And shout out to all the people still saying ± 6 because "the square root of 4 is ± 2". Really simple, x² = 4 then x = ± √4 = ± 2; but √4 = 2 and 2 alone. The ± symbol is always placed before the √ symbol when inverting a square.
@@WombatMan64that’s true if we’re inverting an x^2 term, but from what I understand the radical symbol denotes the principle root which is the positive branch, so -6 should be the only answer in this case. You could probably rewrite the thing with an x^2 term to get two solutions if you changed a bunch of things.
If we are taking the square root of negative numbers, then aren't there only complex solutions? The only square root of -4 is 2i and the only square root of -9 is 3i, correct?
Imaginary numbers aren't mathematically real, but in all fairness, the sets have a terrible naming system. Gauss himself suggested that imaginary numbers be called "lateral numbers" instead. Obviously, in a modern mathematical stance, imaginary numbers are not in the set of real numbers. But from a more debatable, linguistic standpoint, "imaginary numbers are real."
A real number may be defined as a + i*0, where a is a real part and i*0 is an imaginary part. That way there is a connection between real and imaginary numbers. Naming conventions are a bit misleading and don’t mean “real” as in “exist”. Would you call -5 a number that exists? Outside of math negative numbers can signify loss or deficit, but you might as well use positive numbers to measure the size of it.
Imaginary numbers show up in very real ways in electrical engineering. By real, I mean "physical things you can actually show on a meter or display". For example, the impedence of an antenna at various frequencies.
considering thw bias in the poll that mind your decisions conducted, I would say we should omit that data. There is even a bias in the brilliant org poll
@3:06 The rule that sqrt(x) times sqrt(y) = sqrt(xy) - why do NO grade school or high school math teachers EVER bother to mention that this only applies to positive numbers???
It shouldn't be true. Start backwards: sqrt(36)=sqrt(6*6) or sqrt(-6 * -6), so why not sqrt(9*4) or sqrt(-9 * -4)? If this isn't true, then all square roots would be positive, but as we all know the correct answer to any square root is pos or neg
As an electrical engineer (by qualification, if not by trade) I get offended by people referring to j as "i". I is current! And also something in mechanical engineering; because they use j too, and probably not just in solidarity.
I like how you explain all the wrong answers and how people might have gotten to them -- I think it clarifies a lot of important misunderstandings, and even though I got the correct answer, I still learnt a lot (especially from the greater clarity in definitions, which is the most important thing in maths imo). thank you.
None of them are "wrong", just incomplete. +6 and -6 are both completely correct answers. The problem is not in the math itself, but in the ambiguity of the √ symbol having two solutions without a clear indication of which one we want. With positive numbers it's easy to assume we want the positive root because of the obvious association with a literal square from geometry, but with negative numbers and the need to go complex, that "obvious association" is lost. The only answer that's arguably wrong is "undefined". It's taking the square root of an explicitly negative number, so assuming that you're working in the realm of strictly real numbers is kind of silly.
@altrag if they were looking for both the positive and negative answer there would be a +/- before each root. Since that is not there it means that only principle root is to be considered. That is standard math notation.
@@mudfish78 > there would be a +/- before each root That is most definitely not as "standard" as you seem to believe. It may well be what you're most familiar with, but it's hardly universal, especially when you start dealing with complex numbers and "positive" / "negative" isn't as meaningful (is 2-3i positive or negative?) In this case because the result of each root is purely imaginary you can claim "+i is right", but pure imaginary numbers are a rare special case within the entire complex place (as are pure real numbers) so it's kind of questionable to even claim there is a "positive answer". The principal root that you note is the closest we could come to being the "right" answer, but given the simplicity of the question I'd assume this is pretty early on in an introduction to complex numbers and the concept of principal root might not be known by test takers.
@@altragYou should assume that the person solving the problem would know that the radical symbol is only the principal root. Why would you in any universe assume they don't? Suppose you're a teacher making a question. Would you assume the person solving doesn't know how the order of operations works? Would you, as an examiner, mark 1 + 1 x 3 = 6 as correct because there was no rule stating PEMDAS/BODMAS must be followed? Obviously not, because that's common sense. So if not acceptable in that case, then why would you accept a factually wrong mathematical answer because the examinee didn't know an 8th grade rule?
@@interloper9589 > You should assume Why should I assume that? You're asserting that it's "standard", and maybe that's the only way _you've_ seen it used but I've seen it used in both ways on multiple occasions. My first assumption when I saw the question was that it would be a +/- result, just because that's how often it comes up. If I'd taken that test I'd have probably given both answers (knowing what I know today - again, that looks like a fairly introductory level question so I can't say what those students should be expected to know). > no rule stating PEMDAS/BODMAS must be followed The fact that we can't even agree on the acronym tells you what I think of that comparison. > didn't know an 8th grade rule? Because it's not a "rule". PEMDAS can be a "rule" because it's convenient. You could choose any other rule and you'd still only get one right answer (as long as you apply it consistently). It might not be the same answer you get with PEMDAS, but it would be unique under the rule you choose. 1+1*3=6 is not only correct but the _only_ correct answer if you choose the rule "strict left-to-right evaluation". The "principle root" rule is different. The square root operation really does have two (generally) unique answers, and there is no choice of "rule" that can get around that. Picking only one of them to be "right" is a much bigger deal than merely convenience, and as noted that "rule" is not always followed - _and cannot be assumed_ - because it is not and fundamentally cannot be consistent in the same way that PEMDAS can be.
Team plus or minus 6 here. Nowhere did it say that the root should be considered as a function. Restricting the result to the principal branch is an assumption that was never specified.
It's not an assumption. It's literally specified right in front of your eyes. That's what the √ symbol means. If the writer had wanted you to consider both square roots of these numbers they would have written a ± in front of the √ symbol. That's how you say "both square roots" in mathematical notation.
But even if we want to define a principal square root, why is √−̅4̅ = 2𝒊, not √−̅4̅ = −2𝒊? What makes us choose 2𝒊 = 2∠90° and not −2𝒊 = 2∠(−90°), when they are equally valid? If we instead define our principal argument to be −π ≤ φ < π (not −π < φ ≤ π), then √−̅4̅ = (4∠(−180°))¹ᐟ² = 2∠(−90°) = −2𝒊, not 2𝒊.
@@cyberaguaNo, √x = -2 has no solution. You don't need to think about complex numbers for that. As for the cube root, as I said in another comment elsewhere, you'd need to ask someone who plays with higher order roots and complex numbers. There isn't (that I know of, at least) an obvious notational option to indicate which root or roots you're taking about.
@@gavindeane3670 > No, √x = -2 has no solution. You don't need to think about complex numbers for that. It's all about definitions and conventions. You may define √x̅ to have only one (principal) value, or (as some authors do for complex roots) let it be multivalued. In the latter case we have √4̅ = ± 2 = {2; −2} ∋ −2, so x = 4 is a solution to the irrational equation √x̅ = −2 over the field of complex numbers ℂ with the appropriate definition of the multivalued complex square root function. Also equality in this case is replaced with inclusion √4̅ ∋ −2, as it's normally done when solving equations with multivalued functions. But I don't think you've ever heard of it, since only those who solve problems that require multivalued or set-valued functions really need this stuff. Look up Set-valued function on Wikipedia, for example. @gavindeane3670 > you'd need to ask someone who plays with higher order roots and complex numbers It's actually the same for any order roots: ³√−̅8̅ = {−2; 1+𝒊√3̅; 1−𝒊√3̅} ∋ −2.
@@nabibunbillah1839 just 2 the reason +- exists in certain context is to find the roots of functions that are some form 0=ax^2+bx+c (or related, like other polynomials, sin(θ)=0, etc, which have multiple answers), because technically two answers make this true (fundamental theorum of algebra) when you are asked what the sqrt(a) is, the answer is always positive, which is why sqrt(x^2) is |x| for all x, rather than just x saying x^2=4 is a different problem then x=sqrt(4), even if the latter is used in the former
the reason this problem is framed weirdly is because roots and radicals fundamentally work differently for complex numbers, even if the concept is the same for instance i i^(4/4) (i^4)^(1/4) 1^1/4 1 so i=1 obviously this is wrong, and the reason it fails is because there can be multiple answers that technically satisfy 1^(1/4) in the complex plain, including the integer -1, that we dont consider to be valid answers in real-valued computation if you can apply the roots/whatever more diligently, you can do: i^(4/4) (i^(1/4))^4 (exp((π/8)*i))^4 exp((π/2)*i) i i=i
I was in the "-6" camp, but that argument "against" "+-6" really makes me believe in the "+-6" option. As much as I like the idea of the principle square root, I feel like this is akin to "This shouldn't be possible... but according to the math, it exists" (usually referring to the consequences of Einstein's equations)
If the calculation has a purpose other than satisfying a maths teacher (e.g. an engineering problem), you should always consider all possible solutions if you come to a root, arcsin or other inverse function with a "principal solution". Sometimes there is really more than one solution to the problem, very often one of the solutions is obvious nonsense in the real world.
If imaginary numbers, which have no real world meaning, leads to the "correct" answer, then I didn't see why the non-principal square root can't be used too.
@@djinn666 I know, right? He goes "You're only assuming that this is about real numbers, but imaginary numbers can be included" and then he just assumes that it should be a function and that only the positive values should be considered for the square roots.
It's a definition thing. The square root function is defined as principal root always. Thats why you'll find formulas likes the quadratic equations that have "+-sqrt(...", the negative root has to be added back in because the standard is that the symbol means the positive root. It's in a similiar boat as order of operations, you can do addition first but thats a completely different system that will get you different results from how everyone else reads the notation
@@jacobmerrill693 There is no international authority on mathematical definitions. Some would say 1+2+3+4+... is undefined. Others would say it's -1/12. Both are correct if you ask any reputable mathematician. It's the elementary school math teachers who demand that everyone use the same definition.
I think it's disingenuous to go: "we should definitely be considering complex numbers" in one breath and "let's ignore the other half and just consider the principal square root" in the next. Also, who decided Wolfram Alpha was the final authority on how to do math. It's a calculator. An advanced calculator, but a calculator nonetheless. And decisions were made when programming it. Without an understanding of what went into making that decision, I'm not quite ready to say they didn't make a mistake. Edit: All these justifications about established convention and the definition of the square root operator are great and all, but the fact remains: he gave none of them. He gave a reasoning about it not being a function (by the strict definition of only producing one output), but that felt like him deciding on his own that it had to conform to that standard. After all, if you were solving for a value, you wouldn't care that it doesn't match the definition of a function, why would you care here? He said nothing about established conventions.
Wikipedia also defines the √¯ symbol as the principal square root and if you want both halves you'd want to write ±√x̄, I'd guess it's a consensus in the entire math community, so yeah
@@yukimoeI suppose that makes sense in the same way we write "6" rather than "+6" when writing positives, still think that it feels somewhat arbitrary tho
I do agree, somewhat - the root symbol typically means the principal root. Which is what the wolfram alpha function uses too. But I'd suggest if you take the secondary root for one, you should for the other as well, and that's still -6. I do think +/6 would be a valid choice though, with an explanation. But like undefined, you're making an assumption that most people wouldn't make, so you'd want to be upfront about that when you answered. Like, no real solution is a fine answer too. As long as you say there is no real solution, not just undefined.
Maybe you're better at math than you've been giving yourself credit for. I bet if you keep watching this channel, or look for other opportunities to learn math, that you'll get pretty good.
I'm not very good at math either, and I figured that as sqrt(4) is 2 or -2, sqrt(-4) would also be 2 or -2, so I completely ignored the option to multiply the -2 with the 3 or the 2 with the -3 and ended up with 6... Like I said, I'm not very good at math :P
In the original question where both 6 and -6 are options, and +-6 and undefined are not, my answer is -6, as I feel it's the most correct. In your version of the question, where all choices are potentially correct I would say +-6. If +-6 is one of the options, that implies to me that such answers are allowed under the writer's definition of square root.
Using the root symbol (√) is defined as taking the positive root. Taking to a fractional even power (like ½) gives the positive and negative root. 4^(½) = ±2 but √4 = 2
@@pageboysamThat's not true either. 4^(1/2) is the same as √4. It us only the principal square root. No combination of elementary arithmetic symbols produces more than one value without explicitly using ±.
Yeah this is the key here. I too was emotionally attached to my answer of the initial question, too much to realise that the new options for answers changes things, so whilst before the video I chose -6, by the time I was 75% of the way through this video, my choice changed to +/-6
7:47 No. Not everybody. Complex numbers are not ordered, so predicates () are not defined for them, you can't universally choose (i) to be the principal sqrt(-1). In some schools it's taught that if z = r * (cos(t) + i * sin(t)), where -pi < t
"Complex numbers are not ordered, so predicates () are not defined for them, you can't universally choose (i) to be the principal sqrt(-1)" - As it happens this is located on the ordinate of the complex plane, therefore on a straight line with real coordinates, so can't we ?
"To conclude this, math language is not universal." Yes it is, and mathematicians have spent centuries making sure everything has a single and unambiguous definition
I was solidly in camp minus 6 before watching the video, but afterwards you really convinced me that +/-6 is the most principled answer for the same reasons you gave to argue against the Undefined answer. Nothing in the question limits the calculation to the "principal root"... So +/- is the most inclusive way to see the answer... Thanks!
