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: ruclips.net/video/06NqtlmkPK0/видео.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
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 It may not be obvious that the steps are no longer valid!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.
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 😅
The main problem is that i²=(−1) ⇒ i = ±√−1 It’s not technically a separate rule but rather the fact that -i is also a square root of (-1). Since the square root is positive, by separating the terms we should normally have: 2= 1+ (-i)× i = 1+1 =2
@@VeteranVandal well square root is, but that dose not mean there is a eqvivelence between the statements, if you have x^2 = 9 x^2 = 9, but you this is not a eqvivelence ( ) just an implication, (if a then b). so if x is equal to 3, then x^2 is equal to 9. but it dose not mean that if x^2 is equalt to 9 that x MUST be equal to 3. becuase as we all know x can be +-3. But pure math states that sqrt(a^2) = abs(a) meaning that the square root always gives an answer sqrt(a^2) >= 0 for all a in the Real domain.
Yes. Op is picked the wrong line. The i = 1 proof uses the same issue. Fractional exponents are multi valued. The same principle clearly resolves both connundrums
@@BOOMDIGGERwhy should they? exaggeration is a perfectly acceptable method of communicating an idea. Most of us know enough not to take it literally (and if someone did take it literally, I can't see how it would cause any kind of problem)
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?
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)
I saw this and worked it out with pen and paper, and I think it was the first time in my life I actually *understood* why in calc and trig you are able to just "eliminate" results that don't make sense when dealing with functions that have more than one output. This is an amazing example for that, because my teachers all started with sine, basically telling us "well if the result isn't in the quadrant you want, then it's not the right one." Sometimes silly examples are the best examples
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 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.
I don't think your solution is correct. The real issue is step 3 because it is no longer equal to step 2 as the root of 1 has two solutions, one being -1 which gives the final result. I don't think your rule is true, √(ab) is always equal to √a•√b regardless. To circumvent this issue you need to constrain √1 to be positive so step 3 should be 2 = 1 + |√1| Which indeed gives 2 = 2 when you follow through Edit: even though it works out I don't think I am right because √1 really does just mean the principal square root of 1.
sqrt(1) is always positive because sqrt() function cannot be solved for negative values when the inside is positive and a real number. You are confusing this with the function x^2 where x can be negative or positive.
As mentioned in above comment, √1 and roots of x^2 = 1 have completely different meanings. While the square root of 1 is ±1, √1 only considers the principal root. This is why you write roots of x^2 = 3 as ±√3.
@alexwaters4133 someone didn't take any class past algebra, sqrt is always the principal root unless otherwise specified (for example, you derived it by solving for an x^2)
@@alexwaters4133 √1 is not an algebraic expression it's a constant. How you're finding multiple solutions to a constant term is beyond me. I'm assuming you did quadratic equations at some point. The solutions are x=-b±√D/2a. But according to you mathematicians are dumb and just writing -b+√D/2a is adequate since √D is ± inherently?
"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
I contend the error is in step 3. There are in general two solutions for a square root, so the substitution opens the door for picking the "wrong" solution when going the other direction.
The only took the principal value of the square root throughout, so it seems like all the square root computations were consistent (apart from the wrong step bprp said)
@@BryanLu0 for people who don't know why: squaring takes away information ("was it negative?"), so we can only get the absolute value back. So x²=4 resolves to |x|=2 -- and x can be positive or negative inside the abolute sign, so +x=2 or -x=2
problem is: that's how they teach you sqrt in high school they don't tell you that √9 = ±3, they just tell you that it's 3 (and in the quadratic formula they just add the ± outside of the root without any explanation to why it's there instead of a +)
@@PFnove well when you complete the square to derive the quadratic formula and then take the square root you get two values of x. By definition sqrt(x^2)=|x|. like x^2=4 -> |x|=2 -> x=+-2, and you do learn that in high school. in equations you would get two values of x but since the square root gives only the nonnegative result (its a function so it returns only one value) sqrt4=2 not just +-2.
@@hallrules no, in general, √9=±3. We split the √ function into branches so that it can be a union of single valued functions, but there are multiple branches, and hence multiple options for √9.
I got it right for the wrong reason. Lol. Anywho, Im almost more impressed by that marker switching technique you've got going there, than by the math.
If we take ALL 4 roots of 1 we see that the principal root is 1 and the other real root is -1 since (-1)^4=1 but the imaginary roots that make this true is i or -i since i^4 = 1 and (-i)^4=1
The result of ANY even-power root is an absolute value. So no, there is no other "real root" of √1. But the equation of x⁴ = 1 has 4 solutions: 1, -1, i, -i By turning i into i^(4/4), he artificially raised i to the fourth power and than took the 4th root. This creates 3 extraneous solutions, that break the equation.
@@Miftahul_786 The way it was written implies 4 potential solutions since it wasn't written as √x^2 i * i = i^2 = -1 i * -i = -i^2 = 1 -i * i = -i^2 =1 -i * -i = i^2 = -1
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
@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?
