Mathematically, which of these numbers is the largest? (A) Number of hours in a year (B) Number of seconds in a day (C) Number of days in a decade (D) Number of minutes in a week Answer here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-06NqtlmkPK0.html
no of hours in a year- 52*7*24=8736 no of seconds in a day- 60*60*24=86400 no of days in a decade= 365*10= 3650 no of minutes in a week= 60*24*7=10800 hence option b
This guy is talented but the marker thing is pretty common amongst students. I mean I used to use blue and black gel pen by holding in the same hand to save time 😅
It's clear enough when you are dealing with just numbers, but imagine this in some algebra calculation where you have, in the middle of a big expression sqrt(sin(theta)*tan(theta)) which you split into two square roots (maybe planning to use some tan(theta/2) expression) and later on you instantiate theta to an angle where both sin and tan are negative. It may not be obvious that the steps are no longer valid!
That is why it is very important to check your domains. So for example you have equation a/b then automatically your domain is restricted to b≠0. But yes most of the time domain restrictions are forgotten and it can lead to incorrect solution.
Thank you - you (and this video) just fixed a bug in this script I wrote that would output weird numbers when I give it 2 negatives. When I was first figuring out the formula, there's a spot where I ended up doing just that.
@@sandro7 The goal of the script was to take two disjointed arcs where you had start point, end point, and radius and spit out a new single arc giving you the center point instead _(as you already had the start and end point from the initial arcs)_ where the arc goes through all 4 initial points. As a side note: one stipulation was that the arcs had to be in places where they could be connected by a straight line from one end-point to another, and that line was _(mostly)_ tangent to both arcs. While checking that wasn't really part of the script, that situation being the case was part of what decided the script's use in the first place. The bug was in the part of the script that would find the two possible center points of each arc, as it required these trig functions. If the arc was fully in quadrant 1 (Q1), then the script would work flawlessly. If any or all of the 3 points of one of the initial arcs was in Q2 or Q4, then it would do this weird thing where it would _sometimes_ swap one or both signs. I would have to figure by hand which signs are correct, but it at least gave a starting spot. I was in the middle of trying to figure out the pattern based on which quadrant each point of the arc was in and then just writing a switch statement to do the conversion for me as a band-aide solution when I stumbled across this video. If it was in Q3 though, the script was entirely useless and would give completely erroneous outputs. I would have to rotate the arc around the origin point until none of the three points landed in Q3, and then rotate it back by the same amount when moving on to the next step.
By turning i^1 into i^(4/4), you artificially raised it to the 4th power and then took the 4th root. This creates 3 extraneous solutions, that are false (the other 2 false solutions are -1 and -i).
yeah I think maybe the simpler solution in step 3 is that you really shouldn't be taking the root of one side of the equation, even if it is still technically equal, and the second example is just the opposite
I think turning i^1 into i^(4/4) is fine. Order of operations forces you to reduces (4/4) to 1 first before taking the exponent. I was thinking that by putting i^4 in parentheses, such that the right side is now (i^4)^(1/4), you change the order of operations, and therefore change the equation.
The true reason that this doesn't work is that a square root(in Complex numbers) has two different roots. In this problem only one of these roots satisfies the equation(hint: it is not i).
@@drrenwtfrick not exactly. What you are thinking of is the solution to the equation x^2=b. x has two possible solutions; x= squareroot(b) and x= - squareroot(b). In general squareroot(b) is always positive.
@@drrenwtfrick No. Square root (of a real number) is a function defined like this: sqrt(a) is a number b, b≥0, that satisfies b^2=a. As you can notice, that's always one number. The reason why you may think that it should be two numbers probably has to do with the solutions of an equation like x^2=9. Let's look at it: As I assume you know, the first step is to apply sqrt() to both sides sqrt(x^2) = sqrt(9) By the definition, sqrt(9)=3, so sqrt(x^2)=3 Now, what is sqrt(x^2)? It can't be plain x, because the result must be ≥0, if x
@@shadowblue4187 but did you hold the pencil and the pen at the same time with one hand like this guy in the video who is holding two different marker in one hand?
