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]

🅱️

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 It may not be obvious that the steps are no longer valid!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

[@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

[@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

[@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.

[@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.

[@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

[@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.

[@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

[@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

[@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

[@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

[@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

[@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.

[@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

[@syedabid9767]

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

[@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

[@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

[@slightlyscummy]

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

[@marioamro]

that was a smooth color change at 1:20

[@DankeyKang0816]

Let a > r > 0. Consider the r torus of revolution generated by rotating the circle (x-a)^2 + z^2 = r^2, y=0 about the z-axis. The parallels generated by rotating the points (a + r, 0), (a − r, 0), and (a, r) are called the maximum parallel, the minimum parallel, and the upper parallel, respectively. Justify which of these parallels is a geodesic on the torus. Solve this nerds

[@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

[@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!

[@vedarovski4110]

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

[@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)

[@armedcannon]

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.

[@BryanLu0]

Usually if you have a square root you only take the principal value

[@MrJuliancarroll]

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.

[@kingoreo7050]

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]

@@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

[@Engy_Wuck]

@@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

[@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

[@longway5483]

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

[@HorseMessiah]

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

[@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

[@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.

[@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.😊

[@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.

[@FarisYKamal]

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

[@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?

[@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.

[@starhalv2427]

Prediction: the mistake was made at point 5

[@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.

[@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.

[@audioedits-kq7zx]

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

[@sethd2738]

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

[@sameerverma5997]

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

[@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.

[@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.

[@lucasjammal5153]

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

[@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?

[@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

[@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

[@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

[@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.

[@Ozasuke]

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

[@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.

[@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

[@mAny_oThERSs]

"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.

[@mAny_oThERSs]

@@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.

[@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.

[@Asango]

The product rule for square roots sqrt(ab)=sqrt(a)*sqrt(b) explicitly states that both a and b must be positive real numbers.

[@gavindeane3670]

What's the problem with term 3 and term 2?

[@Asango]

​@gavindeane3670 Nothing, just a miscalculation that I forgot to remove from my comment.

[@viCuber]

Credits to the multipen writing with one hand. Insane.

[@Sillimant_]

The way he's switching between the pens is messing with my head more than the maths

[@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!

[@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...

[@dastranjer9274]

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

[@lawrencejelsma8118]

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]

but... square roots are already always positive....

[@dastranjer9274]

@@lawrencejelsma8118 i never knew about the practicle uses of +/- for square roots, i might look into that further in my free time.

[@dastranjer9274]

@@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

[@GeezSus]

@@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

[@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."

[@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.

[@kazagucci]

These comments hurt my brain. So many people that are confidently wrong.

[@seriousbusiness2293]

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.

[@epikherolol8189]

Ur answer is flawed from ur very first statement 😅. Sqrt(1) is 1 and ONLY 1. NOT -1

[@Darkness18641]

@@epikherolol8189 Then how much is (-1)^2?

[@rxuge]

@@Darkness18641its 1 but epik is still right

[@potaatobaked7013]

@@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

[@seriousbusiness2293]

@@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.

[@apoorvasingh1981]

For the solution of the question at the end, reference to the topic 4th root of unity.

[@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.

[@alexeyanokhin6545]

I suppose, step 3 is not correct. Sqrt(1) = 1; -1. So, we can use only arithmetic root.

[@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.

[@aditigunjal4231]

Why is noone talking about how cool he is when he switches the markers in his hand

[@marknieuweboer8099]

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.

[@juancarlos-oc4qi]

the marker changes are so smooooth

[@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

[@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.

[@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

[@MatthewCampbell765]

Before reading I think around 5 is where the mistake was made.

[@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.

[@ShabriMohan]

Sir you are so good at teaching, i mean the best, but how can u be so underrated when u are too good 🔥

[@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?

[@darshans6321]

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

[@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).

[@joshuagillis7513]

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

[@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).

[@josecarlosvidalgherman5373]

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.

[@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.

[@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

[@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.

[@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.

[@Bangaudaala]

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

[@manik5145]

bring back memories, both questins in this video used to be worm in my head during school, but i gave up when my teachers couldnt answer them.... glad that i finally found that its not a valid operation.

[@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.

[@Someone_43]

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

[@Kneegrowww_911]

Any number n when square rooted at the same time...which means √(n²) will always give |n|... learnt this back in first year of high school..and has been helping me ever since

[@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.

[@andrewjknott]

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.

[@JubeiKibagamiFez]

0:20 Where in the hell did 2=1-1 and 2=0 come from??

[@anyaforger844]

i²=-1

[@idrawfurries6136]

steps 1 through 7

[@cprhroaehssneteonr5488]

i^2=-1, therefore 1+i^2=1+(-1)=1-1=0

[@gunpon1803]

The bishop from the thirteenth dimension