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.
You are not wrong, but for me you kinda failed to explain and just said "because that is the rule". Perhaps more insight into "principle value" would be appreciated. So do you mean that the sum of "phases" of multipliers inside the square root should not exceed or be equal to 2*pi?
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.
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.
For clarification of the i = 1 proof: The error lies in the very last step, going from 1^1/4 to 1. Here‘s why: In school, you are always taught to put +- in front of the result of a square root or any even number root for that matter. That’s because these roots have multiple possible solutions. What you aren’t taught, at least not until higher grade maths, is that the number of solutions is actually the same as the number in front of the root. It just so happens that many of these solutions will be complex and irrelevant for working with real numbers. If you display complex numbers in a 2D grid with imaginary on the y axis, the roots of any number can be displayed as a set of points around the origin. Their distance to the origin will be the same all around, so they will lie on a circle around the origin, and they will be arrayed so that if you were to connect each point with the origin, the circle would be divided into parts of equal size all around. Like that, it is easy to imagine what 1^1/4 entails. Yes, 1 and -1 are possible roots, but the last two roots are -i (directly down) and i (directly up). So if you were to consider all possible roots, you would actually get i back, which is true.
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.
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).
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.
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)
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
√1 =±1; however, +1≠‐1 Everything else is just a roundabout way of saying 1 =√1 =-1 This is incorrect. a≥0, a =√(|a|²) =√(|a||a|) =√(|a|)√(|a|) =|a| =a a
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.
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.
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.
Everyone is talking about the problrm being a square root has two solutions without actually discussing the core of the issue at hand. Do people not learn about branch cuts principle value and the fact that the square root is in fact a discontinuous function? Admittedly this leads to square roots having two solutions depending on which branch you are dealing with, but it kind of misses the point
I love these videos. I have children who will benefit from them. Also this one reminds me of my secondary physics teacher who was asked by a student for “extra credit work” to improve her grade. He asked why she wanted extra credit work when she could not do the work he already gave her. 😧
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 look at it as complex numbers spinning around a circle it is very simple to understand why this breaks. Multiplication of two numbers is basically adding their angles together and taking a root is like halfing the angle. So you can either first wind up around the circle by adding the angles and then take half of the angle that you get or you can half both of the angles first and then add them up. Under normal circumstances they both produce the same result. The issue is when you make a full loop around the circle. In this example ✓(-1 * -1) is like (180+180)/2 if you first add them up you get (360=0)/2 = 0/2 = 0 or if you split it up it's 180/2 + 180/2 = 90+90=180. So basically it breaks because after 360 the angles reset to 0. So you can't split up the roots if the sum of the angles of the numbers inside exceed 360.
You first turned 1 into the positive answer of the square root of 1* which is correct, but then truned the one inside into the product lf two negative ones, which wiuld imply using the negative answer of the square root you just used to transform one of the ones. Therefore, you can't just use two definitiones, you gotta stick to one if the values of the kutlivaluated function through all the process with one number.
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.
I knew which step contained the mistake, but I couldn’t say why. I must’ve learned the rule at some point and now it’s only floating in my subconscious.
You cant do this because the result would be wrong is not an explanation. This is math and there is an actual reason why you cant switch from real roots to complex roots like that.
@@bprpmathbasics The actual reason is that the root as a function is only defined for non negative real numbers. In the complex numbers all roots have multiple solutions and it "stops" being a function. So writing sqrt(-1) = i is actually not correct, because -i is also a solution to the equation x² = -1. So technically replacing sqrt(-1) with i is not allowed. Sometimes bending the rules in Math leads to suprisingly correct results (like physicists splitting dy/dx or some infinite sums) however often you get results that are simply wrong (like the famous example of the sum of all natural numbers being -1/12).