Can u help me please Turn on audio option in all the videos please I want to listen it in hindi Sry I can't understand ur English bcoz ( I think u understood) If u can't turn it in all just please turn it in 100 problem series please It's an humble request.............
It's not just square roots. You can't generally distribute exponents over products if the products are < 0 and the exponent is fractional. Generally, (a*b)^N = (a^N)*(b^N) only holds if 'a' and 'b' are positive real numbers or 'N' is an integer or both.
For me I see problems in 5 to 6 transition and in 2 to 3 transition. If we are working in the Complex numbers, square root has two roots. Square root of -1 is equal to { i, -i}, and square root of 1 is equal to { 1, -1}. Which means that every time we introduce square roots we transition from operating over numbers to operating over sets of numbers and each time we go from square root to number we transition from operating with sets of numbers to operating with numbers. Of course it breaks the equality.
I think when we ask for the square root of four, the answer is simply 2. But if we have an equation, we need to find what satisfies the X and then we have solutions 2, -2.
@@websparrow, it works with reals because reals are ordered and we can just pick the biggest one. With complex it's no longer true, there is no difference between i and -i, we can't choose one over the other, there is no more the biggest one. Because of that we are forced to consider all the roots and work with sets of numbers.
This answer sorta feels like a "because I say you can't" kinda answer... like i get it doesn't make sense to allow the split, but it just feels really unsatisfying.
I agree. Just the simple reminder that -1 has two square roots, +i and -i, would have gone a long way for explaining the why. He added the disclaimer that he is using the principal roots here [the roots with the smallest polar angle] but that is a definition that the target audience of this channel might not even know. (not to mention his clickbait OMG WOLFRAM ALPHA IS WRONG videos on the main channel when WolframAlpha does only consider the principal roots as a first result). The breaking point for the exaples he showed is exactly that only principal roots are considered, and is exactly the mistake that the original problem made. There are always two square roots.
@@almscurium That's because when they proved that sqrt(xy) = sqrt(x)sqrt(y) they had to make some assumptions to make that proof. The "rule" isn't an axiom - it doesn't "need" to be true, it was something that was derived from other rules, and when it was derived it was only ever true under certain conditions. iirc. it goes something like this: (sqrt(x)sqrt(y))^2 = xy = sqrt(xy)^2 Therefore, sqrt(x)sqrt(y) = +/- sqrt(xy) if x and y are both positive, then you can rule out the negative solution which is where the "rule" comes from (because obviously sqrt(x) and sqrt(y) are both positive numbers if x and y are both positive so the negative solution is invalid).. but you can only make that assumption when you know that x and y are positive.- otherwise you're just left with sqrt(x)sqrt(y) = +/- sqrt(xy)
One more thing about the first problem 5th Step: √-1 . √-1 = -1 which is different result if we look at previous step which is (√(-1).(-1)) which results to = 1
I guess going from step 3 to step 4 is ok as long as the number at the base is not purely imaginary. Since i is purely imaginary we are not allowed to go from step 3 to step 4. Correct?
i = √-1 by definition i^2 = -1 i^3 = -i i^4 = 1 i = 1^(1/4) up until then, the problem is correct. The error is in assuming that 1^(1/4) = 1 here. Which would be true, were it to be a simple operation. But here, we have i = 1^(1/4). This means that i = ⁴√1, and rewriting the equation with i as x, we get x^4 = 1, which has 4 possible solutions: 1, -1, i, and -i.
this is a good way to direct my media addiction towards something useful. i dont even need to learn this stuff, its just plain interesting and explained well
@@somethingsomething2541 well yeah, but that works for basically everything. most people wont repair their own car, but that doesn't make a video of such "mostly useless". I think to inspire curiosity about math you have to have videos like this that don't just talk about the what but also go into the why, and give you an easy "aha moment" that might inspire to you to seek more of those.
