Hate to be that guy but I need the extra credit! Reddit complex numbers r/theydidthemath

[@bprpmathbasics]

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: kzbin.info/www/bejne/ZmexoqeiotCDgZI

[@thesecretartist327]

B)

[@timecubed]

B

[@vaibhav3238]

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

[@sreekar47807]

b

[@Brocseespec]

🅱️

[@davidecordella9015]

0:25 "We start with 2 = 1 + 1, I agree 👍" the whole of mathematics trembling in fear for his answer

[@Zvxers7]

I disagree.

[@davidecordella9015]

@@Zvxers7 scientists shaken by shocking Zvxers7's disapproval

[@HerrFinsternis]

​@@Zvxers7 Russel and Whitehead are disappointed with you.

[@andreaaazo]

@@Zvxers7nah I agree. Just use Dedekind Cut to demonstrate

[@Dokattak]

It took over 200 pages... but we got there!

[@georgepajeejo7045]

this guy is crazy with the two markers in one hand, pure talent

[@aloksingh-em8cv]

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 😅

[@nitsua6300]

I just noticed lol, he's so smooth with it

[@a_disgruntled_snail]

@@aloksingh-em8cv I should probably learn this trick. I spend a lot of time pen shifting.

[@aloksingh-em8cv]

@@a_disgruntled_snail very easy u will get used to it

[@Milky_tea07]

​@@a_disgruntled_snailsame

[@silver6054]

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 It may not be obvious that the steps are no longer valid!

[@kazedcat]

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.

[@veroxid]

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.

[@MinecraftMasterNo1]

@@veroxid sometimes people forget computer science is a branch of mathematics.

[@sandro7]

I guess you can always just add a plus or minus to be safe until you find the solution but that seems so painful

[@veroxid]

​@@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.

[@sammyjones8279]

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

[@UserT049.]

Leaving a reply so I'll come back here when I know what you're talking about (i haven't even started calc. yes I'm young)

[@collegephysicsforeveryone7744]

Sounds like they got to you too. Just throw away answers right because they dont fit what we were taught

[@gerald2508]

@@collegephysicsforeveryone7744 I am sad that I have to ask you to confirm that this is a joke. Edit: I am sorry, I should have asked before I did sarcasm - is this just about math or are you making a political statement? The former is confusing but only the latter is sad.

[@SGD2718]

@@gerald2508as someone who took all high school math through proof based classes, this has always made sense as a division by zero or a false inversion error

[@nikhildeodhar143]

More than "eliminating" results, it's about defining the correct sets (domains and co-domains) where the mapping will be unique and consistent, at the start of the process itself. Therefore in a lot of maths problems in graduate courses and above, you will see the domain and co-domain explicitly mentioned. Even if it is obvious.

[@C0pium_]

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]

Yep. Square root isn't singly valued. If you write in polar notation you notice the problem immediately.

[@hampustoft2221]

@@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.

[@C0pium_]

@@hampustoft2221 yup

[@NLGeebee]

Yes! Stating that √-1 = i is just sloppy math.

[@ncoderre1]

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

[@taborsmrcna]

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

[@arachnohack9050]

I love how this chanel is called maths "basics" when it never fails to blow my mind. I need to sit down after this

[@alien3200]

It just means you don't know the basics

[@BOOMDIGGER]

stop exagerating...

[@wvoxu]

the education system failed you 🤦

[@MikoRalphino]

No need for judgment guys

[@douglaswolfen7820]

​@@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)

[@psychologienerd7546]

3:40 so the reason its not allowed is because it leads to conflicts? isnt there a proper reason out there?

