Solving x^2=-1, sqrt(x)=-1, abs(x)=-1

[@bprpmathbasics]

Is sqrt(x^2)=abs(x) true for everything? kzbin.info/www/bejne/f3e6qnqGjrl2hqs

[@quzpolkas]

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.

[@_rd_5043]

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.

[@quzpolkas]

@@_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.

[@westy9447]

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.

[@isaacto8761]

​@@westy9447 To properly appreciate this way of thinking, one has to go all the way to Rienman surface. But that is really neat.

[@bprpmathbasics]

Solving x^4=1 kzbin.info/www/bejne/bqLLeK2peN6eorM

[@TheMathManProfundities]

|x| = √(x²) is only generally true for real values of x.

[@chaosredefined3834]

sqrt(x * conj(x)) works though.

[@TheMathManProfundities]

@@chaosredefined3834 OK, I should have said only generally true. I'll edit my post.

[@CalebAnderson-gn3ne]

it is only true if you take the prinipal root which is only defind for real x

[@TheMathManProfundities]

@@CalebAnderson-gn3ne Principal root is defined for complex numbers as well. The radical symbol (√) refers explicitly to the principal root.

[@JJ_TheGreat]

I don't get it: How is the absolute value of x equal to the square root of x²?

[@Eternitycomplex]

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?

[@reizinhodojogo3956]

√(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

[@TheMathManProfundities]

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.

[@TheMathManProfundities]

​@@reizinhodojogo3956x^(y/2) = ±√(x,y) for all values of x and y where x≠0. This holds true for complex values as well.

[@sayonbasak-x3w]

The later statement is correct.

[@WLYLL]

sqrt(i^4) IS NOT i^2 but |i^2| which is 1

[@gamingmonke1269]

last time i was this early Archimedes was still finding the area of a circle

[@ZDTF]

It's 2πr

[@ZDTF]

@ZDTF bro that's the perimeter

[@ZDTF]

Oh my bad G

[@ZDTF]

Nah you are fine

[@Duckz4bucks]

^^ Bro forgot to take his meds 💀

[@kupopo1]

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.

[@pete4416]

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

[@the_osprey314]

What about e^x=-1? What would that be? Probably no solution, but I would like to see if there is anything interesting involved

[@bprpmathbasics]

The answer is iπ (and in fact, i*odd*π). You can see my video here kzbin.info/www/bejne/gKLXZ2qqf6icrbMsi=GJNrBy4D02FC5UjD

[@starleaf-luna]

I could tell |x| = -1 didn't make sense at a glance, because there's no number where its absolute value is negative, right?

[@Ninja20704]

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

[@Anik_Sine]

I like how square rooting a negative number is allowed in a significant part of higher maths but squaring a negative number is not

[@CalebAnderson-gn3ne]

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

[@Steve_Stowers]

"Equation #1 can be solved" is a direct consequence of the Fundamental Theorem of Algebra.

[@mstarsup]

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...

[@JJ_TheGreat]

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?

[@WookieRookie]

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.

[@Freedom-of-Thought]

Please explain further, we will like to hear

[@YourLifeWasting]

Your math kun fu is strong love these videos keep up the good work 💪

[@masteroftheinternetverse1296]

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???

[@dannyyeung8237]

I got a solution to |x|=-1. It is x= | 0 1 | | 1 0 |, which x is a matrix

[@ParnavaMallick]

Can we solve the 2nd equation like this? √x = -1 or, √x = e^(iπ) [from Euler's formula] or, x = e^(2iπ)

[@TacticalAnt420]

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!

[@c42xe]

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.

[@F1r1at]

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.

[@c42xe]

@@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?

[@F1r1at]

@@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.

[@janisir4529]

2. totally has a solution, just need to define sqrt in a way where it has multiple return values.

[@_rd_5043]

​@@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.

[@lool8421]

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

[@westy9447]

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.

[@CarmsCelestia]

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 [...]"

[@phiefer3]

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.

[@westy9447]

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 +/-.

[@corruptconverter2616]

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.

[@blushingmushroom1949]

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?

[@Orenotter]

On #2: Whaaat? How is 1 not a solution? -1 is a valid square root of 1.

[@willywillycow]

I wish they had just made another symbol for principal root and ±√

[@amberthelostsoul]

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?

[@westy9447]

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?

[@Erlisch1337]

oh that was suprisingy easy. finally a video where i got it correct

[@myfyrmadocjones]

Regarding (2), since -1 * -1 = 1 then x = 1 is a solution to sqrt(x) = -1?

[@RealQinnMalloryu4]

(x ➖ 1ix+1i ).(x ➖1ix+1i).(x ➖ 1ix+1i)

[@teelo12000]

Show us a 3d plot of (1) showing it crossing 0 at im = +/- 1?

[@Ninja20704]

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.

[@teelo12000]

@@Ninja20704 not true, we pick one and time shift it

[@justkarl2922]

In C abs(x)=sqrt(xx')

[@aveol7855]

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"?

[@alibouskour4821]

Sqrt(x²)=|x|

[@nfpnone8248]

The absolute value of i^2 equals -1.