"because we are not allowed to do so" to me does not sufficiently explain why you can't split the square root, it just sounds like a random axiom you pull out of thin air.
Actually, that's the opposite. The fact that √ab=√a√b is true for any a or b is an axiom you pull out of thin air. It has been proven for positive a and b only. You can't prove it if both are negative because it's simoly not true. And it's easy to prove it is not true with √1=√((-1)(-1)) but not equal to √(-1)√(-1)
@@afanebrahimi7278 I know it's not true, but the video didn't explain why it's not true. It just said "you can't do that" which adds no insight whatsoever to understanding the problem.
@@StefaanHimpe The reason he didn't explain it is because it gets quite complicated and really requires a university level of understanding. You can't do it because the square root function is discontinuous, owing to the rotational element of the complex system, once you introduce complex numbers. In order to fix that discontinuity you need something called a branch cut which is just a line we say you can't rotate past. Once you choose this branch cut the square root is a nice function with only one solution. By splitting the square root with two negative numbers like in this video you cross the branch and introduce that discontinuity in to the equation which is how you get the weirdness
@4:41 I swear I did not look at any solution to this, but here is my guess. The 4th row where you wrote (i^4)^(1/4) can be written as (i*i*i*i)^(1/4) = i^(1/4)*i^(1/4)*i^(1/4)*i^(1/4) = undefined and *NOT* i since i^(1/4) is undefined. The rules of exponents say that (x^a)^b = (x^b)^a. But in this case the rules of exponents only work for real numbers, not imaginary ones. How'd I do?
sqrt(-1*-1) =/= i*i = -1 This is because the complex function f(z) = z^1/2 with a branch cut on R+ with f(1) = 1 defines the function sqrt(z). U cannot split the product and say (z1z2)^1/2 == (z1^0.5)(z2^0.5), as then u adding the arguments: pi + pi = 2pi, which crosses the branch cut. Rather, sqrt(-1*-1) = (e^2*pi*i)^1/2 = (e^i*0)^1/2 = 1 (in this principal branch), and we don't get nonsense like 1 = -1 NOTE, this branch cut PREVENTS us from saying that: 1 = sqrt(1) = (-1*-1)^1/2 = (e^2*pi*i)^1/2 = e^(i*pi) = -1, which is WRONG as we don't change the 2*pi -> 0 in the exponent. But the above function CAN represent some other branch of f(z) = z^1/2, e.g. say sqt(z), in which sqt(1) = -1 (and this does not mean 1 = -1 either!) Hence, it’s also worth noting that for arg(z1) = k, arg(z2) = m in [0,2*pi); If k+m < 2pi, then: sqrt(z1z2) == sqrt(z1)sqrt(z2) = r1 r2 exp[i(k+m)/2] People say this splitting property holds for 2 positive reals, 1 positive and 1 negative real, but not for 2 negative reals - this is precisely why. I’ve just mentioned the complete version in which case it is appropriate to split the product under the square root for any complex numbers z1 , z2
Bro breaking i^1 into i^(4/4) is the same as breaking (-1)^1 into (-1)^(2/2) and then ((-1)²)½ = 1½ = 1 Yeah the properties of exponents like distribution, squaring/rooting both sides etc. doesn't work with numbers like 0, 1, -1, rt(-1) or i, -rt(-1) or -i, rt(-i) etc.
did I miss something or did he basically just say "yeah this disproves math, but we have this special rule that says you can't do that, therefore math has not been disproved" I feel like there has a to be a more intuitive explanation somewhere
It's "you can't do this otherwise you will end up with a contradiction" For example, to show "can't divide by 0" 1*0=0 and 2*0=0 1*0=2*0 *divide both sides by 0* 1=2
@@bprpmathbasics right but why is it sometimes a contradiction and sometimes not? Why does this specific case result in a contradiction? What about those specific inputs makes the formula not work anymore? What can we learn about math in general from the knowledge of why negative numbers don't work? How do we know that this isn't a problem with the rule itself and that the formula can't be improved? There has to be a better explanation than "it doesn't work because you get a contradiction when you work it out". There has to be a deeper reason that could convince us that it wouldn't work before we even started doing all the math. Something deper down causing negatives to fail.