[@comm1nd]

Because when you multiply 2 negative numbers they become positive, if you wanna split them you make them both positive then split them as you wish You can think about it as 2=sqrt((-1*2)(-1*2)) The -1 cancel out each other and your left with 2 on each side You can also write -1*1

[@prithvibhat1017]

The square root of 1 is either +1 or -1. So if you are taking -1 as the square root, then it is immediately clear that it is not consistent with statement number 2. To write it formally, statement number 3 should be written as : 2 = 1 +(-) sqrt[+(-) 1]. You either consider the operators inside the bracket or the ones outside. The important part about manipulating equations is that the meaning of each statement must be the same as the previous one. Taking the negative root without putting a minus sign does not give you 1+1

[@filip_1432]

yeah, 2 negative numbers multiplied are just the same two numbers multiplied, but positive. √-1=i is a complex number, while √1 is a real number, so you can't say √1=√-1, because no complex number is equal to a real one. therefore, √1×√1 is not equal to √-1×√-1

[@personpeoplepeoplepersons5722]

It probably has to do with order of operations. Multiplication comes before exponents, right?

[@nikhildeodhar143]

The proper reason corresponds to the definition of the functions themselves (square root in this case) and the set of numbers over which the it is defined. In set of real numbers, square root is only defined on zero and positive numbers. So I am pretty sure that splitting a square root like that would itself be an illegal operation for the set of real numbers. Since square root of -1 is not 'i', but is simply undefined. In complex numbers on the other hand, you have multiple roots on the unit circle. So you cannot just substitute one square root with another. It would be more accurate to say that a number "belongs to set of square roots" rather than "equal to square root". The rules will be different for that set. This is something that has been pointed out by others in the comments I believe. I am sure that someone who is more experienced in the subject of Mathematical Analysis will be able to give a more accurate and clearer form of this answer though. I am still a novice with it.

[@michaelbyrd1674]

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]

wait doesnt the square root already have 2 roots by default but we usually ignore the negative roots

[@michaelbyrd1674]

@@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.

[@davidebic]

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.

[@luminessupremacy]

​@@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

[@BakrAli10]

@@luminessupremacy bookmark comment later

[@chungkhang3444]

Love how you switch between blue marker and red marker. So skilled

[@shadowblue4187]

Bruh I did in every exam switching between a pen, pencil and an erases even

[@odintakerprime6595]

​@@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?

[@voidarkyt]

That’s his channel name. Bprp blue pen red pen

[@m.h.6470]

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).

[@klm2558]

100% agreed

[@SirGoP]

This explains the weirdness in the first example aswell. Thank you.

[@fresh_dood]

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

[@sugardude]

Wow, this is the most intuitive and concise explanation so far. Thanks!

[@ernestrobinson8441]

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.

[@TGW757]

I'm impressed at how someone can perfectly and effortlessly write the horizontal crosses of the letters 'f' and 't' before the vertical strokes.

[@naomiparsons462]

I didn't even notice!

[@BrickBreaker21]

Best comment 😂

[@justinhowe3878]

There's a chance that comes from learning Chinese as a first language, because the stroke order of Chinese characters tends to prioritize the horizontal stroke in those cases where a horizontal and a vertical stroke cross. As a result, Chinese writers frequently place horizontal strokes before the vertical strokes that cross them. Moreover, the complexity of Chinese characters means that correct proportions are essential to their easy recognition by a reader, so Chinese writers become very practiced at placing and spacing earlier strokes accurately in relation to future strokes they haven't yet made. Such a background could lead a person to write English letters like f and t crossbar first, with perfect proportions at the end. However, this is just a guess.

[@Gwen_Hemoxia]

​@@justinhowe3878 that is what I thought and it stands as a solid theory

[@glottalstop2080]

@@justinhowe3878thanks chatgpt

[@bprpmathbasics]

Try the problem at 3:48 Here’s the answer: kzbin.info/www/bejne/l6jVmIuOZbahoNUsi=Jzz5j4KLJu14Mv3e

[@anhada.8347]

👆👆👆 This comment was made before the video was uploaded. 🤨🤨🤨

[@samarjitdasIISERkol]

​@@anhada.8347yes probably the video was private

[@jamescollier3]

solved broken sorry. not sorry😅😂

[@dubby_ow]

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

[@onradioactivewaves]

Don't be mean to the complex conjugate

[@davidl.reimer2762]

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

[@DeKevers]

It’s still entertainment though.

