroot(z) is defined as taking the principle nth root of z, and does not always have to be positive, it must simply have the smallest argument in the complex direction, in other words, it is the one where k=0 in the form sqrt(a^2+b^2) x e^(Theta/n + 2k x pi/n). Where a and b are the magnitudes of direction in the real and complex direction.
In other words, it is not entirely true that the square root operation can output any complex number, it would only output the principle square root, even when taken of a complex number.
NUMBER 2 HAS A SOLUTION : It's i^4
IF Sqrt(x) = -1 , and X=1, Therefore Sqrt(i^4) = i^2 = -1
i^4 = 1, because i² = -1, and (-1)² = 1
So, it's not a solution
Sure, so x = 𝒊⁴ = 1 is an absolutely valid solution. Pretty strange Steve keeps denying this obvious fact 🤷♂
@@mxcoyl It *_is_* a solution, since in algebra √1̅ = ±1 = {1; -1} ∋ -1.
@cyberagua it's not obvious: x=exp(i*pi*2) is a solution, and x=exp(0) is not, though both expressions evaluate to one
@@mxcoyl Both are solutions, since for z = exp(𝒊φ) we have:
wₖ = √z̅ = z¹ᐟ² = exp(𝒊(φ+2πk)/2), k ∈ ℤ,
so for φ = 0 we get two root values:
• w₀ = exp(𝒊(0+2π·0)/2) = exp(0) = 1
• w₁ = exp(𝒊(0+2π·1)/2) = exp(𝒊π) = -1
第二題：
√(i^4)=i^(4/2)=i^2=-1，x=i^4=1
老師想問這個是做錯哪部份了？是4次方的問題嗎？
√(a^2) 是等於 |a|而非a
Your math kun fu is strong love these videos keep up the good work 💪
Can we solve the 2nd equation like this?
√x = -1
or, √x = e^(iπ) [from Euler's formula]
or, x = e^(2iπ)
That's tantamount to what he tried; square both sides to get the extraneous answer. I.e. your last step doesn't work.
Why cant i to the power 4 be solution on the second eqn
Because you get 1, not -1
Are you talking about the mod after the root
Second one has solution. √1 = √(cos(2pi*k) + i*sin(2pi*k)) = cos(pi*k) + i*sin(pi*k) = ±1. So, x = 1. In C, n-th root has n values.
Absolutely! 👍
|x| = -1 confused me because I never thought absolute value could be negative
and I was right
So I had to talk about equation 2 but since many people talked about this, I won't tell you.
However, I have something to talk about the third equation because that inspired me to rewrite algebra.
The third equation most definitely has no solutions (neither real, nor complex), while the second equation most definitely has a solution x = 1 with the definition of the multivalued complex square root, so that √1̅= ±1 = {1, -1} ∋ - 1.
@@cyberagua exatly. Like I said, I am going to rewrite algebra.
"Equation #1 can be solved" is a direct consequence of the Fundamental Theorem of Algebra.
Can we define a solution to eq 2. into existence? Like R[j] where sqrt(j) = -1?
that would mean that (-1)^2 is j
@@jacpa2011 Yes... in this Field, squaring a negative number gets you to the j axis. Does that end up in a contradiction?
@lreactor yep, j would appear everywhere and it would be very invasive
Wow, wow, dude! Be simple: √1 = ±1, you remember? The arithmetic root can't, but a root can be negative!
Sqrt(1) is only 1, not - 1, not +/- 1. It is not the same as solving x^2 = 1.
Have you ever heard anyone say
sqrt(2) = +/- 1.4142…
sqrt(3) = +/-1.732…
or etc?