He didn't explain it correctly, that's true, but -6 is the only right answer. What he should've explained is that the radical sign "√" implies the principal square root. It's why we use the "±" sign when we are talking about both roots, like ±√4. Therefore, by the way the question is written, since it uses the radical sign "√", only the principal square root should be considered. In summary, something in the question DOES limit the calculation to the principal root, and that's the radical sign.
@@ric6611 I actually never heard the term "principal square root" before this (I was in the ±6 camp). And it's interesting that the √ symbol implies the principal square root. I'm curious if there's a way to write the question to ask for all solutions. Maybe "Find all solutions for x such that x = (-4)^1/2 * (-9)^1/2" (Edit: Wolfram-Alpha didn't like my 'find all solutions' clause.)
@@paulgruner9644 You can either use ±√(-4) * ±√(-9), which I know is ugly, or something like ±(√(-4) * √(-9)), ask for the solution to the equation x² = -4 * -9, or you could even just be explicit about it. For example, I have no problem if the question said something like "consider both roots when the radical sign is used". My only problem is that by omission, the default should be used, and the default is the principal root. If it makes sense for you to always be talking about both roots, and you don't want to keep using ±, it's entirely valid to say something like "assume both roots from now on", for a paper, or maybe even for an entire class. I think that's what happens when people are learning about both roots, and then they think that the radical sign ALWAYS means both. That's just not the default case.
There is a big difference between : - the function square root, which by definition of all functions has one and only one value where it is defined, eg sqrt(9)=3, and sqrt(-9) is not valid and has no precise meaning. - the solutions of the equation x^2=-9, which has two solutions 3i and -3i. Well defined maths have no logical flaw, you just need to apply right definitions to each concept.
@@verkuilb Glad to be better in Math than in English 🙂 Only one mistake is still very good, as English is not my mother language 🙂 Sometime you should think about assumptions before being ironic 😞
@@StephTBM4yeah he is one of those people who will see that his/her son scoring 95 percent in exams and still complain that spelling of blah blah is wrong in the report card instead on focusing on the result😂😂😂😂
This is a question of definitions, and I do not feel that the definition given here for the square root of a number has been sufficiently justified. Yes, you CAN define the square root as a function that gives back a number with a positive/zero imaginary component, but... why? What reason do we have to discard the other root, if different? Why do we need to find only one answer to the problem given? Isn't it better to find ALL answers to the problem? In the case of Brilliant's question, the choices given make clear that only one answer is desired, and IN THAT CASE, I can get behind -6 being "the correct answer." But the poll question implies, through the presence of the plus-or-minus 6 option, that having multiple answers is valid, and as such, multiple answers should be accepted (if multiple answers exist, which in this case they do).
One can define square root of x as being "whatever number is such that root(x)×root(x) = x." So root(-1)×root(-1)=1 would not be correct because it violates this definition of root. Technically, there are two different complex numbers that, when multiplied by itself, result in -1, and they can booth be considered root(-1), but "root(x)×root(x) = x" implies that root(x) should be a _consistent_ value, so it can be one or the other, but you shouldn't substitute two different values into the two instances of root(x). This is a similar situation: root(-4)×root(-9), this simplifies to 2×3×root(-1)×root(-1). The root(-1) can be either i or -i, and either case gives you -6. One only gets 6 if you substitute i for one and -i for the other.
@@stevenfallinge7149 Okay, yes, that is a way that square roots can be defined, but we still haven't established WHY we should use such a definition, as opposed to any alternatives. We could just as easily define it as "y = sqrt(x) if and only if y^2 = x". Which does NOT require there be only one value for y.
@@NettoTakashi Without supposing sqrt(x) is uniquely defined, it should at least be reasonable to suppose that if sqrt(x) appears twice in the same equation, then it should refer to the same value. So sqrt(x)×sqrt(x) will not equal -x for any x besides 0, no matter which root is chosen.
@@NettoTakashi If sqrt(x) takes on multiple values, then it is not a function, and many familiar manipulations we routinely perform with it are no longer possible. For example, if we subtract sqrt(x) from both sides of the equation, it will not cancel with itself, because we don't know whether they're the same value of sqrt(x) or different values. This makes algebra functionally impossible. It is always possible to express the "I want both roots" sense anyway, by writing something like x² = 9 instead of writing sqrt(9). But the trick is, you need to introduce a variable to represent the unknown value, and once you have a variable, it is routine and familiar to find that there are multiple solutions. We do not normally encounter this behavior when we write things like sqrt(9) - instead, given some function f, we expect that f(9) is exactly one number (or is undefined).
Its not the definition of square root that is up to interpretation here, "square root" only has one definition, but rather its the application of the radical symbol √ and how it is used in context. While it is almost never explicitly stated as such, the radical symbol refers exclusively to the principal square root of a number, therefore it would be incorrect to say that √4 = -2 or √4 = ±2, even though you would say that 2 and -2 are the two square roots of 4. To make this point more clear, consider the square roots of non-perfect squares. You would never say √2 can be positive or negative, it doesn't exist in some sort of numerical superposition, √2 is a single number approximately equal to 1.414. The square roots of 2 are √2 and -√2. Notice how the negative is used in relation to the radical symbol there. If we want to refer to 2's negative square root, the root approximately equal to -1.414, we use a negative sign AND a radical sign in that specific order to make -√2, because √2 is defined as inherently positive. Thus, the problem √-4 · √-9 = ? is solved by finding the principal square roots of -4 and -9, which are 2i and 3i, respectively, then finding their product which is -6. Although if one were to ask this question verbally like "what is the product of the square root of -4 and the square root of -9?" Then it is perfectly reasonable to assume one is allowed to use both the principal square root or its "negative" counterpart to produce ±6. I think a big part of the issue here is that we in math typically ask the question "what is THE square root of x?" because its believed to be simpler for kids to grasp the concept if you teach the principal square root first as THE square root and then introduce the "negative" one later. Which unless the number is zero, there is no "THE" square root, every nonzero number has exactly two square roots (unless you believe in hyper-complex numbers i.e. unless you're completely unhinged) its linguistically incorrect and both unclear and up to interpretation to just say THE square root. Note: I put "negative" in quotes because you don't define a complex number as positive or negative, -2i is not a negative number, but I'm not sure if there is a term that refers to the non-principal square root of a complex number.
I mean, it depends on the agreement on what branch you can take for sqrt, you could even take it as "multivalued function" However, given no context we should take the principal branch, probably the test it was taken assumed principal branches too...
3:14 [citation needed] I think the only basis to say that this works only when x and y are greater than 0 is "we aren't aware of a generalized operation like that". That doesn't mean, however, that it can't be done. Given that imaginary numbers came out of the effort to solve a geometry problem, it might be worth analyzing this question as a geometry problem. 81 = 9*9 = -9*-9 sqrt(81) = sqrt(9)*sqrt(9) = sqrt(-9)*sqrt(-9) or perhaps someone can explain why this doesn't work with something other than "because it doesn't"
You'd be looking for a proof that sqrt(x) distributes under multiplication for non-negative numbers. Sometimes properties like that hold only for a limited domain and this is one of them.
@@isHavvycounter point: the replacement shown in the video of √-9 = i√9 skips the steps in the middle of √-9 = √(-1)(9) = √-1√9 = i√9 which uses the same sorting step in the middle. Sorry for wonky phone symbols
Oh i got it. Creates a logical problem. Take your example and reduce √-9 to i√9. I now have a line saying √9√9 = -√9√9. Or if you take it as a variable X=-X.
3:26 Well, since the rule is in the form of an implication, you can't really say the conclusion of (erroneously applied) implication is definitely false just because the premise of the said implication isn't true. You simply can't decide.
I would start by simplifying: The square root of (a*b*b) should be b*sqrt(a), so this would become 2*sqrt(-1)*3*sqrt(-1) -> 6*(-1) -> -6. OTOH, the sqrt(4) could be either +2 or -2, with +2 being the primary solution, but (-2)*(-2) -> 4 as well.
this is why i don’t like how roots of even numbers are ONLY positive. it just throws various rules out the window. it’s like mathematicians were like “how do we do something that seems reasonable, but actually ends up being annoying?” like you could solve it like: sqrt(-4)sqrt(-9)=x (-4)(-9)=x^2 36=x^2 and then wait a minute. you don’t actually take the square root of 36. what you do is: +-sqrt(36)=x and THEN you have to check both values of x to see if they’re valid
@PopeVancis expecting a grade 1-3(7-9yr old) to compute arithmetic above 50......have adults forgotten this is about learning ...esp mathematics semantics....
Yeah, I don't believe that limitation was ever taught in any class I've taken. I was under the impression that (x^k)*(y^k) always equaled (xy)^k. If it doesn't apply for the specific case of (x< 0 | y < 0) & (-1 < k < 1), that's very strange. I'm doing some cursory googling and can't find it mentioned anywhere.
Its assumed that > Sqrt(x)^2 = x > Sqrt(x)*Sqrt(x) = x = Sqrt(x)^2 Now, if we were to assume that sqrt(x)*sqrt(y) = sqrt(xy) for ALL numbers, we reach this logical contradiction: Sqrt(-9) * Sqrt(-9) = Sqrt(81) = 9 (or maybe ±9, but still not -9, so we contradict that a sqrt of x times itself should give x)
@@HeroOfHyla I think every competent high school level course on the topic teaches that. People just tend to miss and forget things. There are even famous -1 = 1 "proofs" to demonstrate how it does not hold generally. I am also very surprised that you can't find this information online. For example, it is clearly stated e.g. in the Wikipedia article on the square root, under "Properties and uses": "For all _nonnegative_ real numbers x and y, √x√y = √(xy)." Similar limitations can be found in the article about exponents, under "Rational exponents", just to name a few sources immediately found by Google.
I guess by pemdas, sqrt comes before multiplication, meaning -6 is technically correct, but taking x=sqrt(-4)(sqrt-9), x^2=sqrt(-4)^2(sqrt-9)^2, x^2=-4*-9, x= + or - 6 but it's all about definitions
±6 is objectively correct, because there's no such thing as a principal square root in complex numbers. There is n sollutions to n-th root of any complex number and there's absolutely no reason why we should pick one instead of the other
If there is no such thing as a principal square root in complex numbers, why is "i" defined as only +sqrt(-1) and not positive and negative? If it were positive or negative it wouldn't be unique and would vastly change results. Complex numbers are two-parters which makes them a dot on a field and not a position on a string. If you didn't have exact coordinates on a field, you'd end up in the wrong place. That's also why they are vastly different to the remaining number definitions. Imagine it like a geo coordinate. If you try to navigate to europe, e.g. longitude 7, latitude 47, you better hope they are defined as positive. If you miss it, you might end up in the Ivory Coast or somewhere in the middle of the atlantic ocean.
Whomever it was that chose the words “real”, “imaginary”, “rational”, “irrational”, etc., to describe those types of numbers, did a HUGE disservice to mathematicians for the rest of history. How can we convince non-mathematicians of the fundamental correctness of our results, when we tell them that we arrived at our conclusions using “imaginary” and “irrational” numbers?? Or that when you multiply one “imaginary” number by another “imaginary” number, you get a result which is somehow not imaginary??
I think "imaginary" was a term coined by those trying to discredit their use to solve quadratics and cubics I don't think these terms could be any better, they are inconvenient for non-mathematicians sure but they all have a use In their wordings
Every new type of number gets a pejorative name: "negative", "fractional (Latin for 'broken')", "irrational", "imaginary", because the previous generation of mathematicians are uncomfortable with it.
Those successively less useful "numbers" get more and more dismissive labels because they are further and further removed from what numbers are and how numbers behave. It's not pejorative, it's descriptive.
Based on reading the comments, I understand now: we just define √ to mean "the positive square root". I am unsure if x^(0.5) is also supposed to mean the positive square root only. So the square root SYMBOL means only the positive root: if I'm not messing up the terminology, the square root OPERATION, when applied as a transformation, can result in both positive and negative roots. So, if x^2 = 7, then we can apply the square root OPERATION and end up with, "x = √7, or x = -√7". As for not being able to combine (-4)^0.5 * (-9)^0.5 into (36)^0.5, I just refer to Wikipedia on Exponentiation, section Identities and properties, which says that this works "for all integer exponents, provided that the base is non-zero". This restriction to integer exponents might be as fundamental and difficult to prove as 1+1=2, so I'll just accept it.
If you absolutely want √x to mean something for negative numbers, yes, you can choose to give it a value, such as i√(-x). That's totally possible, in mathematics you can define symbols however you want. The US made this choice. But this causes problems like √a√b not always being equal to √(ab), and it has no real benefit when you can just write the transparent and unproblematic i√(|x|) instead. That's why other countries like France decided to make it undefined.
I answered -6 on the poll. No other answer even occurred to me because my background is in physical chemistry and physical chemists deal with wave functions which assign a physical meaning to the imaginary number.
It would be, if it weren't for the fact that the rest of the video is about finding why people might believe the other answers are right and showing why they are wrong.
+6 and -6 both are correct according to the usual definition of the radical sign meaning. For a real positive number its square root is the positive one, but for a complex number the square root of z is defined as the solution of the equation x^2 = z. In a similar way, the cubic root of -1 is the solution of x^3 = -1, i.e. the set ( e^(i*pi/3) , -1, e^(-i*pi/3) )
@angeljimenez3362 I don't think so. The square root symbol is meant to refer to the principal value in all cases. For a negative number, it is the one with positive imaginary part.