[@somethingsomething2541]

Its not usefull for general audience , most people here will probably never use it.

[@dadh-dj8em]

@@somethingsomething2541 It's useful to distract yourself from the distraction xd

[@PinkeySuavo]

Nobody ever will use it besides as an interesting "trick" to know

[@davidl.reimer2762]

@@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.

[@uhlan30]

I’m just impressed someone finally did one of these false proofs by doing something else besides dividing by zero

[@KaikyAlmeida-b8t]

or subtracting

[@РайанКупер-э4о]

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.

[@HellGirl-nw9er]

Thus, proofs that use a such function or notation must make sure it is well defined for the problem approached.

[@websparrow]

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.

[@РайанКупер-э4о]

@@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.

[@biawakstruus]

it also breaks the reality

[@Gezraf]

its really cool you proved i = 1 at the end cuz in the complex plane i actually represents 1 in the imaginary axis

[@aravindmuthu95]

'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

[@Gezraf]

@@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

[@RuleofThehyperbolic]

​​​@@Gezrafif you're talking about the distance from zero, then you should've said abs( *i* )=1 which is true

[@mitchratka3661]

The whole point of that proof was that it was WRONG; he is asking you to find the problem with it lol

[@PhilosophicalNonsense-wy9gy]

Lol

[@jamesnadin3168]

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

[@GIRGHGH]

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.

[@BillyONeal]

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

[@pmnt_]

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]

@@BillyONealyes but why does the rule not apply to two negatives in a square root

[@BillyONeal]

@@almscurium I don't know, complex numbers are weird

[@asdfqwerty14587]

​@@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)

[@bprpmathbasics]

Try the problem at 3:50

[@CrtTrc-n2m]

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

[@phenixorbitall3917]

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?

[@williammarshal4043]

I wonder if a,b

[@CMTRN]

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.

[@glitchy9613]

@@CMTRN this is the correct answer

[@prodyst]

1:14 I paused. I think the wrong one is #5, because sqrt((-1)(-1)) is not equal to sqrt(-1) times sqrt (-1)

[@StefaanHimpe]

"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.

[@afanebrahimi7278]

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)

[@StefaanHimpe]

@@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.

[@joshuagillis7513]

@@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

[@sensey181]

​@@joshuagillis7513Exactly!

[@giantclaw138]

Skill issue

[@syedabid9767]

2:23 Bro the way he changes his markers is dope!🔥

[@user-hk3ej4hk7m]

This is a side effect of taking multivalued functions and making them arbitrarily single valued

[@PFnove]

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 +)

[@kobalt4083]

@@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]

@@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

[@xinpingdonohoe3978]

​@@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.

[@hallrules]

@@xinpingdonohoe3978 sqrt(x) is a function, functions only give either no output or one output. ±√9=±3, √9=3

[@RigoVids]

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)

[@RigoVids]

1:20 I feel so heard right now 😅😂😂

[@ShreyasGaneshs]

Well I don’t think root 1 can be substituted for +1 since it can also be -1

[@bprpmathbasics]

2 things you just don't do in math! kzbin.info/www/bejne/n6HVgGysadeYitU

[@MubinaSultana-l5v]

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.............

[@NauamUwU]

Dude, I hate love you, good job.

[@StanisławŁapiński-n9d]

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?

[@y29k15]

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.

[@derndabest]

​​@@StanisławŁapiński-n9d no this isn't technically wrong the square root of 1 is 1 if you multiply 1 by 1 it is 1 so taking the square root of 1 does not change the equation. Edit I thought about some more and there is an argument either way. The square root of 1 is both 1 and -1 so the answer is right and wrong

[@adw1z]

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

[@lyokol]

Finally someone that explains these strange behaviours using complex analysis and not only some "rule". You should be the comment on the top.😊

[@rcg5317]

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. 😧

[@bprpmathbasics]

Thank you!