i think the end can still be valid if you write it as 2= 1+or - 1,because of the sqaure root, then the default asnwer will be + as that is the only valid solution
That's a great notice in the first problem. I hadn't thought of that because he is a teacher promoting real number primary root answers. Such teachings is hindering innovative thoughts like yours. If we kept ourselves in the Complex numbers world we can easily see how in both problems why only taking a primary root answer can fail. The teachers like this guy assume the +/- answers have to be an "AND" set of two answers rather than an "OR" choice of two choices. That means at a store in the last problem of fourth root of "cherry, pumpkin, apple and lemon" pie I'm not choosing all four to bring home (unless that is what I want). I choose one free of what I think is more satisfying. That is not the case with his fallacy. The same thing with Electrical Engineering and power calculations using voltages, currents and impedances a profession that analyzes both answers of complex numbers math and choose the answer based off "OR" where only one is a better solution than the other. That is why enforcing a voltage and current direction led to instantaneous designs of bridge rectifier circuit electronics because there was that equal opposite direction voltages and currents producing undesirable effects. This also is true in distance measurements in polar coordinates. Forcing an airplane pilot to head North East to fly a distance D makes no sense for any other direction(s) not North East flying that distance D, and especially, for a plane flying Southwest in the "mathematics of vectors in N dimensional space!" By the way in Complex numbers and vector mathematics the square roots of complex numbers come from equivalent domains of 0 to just under 2π then 2π to just under 4π, etc. + or - considerations. That is how we got √(a) in a + ib of b=0 produced the -√a result from 2π bringing it into [0, 2π) range of answers as much as √a
@@GeezSus no because the answer to a square root can be positive or negative because if you square a positive or negative number it will always be positive. so sqaure rooting a number means that there could be 2 possible valid solution
@@dastranjer9274 No?? A square root is ALWAYS positive, this channel has like a thousand videos on it just watch it. Square root is the magnitude, ± are the roots
Squaring both sides of an equation and taking the root of both sides are both problematic, because they change the set of solutions. While squaring gains new solutions, taking the root loses solutions. For example, a = 5 has only one solution, which is a = 5. But if you square, you get a² = 25, which has two solutions (a = 5 and a = -5). If you were to start with a² = 25, an equation that has two solutions, taking the root would lose one of them. In your particular equation, there are no variables, so the initial equation is just true. After taking the root it is just no longer true, you lost the solution that makes it true.
My take is this - whenever you take a square root, you should consider whether you need the principal or secondary root, or both. If step 5 was 2 = 1 +/- root(-1*root(-1) we could still have an equality.
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.
@@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.
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.
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.
With a limited grasp on the concept, basics, and process of algebra, I was pretty sure there was something wrong with the process within sets 4-6. Glad I wasn't completely off.
I'm not satisfied by your solution. Why are you not allowed to separate two negative numbers under a square root? I've never heard of this, and your explanation of "just because" doesn't suit me well. I thought the problem was at 1=sqrt(1) because sqrt(1) has two solutions in the complex number set, i.e. 1 and-1. So it does not surprise me that OP managed to turn sqrt(1) into -1
That's... exactly the issue, but also not. You are allowed to define the sqrt symbol to mean the principal square root if you wish, but the key issue is that the following law stops working, as there's no real preference for one or the other.
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.
Actually the term ✓-1 is "i" which is imaginary, where "i" is only used for theoretical imaginary problems, which can't be used as a practical solution. That's why this is not true
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.
i would say that its wrong in step 3 since root of 1 is also -1 (since were using complex numbers anyway , we can ditch the "positive square root" notation) this diverges us to the other solution , while keeping lhs original
That's my first thought too, but curiously it's step 5 where the equality breaks down? But doing the correct thing in step 3 and putting a ± to diverge when taking a square root seems to fix it? I honestly have no idea what's happening here
@@descuddlebat yes , the incorrect manipulation of square roots does cause the problems. generally you can apply the rules of normal math in the complex world , but the fact is , the foolproof definition of the complex logarithms and exponential functions isnt the same , since complex powers are periodic , a lot of interesting properties are observed. The complex exponential is described as z^w=e^(wln(z)) and logarithms are lnz=ln|z|+i(arg(z)) these definitions should help you steer clear of absurd results and also detect when youre getting a range of answers (of the type n π+k) by interesting properties , i mean landmines that make you lose marks
@@descuddlebat i suggest you check out the nth roots of one i could have done the same thing with 1^(2/3) .which would be one in the real plane , but actually isnt the same in the complex world
Such a lazy explanation. It doesn't work because it doesn't work... especially because it wouldn't have been to complex to explain what actually goes on there (or rather just complex ;))
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 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".
This guy really screwed up. In step 3 he should take the square root of both addends. 4. 2 = sqrt(-1*-1)+sqrt(-1*-1) 5. 2 = sqrt(-1)*sqrt(-1)+sqrt(-1)*sqrt(-1) 6. 2 = i*i+i*i 7. 2 = i^2+i^2 8. 2 = -1+-1 9. 2 = -2 qed
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
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: ruclips.net/video/06NqtlmkPK0/видео.html
B)
B
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
b
🅱️
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.
@@veroxid sometimes people forget computer science is a branch of mathematics.
I guess you can always just add a plus or minus to be safe until you find the solution but that seems so painful
@@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.
this guy is crazy with the two markers in one hand, pure talent
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 😅
I just noticed lol, he's so smooth with it
bprp = black pen red pen
@@aloksingh-em8cv I should probably learn this trick. I spend a lot of time pen shifting.
@@a_disgruntled_snail very easy u will get used to it
I don't agree that 1=√1. Instead, 1=|√1|. From that, you will get 2=2
No. If x is sqrt3
Then x is plus/minus sqrt3 (and sqrt 3 is positive)
the square root function √d is defined as the unique positive root of x^2=d. Thus √1=1 by definition
@@xmasterjkmyou are wrong if square root function defined like that
Sqrt_x^2 should be = x instead , but it is equal to IxI as we know
@@overpredor3412 notice i said positive root
@@xmasterjkm if square root were defined as positive root then
sqrt_x^2 would be equal to only x instead of +-x
The main problem is that i²=(−1) ⇒ i = ±√−1
It’s not technically a separate rule but rather the fact that -i is also a square root of (-1). Since the square root is positive, by separating the terms we should normally have: 2= 1+ (-i)× i = 1+1 =2
Yep. Square root isn't singly valued. If you write in polar notation you notice the problem immediately.