Going from step 2 to 3 is technically also wrong because the root of 1 is +-1 not +1 so it is mathematically not valid as it's not an equal transformation. It also makes sense because if the equation would be 2 = 1 +- 1 then 2 = 0 would be a correct solution (just as an example it obviously is not) and step 2 to 3 would be valid.
I think that the problem is created by square root convention: instead of treating 1^(1/2) as having 2 solutions, we assume that its the positive one, and multiply by -1 if we're using the other one. If you treat 1^(1/2) as equal to 1,-1, the problem disappears. If we ignore this convention, then step 3 introduces an extraneous solution, and if we acknowledge it, step 5 is incorrect. 2=2 2=1+1 2,0=1+1^(1/2) 2,0=1+((-1)(-1))^(1/2) 2,0=1+(-1)^(1/2)(-1)^(1/2) 2,0=1+i*i,1+i*-i,1+-i*i,1*-i*-i 2,0=0,2,2,0
The mistake is really clear when you realize you can effectively remove steps 4-7. It's really clear that, even though 1 and -1 are both square roots of 1, they aren't equal. Jumping through the hoops of complex numbers obfuscates that fact, making it harder to find the error in the overall "proof".
The only thing that jumps out to me as possibly algebraically incorrect is the jump from 4 to 5, as the definition of a square root may not be defined for complex numbers in the same way as it is for rationals. I don’t know the particular algebra rules yet since I haven’t gotten to complex analysis yet, but everything else looks algebraically fine, so that stands out as the only part which may be wrong, leading me to guess that step 5 is the incorrect step. Let’s see (0:28 btw)
Another issue which isn't pointed out but I thought worth mentioning: Sqrt a^2 = a So from line 3 to line 4, they are suggesting that 1=-1 which is not the case. Although I suppose it fits the pattern with 2=0
I am somewhat surprised with myself that I managed to point out this problem right away. It has been over a decade since I worked with complex and imaginary numbers but still saw the problem.
Another way to view this: Technically sqrt(1) = 1 and -1 but the function always picks the positive. The calculation above "forces" the inconsistent -1 to be the answer.
@@Darkness18641 sqrt(a) for a positive number a is defined to be the value x such that x is positive and x^2 = a. So the negative solution to x^2=a is excluded from sqrt() by definition in favor of adding a ± in front. If instead, I were to write a^(1/2) then it would be ambiguous
@@epikherolol8189 Technically roots of order n are defined as solutions of x^n = y But once you expand your view to the complex plane you will always get n different valid roots. In the case of square roots n=2 that's a positive and negative root. But functions need to be well-defined and so the root FUNCTIONS like sqrt(y) always pick the "first main root" and are uniquely defined that way.
I don't get step 3. It works with the number 1, but as a rule you can't take the square root of a single unit within an addition and maintain the equal sign.
It come from the fact that the sqrt function (denoted by the symbol that I don't have easy access to) is a function that picks a single square root out of multiple ones deterministically.
Yeah, but that's how things work. We have to have rules in place to allow for calculations to have meaning. If we didn't, none of it would matter or be useable.
@@DaSquyd yeah but he basically dismissed the crazy idea for being a crazy idea. Setting rules in place and then making new rules that basically say "dont do that" because you dont like the results is just dumb and there is definitly a better way to go about this. For example instead of saying "you are not allowed to do that" they should instead redefine how to calculate a certain thing so stupid results dont come out. Dont get rid of the bad result, get rid of the problem that caused the bad result. Of course that's not all his fault, but its still something i dislike.