[@kaustubhgupta168]

maybe we need to not do that and stay happy

[@CharlesShorts]

yes, peace

[@athrunmoza8996]

I saw this in one of my highschool, gotta say it was so stupid trying to prove 2= 0

[@suspicioussand]

Reject calculus, return to monke

[@yamatanoorochi3149]

3:32 what's the logic behind why we can't do that? You're saying it like it's just an arbitrary rule placed to avoid these kinds of results, but what really is the logic here?

[@texasaggie2378]

Since the product of two negative numbers will always be positive, and the square root of any positive number cannot be complex. We cannot separate then into two square roots since they would always(individually not evaluated) equal a complex value, which would go against the rules we’ve stated in the first place.

[@Krmpfpks]

Sometimes showing that something leads to a contradiction is enough to show it’s incorrect.

[@redubuci537]

Me when axioms:

[@aquarock-fq2lm]

It's frustrating he gave examples of when it's valid but didn't explain an example of when it isn't valid. That seems like the most important part to me...

[@Krmpfpks]

@@aquarock-fq2lm If it was valid, then 2=0 would be true. So the entire left side of the board shows that allowing it would lead to a contradiction, therefor proving that it cannot be allowed.

[@miroslavzderic3192]

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.

[@FarisYKamal]

“You can’t do that” “Why” “Because you can’t”

[@tonywong8134]

@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?

[@audioedits-kq7zx]

4:21 Properties of exponents are invalid for base < 0

[@SeegalMasterPlayz]

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

[@RuleofThehyperbolic]

the root function always takes the absolute value as in: √1 equals only 1 but x²=1 has the four answers

[@m.h.6470]

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]

@@RuleofThehyperbolicx^2=1 only has 2 solutions not 4

[@samkadel8185]

​@@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

[@darshans6321]

Mathematicians saw the reddit guy and went back in time to change the rules of square root

[@BigSources]

"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

[@DaSquyd]

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.

[@BigSources]

@@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.

[@SaiCharithSharma0609]

0:20 It's at step 4 if I'm right. Root over -1 times -1 isn't root over 1, it's negative root over 1.

[@DreamyAileen]

Yes and no. sqrt(-1*-1) == sqrt(1) == +/-1 It's convention to generally only consider the positive root, but the negative root is also a valid root because 1^2 == (-1)^2

[@vedarovski4110]

I knew there was a problem with step 4 or 5 but couldn't tell. "1=|√1|" makes so much sense.

[@Cloud_Seeker]

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.

[@TheVnator]

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_op]

Let's just be honest... The whole concept of imaginary number is like an excuse made by mathematicians to cover up for their mistakes 😂

[@thevorhandener5280]

​@@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

[@Salmacream]

It's because they aren't the same thing. Like apples to oranges. Are you asking for a proof on why they are different?

[@elitrefy_op]

@@thevorhandener5280 ofcourse I just let my true feelings slide for a second there 😂😂

[@austinyun]

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).

[@HorseMessiah]

my father, a college level math teacher, always said "if god were a mathematician the eleventh commandment would be 'respect thy negative'"

[@Peter_Morris]

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.

[@Vansharora9]

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.

[@slightlyscummy]

dude, the way you change between markers is mesmerising, its like they're extensions of your fingers

[@leos.8374]

I'd also say the function square root is only defined on positive or null real number. i is described as a solution of x² = -1 (with -i). It is properly defined after the complex numbers are introduced and it doesn't actually use the square root function. So by stretch, we get x² = (-1)(-1), which doesn't imply x = -1, but x is in {-1;1}.

[@leos.8374]

The last problem also creates this kind of issue, where x to the 1/4 power isn't properly defined in the set of complex numbers. By the same stretch, x^4 = 1 has multiple solutions, which never implies x = 1 in the set of complex numbers.

[@Anduardus]

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.

[@FinesseBTW]

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

[@bprpmathbasics]

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

[@FinesseBTW]

​@@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.