@@VeteranVandal well square root is, but that dose not mean there is a eqvivelence between the statements, if you have x^2 = 9 x^2 = 9, but you this is not a eqvivelence ( ) just an implication, (if a then b). so if x is equal to 3, then x^2 is equal to 9. but it dose not mean that if x^2 is equalt to 9 that x MUST be equal to 3. becuase as we all know x can be +-3.
But pure math states that sqrt(a^2) = abs(a) meaning that the square root always gives an answer sqrt(a^2) >= 0 for all a in the Real domain.
@@hampustoft2221 yup
Yes!
Stating that √-1 = i is just sloppy math.
Yes. Op is picked the wrong line. The i = 1 proof uses the same issue. Fractional exponents are multi valued. The same principle clearly resolves both connundrums
I love how this chanel is called maths "basics" when it never fails to blow my mind.
I need to sit down after this
It just means you don't know the basics
stop exagerating...
the education system failed you 🤦
No need for judgment guys
@@BOOMDIGGERwhy should they? exaggeration is a perfectly acceptable method of communicating an idea. Most of us know enough not to take it literally (and if someone did take it literally, I can't see how it would cause any kind of problem)
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).
100% agreed
This explains the weirdness in the first example aswell. Thank you.
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
Wow, this is the most intuitive and concise explanation so far. Thanks!
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).
wait doesnt the square root already have 2 roots by default but we usually ignore the negative roots
@@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.
Was looking for this comment. The roots of -1 are i and -i. So in reality you could have two possible solutions to sqrt(-1)^2, which are ±i^2 = ±1.
@@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
@@luminessupremacy bookmark comment later
Love how you switch between blue marker and red marker. So skilled
Bruh I did in every exam switching between a pen, pencil and an erases even
@@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?
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.
It isn't "because I said so", it is "because sqrt((-1)(-1)) != sqrt(-1)sqrt(-1)", since the left side is 1 and the right is -1
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.
@@BillyONealyes but why does the rule not apply to two negatives in a square root
@@almscurium I don't know, complex numbers are weird
@@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)
its really cool you proved i = 1 at the end cuz in the complex plane i actually represents 1 in the imaginary axis
'i' does not represent 1 in the imaginary axis. it represents 'i ' in the imaginary axis. just for clarity 1*1 = 1, i*i = - 1. Both are not the same
@@aravindmuthu95 the reason I meant i represents 1 in the imaginary axis is because it's radius in the circle of the complex plane is equal to 1
@@Gezrafif you're talking about the distance from zero, then you should've said abs( *i* )=1 which is true
The whole point of that proof was that it was WRONG; he is asking you to find the problem with it lol
Lol
I saw this and worked it out with pen and paper, and I think it was the first time in my life I actually *understood* why in calc and trig you are able to just "eliminate" results that don't make sense when dealing with functions that have more than one output.
This is an amazing example for that, because my teachers all started with sine, basically telling us "well if the result isn't in the quadrant you want, then it's not the right one."
Sometimes silly examples are the best examples
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.
Thus, proofs that use a such function or notation must make sure it is well defined for the problem approached.
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.
it also breaks the reality
I'm impressed at how someone can perfectly and effortlessly write the horizontal crosses of the letters 'f' and 't' before the vertical strokes.
I didn't even notice!
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
It’s still entertainment though.
Its not usefull for general audience , most people here will probably never use it.
@@somethingsomething2541 It's useful to distract yourself from the distraction xd
Nobody ever will use it besides as an interesting "trick" to know
@@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.
I don't think your solution is correct. The real issue is step 3 because it is no longer equal to step 2 as the root of 1 has two solutions, one being -1 which gives the final result. I don't think your rule is true, √(ab) is always equal to √a•√b regardless.
To circumvent this issue you need to constrain √1 to be positive so step 3 should be
2 = 1 + |√1|
Which indeed gives 2 = 2 when you follow through
Edit: even though it works out I don't think I am right because √1 really does just mean the principal square root of 1.
sqrt(1) is always positive because sqrt() function cannot be solved for negative values when the inside is positive and a real number. You are confusing this with the function x^2 where x can be negative or positive.
As mentioned in above comment, √1 and roots of x^2 = 1 have completely different meanings. While the square root of 1 is ±1, √1 only considers the principal root.
This is why you write roots of x^2 = 3 as ±√3.