I think the main problem is that the square root is separated even though it is undefined, √(ab) =√a ×√b only if a and b are non negative real numbers. You can see this problem for instance in √(-9×-4) if you separate the root you will get -6, but if you don't you will get 6. The correct one is 6 because separating the root is possible when the components it is separated into are defined under real numbers
Problem is between last two steps. 1^(1/4) = 1 yes, but its only one of the solutions (with complex numbers seemingly trivial equations can have more "hidden" solutions), 1^(1/4) can also be -1, i or -i.
The inherent problem with both of these proofs is that roots don't have only one solution, but we treat them like they do, because we want a nice, clean answer. The square root of -1 times the square root of -1 does actually equal the square root of 1, but it's the negative root, and since we started with the positive root, it changes the answer. Same with the end example. i^4 is 1, but 1^(1/4) can be i, -i, -1, or 1. That's why you need to be careful with roots. The problem doesn't lie within the axioms or the notation, but within the way we treat mathematics for our convenience
If you are still wondering "But why is the rule √(a × b) = √a × √b false for a,b < 0?", then consider this: Given a > 0, √a refers to the positive square root of a. But √(-a) refers to 𝒾√a, which is no more "positive" than -𝒾√a. If we use one "negative" and one "positive" square root of our negative numbers, then √(a × b) = √a × √b for a,b < 0.
@@liambohlwhile we do CONVENTIONALLY assume the positive square root when we write √ of a positive number, it doesn't apply when we use it as a tool to solve equations (or, like in the example in the video, complicate it and then simplify it back), because then we either halve or double the number of solutions we get. In case of the video it gets doubled, but only the extraneous one that leads to 2=1 is shown.
@@voomneshka when solving equations, we do see the +-, but thats only because of the rule sqrt(x^2)=|x|. for example: x^2=4 -> |x|=2 -> x=+-2. the equation in the video didnt involve an equation with a variable.
To your last paradox, you don't even need to use imaginary numbers: -1 = -1^1 = -1^(2/2) = ((-1)^2)^1/2 = 1^(1/2) = 1. Honestly this is why I'm not a fan of sqrt or fractional powers. Because people get two questions confused, for example the 4th root of 1 is 1, ie 1^(1/4) = 1 but the answer to what number to the 4th power is 1 ie x^4 = 1 actually has 4 solutions, (1, -1, i, -i). Honestly thinking of numbers as a set of all rotations in the complex plane that result in a value makes this way easier to see 1^1/4 = (e^(2*pi*n*i))^1/4 = e^(1/2*pi*n*i) = [e^(0), e^(1/2*pi*i), e^(pi*i), e^(3/2*pi*i), and equivalent rotations] = [1, i, -1, -i]
From steps 3 to 4 you are multiplying sqrt (-1)(-1) in just one element of the equation. It needs to be applied in the entire equation which will end with -2=-2. The same mistake happens from steps 2 to 3, a sqrt simply appears on one element of the equation.
I would argue the first mistake is from line 2-3, 1 is not equal to sqrt(1). Sqrt (1) is +/- 1, so statement 3 is asserting that 2=2 and that 2= 0 (1 -1) which is clearly not true.
If you write -1 in exponential notation, it is e^(i*pi + i*2*pi*k) where theta = pi + 2*pi*k is the argument, k is an integer denoting the branch So taking the square root gives you sqrt(-1) = e^(i*pi/2)*e^(i*pi*k) = + or - i The -i arrises from a different branch (odd value of k) The new argument, theta_2 = pi/2 + pi*k, has been "compressed in half" compared to theta. This is different from the "convention" that sqrt(-1) = i, which probably was originally defined for the branch k=0.
Hold on, so you state that the note written out at 3:15 is true because otherwise things would get screwy like with the problem in question, but that sounds like an excuse rather than a reasoning. The rule exists due to a consequence, not because there's a solid proof behind it. Can someone explain what this solid proof actually is? I'm honestly really curious.