[@AikaterineG]

In the book "O Algebrista" (lang: PT-BR) says that you cant separate a real number into 2 imaginary, and give examples like that

[@tessiof]

Valeu pela dica!

[@Jerrremy]

Obsessed with his hand dexterity and the way he flips the markers around

[@erin1569]

Does "if a, b < 0 then √(ab)≠√a√b" come from something more fundamental, or does it come directly from this kind of equation?

[@MadocComadrin]

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.

[@randylazer2894]

Finally, somebody who can actually teach math on KZbin, as opposed to those who screw up pre-algebra that a fifth grader should not screw up. Very well done.

[@Ozasuke]

If you work with imaginary numbers, you get imaginary answers.

[@3characterhandlerequired]

I remember one similar case from high school, I was making equations on blackboard and doing it "on the fly", and one of the variables turned out to be zero which made the result weird. I noticed it at about 3/4 equation done that something isn't right, and it took teacher and me a bit time to figure out where the error was. It was way more fun to hunt the error than doing the actual equation.

[@longway5483]

Thanks, now I can finally take my revenge from my maths teacher 😈

[@TuhinRaj-u3y]

1 over any power is one, yes, but the thing is something power 1/4th is the 4th root. so, the fourth root of 1 can have 4 possible solutions. 1, -1, i, and -i. Here, it is strictly equal to i. just like how we ignore negative answers when we factorise quadratic equations for stuff like time and distance, we ignore all other 3 possible answers. Am i right?

[@viCuber]

Credits to the multipen writing with one hand. Insane.

[@yougerloger6824]

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.

[@cinderwolf32]

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.

[@Zevoxian]

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

[@ZipplyZane]

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.

[@sadhanaduttapramanik2663]

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.

[@vladshkurat3005]

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!"

[@etakarinae248]

It is not a loophole🤣. Read some math literature and find the answers yourself.

[@lucasjammal5153]

So basically this rule avoids that any numbers becomes equals to 0

[@sameerverma5997]

7th step: (-1 * -1) gives us 1. 8th step: 2= 1+1 2=2

[@Teknospacepotato]

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

[@nicholashartmann4525]

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.

[@liambohl]

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.

[@voomneshka]

​@@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.

[@kobalt4083]

@@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.

[@sethd2738]

I started watching the video for the math but kept watching for the marker swaps. Very impressive marker skills.

[@MrFirecheese]

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.

[@dutchraider2]

Yea I also didn't quite understand how 1 turned into sqrt(1) without any explanation whatsoever

[@fiprandom3783]

√1 is just 1, not -1 nor ±1

[@zichenghan7585]

⁠@@fiprandom3783-1*-1=1. Square root of 1 is ±1

[@NecroKoopa]

If x^2=1, then x can be 1 or -1. But the square root of 1 is just 1.

[@zichenghan7585]

@@fiprandom3783-1*-1=1

[@tmsniper9229]

To make it clear, it comes from the definition of the square root over complex numbers which is by definition the inverse function f(x)=x², for all x=r*exp(i*theta) notice that f is not bijective if we don't specify that theta is in the interval [0,pi[ and the output of the function should be in the interval [0,2*pi[ , by saying sqrt(1)=sqrt((-1)*(-1)) we are being inaccurate, if we're working with complex numbers it would be more convenient to write it sqrt(exp(i*0))=sqrt(exp(i*pi)*exp(-i*pi)) this way we can clearly see the absurdity that the 2nd -1 is out of the range of the definition of the square root of complex numbers so we can't really separate them in this case. We can only separate them when they are both within the domain of square root. I hope this was helpful

[@kj6221]