Someone didn’t pass algebra… sqrt is always +/-, we usually constrain it to absolute value because a negative solution wouldn’t make sense
@alexwaters4133 someone didn't take any class past algebra, sqrt is always the principal root unless otherwise specified (for example, you derived it by solving for an x^2)
@@alexwaters4133 √1 is not an algebraic expression it's a constant. How you're finding multiple solutions to a constant term is beyond me.
I'm assuming you did quadratic equations at some point. The solutions are x=-b±√D/2a. But according to you mathematicians are dumb and just writing -b+√D/2a is adequate since √D is ± inherently?
maybe we need to not do that and stay happy
yes, peace
I saw this in one of my highschool, gotta say it was so stupid trying to prove 2= 0
"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
@@joshuagillis7513Exactly!
Skill issue
Thanks, now I can finally take my revenge from my maths teacher 😈
I contend the error is in step 3. There are in general two solutions for a square root, so the substitution opens the door for picking the "wrong" solution when going the other direction.
Usually if you have a square root you only take the principal value
I agree with this. There are two possible solutions to the square root of 1. It is false to say 1 and the square root of 1 are equivalent.
The only took the principal value of the square root throughout, so it seems like all the square root computations were consistent (apart from the wrong step bprp said)
@@MrJuliancarroll Sqrt(1)=1 the square root of 1 is 1, or the square root of 9 is 3
It's only necessary to say +/- when it is the inverse of squaring
@@BryanLu0 for people who don't know why: squaring takes away information ("was it negative?"), so we can only get the absolute value back. So x²=4 resolves to |x|=2 -- and x can be positive or negative inside the abolute sign, so +x=2 or -x=2
This is a side effect of taking multivalued functions and making them arbitrarily single valued
problem is: that's how they teach you sqrt in high school
they don't tell you that √9 = ±3, they just tell you that it's 3 (and in the quadratic formula they just add the ± outside of the root without any explanation to why it's there instead of a +)
@@PFnove well when you complete the square to derive the quadratic formula and then take the square root you get two values of x. By definition sqrt(x^2)=|x|. like x^2=4 -> |x|=2 -> x=+-2, and you do learn that in high school. in equations you would get two values of x but since the square root gives only the nonnegative result (its a function so it returns only one value) sqrt4=2 not just +-2.
@@PFnove they don't tell you that √9 = ±3 cuz its not ±3 (its just 3), unless i dont understand what ur trying to say
@@hallrules no, in general, √9=±3. We split the √ function into branches so that it can be a union of single valued functions, but there are multiple branches, and hence multiple options for √9.
@@xinpingdonohoe3978 sqrt(x) is a function, functions only give either no output or one output. ±√9=±3, √9=3
I got it right for the wrong reason. Lol. Anywho, Im almost more impressed by that marker switching technique you've got going there, than by the math.
Try the problem at 3:48
Here’s the answer: ruclips.net/video/awrgXX0Qnjs/видео.htmlsi=Jzz5j4KLJu14Mv3e
👆👆👆 This comment was made before the video was uploaded. 🤨🤨🤨
@@anhada.8347yes probably the video was private
solved
broken
sorry. not sorry😅😂
he probably uploaded video and published it with a delay, maybe a scheduler and the uploader can already comment on it as soon as its uploaded
Don't be mean to the complex conjugate
2:23 Bro the way he changes his markers is dope!🔥
If we take ALL 4 roots of 1 we see that the principal root is 1 and the other real root is -1 since (-1)^4=1 but the imaginary roots that make this true is i or -i since i^4 = 1 and (-i)^4=1
the root function always takes the absolute value as in:
√1 equals only 1
but x²=1 has the four answers
The result of ANY even-power root is an absolute value. So no, there is no other "real root" of √1.
But the equation of x⁴ = 1 has 4 solutions: 1, -1, i, -i
By turning i into i^(4/4), he artificially raised i to the fourth power and than took the 4th root. This creates 3 extraneous solutions, that break the equation.
@@RuleofThehyperbolicx^2=1 only has 2 solutions not 4
@@Miftahul_786
The way it was written implies 4 potential solutions since it wasn't written as √x^2
i * i = i^2 = -1
i * -i = -i^2 = 1
-i * i = -i^2 =1
-i * -i = i^2 = -1
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
Finally someone that explains these strange behaviours using complex analysis and not only some "rule".
You should be the comment on the top.😊
@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?
Try the problem at 3:50
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 wonder if a,b
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.
@@CMTRN this is the correct answer
You are not wrong, but for me you kinda failed to explain and just said "because that is the rule". Perhaps more insight into "principle value" would be appreciated. So do you mean that the sum of "phases" of multipliers inside the square root should not exceed or be equal to 2*pi?
2 things you just don't do in math! ruclips.net/video/iprO9v4reTs/видео.html
I don't do math at all 💀
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.............
Dude, I hate love you, good job.
I'm sorry, but isn't step 3 already a mistake? Shouldn't everything become a square root? Can we square root just one number in the equation?
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.
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.