Mathematician here. Complex numbers are based on the imaginary unit i, which is defined by i^2 = -1. If you, on the other hand, define i = sqrt(-1), you get into trouble: i^2 = sqrt(-1) * sqrt(-1) = sqrt(-1 * -1) = sqrt(1) = 1. That's incorrect! So suddently there are rules about sqrt() and other functions that you cannot apply. Rather than trying to memorize what you can and cannot do, it's better to never write sqrt(-k) with some positive integer k in the first place. Complex numbers don't allow you to write sqrt(-5), they simply give you i, the number when squared gives -1. That's it, and it turns out that's enough, too, to do all the complex magic.
Since complex numbers DO allow -5 to have square roots, it seems odd to conclude that they don't allow you to write √(-5). I know that you can write it as i√5, but your argument seems to say that being able to write √(-5) is redundant, not that it's prohibited.
@@gavindeane3670 I see where you're coming from. The complex square root function does indeed exist, but is very tricky! For one, it's multi-valued (exept for z=0), so sqrt(-5) isn't just "a number", but can be one of two (sqrt(5)i and -sqrt(5)i in this case). Likewise, sqrt(-1) is i and -i. It's not useful to do any calculation with that. Would you agree?
@@nschloeObviously numbers have two square roots. But the principal root is defined for complex numbers (it wouldn't need to be called "principal" if it was only defined for real numbers - we could just call it the positive one) so I don't see how √(-5) is problematic notation. I can see how there would be a problem if the notation √(-5) was ambiguous as to whether it meant 5i or -5i, but that's no different to the problem we'd have if √4 was ambiguous as to whether it meant -2 or 2. We invented all these squiggles and shapes that we call "mathematical notation" so we can choose what it means.
@@gavindeane3670 no, non-negative numbers only have one square root. Quadratic equations however have 2 solutions. And that is what I think is the greatest misconception. Even Presh here goed from calculating square roots to solving quadratic equations. Jumping from calculating √-4 to solving the equation x² = -4 does not justify the use of i.
@@gavindeane3670no, it doesn't allow that. Complex numbers are used to solve equations like x² = - 5 which results in x = +/- i√5. So 2 answers/solutions/roots.
I'm under the impression that the result of a function can also be a set of values, rather than a singular one. So I prefer the consideration of all roots, rather than just the principal root.
I'm perfectly happy with ±6 as an answer. I have no reason to assume the root here is a principal root, as the question has nothing to do with functions, so I don't have to handwave away half the answers bc they're inconvenient.
square root is defined to only have one output so everyone who says +- is wrong. that only happens with the absolute value operation and a lot of people like to pretend theres an absolute value around square roots when there shouldn't be.
When you're using the quadratic equation you're essentially completing the square. And in completing the square you need to take a square root of a variable squared. A variable squared has a value that could be traced back to two original values for that variable. We use the plus minus because we are *dismantling* something that could be either option.
@@Philip-qq7ql That isn't a flawed argument though, since the square root function is specifically defined that way, the reason being you want to be able to refer to only the positive or only the negative numbers that when squared equal another number as opposed to always referring to both. That said, we could have created a different notation to specify when a square root was positive or negative and kept the base square root as implying both, but it just didn't turn out that way.
Have you ever seen a thing? Or have you only seen one particular object considered to be a thing? Numbers are definitions, such as "thing" "word" "thoughts" "happiness"
But they are distinguishable. For instance, the principle value of the argument of i is pi/2, while for -i it is -pi/2. Or, -i is a root of the polynomial z+i, while i is not. Etc.
@@johnreid5321 The principal value of i being pi/2 simply means that exp(i.pi/2) = i. But since exp(-i.pi/2) = -i, that fails to distinguish i from -i: that is, if you replace all occurrences of i by -i you get the same thing. Similarly -i's being a zero of the polynomial z+i: replace i by -i throughout, and get the equally true statement that -(-i) is a zero of z+(-i). Again i and -i can be switched and true statements come out true again, and once more nothing serves to distinguish i from -i. (But you have to switch ALL occurrences of i of course)
@@russellsharpe288Seems as though you have put it through a specific case though. Would this not be the same as saying because cos(0) = cos(2*pi), 0 and pi are indistinguishable? As terms at their face value i is the sqrt(-1) and -i is -sqrt(-1).
@@johnreid5321 1 is defined as the multiplicative identity. It is easy to show that there can be only one of those (if a,b were two of them then we would have, just using the definition, a = ab = b) -1 is defined as the additive inverse of 1 ie that number which when added to 1 gives the additive identity 0. Again it is easy to show there can only be one (if a,b were two then a = a + 0 = a + (1+b) = (a+1) + b = 0 + b = b. (This uses associativity of addition of course) Could we have 1 = -1? We can in mathematical structures known as fields of characteristic 2, because there 1 + 1 = 0. But for the standard number systems we are all familiar with (Q,R,C...) 1 and -1 are very different. Look at it this way: imaginary numbers are first introduced something like this: we are going to introduce a new number which is a solution to x²+1 = 0 and we are going to call it i. Very quickly it turns out that there is then another solution to x²+1 = 0, namely -i. But how do we know the original solution we introduced was not in fact -i and not i after all? Obviously just by prescription or convention: we have simply *named* the first one i and then the other one must be -i (rather than the other way around). But if somebody else goes through the same procedure, how do we know that what we are calling i is not what they are calling -i ie how do we know that we and they have originally picked the "same" solution to x²+1 = 0? This is pretty clearly a fake question, just because it doesn't matter which one is meant when we label one i and the other -i. And this is because the situation is perfectly symmetrical: there is nothing to distinguish i from -i at all, other than an arbitrary initial act of naming. The only defining property these numbers have is that they are solution of x²+1 = 0, and everything works perfectly fine if we swap all occurrences of i with -i and vice versa. So pretty clearly there is not going to be a way to distinguish i from -i: this is guaranteed by the very way they have been introduced (and indeed it could not be otherwise).
for real number, you can choose the principal square root because IR have an order, so you say sqrt(p) is the greater or equal to 0 solution of x^2=p but for complexe number, there is no order, so no good way to choose a principal square root, so there is 2 square root in the complexe set last argument : we all know that 1 = e^i2kpi where k is an integer by the law of exponent, sqrt(1) = e^ikpi with k is an integer so it’s both 1 and -1
Finally someone in the comments who gets the difference between a mathematical object (the complex field) and some representation of that object using non-unique (re+im) components
> but for complexe number, there is no order, so no good way to choose a principal square root That's just false. Principal square root is perfectly defined for all C. And √ symbol defines specifically principal square root. There is no room for different answers in this problem.
@@lerarosalene Not so simple. I think your confusion arises from the way complex numbers are introduced in highschool and engineering classes as "pair of two real numbers". But those pairs are NOT complex numbers, those are just some representation of the complex number field that you can do calculations on. Now of course one can define a principal square root as an operation on your specific representation, which may be handy for some engineering problems or whatever. But that definition is non-mathematical. The reason is that you can map Re and In in many different ways onto the complex field and get the same mathematics back (but your number pairs would look very differently). In fact, you cannot even tell the difference between i and -i by any equation involving field operations only. And don't say that i is the squareroot of -1, because that would be cyclic reasoning.
@@valentinziegler1649 stop smoking whatever you are smoking. √ symbol is defined to be principal square root and principal square root is also precisely defined. This whole problem is about notation and people like you not understanding it.
So the equation could be resolved and expressed as 6*e^(π+2kπ) with k: 0;1 considering the solutions of the square roots synchronized which both result on -6 . Or you could consider the roots being able to be out of sync which leaves you with the equation as 6*e^j[(π+2kπ)/2 + (π+2hπ)/2] with k,h: 0,0; 0,1; 1,0; 1,1 and then you'll have 4 solutions, 2 being -6 and the other 2 being 6.
I am on the -6 camp never understood why 7:00 - 7:22, finally someone explained to me why only positive roots are used. My HS teacher only told me "just its the rule" but didnt expound more. Thanks!
Edit: thought you were asking someone to explain, I sort of skipped that part of the video. Sorry! Original comment: Based on reading the comments, I understand now: we just define √ to mean "the positive square root". I am unsure if x^(0.5) is also supposed to mean the positive square root only. So the square root SYMBOL means only the positive root: if I'm not messing up the terminology, the square root OPERATION, when applied as a transformation, can result in both positive and negative roots. So, if x^2 = 7, then we can apply the square root OPERATION and end up with, "x = √7, or x = -√7".
I don’t fully buy the argument against undefined (although my answer is -6). First of all, x usually denotes a real variable (compare to z). Second, not all problems imply that you should extend the domain, for example if I say “can you factor x^2 + 1” then “no” seems like the answer you would usually expect unless you had already established that you are working over C.
But even then the correct answer would be to specify "if real: undefined" which would be correct. You are doing a disservice to yourself by assuming properties without writing them down. Then when validating your work nobody can know why you got the answers wrong (because not showing which assumptions you made).
06:39 The information "sqrt(x) is not a function!" is wrong. sqrt(x) IS A FUNCTION. The graph show at 06:39 is NOT a graph of sqrt(x). It is a graph of+-sqrt(x). And this +-sqrt(x) is NOT a function.
He isn't saying that the accepted definition of sqrt(x) is not a function, nor is he claiming the original graph (pre-pruning) to be a graph of the acceptedly-defined sqrt(x). He is saying that if instead sqrt(x) was alternatively defined to included both possible roots, and not just the principal root, then the alternatively-defined sqrt(x) would not be a function.
@@nwoDekaTsyawlA By convention, the √ symbol defines a single-valued function. √a refers only to the principal branch solution of the equation x^2=a. By this convention, √(-4)√(-9) has a single value of -6. This would not apply to other ways of writing. (-4)^0.5 * (-9)^0.5 = ±6
When taking the square root of -4 as a complex number then both 2i and -2i are the two possible values and as far as I can tell, equally valid. So why is -2i discounted? If we allow both square roots for -4 and -9 then +/-6 should be correct. As well, 2i is not a positive number so how does it get the status of "principle"?
That is not the issue. The issue is much more complicated. If you take your domain as complex numbers for the relation f(z) = sqrt(z) it is multivalued. To make it a nice function you restrict the domain and that restriction tells that -1 has an unique square root and it is i. If you need more answers you are talking about roots of unity and that is totally different thing than just taking a square root.
@@SkegAudio Because it isn't a function if you don't do that. It's double valued relation and generally n:th root is n-valued relation and as function you take the first branch. If you want to study the double valued relation it's done by forgetting the 2 and just think how n-valued root handles. And for that you only need to know how the roots of unity work. Remember that any complex number is ae^bi, where a and b are any two real numbers. Then we can discard the a as it is just a multiplicative factor and focus on the e^bi and then n-roots of that are the roots of unity offsetted by e^(bi/n).
@@topilinkala1594 Look while OP is indicating that complex numbers have n nth roots, the √ symbol itself has a specific conventional meaning in complex analysis - it denotes the principal square root. When we write √(-4), we're specifically referring to 2i, not ±2i. It's similar to how √4 means +2, not ±2, even though both ±2 are solutions to x²=4. If we wanted to indicate both possible roots, we'd need to explicitly write ±√(-4)⋅±√(-9). The √ notation alone has a well-defined meaning that gives us one specific result: (2i)(3i) = -6. You're absolutely right that both roots exist mathematically! But the question is about what the specific notation √(-4)√(-9) evaluates to, given standard mathematical conventions.
The actual thing here is, that the task solving sqrt(-4) within complex numbers can easily be rewritten to sqrt((i^2)*4), which then translates to | i | * sqrt(4)
I was learned in school that i has definition i²=-1 and √(negative number) is not defined
2 месяца назад+15
The problem is one of definition, how do you choose which root is the principal root of a number? Easy to do for the positive reals, but why choose i as root(-1) and not -i? And what about root(i) or root(-i) or any other complex number, which of the two "roots" count as the principal one? Choosing arbitrarily is fine, of course, but there would be a continuity problem in the resulting function (from C to C). This is typically why I try to avoid using the square root symbol with negative or complex numbers, to avoid ambiguity.
@@TonyFisher-lo8hh I didn't give a strict definition. Why would you assume that I did? Obviously I was talking about an implied range of positive real numbers. A more technical definition of √x is the principle square root of x. When x is a positive real number, the principle square root of x is defined to be the positive square root. When x is -1, the principle square root is defined to be the complex number "i". So √(-1) does exist.
I do love the argument about the domain... Don't limit the range of 'x' in f(x), but do limit the number of f(x) used to only the principal root function.
I come from a country where conventions were maybe a little different. when writing the Radical sign ("Square root sign"), the convention is that you work in R. If you want to work in C, we normally write ^(1/2) And to avoid any confusion, always, always write the definition domain. I assume that if this is an interview, the very first thing to do is clarify the definition domain. The last thing you want is to be "right" but misunderstood by half of your co-workers, or by authorities.
I solved it as -6 but I have an argument for undefined which you didn’t mention. One can say it’s undefined as there was no domain of x mentioned so I can not be sure what was meant. The reason for this is that most of the time real numbers are used as domain and that is the reason for fair assumption that this is also the case.