@@Ninja20704, yeah, I know that a root usually is understand as positive only value. But in common way its negative value can't be ignored otherwise we would be loss some solutions. (But sometimes we get extra.) It is known that root of N power is an N-arity function. So square root has 2 values, + and -. If you image 1 and -1 on complex plane, you can see that the 1+0i possible to turn on π angle twice to get 1+0i again. Two sequential equal turns is same that squaring thus exp(iπ) aka -1+0i is possible solution of √1
@@Qraizer we say one solution is sqrt(1) and the other is -sqrt(1). We write the negative sign infront of the sqrt to get the negative answer. Just think about how we solve x^2 = 2. We say one answer is sqrt(2) and the other is -sqrt(2).
If you are talking about complex numbers, then the sqrt will only return the principal answer, which is when we write the inside using the principal argument, the argument that lies in (-pi,pi]. Again, no one ever says sqrt(-1) = +/- i
@@Ninja20704 , yes, we use only the principal values of roots. Because it is convenient for us when function has just one value. But this don't change the fact that root of N power is N-arity function.
When we consider ℝ instead of ℂ, for odd N we get only 1 real value due to rest are complex. But for even N we can get 2 real values. If we want. I think that strict solution of √x = -1 is not question of our desires but is question of exact properties of square root as a function.
In the first example, you indicate that the sqrt is (+/-)... but in the second example you say -1 is an extraneous solution, but if you indicate the sqrt as (+/-) as in the first example, then 1 and -1 are both valid solutions. Why do we have (+/-) i as two valid answers in the first case, but not (+/-) 1 as two similarly valid answers in the second case?
For question 2 can you do, e^(2ipi)?
e^2ipi = 1
sqrt(1) is still 1 (a function has 1 output)
You can't always rely on (a^n)^m=a^nm with complex exponentials as they are more restrictive
@@_rd_5043 The real issue is that sqrt(x) = -1 for real numbers has no solutions because sqrt(x) simply IS the principal (positive) square root. sqrt(z) = -1 as a single-valued complex function only has a solution if you pick the appropriate branch (and it is 1 in that case).
You can't always rely on exponential rules to work in complex numbers when one of the exponents has a fractional part, but I believe you can choose a branch of the function z^(fractional part) so that it does work in whatever specific case you're dealing with. I don't know if it works with irrational fractional parts, but I'm 99% sure it does with rational ones.
Could be wrong though, I was never a complex analysis guy.
What if we take (-1)^2 in second one? Can't we just cross these sqrt and ^2? And left with -1? "sqrt((-1)^2)=-1"?
Sqrt(x²)=|x|
I disagree on question 2, as 1 is only an extraneous solution if you consider the square-root function to be single-valued, which I don't think is a fair assumption to make in complex land, beyond computation, when it provides a different result than considering multivaluedness. Considering sqrt(x) as the inverse to x², such that (sqrt(x))² = x, we want sqrt(x) = -1. Square both sides, x = 1, and this is fine, as sqrt(1) = ±1 through consideration of the other branches of the square-root function.
What about sqrt(i^4), isn't that equal to i^2 and so to -1?
Yes, it is.
Vi⁴=±1.
Regarding (2), since -1 * -1 = 1 then x = 1 is a solution to sqrt(x) = -1?
It implies the principal root of one, which is 1.
1 ≠ -1
First! Loving the video so far ❤
I got a solution to |x|=-1. It is x=
| 0 1 |
| 1 0 |, which x is a matrix
Here x is a number.
for (2), x = e^2kpi ?
that's just 1, you can't just cancel out the 2 when using complex exponentials
3:58
sqrt of x = -1
its plus or minus 1
Sqrt(1) = -1?
Sqrt(-1) = -1?
Maybe I'm unpopular with this, but in my opinion the idea that √x = |x¹/²| needs explicit assertion, it is not implicitly true. It is sometimes true, like in geometric context where negative lengths can't exist directly it can be taken as true, and it is used in the definition of standard deviation, but it's not generally true and therefore needs an assertion. Maths should not depend on pedantry, and taking a root is the inverse operation of exponentiation, and should be treated as such.
taking the square root of a number should return 2 possible values, right? A positive and a negative? So sqrt(x)=-1 would absolutely have the solution of x=1 because sqrt(1)=±1, and one of those sides is -1. It works???