This guy completely missed the point. He doesn't explain at all why square root of two negative numbers can't be separated, he just states they can't. Real reason: n^2 has two roots: +n and -n. Thus, sqrt(1) = +1 or -1, thus, 1 + sqrt(1) = 2 or 0, which is not equal to 2. Thus, we cannot simply replace 1 by sqrt(1) in an equality. Now, we can define sqrt as just the positive root, i.e., sqrt(n^2) = |n| for real n. Then we can write 2 = 1 + sqrt(1). But note that imaginary numbers don't classify as either positive or negative, thus with this definition sqrt(-1) is not i, and in fact cannot be computed. Thus, to compute the answer, we would have to combine the two s.t. sqrt(-1).sqrt(-1) = sqrt(-1.-1) = 1 Finally, we can define sqrt such that sqrt(n^2) = |n| and sqrt(-n^2) = i*|n| for real n. This is what this video assumes sqrt to mean (without explicitly stating it). Then, with this definition, sqrt(-n)*sqrt(-n) = -|n|, which is not equal to sqrt(-n*-n) = |n|. This difference arises because with this particular way of defining sqrt, we select only one of the two possible roots. This needs to be conserved across computation steps, which is the reason why we can't separate the sqrt of two negative numbers (with this particular definition of sqrt).

[@licharey]

La raíz de un número siempre es positiva. Sqrt(1)=1 no a 1 y -1.

[@iTeachMyToast]

Your marker technique is impeccable

[@marioamro]

that was a smooth color change at 1:20

[@Extreme-PCIe-Cable]

??

[@Brainmush635]

He’s a professional teacher

[@yeetpathak639]

The problem is in the last step, i believe, coz 1^1/4 has more than one solution, one of them being i, if we start with i we are to assume we are dealing in complex plane, hence, there are bound to be more roots. Edit: just saw bro solve the last q in a vid 3yrs ago. GGs

[@Qermaq]

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.

[@DqwertyC]

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".

[@chaost11]

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

[@bretsheeley4034]

Full agreement. That’s where I saw the problem, and for that exact reason.

[@kobalt4083]

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.

[@the4spaceconstantstetraqua886]

(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

[@RhiboNuclicAcidゅ]

2:49 you didn't explain why we're not supposed to do that or why it's not possible. Or perhaps I didn't give it enough thought.

[@georgewalker3702]

I don't know if you figured it out at some point, but he literally explains it at 3:05.

[@juancarlos-oc4qi]

the marker changes are so smooooth

[@epikherolol8189]

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

[@johnyang799]

Issue is the first step. i^1 is not i^(4/4).

[@kobalt4083]

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.

[@kobalt4083]

@@johnyang799 you are also incorrect. 1 is equal to 4/4, so it is indeed equivalent.

[@johnyang799]

@@kobalt4083 Then the op is correct. It's either when you introduce the 1/4 part or when you execute it.

[@kobalt4083]

@@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.

[@nicerknifes9465]

Not bad... Very nice explanation. Made me remember the roots of maths and why some shortcuts even work the way they do. The writing allmost had me think "maybe negative numbers don't really exist if they can be set equal with their positive numbers (because with that Reddit thing you could write 'x = -x' for any number basically). So maybe negative numbers are an invention of banks to create debt."

[@diamondmemer9754]

Discussing the rules of a number that doesn't exist is peak human development

[@rayaneferouni8658]

They may not exist but you wouldn't be watching this video without them

[@diamondmemer9754]

@@rayaneferouni8658 ok so?

[@shadowyt376]

They have been very useful.

[@diamondmemer9754]

@@shadowyt376 for what

[@HailAnts]

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.

[@antonyandraws9148]

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.

[@themarketgardener]

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.

[@zwxyer]

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 pass algebra… sqrt is always +/-, we usually constrain it to absolute value because a negative solution wouldn’t make sense

[@Fenrakk101]

​@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)

[@zwxyer]

@@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?

[@itzslopchaosz7108]

Ok, can someone explain why you can't separate them? I'm obviously not doubting it, it would just be nice to have a more detailed explanation on the "whys"

[@kielbasaman2824]

try to compute both of them and you will see both outputs are different sqrt(-1*-1) = 1 sqrt(-1) * sqrt(-1) = -1 this is why they can’t be separated like other roots

[@gavindeane3670]

Because otherwise 2 would equal 0.