Exactly. I stopped watching at the half-baked integer mathematics analogy. The domain is the real numbers unless otherwise specified. If this question came up in a textbook in a chapter about complex numbers I wouldn't expect them to state it again in each problem.. but in isolation it's intentionally ambiguous about what domain to use -- there is no right answer because you can't know which of the two right answers it wants.
@@seedmole I disagree, the domain is all numbers unless otherwise specified. Without that assumption the fundamental theorem of algebra breaks. Unless there is something that specifically prohibits them. The characteristic equations of second order differential equations wind up with complex roots even if you can use them to work your way to purely real valued functions.
That makes no sense. If you know about complex numbers then you know how to evaluate this expression. You do not need the author's permission to use that knowledge.
because them being the root of a negative number changes how the maths works it is more that the product of 2 roots is more a guide and a quick shortcut that can be used under some circumstances rather than a good rule
Pretty much just to avoid this specific problem having the answer 6. If you're using the version of square root with both branches present then the rule is fine, if you use the one with only one branch, sometimes the rule gives you the other branch. Since 6 would be the positive and thus correct branch for sqrt(36) but not for sqrt(-4)*sqrt(-9).
@@Stereomoo It would also end up working if there was only one negative number, TBF. Like radical(-9)*radical(4) = 3i*2 = 6i And radical(-36) = 6i Same answer. It's only deceptive when there are multiple negatives, because you'll end up on the non-principal branch. Actually, going to more than 2 numbers... Like, 4 negatives and 5 negatives both work. i^4 = 1, i^5 = i, both on the principal path if you multiplied everything together. 0,1 work; 2,3 don't; 4,5 work; 6,7 don't; etc.
@@durandle9226 in my limited knowledge of pure math, i consider the whole explanation unnecessarily complicated, because sqrt(-4)*sqrt(-9) could be simplified to sqrt ((-4)*(-9)) which simplifies to sqrt (36) = 6
Basically it's an agreement that the square root of a number is _always_ positive; for the x² = c, the answer is plus-minus root c, which means that root c must have a fixed sign.
I think Presh's observation that the (±2i)(±3i) camp would be the hardest to convince is exceptionally astute, as I understand but continue to disagree with his belief that the principal square root bears on the evaluation of this expression. Given the PEMDAS stipulation that exponentiation supersedes multiplication, the ± has to be produced twice and therefore winds up in the final answer, as well. Denying half the complete answer in order to solve an irrelevant manufactured issue (the not-a-function issue) is logically fallacious. The expression has two valid values, not one. // Do I sound terribly opinionated? Well, I probably am -- except for the *terribly*. Cheers. :)
@@jim55price the expression has one value. ALL arithmetic expressions involving addition, subtraction, multiplication, division and powers (including the √ symbol) ALWAYS produce a single value. Even in the context of complex arithmetic no simple expressions ever has more than a single value. There is no equasion here. There are no functions. √-4*√-9 is a simple arithmatic expression with a SINGLE value of -6.
@@nbooth Its not just arithmetic operators, _all functions_ should return a single value, since by definition a function is a one-to-one (or many-to-one, but never one-to-many) mapping between two sets
@@nbooth Your comment is preposterous, given that far simpler expressions, e.g. ±6, ±x, ±i, have two values. I've no idea what you're going on about with your claims of "ALL" and "ALWAYS". All it takes to illustrate your error is a single "±".
@jim55price yes the ± symbol allows you to write two values with a single expression. Congratulations. {-1, 1} is also an expression with more than a single value and [0, 1) is uncountably many. Those aren't arithmetic operations however. They're ways of writing sets, which is what the ± symbol does as well. Raising a number to a power is an arithmatic operation that produces a SINGLE value. That's why we need the ± symbol in the first place.
@@jim55price you didn't read what I wrote. I said all expressions involving the following operations (+, -, /, *, ^) yield a single value and you're claiming some other operation is a counterexample to that.
Regardless of what the correct answer is for this question, it is important to remember that the exponent x^a y^a = (xy)^a only applies for positive real numbers *x* and *y* and real exponents *a* where here, a=1/2.
@@pelasgeuspelasgeus4634 Usually the square root of any complex number is defined as x^(1/2) = e^(1/2 * log(x)) but the important part is choosing which value of log to use.
I was of the opinion that it could be +/-6, but having looked up mathematical notation, the √ sign implicitly means that one should take only the principal root. It's a matter of convention. I don't think the explanation involving functions was convincing to me (or others in the comments I have noticed), it was probably better to just say that the notation implies we should discard the non-principal roots.
• When dealing with square roots of negative numbers in the complex number system, each square root operation returns a single principal value. • The principal square roots of -4 and -9 are 2i and 3i, respectively. • Multiplying these roots results in 6i^2, which simplifies to -6. Therefore, in the complex number system, the answer to -4 x -9 is -6, not ±6. There should be no ambiguity here.
who decided complex numbers must be principal roots? If so, would (-1)^(1/2) be i, or +-i? I really don't think we should always ONLY talk about principal roots. Now if the square root *symbol* is about principle, then that should have been mentioned. P.S. not to mention -i^2 is in fact -1
@entityredstoneonyt to express the fact that the principal square root of 9 is 3, we write √9=3. X²=9 is a different thing. The same applies to negative numbers. That's what I've learned in school, and that's what wikipedia says.
I feel like there should be a new mathematical function where it returns not just the principal root but all the roots: (I'm not a mathematician so the following might sound a bit wierd) let root(a,b) be all solutions to x^b = a For example: root(25, 2) = {5, -5} because 5^2 = (-5)^2 = 25. Now we can find all "solutions" to (sqrt(-4))(sqrt(-9)). root(-4, 2) = {2i, (-2)i} root(-9, 2) = {3i, (-3)i} Multiply them together means take all the possible combinations and multiply them: {-6, 6, 6, -6}. Remove the duplicates, we get: {-6, 6}
On the other hand, if √4̅ = ±2 = {2; −2}, then for √4̅ + √4̅ we have four options: • 2 + 2 = 4 • 2 + (−2) = 0 • (−2) + 2 = 0 • (−2) + (−2) = −4 So it turns out that there are three different possible outcomes for a simple sum, and also √4̅ + √4̅ is no longer equal to 2√4̅ = ±4 ¯\_(ツ)_/¯
@@cyberaguaoh yikes that’s something I’d never considered. If you expand the definition of √x to be the set {+√x, -√x}, then yeah √x + √x ≠ 2√x. That’s bound to mess some stuff up…
@@121DSpCe-Tile In fact, it makes sense when solving irrational equations over the field of complex numbers ℂ - there you have to take into account all possible combinations of the root values.
The problem is that you considered i as sqrt(-1), it ain't. i is the number such as i² is -1, while the answer for sqrt(-1) is i and -i , both branches should always be considered, unless the question amde has a context and only one branch makes sense. is nice that you talked about functions and all, but this isn't a question about functions, is a question about an equation. I don't know how you gringos learn equations, but when i did in primary school, if we had to answer a problem of the kind: "what are the roots of 3x² + 4 - 2" we would solve 3x² + 4 - 2 = 0 and we knew we could put that on the in the square equation formula (A.K.A. Bhaskara) and we could find up to two roots, unless they coincided or one or both couldn't be calculated in real numbers (it was primary school after all). So, it is + or - 6. when we looks at sqrt(-4)*sqrt(-9) we immediately know that we could have four answers because each square root can have two answers and we know that two of hose answers would coincide in the same number, and the other two as well therefore we can have up to two unique answers. then when we expand that to sqrt(-1)*sqrt(+4)*sqrt(-1)*sqrt(+9) we know this expression, written as it is can give us eight answers and we know we will have, at most, two unique answers because we know that (+or-)i*(+or-)2*(+or-)i*(+or-)9 have eight anwers.
@@noname_atall there is no equasion here! "√-9*√-4" is a simple arithmatic expression. There are no equasions or functions involved. Even in the realm of complex numbers, simple arithmatic expressions have SINGLE values.
But the "√" is _not_ defined as the solution to a quadratic equation. It's defined as the principle square root. This symbol always denotes a single positive number. Recall the quadratic formula that you referenced. Note that it includes the construction "±√...". If "√" could be either a positive or negative number, why would the quadratic formula have to include the ± here? If it was already implicit in the √ symbol, it wouldn't be a necessary to include it in the formula.
The principal value of the root can be expressed and dissuaded from debate by noting that we start with the root already taken. The plus/minus is only necessary to solve when applying the root as a solution to the quadratic equation.
There is no such limit to negative numbers under the root. We all have happily used this with variables, not even wasting a single thought about signs.
The undefined solution is largely given to students learning square roots initially. In order to not overwhelm students, the teachers simply want students to say it is undefined. But, once students learn of complex numbers, negative square roots can be taught in their entirety and should be viewed as the actual correct answer.
many people might see 2 negetives in multiplication and get the answer as +6 but it actually goes like this: since i=sqrt(-1) we get, -> [{i*sqrt(4)}{i*sqrt(9)}] -> (i^2)*(2)*(3) and since {sqrt(-1)}^2=-1 the answer is -6
-2i squared equals -4, so I can understand why they might say the original problem's answer is plus or minus 6. I confirmed this approach by trying it with apples, and ended up with plus or minus 6 apples, which did kind of mess up my bookkeeping, to say nothing of determining how much space is left in my refrigerator!
Even if we extend the principal branch of the square root to complex numbers, it is usually not defined on a ray. That ray is typically negative real numbers. If we extend it continuously to negative numbers (and remove another ray instead), a choice must be made as to whether sqrt(-1) is i or -i. Fortunately, for either choice, sqrt(-4) * sqrt(-9) evaluates to -6.
All you need is the definition of the radical symbol. From Wikipedia - The two square roots of a negative number are both imaginary numbers, and the square root symbol refers to the principal square root, the one with a positive imaginary part. Additionally, I don’t think the tree pruning for functions explanation is relevant. There’s nothing in the original question that requires a function. We’re simply asked to evaluate an expression. -6 😎
I’m a bit closer to agree with the Principal Root argument. However, as a counter example in the quadratic formula +/- is part of it even when the radicand is negative. There is no enough information in the problem to exclude the positive answer. What am I missing?
It’s just the definition of the radical symbol. The two square roots of a negative number are both imaginary numbers, and the square root symbol refers to the principal square root, the one with a positive imaginary part. So when evaluating the expression, we use the positive imaginary part for both radicals and the answer is -6. The answer is very straightforward. The quadratic formula is a different situation. In this case we’re solving an equation (not simply evaluating an expression) and it will have two answers (unless the radicand is 0). Here we want to include both roots, positive and negative, no matter if the radicand is real or imaginary. Since the radical symbol only refers to the positive square root, we must add the plus minus sign to specify both roots and find both solutions.
@@johnhoslett6732 Interesting. So you are saying "Simplify the following term" and "x equals the following term, solve for x" are two different things. But who says that I am not allowed to pick the second option to solve this problem?
@@manfredlemke4671they are different things. If you take a square root of something (call it a) you get two possibilities: ±√a. If you're given a square root √a, that *always* refers to just one of them.
The fact that the square root symbol refers to only the principal square root is similar to the fact that 1 is not a prime number. It might not fit with your world view, but mathematicians have adopted is as a standard because it makes mathematics easier, and you should too.
Neglecting all but the principle root is a good way to pass the vertical line test. Just like kicking the ball into the hole is a good way to make par. If you were given more than just an equation, some real world situation, for which negative values could not be valid, then you would be justified in ignoring the negative roots. But, this problem did not give details of any particular situation, we don't know that x represents a real value. Since the notation given does not indicate a function ( it is ?, not f(x)) and doesn't say to graph the result, the vertical line test for a function is also irrelevant. So it all comes down to your assumptions. If you assume a real value function, then you would have undefined, because you would not have i to deal with the negative square. If you assume complex numbers, then it makes sense to use the complex plane for a graph if you want to graph it, in the complex plane using both positive and negative roots is just fine, then you get ±6. You only have the answer -6 if you assume complex numbers but choose to graph the answer in rectangular coordinates as a function of x. I always used to tell my students that they could use reasonable assumptions, but they must identify the assumptions they make.
In all honesty, ±6 is still incorrect, all complex functions have infinitely many values depending on what branch you evaluate the function on. In the complex plane there is simply isn't really an answer, so the assumption about where branch cuts are and the which branch is being chosen for your function is explicitly stated
I think the assumption of a principal square root is the best assumption to make since that's literally what the radical sign means. Yes, numbers have two square roots, but as soon as you use "√", you're automatically referring to the positive one. It's just what the notation means.
@@ric6611 Unfortunately the radical symbol is in and of itself, ambiguous. As the same radical symbol itself is used for the square-root function over the positive reals as well as the square-root over the entire complex plane. i.e. f: ℝ⁺ → ℝ⁺ is f(x) = √x, and pertains to the positive square root of a non-negative real number. while, f: ℂ → ℂ is f(z) = √z, and may pertain to any of the square roots of any complex number, on any branch of the function. I think personally, the original square root function itself, f: ℝ⁺ → ℝ⁺ is f(x) = √x does not apply here, as the second you put a negative real number in there, we have to expand our definition of f, to that of f: ℂ → ℂ is f(z) = √z, wherein we must define which branch of the square function we are working on, that usually being the principal branch. I think that √-3 itself implies we are talking f(z) here, since if we used f(x), -3 is simply not in the domain of the function.