"taking the square root of a number should return 2 possible values, right? A positive and a negative? "
No. The square root of a (non-negative) number is _defined_ as the non-negative number whose square is the original number.
"sqrt(1)=±1"
No. sqrt(1) = 1 only. sqrt(1) = -1 is wrong. Look into a math textbooks how the "square root" of a number is defined, if you don't believe me.
E.g., V9=±3.
@@bjornfeuerbacher5514E.g., V9=±3.
@@Misha-g3b No again. Read my comment above.
@@bjornfeuerbacher5514 V1=±1, because (±1)²=1, but 1(=V1) is the main, arithmetic, principal value. E.g., 2+V9=2±3=5 or -1; V1-V(16)=1-4=-3. Remember that.
What about exp(2pi*i)? It's square root should be exp(pi*i)=-1
Sure
Isnt absolute value only equal to square root of x squared in the real world?
solve sqrt(x)=-1 has no silution at all?
I'm stuck somehow:
Simplifying: sqrt(x) =-1
-1 =e^(i(pi+2npi))
Prinzipal value n=0
-1 = sqrt[(e^i(pi+2npi))²]
= sqrt[e^(i(2pi+2×2npi))]
For sqrt(e^(i0pi)) it doesn't work, since its =1
but e^(i2pi), e^(i6pi) and so on?
For x^2 = -1, x = i. For sqrt(x) = -1, x = no
is x=i^2 can't be the solution for the 2nd?
if x=i^2 then x=i^2=-1
sqrt(-1)≠-1
Wait a minute... 5:28
|i|=1
sqrt(i^2)=1
sqrt(-1)=1
(x ➖ 1ix+1i ).(x ➖1ix+1i).(x ➖ 1ix+1i)
When you are new to the complex world and realise that (-i)² is a triple negation.
X^2 = -1 had no solution before a formular to solve cubic equation which cause the founding of complex number
So I believe sqrt(x) = -1 and abs(x) = -1 will be solvable oneday.
Seems like both are solvable already - depends upon the definitions and number sets being used.
1) x = ±i (i² = -1 by definition)
2) No solution (principal roots are never negative)
3) No solution (absolute values are never negative)
The so-called "principal" roots are defined arbitrarily and are generally useless in complex algebra, since, for example, the cube root ³√−̅8̅ has three distinct values: ³√−̅8̅ = {−2; 1+𝒊√3̅; 1-𝒊√3̅}, and it's hard to decide which of them is more "principal" than the others 🤷♂️ And the same goes for the complex square root as well.
@@cyberagua Nope. √−8 = −2 is the only correct solution, and it isn't arbitrary at all.
However, x³ = −8 has three solutions: x₁ = −2 ∨ x₂ = 1 − i√3 ∨ x₃ = 1 + i√3, but that's a completely different question.
@@Nikioko Wolfram Alpha says ³√−̅8̅ = 1 + 𝒊√3̅ 🤷♂
@@Nikioko What grade are you in?
@@cyberagua WolframAlpha gives the solution which lies in the first quadrant of the complex plane, i.e. non-negative values on both real and imaginary axis. Other calculators return -2 for ³√-8.
What do you mean, what grade am I in?
fun fact: on the first equation he wrote, and proved, that the solution for X² is ±i but, for the second equation ( √x=-1) he claimed that √1=1 instead of √1=±1.
as a psychological trick that some people may not notice, he gave 3 equations, one that only has solutions under certain conditions, one that has a similar solution but claimed it doesn't has any and an absurd equation that, clearly, has no solution. this last equation seems to validate that the second also doesn't has a solution despite having one that is coherent with the way he solved the first.