For clarification of the i = 1 proof: The error lies in the very last step, going from 1^1/4 to 1. Here‘s why:
In school, you are always taught to put +- in front of the result of a square root or any even number root for that matter. That’s because these roots have multiple possible solutions. What you aren’t taught, at least not until higher grade maths, is that the number of solutions is actually the same as the number in front of the root. It just so happens that many of these solutions will be complex and irrelevant for working with real numbers. If you display complex numbers in a 2D grid with imaginary on the y axis, the roots of any number can be displayed as a set of points around the origin. Their distance to the origin will be the same all around, so they will lie on a circle around the origin, and they will be arrayed so that if you were to connect each point with the origin, the circle would be divided into parts of equal size all around.
Like that, it is easy to imagine what 1^1/4 entails. Yes, 1 and -1 are possible roots, but the last two roots are -i (directly down) and i (directly up). So if you were to consider all possible roots, you would actually get i back, which is true.
Yes the concept from n roots of unity.
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.
In step 3 there was a rule that if 2 numbers are negative in under root and they are in product then they can't separate.
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.
Let's just be honest... The whole concept of imaginary number is like an excuse made by mathematicians to cover up for their mistakes 😂
@@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
It's because they aren't the same thing. Like apples to oranges. Are you asking for a proof on why they are different?
@@thevorhandener5280 ofcourse I just let my true feelings slide for a second there 😂😂
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).
These comments hurt my brain. So many people that are confidently wrong.
Funny how every time somebody finds a loophole in math, math just says "oh, this is actually an exception of math, you cant do that!"
It is not a loophole🤣. Read some math literature and find the answers yourself.
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.
4:21
Properties of exponents are invalid for base < 0
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.
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)
1:20 I feel so heard right now 😅😂😂
Well I don’t think root 1 can be substituted for +1 since it can also be -1
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
Also x+sqrt(x)≠ x+x
I’m just impressed someone finally did one of these false proofs by doing something else besides dividing by zero
or subtracting
√1 =±1; however, +1≠‐1
Everything else is just a roundabout way of saying
1 =√1 =-1
This is incorrect.
a≥0, a =√(|a|²) =√(|a||a|)
=√(|a|)√(|a|) =|a| =a
a
If you work with imaginary numbers, you get imaginary answers.
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.
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.
Ur answer is flawed from ur very first statement 😅.
Sqrt(1) is 1 and ONLY 1.
NOT -1
@@epikherolol8189 Then how much is (-1)^2?
@@Darkness18641its 1 but epik is still right
@@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.
Reddit guys maybe didn't study class 11 complex number chapter.(NCERT book)
Does "if a, b < 0 then √(ab)≠√a√b" come from something more fundamental, or does it come directly from this kind of equation?
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.
Dude went through complex numbers just to get to the point where sqrt(1) =-1
In the book "O Algebrista" (lang: PT-BR) says that you cant separate
a real number into 2 imaginary, and give examples like that
Valeu pela dica!
Everyone is talking about the problrm being a square root has two solutions without actually discussing the core of the issue at hand. Do people not learn about branch cuts principle value and the fact that the square root is in fact a discontinuous function?
Admittedly this leads to square roots having two solutions depending on which branch you are dealing with, but it kind of misses the point
I love these videos. I have children who will benefit from them. Also this one reminds me of my secondary physics teacher who was asked by a student for “extra credit work” to improve her grade. He asked why she wanted extra credit work when she could not do the work he already gave her. 😧
Thank you!
314,735 views, 8.2k likes, 534 comments, 98.1k subscribers. Nice!!
I think it is step 5 to 6. the imaginary unit (i) is defined by its property i^2 = −1. NOT that sqrt(-1) = i.
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.
Yea I also didn't quite understand how 1 turned into sqrt(1) without any explanation whatsoever
√1 is just 1, not -1 nor ±1
@@fiprandom3783-1*-1=1. Square root of 1 is ±1
If x^2=1, then x can be 1 or -1.
But the square root of 1 is just 1.
@@fiprandom3783-1*-1=1
I suppose, step 3 is not correct.
Sqrt(1) = 1; -1. So, we can use only arithmetic root.
If you look at it as complex numbers spinning around a circle it is very simple to understand why this breaks. Multiplication of two numbers is basically adding their angles together and taking a root is like halfing the angle. So you can either first wind up around the circle by adding the angles and then take half of the angle that you get or you can half both of the angles first and then add them up. Under normal circumstances they both produce the same result. The issue is when you make a full loop around the circle. In this example ✓(-1 * -1) is like (180+180)/2 if you first add them up you get (360=0)/2 = 0/2 = 0 or if you split it up it's 180/2 + 180/2 = 90+90=180. So basically it breaks because after 360 the angles reset to 0. So you can't split up the roots if the sum of the angles of the numbers inside exceed 360.
You first turned 1 into the positive answer of the square root of 1* which is correct, but then truned the one inside into the product lf two negative ones, which wiuld imply using the negative answer of the square root you just used to transform one of the ones. Therefore, you can't just use two definitiones, you gotta stick to one if the values of the kutlivaluated function through all the process with one number.