An important distinction he made is that although a number has more than one square root, the square root function has only one answer which is why we are careful to write +/- in front of the square root symbol in the quadratic formula for example.
They way I like to think about it is the sqrt(x) is really a shorthand for (x)^(1/2). That square root symbol implies powers and parentheses, so you have to do them before you perform any multiplication. Unfortunately, many people are taught in school that sqrt is "fundamental".
Since square root is multivalue function and we have 2 roots in the original question, we have 4 ways of choosing branches (sqrt(-4) = 2i or = -2i). So my answer was undef. until branches are chosen (+-6 is not proper way to write uncertainty in my opinion) Upd: Also, your argument about +-6 is ruined by your previous one about undef. Why you has limited the domain of the function to C, why not to choose the Rieaman surface for square root, so both branches could exist simultaneously.
@@MrSummitvillewdym "for this problem". Math isn't working like this. We need domain to consider whether function is multivalued or not. If it is, we need either specified the leaf on which we are, or use structure called Rieman surface to make a well-defined term "square root". you don't know the complex analysis doesn't mean complex analysis disappears...
@@MrSummitvilleDifferent areas of mathematics may use diffrent notation. Different countries and mathematical schools may use different notation. Diffrent contexts require different approaches. If you assume that the radical sign denotes the principal square root of complex number, this does not imply that this is the convention that is always used by everyone else. In complex analysis it is convenient to view the square root as a multifalued function and so we usually use the radical sign to denote the *set* of all square roots of a complex number. As it was mentioned by @MercuriusCh, you could even consider the square root as a nice and well-defined function on the Riemann surface if that works for a particular problem. It is all the matter of what is convenient in any given situation
for the +/- argument it can't be the case because we have the square root of a negative number it is only valid when taking the square root of a positive number (or more broadly the even root of a positive number) because the number was defined as negative we know that were are evaluating a product of positive number in the sub step of "i root 4 and i root 9"
The core of the discussion between -6 and +-6 is the square root. The trouble is the numeric answer, and that is a "bad manner" in math. Just write -sqrt(36). If the root was sqrt(2) nobody would calculate the numeric answer and would let the reader to interpret the meaning depending on the context. Only 10-14 years olds care about giving a precise numeric anwer with 987 decimals. Adult, math-educated, people are not afraid of amswering with the simplest roots or fractions. It is the same if the answer was 22/7, only children would calculate the numeric answer, normal and functional people would let the answer be 22/7 since it retains all the information (all the decimals with 0 error). So the answer should be -sqrt(36) and it would not delete any information.
When you evaluate an expression, you take all possible results. The correct answer should be +/-6 as it takes all possible outcomes into picture. I like your logic of principal square roots but that falls short of a complete answer.
The symbol √ has a meaning, it will represent one number. For example √ 2 is the exact value of 1.414….. It is a number, not several numbers. If √ 2 could mean both 1.414… and -1.414… (± √ 2) then how could you possibly write just one of them. You wouldn’t be able to. X=√ 2 is not the same as x^2=2
I disagree, we are simply performing a binary operation on two complex numbers, which by definition has one output. We define the square root on positive reals as the principal square root so there is no reason to assume otherwise for the complex numbers
@@bobh6728when you do calculations on physical objects, value or principal roots are considered. However, when it is asked as a general question to evaluate, all possible results are to be used.
It isn't correct. There's no principal square root in complex world. In this case you must consider it as a power 1/2. Power has multiple values in complex world.
There is. Source: wikipedia this is my understanding Put a complex number in polar form: r * e^(i * θ) With a restriction on θ: -π < θ ≤ π Now, the principal square root would be defined as: sqrt(r * e^(i * θ)) = sqrt(r) * e^(i * θ/2) With this definition sqrt(-1) = 1 * e^(i * π/2) = = i
How stupidly inconsistent. It makes sense if the radical symbol to be the positive square root only for all the reasons they give then the exact same reasoning must apply to the imaginary verson.
The thing is, ±6 comes with the assumption that we can take both square root values both times. That's (usually) a bigger assumption than "there's a square root that is explicitly and obviously that of a negative number, we can use i", which makes ±6 feel still correct, but not quite as correct as just -6. Especially when you can tell that the i's are gonna "cancel each other out" into -1, it feels much neater to remove that uncertainty even if it takes two weird steps, than if you were to take two more uncommon weird steps and still have this "uncertain" result in the end.
it is not correct to use an ambiguity like that in the first place solving the sub step of " i root 4" is not something to consider the 4 might be the product of 2 negative numbers we KNOW it is a product of positive numbers in that state because we started with the square root of a negative number
Limiting to the principal root is the definition, not the assumption. Also, "wolfram alpha said so" is a valid argument. WolframAlpha just follows the definitions we gave it. 1+1=2 is defined. We follow that definition. So do we follow the definition of a square root.
Its odd to me that the segment that went off on people who only used Real numbers is the followed by an assumption that we should take the positive values for both roots. sqrt(4) =±2 is not a function, true, but sqrt(4) = -2 is. Its is completely possible to get sqrt(4)i * sqrt(9)i -2i * 3i -6i^2 6 There are 4 potential routes for the equation. 2 give you -6 and 2 give you 6. Therefore the general solution is ±6. The answer is only -6 if you add a qualifier of using principle roots.
The question is not to find solutions for a quadratic function, where considering the negative root part is required, but it is about calculating the root, which is defined to be positive only.
Mathematician have agreed that whenever your are talking about "the square root" your are talking about "the principle square root." And especially in this case, as when that sign (radical sign/radix) is used, it specifically denotes the principle square root.
@@arejaybee there's no equasion here. We just have arithmetic operations on complex numbers. The expression has a single value: -6. (Yes the √ does mean principal square root).
@@nbooth The question for the second poll was "What is the product of the square root of -4 and the square root of -9 equal to" and at 1:34 (where I'm currently paused at) it is even shown with an equal sign between the formula and the possible solutions. So there is an equasion of root(-4)root(-9)=? implied.
@@TyroRNG but the notation was not in the original question (for the second poll, 1:34), so that argument does not apply. And based on the phrasing of the poll, it is root(-4)×root(-9)=? So it comes down to finding ALL solutions for ?, which would mean, unless extracting root(-1)=i and then squaring root(4)i×root(9)i is forbidden by some rule ?=6 needs to be considered
Either we operate in the real domain, in which case it's undefined, or in the complex domain, where each non-zero number has exactly two square roots (yes, even if it is a positive real number). In the latter case, the expression has two values: 6 and -6. But this is why the square root symbol is better avoided in complex arithmetic, the usual assumption that an expression evaluates to one value breaks down and one has to be careful about the consequences. In typical uses it's better to start by saying that -1 has two square roots, we call one of them i, and then express everything in terms of i rather than using the tricky square root symbol.
If the rule √x•√y = √(x•y) only applies to x,y≥0 then you can't split √-4 to i•√4 Once we start working with complex numbers we need to remember that roots have two results. So, the result of the product now depends on which value you get for √-4 and √-9. It turns out that all the combinations ends up giving you 6 or -6. The result is ±6
In the survey run by Brilliant, less than 50% got it right, so "most people get it wrong" is technically correct. When the video maker re-ran the survey, 55% got it right. Presumably, this says something about the math knowledge of folks who watch this channel vs. folks who follow Brilliant.
@jmr5125, and John. Generally speaking while something over 50% is a majority, that does not mean that it is most as most requires nearly all. I'd be looking in the 80s or 90s for most. However, if we are talking comparatives, many, more, most and comparing groups containing different numbers then many (of a large group) might require only 20%, with a group containing 35% being more and a remaining group containing 45% of the total might indeed represent the most. Ain't words fun too.
The concept of the principal value is a standard that we invented and decide to use to make maths make more sense for us. I think the fact that when we don't use the principal value standard we get the same answer whether we take the square root of the numbers then multiply as we do when we multiply the numbers and then take the square root means that ditching the principal value standard would be a more consistent model. Taking the principal value standard is like saying the answer is undefined because we're restricting ourselves to the real number line. It's a standard we can impose, not the actual answer.
Yes, when using only the principal values, we cannot solve irrational equations like √x̅ = -2 or ³√x̅ = -2 over the field of complex numbers ℂ, but if we let our complex nth roots be multivalued, then these equations have solutions.
At a guess, I think the reason why someone might go for +6 (which was my first instinct, before I thought that it could also be -6) is that when they're taught that √x√y = √(xy), complex numbers haven't been introduced yet, so the teacher doesn't need to specify that x and y have to be ≥ 0 for the rule to work - and then by the time complex numbers *are* introduced, this particular subtlety gets swept under the rug because we've got a lot of material to get through on these things! So possibly people might think that the rule still holds for negative numbers.
Shout out to the person who said "Undef. if real, -6 if complex"
And shout out to all the people still saying ± 6 because "the square root of 4 is ± 2".
Really simple, x² = 4 then x = ± √4 = ± 2; but √4 = 2 and 2 alone. The ± symbol is always placed before the √ symbol when inverting a square.
Oh i figured if real it would be 6 unreal would be -6
@@WombatMan64that’s true if we’re inverting an x^2 term, but from what I understand the radical symbol denotes the principle root which is the positive branch, so -6 should be the only answer in this case. You could probably rewrite the thing with an x^2 term to get two solutions if you changed a bunch of things.
If we are taking the square root of negative numbers, then aren't there only complex solutions? The only square root of -4 is 2i and the only square root of -9 is 3i, correct?
But -6 is real.
i would never say "imaginary numbers are real," I say "imaginary numbers exist."
Imaginary numbers aren't mathematically real, but in all fairness, the sets have a terrible naming system. Gauss himself suggested that imaginary numbers be called "lateral numbers" instead.
Obviously, in a modern mathematical stance, imaginary numbers are not in the set of real numbers. But from a more debatable, linguistic standpoint, "imaginary numbers are real."
A real number may be defined as a + i*0, where a is a real part and i*0 is an imaginary part. That way there is a connection between real and imaginary numbers.
Naming conventions are a bit misleading and don’t mean “real” as in “exist”. Would you call -5 a number that exists? Outside of math negative numbers can signify loss or deficit, but you might as well use positive numbers to measure the size of it.
But that's not funny
Imaginary numbers show up in very real ways in electrical engineering. By real, I mean "physical things you can actually show on a meter or display". For example, the impedence of an antenna at various frequencies.
well on that.. numbers dont really exist, they are supposed to be an abstraction
If 55% of people got it right, then didn't "most people" get it right? As in the majority?
42% got it right on the original poll.
Do you mean the principal majority or unprincipled majority? 😊
considering thw bias in the poll that mind your decisions conducted, I would say we should omit that data. There is even a bias in the brilliant org poll
On RUclips poll you can change your vote to the one with most % / maths specific audience
55% of mathematical interested people got it right, 42% of a generic audience.
@3:06 The rule that sqrt(x) times sqrt(y) = sqrt(xy) - why do NO grade school or high school math teachers EVER bother to mention that this only applies to positive numbers???
Because in highschool students will never encounter roots of non positive numbers.
I was searching for this, so that's why it's only -6, thanks
Because it doesn't only apply to positive numbers.
@@gabrielbarrantes6946 Complex numbers are introduced as early as 10th grade in many schools.
It shouldn't be true. Start backwards: sqrt(36)=sqrt(6*6) or sqrt(-6 * -6), so why not sqrt(9*4) or sqrt(-9 * -4)? If this isn't true, then all square roots would be positive, but as we all know the correct answer to any square root is pos or neg
There's a difference between arguing about math and standardization.
Agreed. This is an issue of definitions.
Maths!
(Sorry, i know i am being a bafoon, i just think it is funny to argue "standardising" linguistics out of nowhere)
@@volodyanarchisthe wrote in US English and used correct spelling.
@@ronald3836 r/woooosh
@@ronald3836 A contradictory statement
As an electrical engineer, I feel offended when people tell me imaginary numbers aren't real.
In electrical engineering the square root of -1 is an jmaginary number.😁
@@nielshoogev1 That is funny! Thank you for that.
I agree. How could sine exist without j?
As an electrical engineer (by qualification, if not by trade) I get offended by people referring to j as "i". I is current! And also something in mechanical engineering; because they use j too, and probably not just in solidarity.
Imaginary numbers don't have feelings, you don't need to defend them.
I like how you explain all the wrong answers and how people might have gotten to them -- I think it clarifies a lot of important misunderstandings, and even though I got the correct answer, I still learnt a lot (especially from the greater clarity in definitions, which is the most important thing in maths imo). thank you.
None of them are "wrong", just incomplete. +6 and -6 are both completely correct answers. The problem is not in the math itself, but in the ambiguity of the √ symbol having two solutions without a clear indication of which one we want. With positive numbers it's easy to assume we want the positive root because of the obvious association with a literal square from geometry, but with negative numbers and the need to go complex, that "obvious association" is lost.
The only answer that's arguably wrong is "undefined". It's taking the square root of an explicitly negative number, so assuming that you're working in the realm of strictly real numbers is kind of silly.
@altrag if they were looking for both the positive and negative answer there would be a +/- before each root. Since that is not there it means that only principle root is to be considered. That is standard math notation.