A rather deep analysis! 👍 And sure the second equation has a solution x = 1, since √1̅ = ±1 = {1; -1} ∋ -1
In C abs(x)=sqrt(xx')
Cant you use √i^4
√i⁴ = √1 = 1 ≠ -1
∴ √x = -1 has no solutions
oh that was suprisingy easy. finally a video where i got it correct
We need a square function that instead of being x*x it is x*|x|
On #2: Whaaat? How is 1 not a solution? -1 is a valid square root of 1.
however -1 is not a valid principal root of 1, when theres a sqrt in an expression, unless specified otherwise is always the principal root
The absolute value of i^2 equals -1.
If x=e^(2i*pi), then sqrt(e^(2i*pi))= e^(i*pi)=-1. Where is the mistake?
That's correct.
On (2) -1 is a square root of 1, but it is not the square root of 1.
it's a root of polynomial x²=1 but it's not a square root, as that is a function
@@_rd_5043 Ok it's a -2 power then. You know what I'm saying.
Isnt sqrt(i^4)=i^2=1?
Sqrt(i⁴)=(i⁴)^½=i^⁴/²=i²=-1
The rule z^(m/n) = [z^m]^(1/n) does not hold if m/n is not the simplest form of the fraction and z is non-real (or real but negative). So (i^4)^(1/2) ≠ i^(4/2)
Isn't x=i when √x=-1
Someone tell me, and If not why?
Sqrt(i^4)=i^2=-1
For equation 3, you should also mention, for the equation abs(x)=sqrt(x^2), that this works as long as x is a real number.
Otherwise, you would get the following for x=i,
abs(i) = sqrt(i^2) = sqrt(-1), but sqrt(-1) = i
therefore, abs(i) = i, which is not true at all since abs(i) = 1
Why doesn't √(i)^4 work?
I think order of operations. (i)^4 = 1 then sqrt(1) = 1
Because sqrt(x^2) = abs(x), not x, as he explained near the end of the video. It is often treated as if sqrt(x^2) = x because it works when x >= 0, but if x < 0 or is imaginary it is important that the abs(x) is used
The rule of exponent z^(m/n)=[z^m]^(1/n) does not apply if m/n is not the simplest form of the fraction and z is non-real (or even negative real numbers).
So [i^4]^(1/2) ≠ i^(4/2)
Because x^(1/2) =/= sqrt(x)
Really, the fundamental operation as described with real values, as you see here, is poorly defined in complex values. It results in multiple 'branches' of possibilities - as 1=i^4=i^8 etc so the 'general square root of 1' (vaguely defined) can equally be 1, i^2, i^4 etc.
The square root for complex values is defined to be the *principal* result of all such equivalences. That is, effectively, the value being square rooted is first reduced to 'simplest' form before the operation proceeds, therefore giving a consistent result for equivalent inputs.
The language I use here is deliberately vague as a lot of the precision makes more sense using polar form, where i=e^(pi/2) [or, technically, i=e^(pi/2+2*n*pi) where n is an integer]
The same reason why 1=sqrt(1)=sqrt(-1 * -1)=sqrt(-1)sqrt(-1)=i*i=-1 is not valid. Many of the exponential laws we're so attached to simply don't work anymore with complex numbers
Why does sqrt(1) not equal one? Look, If we write 1 as e^(2i*pi) and then take the sqare root of it which is by definition: e^(i * pi), because you have to devide the exponent by 2, you get -1 as an answer! So where is my mistake?
No mistake. You are correct.
@@cyberagua But sqrt(1) isnt 1 and -1 he said in the video ( i wrote the first sentence wrong)
You are right and Steve is wrong here. Over the complexes, the square root of a nonzero complex number is normally defined to have two distinct values:
√−̅4̅ = ±2𝒊 = {2𝒊; − 2𝒊}
and the cube root of a nonzero complex number has three distinct values:
³√−̅8̅ = {−2; 1+𝒊√3̅; 1−𝒊√3̅}
While over the non-negative reals we can define a so-called _principal_ square root, it makes very little sense (or rather no sense at all) over the set of complex numbers ℂ.
@@cyberagua Okay thank you :) you ment to write sqrt(4) and not sqrt(-4) tho, right?