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.
Appreciate that the guy used something a bit more sophisticated than division by zero that is typically used to "derive" these types of 1=2 results
I knew which step contained the mistake, but I couldn’t say why. I must’ve learned the rule at some point and now it’s only floating in my subconscious.
How do we know its not the entire imaginary realm that is wrong 🤔🤔🤔
You cant do this because the result would be wrong is not an explanation. This is math and there is an actual reason why you cant switch from real roots to complex roots like that.
What’s the actual reason?
@@bprpmathbasics
The actual reason is that the root as a function is only defined for non negative real numbers. In the complex numbers all roots have multiple solutions and it "stops" being a function. So writing sqrt(-1) = i is actually not correct, because -i is also a solution to the equation x² = -1. So technically replacing sqrt(-1) with i is not allowed.
Sometimes bending the rules in Math leads to suprisingly correct results (like physicists splitting dy/dx or some infinite sums) however often you get results that are simply wrong (like the famous example of the sum of all natural numbers being -1/12).
i think the end can still be valid if you write it as 2= 1+or - 1,because of the sqaure root, then the default asnwer will be + as that is the only valid solution
That's a great notice in the first problem. I hadn't thought of that because he is a teacher promoting real number primary root answers. Such teachings is hindering innovative thoughts like yours. If we kept ourselves in the Complex numbers world we can easily see how in both problems why only taking a primary root answer can fail. The teachers like this guy assume the +/- answers have to be an "AND" set of two answers rather than an "OR" choice of two choices. That means at a store in the last problem of fourth root of "cherry, pumpkin, apple and lemon" pie I'm not choosing all four to bring home (unless that is what I want). I choose one free of what I think is more satisfying. That is not the case with his fallacy.
The same thing with Electrical Engineering and power calculations using voltages, currents and impedances a profession that analyzes both answers of complex numbers math and choose the answer based off "OR" where only one is a better solution than the other. That is why enforcing a voltage and current direction led to instantaneous designs of bridge rectifier circuit electronics because there was that equal opposite direction voltages and currents producing undesirable effects.
This also is true in distance measurements in polar coordinates. Forcing an airplane pilot to head North East to fly a distance D makes no sense for any other direction(s) not North East flying that distance D, and especially, for a plane flying Southwest in the "mathematics of vectors in N dimensional space!"
By the way in Complex numbers and vector mathematics the square roots of complex numbers come from equivalent domains of 0 to just under 2π then 2π to just under 4π, etc. + or - considerations. That is how we got √(a) in a + ib of b=0 produced the -√a result from 2π bringing it into [0, 2π) range of answers as much as √a
but... square roots are already always positive....
@@lawrencejelsma8118 i never knew about the practicle uses of +/- for square roots, i might look into that further in my free time.
@@GeezSus no because the answer to a square root can be positive or negative because if you square a positive or negative number it will always be positive. so sqaure rooting a number means that there could be 2 possible valid solution
@@dastranjer9274 No?? A square root is ALWAYS positive, this channel has like a thousand videos on it just watch it. Square root is the magnitude, ± are the roots
Why is noone talking about how cool he is when he switches the markers in his hand
7th step: (-1 * -1) gives us 1.
8th step: 2= 1+1
2=2
1=1
-1 = -1
-1/1 = 1/(-1) (take root both sides)
i=1/i
i²=1
=> 1= -1
Can anyone tell me what is wrong in here
Squaring both sides of an equation and taking the root of both sides are both problematic, because they change the set of solutions. While squaring gains new solutions, taking the root loses solutions. For example, a = 5 has only one solution, which is a = 5. But if you square, you get a² = 25, which has two solutions (a = 5 and a = -5). If you were to start with a² = 25, an equation that has two solutions, taking the root would lose one of them. In your particular equation, there are no variables, so the initial equation is just true. After taking the root it is just no longer true, you lost the solution that makes it true.
My take is this - whenever you take a square root, you should consider whether you need the principal or secondary root, or both. If step 5 was 2 = 1 +/- root(-1*root(-1) we could still have an equality.
Prediction: the mistake was made at point 5
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
Full agreement. That’s where I saw the problem, and for that exact reason.
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.
Square roots don't expand like that in step 5 if there is a negative number inside(or in case an imaginary no.)
0:05 step 3 is wrong
Nope, it's step 5
@@neerav10i think he refers to √1 = |1| not 1
@@appmeurtreThe absolute value of 1 is 1 though. Step 3 is fine because it’s just saying the principal root of 1 is 1
@@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.
Nah i was mesmerized by the way you handle your markers
Credits to the multipen writing with one hand. Insane.
"this ridiculous proof is wrong because if it would be true then ridiculous things would come out so its wrong" is basically what you said
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.
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.