@@elitrefy_opim sure you are mostly kidding but imaginary numbers describe how our universe function in various formulas, in fact its less that the numbers are imaginary and mire that they are beyond what we can see
Because sqrt(x) is undefined on negative numbers. It can be extended to the complex plane, which is where we get sqrt(-1)=i, but that introduces periodicity. e^i*pi = -1 but also e^3*i*pi = -1 etc. After some complex analysis you see that you can't split the negative radicals because you end up hitting the line where there is a discontinuity in the real part of the function (although the complex function remains smooth) as you have made a full rotation around the complex plane (at 2*pi).
1 + √1 = 0 is a valid and correct equation when we choose to define √1 = -1. The fact that the square root "function" in fact has two values at each positive real number is the true problem with the derivation. In practice, we always implicitly mean the square root that chooses the positive value for the square root of a positive number. In terms of complex numbers this is equivalent to the following :(we just treat unit norm complex numbers as it's sufficient here and simpler) let z = exp(ia) where 0
I didn't learn before you can't sperate negative square roots but I have to ask. What do we do if there are two negative numbers under the root multiplied with another postive one or a negitave one.
I think you would make them no longer both negative. Like if you had sqrt(-3*-5*-20) I would evaluate that as sqrt(-300) in which case there is only one negative. I'm curious if this gets messy when considering something like a square root of a polynomial with various negative terms.
It doesn’t really matter how many numbers you multiply together. If it’s positive inside it’ll be positive and real (no imaginary component) if it’s negative it’ll be the possible root * i, and that’s really the way to define the square root function
Basically, you need to resolve the negative signs first. It doesn't matter how many negative numbers are under the square root. If the final result of everything under the square root sign is positive, you get a real number. If the result is negative, you get an imaginary number. Or another way to put it: never split the square root into more than one negative number.
The n'th root will always return n results. The square root has two solutions, the cubed root has three, etc.. By assuming that it only has a scalar solution, we introduce the ambiguity that is causing this issue.
Isn't the problem occuring before the 4 -> 5 rewrite? I would argue it occurs as early as the 2 -> 3 rewrite (due to the (false?) assumption that √1 = 1, when in reality it is more accurate that |√1| = 1 (or √1 = ±1) thereby our 2 -> 3 rewrite introduces the ambiguity resulting in the false proof) Another way to demonstrate this, while also avoiding what you already addressed is: > 2 = 1 + √1 > 2 = 1 + √((-1)*(-1)) > 2 = 1 + √((-1)²) > 2 = 1 + ((-1)²)^(1/2) > 2 = 1 + (-1)^(2 * 1/2) > 2 = 1 + (-1)^(2/2) = 1 + (-1) = 1 - 1 > 2 = 0 I might be way off, just a lousy engineer after all :^) Interested in seeing the responses to this
no. it is not an assumption that sqrt1=1, and it is true. you learned the misconception that sqrt1=+-1, but the square root is a function and therefore returns only one value. you can test this by searching up sqrt1 or on a calculator, and verifying that y=sqrtx does indeed have exactly one corresponding value/output for each input on a graph.
Any number n when square rooted at the same time...which means √(n²) will always give |n|... learnt this back in first year of high school..and has been helping me ever since
What I learned was that i does not equal sqrt(-1) but i² = -1 and the two expressions are not equivalent. Compare 2 = sqrt(4) not being equivalent to 2² = 4, because (-2)² = 4 too.
The answer to the last question: The (1)^(1/4) part is correct. But the last step isn't. In this case it would form 4 roots of unity and i will be one of them. All the 4th roots of unity are ±1,±i
I disagree. the square root, cube root, etc. are functions, so they can only return one nonnegative value. sqrt4 is not equal to +-2 (you mustve learned the misconception). sqrt(x^2)=|x|. for example sqrt((-2)^2) is not -2. however, x^2=4 -> |x|=2 -> x=+-2. the 4th roots of 1 are indeed 1, -1, i, and -i, because those are the solutions to x^4=1, but 1^(1/4) is essentially the 4th root of 1, which of course 1. even typing it in a calculator, youll see it works.