@@mudfish78 > there would be a +/- before each root
That is most definitely not as "standard" as you seem to believe. It may well be what you're most familiar with, but it's hardly universal, especially when you start dealing with complex numbers and "positive" / "negative" isn't as meaningful (is 2-3i positive or negative?)
In this case because the result of each root is purely imaginary you can claim "+i is right", but pure imaginary numbers are a rare special case within the entire complex place (as are pure real numbers) so it's kind of questionable to even claim there is a "positive answer". The principal root that you note is the closest we could come to being the "right" answer, but given the simplicity of the question I'd assume this is pretty early on in an introduction to complex numbers and the concept of principal root might not be known by test takers.
@@altragYou should assume that the person solving the problem would know that the radical symbol is only the principal root. Why would you in any universe assume they don't?
Suppose you're a teacher making a question. Would you assume the person solving doesn't know how the order of operations works? Would you, as an examiner, mark 1 + 1 x 3 = 6 as correct because there was no rule stating PEMDAS/BODMAS must be followed?
Obviously not, because that's common sense. So if not acceptable in that case, then why would you accept a factually wrong mathematical answer because the examinee didn't know an 8th grade rule?
@@interloper9589 > You should assume
Why should I assume that? You're asserting that it's "standard", and maybe that's the only way _you've_ seen it used but I've seen it used in both ways on multiple occasions. My first assumption when I saw the question was that it would be a +/- result, just because that's how often it comes up. If I'd taken that test I'd have probably given both answers (knowing what I know today - again, that looks like a fairly introductory level question so I can't say what those students should be expected to know).
> no rule stating PEMDAS/BODMAS must be followed
The fact that we can't even agree on the acronym tells you what I think of that comparison.
> didn't know an 8th grade rule?
Because it's not a "rule".
PEMDAS can be a "rule" because it's convenient. You could choose any other rule and you'd still only get one right answer (as long as you apply it consistently). It might not be the same answer you get with PEMDAS, but it would be unique under the rule you choose. 1+1*3=6 is not only correct but the _only_ correct answer if you choose the rule "strict left-to-right evaluation".
The "principle root" rule is different. The square root operation really does have two (generally) unique answers, and there is no choice of "rule" that can get around that. Picking only one of them to be "right" is a much bigger deal than merely convenience, and as noted that "rule" is not always followed - _and cannot be assumed_ - because it is not and fundamentally cannot be consistent in the same way that PEMDAS can be.
Team plus or minus 6 here. Nowhere did it say that the root should be considered as a function. Restricting the result to the principal branch is an assumption that was never specified.
It's not an assumption. It's literally specified right in front of your eyes. That's what the √ symbol means.
If the writer had wanted you to consider both square roots of these numbers they would have written a ± in front of the √ symbol. That's how you say "both square roots" in mathematical notation.
But even if we want to define a principal square root, why is √−̅4̅ = 2𝒊, not √−̅4̅ = −2𝒊? What makes us choose 2𝒊 = 2∠90° and not −2𝒊 = 2∠(−90°), when they are equally valid? If we instead define our principal argument to be −π ≤ φ < π (not −π < φ ≤ π), then √−̅4̅ = (4∠(−180°))¹ᐟ² = 2∠(−90°) = −2𝒊, not 2𝒊.
@@gavindeane3670 Does the equation √x̅ = −2 have a solution (real or complex)? And what about ³√x̅ = −2?
@@cyberaguaNo, √x = -2 has no solution. You don't need to think about complex numbers for that.
As for the cube root, as I said in another comment elsewhere, you'd need to ask someone who plays with higher order roots and complex numbers. There isn't (that I know of, at least) an obvious notational option to indicate which root or roots you're taking about.
@@gavindeane3670 > No, √x = -2 has no solution. You don't need to think about complex numbers for that.
It's all about definitions and conventions. You may define √x̅ to have only one (principal) value, or (as some authors do for complex roots) let it be multivalued. In the latter case we have √4̅ = ± 2 = {2; −2} ∋ −2, so x = 4 is a solution to the irrational equation √x̅ = −2 over the field of complex numbers ℂ with the appropriate definition of the multivalued complex square root function. Also equality in this case is replaced with inclusion √4̅ ∋ −2, as it's normally done when solving equations with multivalued functions. But I don't think you've ever heard of it, since only those who solve problems that require multivalued or set-valued functions really need this stuff. Look up Set-valued function on Wikipedia, for example.
@gavindeane3670 > you'd need to ask someone who plays with higher order roots and complex numbers
It's actually the same for any order roots: ³√−̅8̅ = {−2; 1+𝒊√3̅; 1−𝒊√3̅} ∋ −2.
Good video, but please realize that "because Wolfram Alpha said so" is not a convincing argument.
has anyone ever found an error in Wolfram-Alpha? For me, that makes it an authoritative source.
It's not an argument, sure, but Wolfram Alpha is an extremely reliable source. You may notice that he gave an argument, though.
sqrt(4) = +-2 or sqrt(4) = 2 which one is it?? 😑
@@nabibunbillah1839
just 2
the reason +- exists in certain context is to find the roots of functions that are some form 0=ax^2+bx+c (or related, like other polynomials, sin(θ)=0, etc, which have multiple answers), because technically two answers make this true (fundamental theorum of algebra)
when you are asked what the sqrt(a) is, the answer is always positive, which is why sqrt(x^2) is |x| for all x, rather than just x
saying x^2=4 is a different problem then x=sqrt(4), even if the latter is used in the former
the reason this problem is framed weirdly is because roots and radicals fundamentally work differently for complex numbers, even if the concept is the same
for instance
i
i^(4/4)
(i^4)^(1/4)
1^1/4
1
so i=1
obviously this is wrong, and the reason it fails is because there can be multiple answers that technically satisfy 1^(1/4) in the complex plain, including the integer -1, that we dont consider to be valid answers in real-valued
computation
if you can apply the roots/whatever more diligently, you can do:
i^(4/4)
(i^(1/4))^4
(exp((π/8)*i))^4
exp((π/2)*i)
i
i=i
I was in the "-6" camp, but that argument "against" "+-6" really makes me believe in the "+-6" option. As much as I like the idea of the principle square root, I feel like this is akin to "This shouldn't be possible... but according to the math, it exists" (usually referring to the consequences of Einstein's equations)
If the calculation has a purpose other than satisfying a maths teacher (e.g. an engineering problem), you should always consider all possible solutions if you come to a root, arcsin or other inverse function with a "principal solution". Sometimes there is really more than one solution to the problem, very often one of the solutions is obvious nonsense in the real world.
If imaginary numbers, which have no real world meaning, leads to the "correct" answer, then I didn't see why the non-principal square root can't be used too.
@@djinn666 I know, right? He goes "You're only assuming that this is about real numbers, but imaginary numbers can be included" and then he just assumes that it should be a function and that only the positive values should be considered for the square roots.
It's a definition thing. The square root function is defined as principal root always. Thats why you'll find formulas likes the quadratic equations that have "+-sqrt(...", the negative root has to be added back in because the standard is that the symbol means the positive root.
It's in a similiar boat as order of operations, you can do addition first but thats a completely different system that will get you different results from how everyone else reads the notation
@@jacobmerrill693 There is no international authority on mathematical definitions. Some would say 1+2+3+4+... is undefined. Others would say it's -1/12. Both are correct if you ask any reputable mathematician. It's the elementary school math teachers who demand that everyone use the same definition.
I think it's disingenuous to go: "we should definitely be considering complex numbers" in one breath and "let's ignore the other half and just consider the principal square root" in the next.
Also, who decided Wolfram Alpha was the final authority on how to do math. It's a calculator. An advanced calculator, but a calculator nonetheless. And decisions were made when programming it. Without an understanding of what went into making that decision, I'm not quite ready to say they didn't make a mistake.
Edit: All these justifications about established convention and the definition of the square root operator are great and all, but the fact remains: he gave none of them. He gave a reasoning about it not being a function (by the strict definition of only producing one output), but that felt like him deciding on his own that it had to conform to that standard. After all, if you were solving for a value, you wouldn't care that it doesn't match the definition of a function, why would you care here? He said nothing about established conventions.
I was thinking the exact same thing when I saw him pull out Wolfram alpha to “justify” against that answer hahah
Wikipedia also defines the √¯ symbol as the principal square root and if you want both halves you'd want to write ±√x̄, I'd guess it's a consensus in the entire math community, so yeah
@@yukimoeI suppose that makes sense in the same way we write "6" rather than "+6" when writing positives, still think that it feels somewhat arbitrary tho
I do agree, somewhat - the root symbol typically means the principal root. Which is what the wolfram alpha function uses too. But I'd suggest if you take the secondary root for one, you should for the other as well, and that's still -6. I do think +/6 would be a valid choice though, with an explanation. But like undefined, you're making an assumption that most people wouldn't make, so you'd want to be upfront about that when you answered. Like, no real solution is a fine answer too. As long as you say there is no real solution, not just undefined.
@@MarkEmerAndersonII That's not how square root works. It is not that you take one or the other. You take both.
I'm not very good at math. I quickly came up with -6, so I figured it was wrong.
Maybe you're better at math than you've been giving yourself credit for. I bet if you keep watching this channel, or look for other opportunities to learn math, that you'll get pretty good.
Well obviously you must be decent since you got the answer right.
i also came up with -6 and thought it was wrong
I'm not very good at math either, and I figured that as sqrt(4) is 2 or -2, sqrt(-4) would also be 2 or -2, so I completely ignored the option to multiply the -2 with the 3 or the 2 with the -3 and ended up with 6... Like I said, I'm not very good at math :P
@@helaluddin-bo9kr Well the title of the video set us up to think we are wrong, same thing happened to me.
In the original question where both 6 and -6 are options, and +-6 and undefined are not, my answer is -6, as I feel it's the most correct. In your version of the question, where all choices are potentially correct I would say +-6. If +-6 is one of the options, that implies to me that such answers are allowed under the writer's definition of square root.
Using the root symbol (√) is defined as taking the positive root. Taking to a fractional even power (like ½) gives the positive and negative root.
4^(½) = ±2
but
√4 = 2
@@pageboysam Not universally.
@@TheFinalChapters I’d be interested to hear which culture’s math system doesn’t.
@@pageboysamThat's not true either. 4^(1/2) is the same as √4. It us only the principal square root. No combination of elementary arithmetic symbols produces more than one value without explicitly using ±.
Yeah this is the key here. I too was emotionally attached to my answer of the initial question, too much to realise that the new options for answers changes things, so whilst before the video I chose -6, by the time I was 75% of the way through this video, my choice changed to +/-6
Every time you evaluate the square root of a negative number, dark matter is created somewhere in the universe
7:47 No. Not everybody.
Complex numbers are not ordered, so predicates () are not defined for them, you can't universally choose (i) to be the principal sqrt(-1). In some schools it's taught that if z = r * (cos(t) + i * sin(t)), where -pi < t
"Complex numbers are not ordered, so predicates () are not defined for them, you can't universally choose (i) to be the principal sqrt(-1)" - As it happens this is located on the ordinate of the complex plane, therefore on a straight line with real coordinates, so can't we ?
@@jige1225 This does not generalize well. What if we need sqrt(1+i)? Which branch are we going to choose?
"To conclude this, math language is not universal."
Yes it is, and mathematicians have spent centuries making sure everything has a single and unambiguous definition
@@КонстантинАртем quite straight forward, the length of the vector would be sqrt(sqrt(2)) and the angle around 30 degrees. One solution
@@salerio61 The angle would be 22.5 + 180 * n degrees.
I was solidly in camp minus 6 before watching the video, but afterwards you really convinced me that +/-6 is the most principled answer for the same reasons you gave to argue against the Undefined answer. Nothing in the question limits the calculation to the "principal root"... So +/- is the most inclusive way to see the answer... Thanks!
He didn't explain it correctly, that's true, but -6 is the only right answer. What he should've explained is that the radical sign "√" implies the principal square root. It's why we use the "±" sign when we are talking about both roots, like ±√4. Therefore, by the way the question is written, since it uses the radical sign "√", only the principal square root should be considered.
In summary, something in the question DOES limit the calculation to the principal root, and that's the radical sign.
@@ric6611 I actually never heard the term "principal square root" before this (I was in the ±6 camp). And it's interesting that the √ symbol implies the principal square root. I'm curious if there's a way to write the question to ask for all solutions. Maybe "Find all solutions for x such that x = (-4)^1/2 * (-9)^1/2" (Edit: Wolfram-Alpha didn't like my 'find all solutions' clause.)
@@paulgruner9644 You can either use ±√(-4) * ±√(-9), which I know is ugly, or something like ±(√(-4) * √(-9)), ask for the solution to the equation x² = -4 * -9, or you could even just be explicit about it.
For example, I have no problem if the question said something like "consider both roots when the radical sign is used". My only problem is that by omission, the default should be used, and the default is the principal root.
If it makes sense for you to always be talking about both roots, and you don't want to keep using ±, it's entirely valid to say something like "assume both roots from now on", for a paper, or maybe even for an entire class. I think that's what happens when people are learning about both roots, and then they think that the radical sign ALWAYS means both. That's just not the default case.
There is a big difference between :
- the function square root, which by definition of all functions has one and only one value where it is defined, eg sqrt(9)=3, and sqrt(-9) is not valid and has no precise meaning.