Bro used imaginary numbers 💀💀
Case solved, 2 does NOT equal 0 😂😂😂
imaginary numbers are used in quantum mechanics which describes the movement of subatomic particles, they are very real
With a limited grasp on the concept, basics, and process of algebra, I was pretty sure there was something wrong with the process within sets 4-6. Glad I wasn't completely off.
I'm not satisfied by your solution. Why are you not allowed to separate two negative numbers under a square root? I've never heard of this, and your explanation of "just because" doesn't suit me well.
I thought the problem was at 1=sqrt(1) because sqrt(1) has two solutions in the complex number set, i.e. 1 and-1. So it does not surprise me that OP managed to turn sqrt(1) into -1
That's... exactly the issue, but also not. You are allowed to define the sqrt symbol to mean the principal square root if you wish, but the key issue is that the following law stops working, as there's no real preference for one or the other.
0:20 Where in the hell did 2=1-1 and 2=0 come from??
i²=-1
steps 1 through 7
i^2=-1, therefore 1+i^2=1+(-1)=1-1=0
The bishop from the thirteenth dimension
If paused the video and after my thorough analysis I have determined the issue is somewhere between step 3 and 7
Yep, 5
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
Issue is the first step. i^1 is not i^(4/4).
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 you are also incorrect. 1 is equal to 4/4, so it is indeed equivalent.
@@kobalt4083 Then the op is correct. It's either when you introduce the 1/4 part or when you execute it.
@@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.
Actually the term ✓-1 is "i" which is imaginary, where "i" is only used for theoretical imaginary problems, which can't be used as a practical solution. That's why this is not true
i is used all the time bud
Thats why in french, sqrt is only defined with real numbers. Makes a lot more sense then inventing random rules.
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.
Two apples aren't the same as no apples. Boom solved! 💪
i would say that its wrong in step 3
since root of 1 is also -1 (since were using complex numbers anyway , we can ditch the "positive square root" notation)
this diverges us to the other solution , while keeping lhs original
That's my first thought too, but curiously it's step 5 where the equality breaks down?
But doing the correct thing in step 3 and putting a ± to diverge when taking a square root seems to fix it?
I honestly have no idea what's happening here
Does the wrong application of sqrt(ab) somehow "jump between the roots" ? And what's the mechanism? I'm really curious about this now
@@descuddlebat yes , the incorrect manipulation of square roots does cause the problems. generally you can apply the rules of normal math in the complex world , but the fact is , the foolproof definition of the complex logarithms and exponential functions isnt the same , since complex powers are periodic , a lot of interesting properties are observed. The complex exponential is described as z^w=e^(wln(z))
and logarithms are lnz=ln|z|+i(arg(z))
these definitions should help you steer clear of absurd results and also detect when youre getting a range of answers
(of the type n π+k)
by interesting properties , i mean landmines that make you lose marks
@@descuddlebat i suggest you check out the nth roots of one
i could have done the same thing with 1^(2/3) .which would be one in the real plane , but actually isnt the same in the complex world
No the square root of 1 is always 1 its never -1!
Such a lazy explanation. It doesn't work because it doesn't work... especially because it wouldn't have been to complex to explain what actually goes on there (or rather just complex ;))
But then I guess I'm not the target audience
Discussing the rules of a number that doesn't exist is peak human development
They may not exist but you wouldn't be watching this video without them
@@rayaneferouni8658 ok so?
They have been very useful.
@@shadowyt376 for what
wait so can you explain why sqrt(a*b) != sqrt(a)*sqrt(b) when a, b, < 0? I know it is a rule but why?
It’s “if you do that, you get a contradiction”. Here’s a video for that ruclips.net/video/iprO9v4reTs/видео.htmlsi=0Zh8iMKA4K7Q3xuP
I'm impressed by your ability to hold two markers and seamlessly switch between them like that
That is actually an incredibly easy thing to do. Next time you’re at a whiteboard try it. Can literally be done by anyone with zero practice.
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 question is WHY?
(At 1:11 in the video)
I think I can replicate this in less steps
2 = 2
2 = 1 + 1
2 = 1 + x (given x² = 1)
2 = 1 + -1 because -1² = 1
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".
For the solution of the question at the end, reference to the topic 4th root of unity.
So basically this rule avoids that any numbers becomes equals to 0
This guy went from Lawful to Chaotic in less than 5 minutes
So, it doesn't work because Algebra made a balance patch, when someone found a bug.
This guy really screwed up. In step 3 he should take the square root of both addends.
4. 2 = sqrt(-1*-1)+sqrt(-1*-1)
5. 2 = sqrt(-1)*sqrt(-1)+sqrt(-1)*sqrt(-1)
6. 2 = i*i+i*i
7. 2 = i^2+i^2
8. 2 = -1+-1
9. 2 = -2
qed
Just to be clear, by "this guy" I mean the guy who sent in the problem, not bprp.
😄
I have a doubt. Instead of splitting it as √-1.√-1, why dont we write it as √(-1)^2 and cancel the sq and sqrt. We'll still have 1-1 🫠
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