[@marcelbrown2174]

0:05 step 3 is wrong

[@neerav10]

Nope, it's step 5

[@appmeurtre]

​@@neerav10i think he refers to √1 = |1| not 1

[@bman5257]

@@appmeurtreThe absolute value of 1 is 1 though. Step 3 is fine because it’s just saying the principal root of 1 is 1

[@appmeurtre]

@@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

[@bman5257]

@@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.

[@NerfPsycho]

4:55 The mistake is from step 3 to step 4. This rule: a^(bc) = (a^(b))^(c) does not hold for all the values of a, b and c; a should be a positive real number, and b and c are real numbers.

[@Iam1Person]

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.

[@potatoheadpokemario1931]

I'm not exactly satisfied with the awnser that it's wrong because it leads to conflicts

[@entcraft44]

There are multiple ways to look at this problem and depending on your viewpoint the reason is different. The equation z^2 = 1 has two solutions +1 and -1. The square root is supposed to be the inverse operation of z^2 but of course that is not uniquely defined. There are those who think that sqrt(z) should be a multivalued function, i.e. sqrt(1) is both -1 and 1 at the same time. Then the steps that replace 1 = sqrt(1) and i = sqrt(-1) are both wrong, since those roots have more than one value. The other way is to define sqrt(z) as the "principal branch of the square root". In this case sqrt(1) = 1 and sqrt(-1) = i so those steps are fine, but the property sqrt(uv) = sqrt(u) * sqrt(v) no longer holds for all complex numbers u and v. You can kind of see this if you look at the substitution 1 = (-1)*(-1) = (-1)^2. This suggests that sqrt(1) = -1, but we use sqrt(1) = 1 in the other steps! If you define the root to have a single value, you better stay consistent! (This is just an idea to get some intuition, the problem is not in 1 = (-1)*(-1), it is in sqrt((-1)*(-1)) = sqrt(-1)*sqrt(-1)). The exact reason requires a more detailed analysis of how the branch cut definition works.) PS: There are a lot of comment treads arguing about which definition/approach is right, but of course both are correct. And which is better? It depends on the application. Multivalued functions have their own problems...

[@Bangaudaala]

How do we know its not the entire imaginary realm that is wrong 🤔🤔🤔

[@MrMikhailAsimov]

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.

[@astromos]

Thats why in french, sqrt is only defined with real numbers. Makes a lot more sense then inventing random rules.

[@antoinebugnicourt808]

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.

[@GameJam230]

This is why the square root function is always used as the principal square root unless specifically stated otherwise.

[@L3monsta]

If paused the video and after my thorough analysis I have determined the issue is somewhere between step 3 and 7

[@the_linguist_ll]

Yep, 5

[@TosGD]

you can look at this for 15 seconds and understand that the problem is where 2 = 1 + 1^2 being "simplified" into 2 = 1 - 1 is the problem

[@TosGD]

meant to say 2 = 1 + i^2 mb, but anyways getting i^2 is the mistake

[@TosGD]

oops another edit all of that was wrong

[@suraj_rodge]

Problem is with the statement 3 , if you write 1 as √1 , then it give √1 two values that is +1 and -1. So the whole calculation after this will have two solutions one being the real and one imaginary.

[@Friendly_Neighborhood_Dozer]

I feel so smart for finding the problem by just looking at the thumbnail for like 5-10 minutes

[@Someone_43]

When you think of it, math as a whole is the most elaborate and complicated puzzle in the universe

[@purplelord8531]

had to watch twice too obsessed with the incredible marker switch

[@paulh4828]

The problem is in fact that in C, the two opposite solutions of z²-a=0 must be considered as "square roots". You can't just take 3i as "the square root of -9", you have to be more careful, and the √ notation (as well as the ^(1/2) thing) clearly doesn't help. So let's just: • ban this notation on C\lR+ • be careful when we write (or just not write at all!) things like: (e^(it))^k = e^(ikt) when k is in C\Z because the result might not be well-defined since not unique, and we might select the wrong solution!