- the solutions of the equation x^2=-9, which has two solutions 3i and -3i.
Well defined maths have no logical flaw, you just need to apply right definitions to each concept.
I agree-but I do find it ironically humorous that “well-defined math” doesn’t even have a well-defined spelling. (“math” vs. “maths”) 😂
@@verkuilb Glad to be better in Math than in English 🙂
Only one mistake is still very good, as English is not my mother language 🙂
Sometime you should think about assumptions before being ironic 😞
@@StephTBM4yeah he is one of those people who will see that his/her son scoring 95 percent in exams and still complain that spelling of blah blah is wrong in the report card instead on focusing on the result😂😂😂😂
@@verkuilb Both are an abbreviation of "mathematics."
@@verkuilb"maths" is used in UK, Australia, etc...
This is a question of definitions, and I do not feel that the definition given here for the square root of a number has been sufficiently justified. Yes, you CAN define the square root as a function that gives back a number with a positive/zero imaginary component, but... why? What reason do we have to discard the other root, if different? Why do we need to find only one answer to the problem given? Isn't it better to find ALL answers to the problem?
In the case of Brilliant's question, the choices given make clear that only one answer is desired, and IN THAT CASE, I can get behind -6 being "the correct answer." But the poll question implies, through the presence of the plus-or-minus 6 option, that having multiple answers is valid, and as such, multiple answers should be accepted (if multiple answers exist, which in this case they do).
One can define square root of x as being "whatever number is such that root(x)×root(x) = x." So root(-1)×root(-1)=1 would not be correct because it violates this definition of root. Technically, there are two different complex numbers that, when multiplied by itself, result in -1, and they can booth be considered root(-1), but "root(x)×root(x) = x" implies that root(x) should be a _consistent_ value, so it can be one or the other, but you shouldn't substitute two different values into the two instances of root(x). This is a similar situation: root(-4)×root(-9), this simplifies to 2×3×root(-1)×root(-1). The root(-1) can be either i or -i, and either case gives you -6. One only gets 6 if you substitute i for one and -i for the other.
@@stevenfallinge7149 Okay, yes, that is a way that square roots can be defined, but we still haven't established WHY we should use such a definition, as opposed to any alternatives. We could just as easily define it as "y = sqrt(x) if and only if y^2 = x". Which does NOT require there be only one value for y.
@@NettoTakashi Without supposing sqrt(x) is uniquely defined, it should at least be reasonable to suppose that if sqrt(x) appears twice in the same equation, then it should refer to the same value. So sqrt(x)×sqrt(x) will not equal -x for any x besides 0, no matter which root is chosen.
@@NettoTakashi If sqrt(x) takes on multiple values, then it is not a function, and many familiar manipulations we routinely perform with it are no longer possible. For example, if we subtract sqrt(x) from both sides of the equation, it will not cancel with itself, because we don't know whether they're the same value of sqrt(x) or different values. This makes algebra functionally impossible.
It is always possible to express the "I want both roots" sense anyway, by writing something like x² = 9 instead of writing sqrt(9). But the trick is, you need to introduce a variable to represent the unknown value, and once you have a variable, it is routine and familiar to find that there are multiple solutions. We do not normally encounter this behavior when we write things like sqrt(9) - instead, given some function f, we expect that f(9) is exactly one number (or is undefined).
Its not the definition of square root that is up to interpretation here, "square root" only has one definition, but rather its the application of the radical symbol √ and how it is used in context. While it is almost never explicitly stated as such, the radical symbol refers exclusively to the principal square root of a number, therefore it would be incorrect to say that √4 = -2 or √4 = ±2, even though you would say that 2 and -2 are the two square roots of 4. To make this point more clear, consider the square roots of non-perfect squares. You would never say √2 can be positive or negative, it doesn't exist in some sort of numerical superposition, √2 is a single number approximately equal to 1.414. The square roots of 2 are √2 and -√2. Notice how the negative is used in relation to the radical symbol there. If we want to refer to 2's negative square root, the root approximately equal to -1.414, we use a negative sign AND a radical sign in that specific order to make -√2, because √2 is defined as inherently positive. Thus, the problem √-4 · √-9 = ? is solved by finding the principal square roots of -4 and -9, which are 2i and 3i, respectively, then finding their product which is -6.
Although if one were to ask this question verbally like "what is the product of the square root of -4 and the square root of -9?" Then it is perfectly reasonable to assume one is allowed to use both the principal square root or its "negative" counterpart to produce ±6. I think a big part of the issue here is that we in math typically ask the question "what is THE square root of x?" because its believed to be simpler for kids to grasp the concept if you teach the principal square root first as THE square root and then introduce the "negative" one later. Which unless the number is zero, there is no "THE" square root, every nonzero number has exactly two square roots (unless you believe in hyper-complex numbers i.e. unless you're completely unhinged) its linguistically incorrect and both unclear and up to interpretation to just say THE square root.
Note: I put "negative" in quotes because you don't define a complex number as positive or negative, -2i is not a negative number, but I'm not sure if there is a term that refers to the non-principal square root of a complex number.
You explanations for "undefined" and "+/-6" are at odds with each other
I mean, it depends on the agreement on what branch you can take for sqrt, you could even take it as "multivalued function"
However, given no context we should take the principal branch, probably the test it was taken assumed principal branches too...
5:05 "imaginary numbers are real" this breaks my brain
3:14 [citation needed]
I think the only basis to say that this works only when x and y are greater than 0 is "we aren't aware of a generalized operation like that". That doesn't mean, however, that it can't be done.
Given that imaginary numbers came out of the effort to solve a geometry problem, it might be worth analyzing this question as a geometry problem.
81 = 9*9 = -9*-9
sqrt(81) = sqrt(9)*sqrt(9) = sqrt(-9)*sqrt(-9)
or perhaps someone can explain why this doesn't work with something other than "because it doesn't"
You'd be looking for a proof that sqrt(x) distributes under multiplication for non-negative numbers. Sometimes properties like that hold only for a limited domain and this is one of them.
@@isHavvycounter point: the replacement shown in the video of √-9 = i√9 skips the steps in the middle of √-9 = √(-1)(9) = √-1√9 = i√9 which uses the same sorting step in the middle. Sorry for wonky phone symbols
Oh i got it. Creates a logical problem. Take your example and reduce √-9 to i√9.
I now have a line saying √9√9 = -√9√9. Or if you take it as a variable X=-X.
Holds up for X OR Y is negative but not both. But try explaining that detail to kids at school
3:26 Well, since the rule is in the form of an implication, you can't really say the conclusion of (erroneously applied) implication is definitely false just because the premise of the said implication isn't true. You simply can't decide.
I would start by simplifying: The square root of (a*b*b) should be b*sqrt(a), so this would become 2*sqrt(-1)*3*sqrt(-1) -> 6*(-1) -> -6.
OTOH, the sqrt(4) could be either +2 or -2, with +2 being the primary solution, but (-2)*(-2) -> 4 as well.
this is why i don’t like how roots of even numbers are ONLY positive. it just throws various rules out the window. it’s like mathematicians were like “how do we do something that seems reasonable, but actually ends up being annoying?”
like you could solve it like:
sqrt(-4)sqrt(-9)=x
(-4)(-9)=x^2
36=x^2
and then wait a minute. you don’t actually take the square root of 36. what you do is:
+-sqrt(36)=x
and THEN you have to check both values of x to see if they’re valid
Sounds like the 5*3 debate for highschool students.
The answer's 53, obviously :D
but is it 5 + 5 + 5 or 3 + 3 + 3 + 3 + 3? /s
@hampustoft2221 it's what ur taught at the time u take the q?
It's 5*3, just do it like one problem 5*3 is 15
Addition is too slow, what are you gonna do when it's 99*768
@PopeVancis expecting a grade 1-3(7-9yr old) to compute arithmetic above 50......have adults forgotten this is about learning ...esp mathematics semantics....
I multiplied the roots and ended up with the answer 6, but I was unaware that the multiplication rule was only valid for positive numbers.
Yeah, I don't believe that limitation was ever taught in any class I've taken. I was under the impression that (x^k)*(y^k) always equaled (xy)^k. If it doesn't apply for the specific case of (x< 0 | y < 0) & (-1 < k < 1), that's very strange. I'm doing some cursory googling and can't find it mentioned anywhere.
Its assumed that
> Sqrt(x)^2 = x
> Sqrt(x)*Sqrt(x) = x = Sqrt(x)^2
Now, if we were to assume that sqrt(x)*sqrt(y) = sqrt(xy) for ALL numbers, we reach this logical contradiction:
Sqrt(-9) * Sqrt(-9) =
Sqrt(81) =
9 (or maybe ±9, but still not -9, so we contradict that a sqrt of x times itself should give x)
@@HeroOfHyla I think every competent high school level course on the topic teaches that. People just tend to miss and forget things. There are even famous -1 = 1 "proofs" to demonstrate how it does not hold generally.
I am also very surprised that you can't find this information online. For example, it is clearly stated e.g. in the Wikipedia article on the square root, under "Properties and uses":
"For all _nonnegative_ real numbers x and y, √x√y = √(xy)."
Similar limitations can be found in the article about exponents, under "Rational exponents", just to name a few sources immediately found by Google.
I asked my maths teacher “how many grooves are there on a record?”, he replied “how should I know?” I told him there is only one groove on each side.
I guess by pemdas, sqrt comes before multiplication, meaning -6 is technically correct, but taking x=sqrt(-4)(sqrt-9), x^2=sqrt(-4)^2(sqrt-9)^2, x^2=-4*-9, x= + or - 6 but it's all about definitions
x=6 is an extraneous solution created by the initial squaring step.
As stated, the problem is ambiguous. It all hinges on which square root you use. The interpretation is for the principal square root.
±6 is objectively correct, because there's no such thing as a principal square root in complex numbers. There is n sollutions to n-th root of any complex number and there's absolutely no reason why we should pick one instead of the other
Amen.
I agree 100%!
If there is no such thing as a principal square root in complex numbers, why is "i" defined as only +sqrt(-1) and not positive and negative? If it were positive or negative it wouldn't be unique and would vastly change results. Complex numbers are two-parters which makes them a dot on a field and not a position on a string. If you didn't have exact coordinates on a field, you'd end up in the wrong place. That's also why they are vastly different to the remaining number definitions.
Imagine it like a geo coordinate. If you try to navigate to europe, e.g. longitude 7, latitude 47, you better hope they are defined as positive. If you miss it, you might end up in the Ivory Coast or somewhere in the middle of the atlantic ocean.
That is incorrect. You are not solving an equation, you are applying a function. There can only be one answer.
@@migga86 I think i = sqrt(-1) is a bad definition, my teachers agree on this. A better definition is i^2 = -1
Whomever it was that chose the words “real”, “imaginary”, “rational”, “irrational”, etc., to describe those types of numbers, did a HUGE disservice to mathematicians for the rest of history. How can we convince non-mathematicians of the fundamental correctness of our results, when we tell them that we arrived at our conclusions using “imaginary” and “irrational” numbers?? Or that when you multiply one “imaginary” number by another “imaginary” number, you get a result which is somehow not imaginary??
I think "imaginary" was a term coined by those trying to discredit their use to solve quadratics and cubics
I don't think these terms could be any better, they are inconvenient for non-mathematicians sure but they all have a use In their wordings
Every new type of number gets a pejorative name: "negative", "fractional (Latin for 'broken')", "irrational", "imaginary", because the previous generation of mathematicians are uncomfortable with it.
Those successively less useful "numbers" get more and more dismissive labels because they are further and further removed from what numbers are and how numbers behave. It's not pejorative, it's descriptive.
Whoever *
There's no such word as Whomever even the objective case.
It's not even imaginary.
@@neuralwarp If you want to be pedantic, there is no objective case: its the accusative case
I would have said -6, but after watching the video I think the people who said +/-6 are more accurate.
You were right the first time. Those symbols will never be equal to positive 6.
no, they're not more accurate 😂
Based on reading the comments, I understand now: we just define √ to mean "the positive square root". I am unsure if x^(0.5) is also supposed to mean the positive square root only. So the square root SYMBOL means only the positive root: if I'm not messing up the terminology, the square root OPERATION, when applied as a transformation, can result in both positive and negative roots.
So, if x^2 = 7, then we can apply the square root OPERATION and end up with,
"x = √7, or x = -√7".
As for not being able to combine (-4)^0.5 * (-9)^0.5 into (36)^0.5, I just refer to Wikipedia on Exponentiation, section Identities and properties, which says that this works "for all integer exponents, provided that the base is non-zero". This restriction to integer exponents might be as fundamental and difficult to prove as 1+1=2, so I'll just accept it.
If you absolutely want √x to mean something for negative numbers, yes, you can choose to give it a value, such as i√(-x). That's totally possible, in mathematics you can define symbols however you want. The US made this choice. But this causes problems like √a√b not always being equal to √(ab), and it has no real benefit when you can just write the transparent and unproblematic i√(|x|) instead. That's why other countries like France decided to make it undefined.
I answered -6 on the poll. No other answer even occurred to me because my background is in physical chemistry and physical chemists deal with wave functions which assign a physical meaning to the imaginary number.
'I believe this is the correct answer' is probably the last phrase you want to hear from the author of a math channel on a simple math question.
