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.
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.
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...
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
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.
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
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
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.
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.
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.
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)
@@_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.
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.
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
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
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
-1 is a possible solution to the first one. Sure, if you RE-DEFINE a (square root of )a equals b not as b^2 = a, (in this case, -1 is a possible solution of sqrt(1)), but strictly as the positive value for b such as b^2 = a, THEN, and only then, there is no solution. *Similar* things happen in residue modulo. Since, in modulo 4, 2*2 = 0, one can say that sqrt(0) = 2 (mod 4) . In addition to 0, for sure. Note that 2, and 0, are then "representant" of the "class" of numbers, generally noted as sets: [0] = { ..., -8, -4, 0, 4, 8 ...} and [2] = { ..., -6, -2, 2, 6, .... } under the "context" of mod 4 and indeed, ANY element of [2], squared, gives 0 (mod 4) , so any element of [0] and of [2] is a candidate for the solution of sqrt(0) (mod 4).
When you're using complex numbers, you can't just change around the order of exponents. In this case, i^4 = i*i*i*i = (-1)*(-1) = 1 so (i^4)^(1/2) = 1^(1/2) = 1
There is a difference between the question, "what number(s) squared yield x," and the solution to the function, sqrt(x). For every complex number, c, there are two complex numbers, r_1 and r_2, such that (r_1)^2 =c and (r_2)^2 = c. We can refer to these as the square roots of c, but a function must have a unique output for each input, so for sqrt(x), we have to pick one. This is why, when solving quadratics, we will will write (+-)sqrt(x) when taking the square root of both sides of the equation. In this case, we have to consider both possibilities. That is different from an equation that starts with a simple square root in it.
Square root is commonly used in geometry where you obviously cannot have negative side lengths, so it makes sense to define a square root function that outputs only positive numbers, which is the way its being used in the video. But you are correct, when talking about numbers on the complex plane it makes more sense to consider root functions as multivalued
@@Samir-zb3xk Square root could be multivalued on real numbers too, it's mostly convention that it doesn't, also it looks stupid for any larger roots that odd ones have 1 and even ones 2...
Is sqrt(x^2)=abs(x) true for everything?
ruclips.net/video/IFWyGPYTCPI/видео.html
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.
|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.
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.
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...
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
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.
Solving x^4=1
ruclips.net/video/8qhGzsCyklQ/видео.html
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
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
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.
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.
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.
"Equation #1 can be solved" is a direct consequence of the Fundamental Theorem of Algebra.
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.
(x ➖ 1ix+1i ).(x ➖1ix+1i).(x ➖ 1ix+1i)
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?
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
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)
Your math kun fu is strong love these videos keep up the good work 💪
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.
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.
oh that was suprisingy easy. finally a video where i got it correct
In C abs(x)=sqrt(xx')
Why cant i to the power 4 be solution on the second eqn
1: {i, -i}
2: x ∉ ℂ note {-1} ⊂ √1
3: x ∉ ℂ
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
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
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.
for (2), x = e^2kpi ?
that's just 1, you can't just cancel out the 2 when using complex exponentials
_The Little 'i' That Could_
for number 3, just invent a new number, let’s call it A
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
-1 is a possible solution to the first one. Sure, if you RE-DEFINE a (square root of )a equals b
not as b^2 = a, (in this case, -1 is a possible solution of sqrt(1)),
but strictly as the positive value for b such as b^2 = a, THEN, and only then, there is no solution.
*Similar* things happen in residue modulo. Since, in modulo 4, 2*2 = 0, one can say that sqrt(0) = 2 (mod 4) . In addition to 0, for sure.
Note that 2, and 0, are then "representant" of the "class" of numbers, generally noted as sets: [0] = { ..., -8, -4, 0, 4, 8 ...} and [2] = { ..., -6, -2, 2, 6, .... } under the "context" of mod 4 and indeed, ANY element of [2], squared, gives 0 (mod 4) , so any element of [0] and of [2] is a candidate for the solution of sqrt(0) (mod 4).
yes i can!
In equation 2 ... Why can x not be -1?
Because √1 can also be -1..... -1*-1=1
What about i⁴? √i⁴=i²=⁻1
When we take a square root, it gives the positive value of the root, that's why.
@@Taric25sqrt(i^4) is not i^2 because the rule that z^(m/n) = [z^m]^(1/n) does not hold if m/n is reducible and z is complex.
@@Ninja20704 Ah, what is the rule if z is complex?
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
Cant you use √i^4
√i⁴ = √1 = 1 ≠ -1
∴ √x = -1 has no solutions
First! Loving the video so far ❤
Came from insta 😂
what if we factor the second answer, 1=i^4
√i^4=-1
(i^4)^1/2=-1
i^4•1/2=-1
i^2=-1
Is it make sense?
When you're using complex numbers, you can't just change around the order of exponents.
In this case, i^4 = i*i*i*i = (-1)*(-1) = 1
so (i^4)^(1/2) = 1^(1/2) = 1
You can’t use the rules of exponents for imaginary numbers from my understanding. Because then you’d be able to prove a bunch of nonsense such as i=1.
x=i⁴ therefore, sqroot of i⁴ is (i)⁴½ = i²=-1
No, i^4=1
You can’t use exponent laws for non-real complex numbers from my understanding.
@@chonkeboiexactly
Imaginary numbers... I always hated imaginary numbers.
I always loved "imaginary" numbers. Turns the graph 3D with the lines popping up out of the page!
Square root has 2 values on the complex plane for any input, sqrt(1) is 1 and -1, so it's a perfectly valid equation in some sense of square root.
There is a difference between the question, "what number(s) squared yield x," and the solution to the function, sqrt(x). For every complex number, c, there are two complex numbers, r_1 and r_2, such that (r_1)^2 =c and (r_2)^2 = c. We can refer to these as the square roots of c, but a function must have a unique output for each input, so for sqrt(x), we have to pick one. This is why, when solving quadratics, we will will write (+-)sqrt(x) when taking the square root of both sides of the equation. In this case, we have to consider both possibilities. That is different from an equation that starts with a simple square root in it.
@@zaelgreen1670 sqrt is a strong independent function that can have 2 values as an output on the complex plane.
Square root is commonly used in geometry where you obviously cannot have negative side lengths, so it makes sense to define a square root function that outputs only positive numbers, which is the way its being used in the video. But you are correct, when talking about numbers on the complex plane it makes more sense to consider root functions as multivalued
@@Samir-zb3xk Square root could be multivalued on real numbers too, it's mostly convention that it doesn't, also it looks stupid for any larger roots that odd ones have 1 and even ones 2...
Eq 1: x = i
What on earth is Susan?😅
2) √x=-1 , here is a solution , x=i^4