@@johnyang799 please read my full reply to the op. you can even type 1^(1/4) on a calculator or search it up, and it will return 1. i understand the roots of unity, but that is irrelevant considering 1^(1/4) isnt equivalent to the 4th roots, which are indeed 1, -1, i, and -i, but the 4th root of 1, which can only return one value as a function: 1.
In complex domain, x^4 is a function, x^(1/4) is not a function because it has always 4 solutions. 1^(1/4) is not necessarily 1. What you are doing is equivalent to this: x^2 = 4. Then x = 2 and x = -2. So, 2 = -2.
@@bman5257 do you know the difference between = | ≈ | == ? √1 = 1 is a true statement but √1 == 1 is false it needs to be |1|, it's been a while since I graduated from highschool but I can clearly remember the basics of absolute numbers and roots
@@appmeurtre But |1| = 1. I think this ultimately boils down to nomenclature. I guess I’m just skipping the step of going to |1| because I immediately just evaluate the absolute value.
The way I've always learned this is that it's technically _not correct_ to say that i=sqrt(-1) Yes, i²=-1, but that's not the same thing. sqrt(a) is defined as the positive solution x to the problem x² = a. This does not exist for a=-1 -1 simply does not have a square root. So, in my opinion, the step from 5 to 6 is really the critical part.
Thanks for this, I was feeling crazy looking at the comments, I didn't consider that could be a French education thing. Everyone seems to confuse the sqrt function (defined only on reals and giving only positive roots) with the idea to look for all the roots. sqrt(1) is always 1 even though (-1)²=1, and sqrt(-1) isn't defined even though i² = -1. A function can't associate multiple outputs to a single input.
The accurate answer is the following: you do not mix complex value* root function and real root function. The examples are fine because each example use one root function, not both. To illustrate, let's respectively rename the real and complex root function f_1 and f_2. So we wrote f_1((-1) × (-1)) = f_2(-1) × f_2(-1). This formula doesn't exist. f_1 = f_2 on positive number, but cannot be applied for negative one. *: the function that associates a complexe number to any real number, with expression √•.
Also when 1 was written as sq root 1 , possible results became +-1 and we needed to specify that it should be |√1| and then proceed as |√-1|×|√-1| or whole as |√-1×-1|
I like to think about it like this: Sqrt(x) actually gives a positive and a negative number. 1 = Sqrt(1) only if you look at the positive root of 1. It doesn't count if sqrt(1) = -1. So if that's the case, then you can't say that 1 = sqrt(1) = sqrt(-1 * -1) because you're essentially just saying 'look at the bottom (negative root) Y value instead of the top (positive root) Y value,' which isn't actually equal to 1.
This is basically just a sneaky way of pulling the negative solution of a square root instead of the positive one. It could also be written as 2 = 2 2 = 1 + 1 2 = 1 + sqrt(1) 2 = 1 + sqrt((-1)^2) 2 = 1 - 1 2 = 0 the mistake here being that sqrt((-1)^2) = | -1 | = 1
With that logic, then that means sqrt(9) = -3...it's true in a pure conceptual sense, but it causes problems when using imaginary numbers. Try going backwards, (-3)^2 = 9? Yes, but sqrt(9) is sqrt(-1*-9) yet somehow is ALWAYS = +3 when we are simplifying expressions, when it should be = 3*i^2, or -3. It's not consistent.
when our teacher taught us about complex numbers, he used √-1 to explain the concept, and immediately warned us that we cannot write it down because it allows you to say anything is equal to anything. you just cannot write it down if doing proper mathematics. I find it funny how so many people can get it wrong.
1 variable complex equations have as many solutions as the highest power of the variable in the equation. (1)^(1/4)= z is the same as z^(4)= 1. And this has 4 solutions, 1, -1, i, -i.
To me, it's just that the square root function was defined as the only positive root. But it's only defined on positive real numbers. When you're not on this interval, it's not really the square root function and you should not assume it works the same
Using the way you did the first 2 examples also shows why. You would have sqrt(-1*-1)=sqrt(-1)*sqrt(-1). Except the left side is then sqrt(1).....which is plus or minus one. Since there are two possible answers, you cannot say that the right side equals the left side.