@@muuke852 Yeah, you are right: it should have been √−̅4̅ = ±2𝒊 = {2𝒊; − 2𝒊}
The real number -1 is a completely valid square root of 1, I have no idea why there's so much racism between principal and non-principal roots when -1 and 1 have the same square. Unless the square root is considered a function and not an operator, but it's still misleading.
As for |x| = -1, I don't even know why that needed to be explained with anything other than "The absolute value of any number is strictly non-negative", because that is (practically) the whole truth. Obviously there are complicated reasons as to why it's true, but that's a whole other topic.
X^2=1x=± i; (x)^1/2=-1 x= i;
|x|=-1 x={-1; i}
|X|= -1 ... What if: |-(-1)|= -1 ?
How about sqrt(i^3) = i^2 = -1
√(i^3) = √-i
i² = -1
√-i = ±√2 * (1 + i)
∴ √(i^3) ≠ i²
@@FiveSixEP Also, sqrt(x^3) = x^(3/2) which is not the same as x^2
Sqrt(x) is range-restricted. You used the exact principle that means sqrt(x)=-1 has a solution in your video. Sqrt(-1) is + OR - i. You said that, for reference, and a rule like that about an operation always applies, or never applies. Therefore, sqrt(1) is +/-1, or x=-i is not a valid solution for x^2=-1. The reason sqrt(x) the function is range-restricted is because otherwise it wouldn’t be a function. However, if you make it a relation instead, and use its maximal range, it is just a sideways parabola. I assume you knew all the info I just said, I just wanted to show my reasoning for why sqrt(x)=-1 has a solution that isn’t just pedantic. (Or at least, no more pedantic than mathematics must be)
Para segunda respuesta, tengo 1, 0i, 2pi(z axis)
Equation 2 has solution x=e^(2πi) check it out
Why is x≠-1 in equation 2? After all if x = 1, -1^2=1, but so does 1^2… To me, in equation 2, x=±1, But, as some others pointed out, [e^2i(pi]=1 On top of this, if you take a square root of a number, you are basically doing x^1/2. Back to that problem, if we take the square root [e^2i(pi)] it’s equal to -1.which is the same solution, just in a different form, which would reinforce the solution of x=-1. On top of this, if we were to consider that the square root of i is i/±1. Neither of these are equal to -1, because it’s either -i or just i. However, we know that i^2=-1 and -i^2=-1. That doesn’t matter though, even if it does restate the equation.
Because, unless specified otherwise, sqrt(x) means the principal square root, and that is defined to be the positive root over the real numbers. There are lots of good reasons for that mainly related to making the square root function single-valued and making sqrt(1) = 1 true.
sqrt(i^4) = ?
Sure, √(𝒊⁴) = 𝒊² = -1 (as one of the values)
±1.
Hold on. For the second case, since you are taking a square root of 1, you have two possible answers, 1 or -1, don't you. I thought square root give two possible answers. One answer may work or two answers may work. So, I am confused.
Yeah, exactly: the algebraic complex multivalued nth root returns n distinct values (for a nonzero argument), so √-̅4̅ = ±2𝒊 and √1̅ = ±1
الجبر
الجبر
Trivialities. This is trying to conjure up something from nothing. It more than likely confuses young students rather than helps them.
In the second one could be x= i^4 I think
Sure 👍
If sqrt(1)=±1, then why does sqrt(x)=-1 have no solutions when sqrt(x)=1 does?
sqrt(1)=±1 only if sqrt isnt a principal root, a principal root is the same as a root but only uses the positive outcome.
0:29 Technically, there is a solution: i
You're a genius
@@Alex1996-v6w Thank you! 😀
@@JJ_TheGreatit was sarcasm, bprp goes over that solution later in the video
for number 3, just invent a new number, let’s call it A
but that doesnt make any sense, the absolute value of something is always positive, if |A| = -1 then it would literally defy the entire purpose of the "absolute value"
@ Oh
Then judt invent a jew number for number2
The first cannot equal -1.