[@michodome9342]

We need a square function that instead of being x*x it is x*|x|

[@GerryCatLord]

Sqrt(i⁴)=(i⁴)^½=i^⁴/²=i²=-1

[@Ninja20704]

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)

[@JJ_TheGreat]

0:29 Technically, there is a solution: i

[@Alex1996-v6w]

You're a genius

[@itslejogd]

quick correction here the square root of x = -1 DOES have a solution, and it is 1, since -1 * -1 is 1, but also because square roots give you negative AND positive values. Its not estraneous. Also, an absolute value does not equal to the square root of a squared number, since like i said, square roots give negative solutions too. Absolute values are simply that value taking to the positive side.

[@DeJay7]

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.

[@Wsp307]

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]

Do you have a problem that you are working on?

[@Wsp307]

@@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?

[@Wsp307]

@@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.

[@Wsp307]

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.

[@owenkasif8937]

For question 2 can you do, e^(2ipi)?

[@_rd_5043]

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

[@Mystic998]

​@@_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.

[@federicopagano6590]

Equation 2 has solution x=e^(2πi) check it out

[@luispinto2643]

Isnt absolute value only equal to square root of x squared in the real world?

[@GrenoGuy]

yes i can!

[@Qermaq]

On (2) -1 is a square root of 1, but it is not the square root of 1.

[@_rd_5043]

it's a root of polynomial x²=1 but it's not a square root, as that is a function

[@Qermaq]

@@_rd_5043 Ok it's a -2 power then. You know what I'm saying.

[@Qraizer]

Wow, wow, dude! Be simple: √1 = ±1, you remember? The arithmetic root can't, but a root can be negative!

[@Ninja20704]

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?

[@Qraizer]

@@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

[@Ninja20704]

@@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

[@Qraizer]

@@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.

[@SeegalMasterPlayz]

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

[@hikari1690]

Stahp~ I don't need my brain to grow~

[@mikeyu3243]

for (2), x = e^2kpi ?

[@_rd_5043]

that's just 1, you can't just cancel out the 2 when using complex exponentials

[@ronjharedrobles2960]

3:58 sqrt of x = -1 its plus or minus 1

[@yodaas7902]

Sqrt(1) = -1? Sqrt(-1) = -1?

[@maverickthe13th52]

Why doesn't √(i)^4 work?

[@rootaccess8331]

I think order of operations. (i)^4 = 1 then sqrt(1) = 1

[@incendiohawk1725]

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

[@Ninja20704]

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)

[@bluerendar2194]

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]

[@_rd_5043]

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

[@cat-tl2kk]

1: {i, -i} 2: x ∉ ℂ note {-1} ⊂ √1 3: x ∉ ℂ

[@williamreynolds6680]

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.

[@Mystic998]

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.

[@prathameshchoudhari8165]

Why cant i to the power 4 be solution on the second eqn

[@yodaas7902]

Because you get 1, not -1

[@prathameshchoudhari8165]

Are you talking about the mod after the root

[@andreiinthedesktopworld1178]

for number 3, just invent a new number, let’s call it A

[@FiveSixEP]

How about sqrt(i^3) = i^2 = -1

[@Exvixcity]

√(i^3) = √-i i² = -1 √-i = ±√2 * (1 + i) ∴ √(i^3) ≠ i²

[@Exvixcity]

@@FiveSixEP Also, sqrt(x^3) = x^(3/2) which is not the same as x^2

[@floofybear]

In equation 2 ... Why can x not be -1? Because √1 can also be -1..... -1*-1=1

[@Taric25]

What about i⁴? √i⁴=i²=⁻1

[@gaganzanwar]

When we take a square root, it gives the positive value of the root, that's why.

[@Ninja20704]

@@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.

[@Taric25]

@@Ninja20704 Ah, what is the rule if z is complex?

[@flatii]

Cant you use √i^4

[@Exvixcity]

√i⁴ = √1 = 1 ≠ -1 ∴ √x = -1 has no solutions

[@vikramgupta7618]

First! Loving the video so far ❤

[@someone_random1283]

Came from insta 😂

[@etrisplayer6169]

nvm

[@xgsan]

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?

[@ttmfndng201]

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

[@chonkeboi]

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.

[@JossoJJossoJ]

When I saw the thumbnail, I was like Davy Jones seeing his hidden chest

[@snnwstt]

-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).

[@holyhayn]

Imaginary numbers... I always hated imaginary numbers.

[@hrayz]

I always loved "imaginary" numbers. Turns the graph 3D with the lines popping up out of the page!

[@netneo4038]

They're very important in practical applications as well, ironically.

[@jamesharmon4994]

Eq 1: x = i

[@janisir4529]

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.

[@zaelgreen1670]

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.

[@janisir4529]

@@zaelgreen1670 sqrt is a strong independent function that can have 2 values as an output on the complex plane.

[@Samir-zb3xk]

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

[@janisir4529]

@@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...

[@chetan9101]

What on earth is Susan?😅

[@uchiha_itach1-t4g]

2) √x=-1 , here is a solution , x=i^4

[@vaib223]

x=i⁴ therefore, sqroot of i⁴ is (i)⁴½ = i²=-1

[@skagna]

No, i^4=1

[@chonkeboi]

You can’t use exponent laws for non-real complex numbers from my understanding.

[@_rd_5043]

​@@chonkeboiexactly