Whats funny is im pretty sure he covered the answer to this videos bonus problem in one of his other videos Iirc its somrthing to do with either thr principle root, or thr losing od roots For example Sqrt((-2)^2) Is 2, and ONLY 2 (the square root function is also known as the principle root function, only defined in positive real numbers) Whereas (sqrt(-2))^2 = -2 (sqrt of -2 is sqrt(2)i, which squared is 2 * -1, or -2) So somrthing something take thr quartic and quartic root and it does the same thing as what i just showed Note that the quartic root of 1 is i, -i, 1, and -1, but since its ALSO the principle quartic root, defined only in positive real numbers, the only root you see is 1. Idk Its something along those lines Powers and roots not being interchangable if the degree is even
squareroot with a negative number simply is undefined. You can say that -1 = x^2 has the solution i or -i Since -1 = 1*e^(pi + k2pi)i, if you halve the angle and square root the norm, you get x= 1 *e^(pi/2 + k*pi)i which give you 2 solutions in complex space, x =i or x= -i but sqrt(-x) does not exist. for x> 0
the thing done at last isint entirely wrong but only a bit The fourth root of 1 has 4 possible answers, 1, -1, i and -i. And whenever we square any number we have chances of getting extra solutions, so at end we write fourth root of 1 as i only as it is the only thing that satisfies the eqn. +1, -1 and -i are extra solutions
Step 5 got me. Even while it's reasonable to think that given i²= -1, then surely sqrt(-1)=i right? Well that's not how square root works. It can't take a negative number since there's no way two equal numbers will be negative, their signs would counter each other into positive anyhow you wanna think about them. There's the magic behind complex numbers, too
I love it how people try to find a loophole in math rules, just to be unproven by the fact that they made something wrong. This is the same thing as the proof that 2 equals 1. But in fact, you have to divide by 0 to get that.
1^(1/4) has four roots: 1, -1, i, and -i. Thus the last step is incorrect, 1 is not the only solution. This is the same as squaring (-1) and then writing square root of 1 as +1, and saying that (-1)=(+1)
The definition is i²= -1. Don't mess up with equivalence transformations. From there it is i = sqrt(-1) OR i = -sqrt(-1) !!! You end up to check two possibilities. Only one has to be true.
Root of product of two numbers say a and b can be broken into two separate roots only when both are NOT (-)ve. Infact √(a)(-b) = √a . √(-b) is correct while both √(-a)(-b) = √(-a) . √(-b) is incorrect. Thus step 5 is incorrect.
The problem is you have to differentiate between the "positive" square-root which is a function only defined for non-negative real numbers and the more general square root which is defined for all complex numbers but always returns two numbers for non-zero input. The "positive" square root is not the same as the more general complex square root even when limiting the input to non-negative real numbers since it will only find one of the two complex square roots. Now let's look at the problem. We can't be using the "positive" square root since √-1 would be undefined, so we must be using the complex square root. But then the transition from the second to the third line becomes invalid since √1 = ±1. But even after that there are more mistakes. (5) -> (6) √-1 = ±i, where the ± for each root are independent so there are four cases which simplify to "same sign" and "different sign". If you are always clear about which square root you are using you don't need to remember any weird rules about how you can and cannot factor square roots.
I remember my math teacher did this in highschool. Well at least the part of 2=0. I also remember it as the day I stopped giving a f about math. Went from 10 mark to 3 and went to literature class rather than maths.
i think the error is not in the rule for the roots but in the definition of i, i is defined as i^2=-1 which is not the same as i= sqrt(-1) in the example the distributions of the roots is wrong. an extra sign is missing
I haven’t watched the video but I’m guessing it’s step 3 where they forgot to make it two quart roots of 1 since you need it to multiple back together to make 1
It is correct to write "i^2 = -1", however you can't write "sqrt(-1) = i", Because the sqrt function is not defined for negative numbers (domain is real non negative numbers)