@@akasyan He's just being modest and polite. The (only) correct answer is -6 and the people saying it should be ±6 are wrong.
Idk, the deeper you get into math, the more you realize how much ambiguity there is based on what assumptions you bring to a math problem.
It would be, if it weren't for the fact that the rest of the video is about finding why people might believe the other answers are right and showing why they are wrong.
+6 and -6 both are correct according to the usual definition of the radical sign meaning. For a real positive number its square root is the positive one, but for a complex number the square root of z is defined as the solution of the equation x^2 = z. In a similar way, the cubic root of -1 is the solution of x^3 = -1, i.e. the set ( e^(i*pi/3) , -1, e^(-i*pi/3) )
@angeljimenez3362 I don't think so. The square root symbol is meant to refer to the principal value in all cases. For a negative number, it is the one with positive imaginary part.
Mathematician here.
Complex numbers are based on the imaginary unit i, which is defined by i^2 = -1. If you, on the other hand, define i = sqrt(-1), you get into trouble: i^2 = sqrt(-1) * sqrt(-1) = sqrt(-1 * -1) = sqrt(1) = 1. That's incorrect! So suddently there are rules about sqrt() and other functions that you cannot apply.
Rather than trying to memorize what you can and cannot do, it's better to never write sqrt(-k) with some positive integer k in the first place. Complex numbers don't allow you to write sqrt(-5), they simply give you i, the number when squared gives -1. That's it, and it turns out that's enough, too, to do all the complex magic.
Since complex numbers DO allow -5 to have square roots, it seems odd to conclude that they don't allow you to write √(-5).
I know that you can write it as i√5, but your argument seems to say that being able to write √(-5) is redundant, not that it's prohibited.
@@gavindeane3670 I see where you're coming from. The complex square root function does indeed exist, but is very tricky! For one, it's multi-valued (exept for z=0), so sqrt(-5) isn't just "a number", but can be one of two (sqrt(5)i and -sqrt(5)i in this case). Likewise, sqrt(-1) is i and -i. It's not useful to do any calculation with that. Would you agree?
@@nschloeObviously numbers have two square roots. But the principal root is defined for complex numbers (it wouldn't need to be called "principal" if it was only defined for real numbers - we could just call it the positive one) so I don't see how √(-5) is problematic notation.
I can see how there would be a problem if the notation √(-5) was ambiguous as to whether it meant 5i or -5i, but that's no different to the problem we'd have if √4 was ambiguous as to whether it meant -2 or 2.
We invented all these squiggles and shapes that we call "mathematical notation" so we can choose what it means.
@@gavindeane3670 no, non-negative numbers only have one square root. Quadratic equations however have 2 solutions.
And that is what I think is the greatest misconception. Even Presh here goed from calculating square roots to solving quadratic equations.
Jumping from calculating √-4 to solving the equation x² = -4 does not justify the use of i.
@@gavindeane3670no, it doesn't allow that. Complex numbers are used to solve equations like x² = - 5 which results in x = +/- i√5. So 2 answers/solutions/roots.
The answer undefined is fine when the one who answer is at pre-calculus class.
Nope, at pre-calculus you learn that √x*√x=x and hopefully that √x*√y=√xy.
I'm under the impression that the result of a function can also be a set of values, rather than a singular one. So I prefer the consideration of all roots, rather than just the principal root.
I'm perfectly happy with ±6 as an answer. I have no reason to assume the root here is a principal root, as the question has nothing to do with functions, so I don't have to handwave away half the answers bc they're inconvenient.
If square roots have multiple answers by default we would not write +- on every quadratic equation
Thats the only argument for roots not giving out multiple solutions, and even that is flawed because theres no standard way to write formulas
@@Philip-qq7qlOf course there is. Some of them are even CALLED "Standard Form" (like the standard form for the equation of a circle).
square root is defined to only have one output so everyone who says +- is wrong. that only happens with the absolute value operation and a lot of people like to pretend theres an absolute value around square roots when there shouldn't be.
When you're using the quadratic equation you're essentially completing the square. And in completing the square you need to take a square root of a variable squared. A variable squared has a value that could be traced back to two original values for that variable. We use the plus minus because we are *dismantling* something that could be either option.
@@Philip-qq7ql That isn't a flawed argument though, since the square root function is specifically defined that way, the reason being you want to be able to refer to only the positive or only the negative numbers that when squared equal another number as opposed to always referring to both. That said, we could have created a different notation to specify when a square root was positive or negative and kept the base square root as implying both, but it just didn't turn out that way.
Have you ever seen a number? Or have you only seen the symbol representing the idea?
Have you ever seen an object, or have you only seen light reflecting off of the object?
Have you ever seen a thing? Or have you only seen one particular object considered to be a thing?
Numbers are definitions, such as "thing" "word" "thoughts" "happiness"
The root of the trouble is that there is no way to distinguish i from -i.
But they are distinguishable. For instance, the principle value of the argument of i is pi/2, while for -i it is -pi/2. Or, -i is a root of the polynomial z+i, while i is not. Etc.
@@johnreid5321 The principal value of i being pi/2 simply means that exp(i.pi/2) = i. But since exp(-i.pi/2) = -i, that fails to distinguish i from -i: that is, if you replace all occurrences of i by -i you get the same thing. Similarly -i's being a zero of the polynomial z+i: replace i by -i throughout, and get the equally true statement that -(-i) is a zero of z+(-i). Again i and -i can be switched and true statements come out true again, and once more nothing serves to distinguish i from -i. (But you have to switch ALL occurrences of i of course)
Well then, how is 1 distinguisble from -1?
@@russellsharpe288Seems as though you have put it through a specific case though. Would this not be the same as saying because cos(0) = cos(2*pi), 0 and pi are indistinguishable? As terms at their face value i is the sqrt(-1) and -i is -sqrt(-1).
@@johnreid5321 1 is defined as the multiplicative identity. It is easy to show that there can be only one of those (if a,b were two of them then we would have, just using the definition, a = ab = b)
-1 is defined as the additive inverse of 1 ie that number which when added to 1 gives the additive identity 0. Again it is easy to show there can only be one (if a,b were two then a = a + 0 = a + (1+b) = (a+1) + b = 0 + b = b. (This uses associativity of addition of course)
Could we have 1 = -1? We can in mathematical structures known as fields of characteristic 2, because there 1 + 1 = 0. But for the standard number systems we are all familiar with (Q,R,C...) 1 and -1 are very different.
Look at it this way: imaginary numbers are first introduced something like this: we are going to introduce a new number which is a solution to x²+1 = 0 and we are going to call it i. Very quickly it turns out that there is then another solution to x²+1 = 0, namely -i. But how do we know the original solution we introduced was not in fact -i and not i after all? Obviously just by prescription or convention: we have simply *named* the first one i and then the other one must be -i (rather than the other way around). But if somebody else goes through the same procedure, how do we know that what we are calling i is not what they are calling -i ie how do we know that we and they have originally picked the "same" solution to x²+1 = 0? This is pretty clearly a fake question, just because it doesn't matter which one is meant when we label one i and the other -i. And this is because the situation is perfectly symmetrical: there is nothing to distinguish i from -i at all, other than an arbitrary initial act of naming. The only defining property these numbers have is that they are solution of x²+1 = 0, and everything works perfectly fine if we swap all occurrences of i with -i and vice versa. So pretty clearly there is not going to be a way to distinguish i from -i: this is guaranteed by the very way they have been introduced (and indeed it could not be otherwise).
for real number, you can choose the principal square root because IR have an order, so you say sqrt(p) is the greater or equal to 0 solution of x^2=p
but for complexe number, there is no order, so no good way to choose a principal square root, so there is 2 square root in the complexe set
last argument : we all know that 1 = e^i2kpi where k is an integer
by the law of exponent, sqrt(1) = e^ikpi with k is an integer so it’s both 1 and -1
Finally someone in the comments who gets the difference between a mathematical object (the complex field) and some representation of that object using non-unique (re+im) components
> but for complexe number, there is no order, so no good way to choose a principal square root
That's just false. Principal square root is perfectly defined for all C. And √ symbol defines specifically principal square root. There is no room for different answers in this problem.
@@lerarosalene Not so simple. I think your confusion arises from the way complex numbers are introduced in highschool and engineering classes as "pair of two real numbers". But those pairs are NOT complex numbers, those are just some representation of the complex number field that you can do calculations on. Now of course one can define a principal square root as an operation on your specific representation, which may be handy for some engineering problems or whatever. But that definition is non-mathematical. The reason is that you can map Re and In in many different ways onto the complex field and get the same mathematics back (but your number pairs would look very differently). In fact, you cannot even tell the difference between i and -i by any equation involving field operations only. And don't say that i is the squareroot of -1, because that would be cyclic reasoning.
Who told you that??? Principal root is perfectly well defined for complex numbers.
@@valentinziegler1649 stop smoking whatever you are smoking. √ symbol is defined to be principal square root and principal square root is also precisely defined. This whole problem is about notation and people like you not understanding it.
So the equation could be resolved and expressed as 6*e^(π+2kπ) with k: 0;1 considering the solutions of the square roots synchronized which both result on -6 .
Or you could consider the roots being able to be out of sync which leaves you with the equation as 6*e^j[(π+2kπ)/2 + (π+2hπ)/2] with k,h: 0,0; 0,1; 1,0; 1,1 and then you'll have 4 solutions, 2 being -6 and the other 2 being 6.
I am on the -6 camp never understood why 7:00 - 7:22, finally someone explained to me why only positive roots are used. My HS teacher only told me "just its the rule" but didnt expound more. Thanks!
Edit: thought you were asking someone to explain, I sort of skipped that part of the video. Sorry!
Original comment: Based on reading the comments, I understand now: we just define √ to mean "the positive square root". I am unsure if x^(0.5) is also supposed to mean the positive square root only. So the square root SYMBOL means only the positive root: if I'm not messing up the terminology, the square root OPERATION, when applied as a transformation, can result in both positive and negative roots.
So, if x^2 = 7, then we can apply the square root OPERATION and end up with,
"x = √7, or x = -√7".
I don’t fully buy the argument against undefined (although my answer is -6). First of all, x usually denotes a real variable (compare to z). Second, not all problems imply that you should extend the domain, for example if I say “can you factor x^2 + 1” then “no” seems like the answer you would usually expect unless you had already established that you are working over C.
Exactly, that argument doesn't even make sense. It just reinforces that we're always talking about the reals unless otherwise specified.
But even then the correct answer would be to specify "if real: undefined" which would be correct.
You are doing a disservice to yourself by assuming properties without writing them down.
Then when validating your work nobody can know why you got the answers wrong (because not showing which assumptions you made).
06:39 The information "sqrt(x) is not a function!" is wrong.
sqrt(x) IS A FUNCTION.
The graph show at 06:39 is NOT a graph of sqrt(x).
It is a graph of+-sqrt(x).
And this +-sqrt(x) is NOT a function.
I hope that we get a clarification / correction on this comment, I think you are right.
He isn't saying that the accepted definition of sqrt(x) is not a function, nor is he claiming the original graph (pre-pruning) to be a graph of the acceptedly-defined sqrt(x). He is saying that if instead sqrt(x) was alternatively defined to included both possible roots, and not just the principal root, then the alternatively-defined sqrt(x) would not be a function.
@@nwoDekaTsyawlA By convention, the √ symbol defines a single-valued function. √a refers only to the principal branch solution of the equation x^2=a.
By this convention, √(-4)√(-9) has a single value of -6. This would not apply to other ways of writing. (-4)^0.5 * (-9)^0.5 = ±6
@oliviervancantfort5327 I agree with everything you said. I don't agree with the image at 06:39 stating "sqrt(x) is not a function".
It's just not an invertible function. :)
I started as a -6 answer myself. But going through the ±6 answer actually convinced me that ±6 is the actual, correct answer.
When taking the square root of -4 as a complex number then both 2i and -2i are the two possible values and as far as I can tell, equally valid. So why is -2i discounted? If we allow both square roots for -4 and -9 then +/-6 should be correct.
As well, 2i is not a positive number so how does it get the status of "principle"?
The issue here is, that (-1) is also (-i)^2, you said it yourself, it is a limit you are imposing upon yourself…
That is not the issue. The issue is much more complicated. If you take your domain as complex numbers for the relation f(z) = sqrt(z) it is multivalued. To make it a nice function you restrict the domain and that restriction tells that -1 has an unique square root and it is i. If you need more answers you are talking about roots of unity and that is totally different thing than just taking a square root.
you're not cooking here, bro. it's literally a principle as to why the branch cutting is necessary
@@SkegAudio Because it isn't a function if you don't do that. It's double valued relation and generally n:th root is n-valued relation and as function you take the first branch.
If you want to study the double valued relation it's done by forgetting the 2 and just think how n-valued root handles. And for that you only need to know how the roots of unity work. Remember that any complex number is ae^bi, where a and b are any two real numbers. Then we can discard the a as it is just a multiplicative factor and focus on the e^bi and then n-roots of that are the roots of unity offsetted by e^(bi/n).
@@topilinkala1594 Look while OP is indicating that complex numbers have n nth roots, the √ symbol itself has a specific conventional meaning in complex analysis - it denotes the principal square root. When we write √(-4), we're specifically referring to 2i, not ±2i. It's similar to how √4 means +2, not ±2, even though both ±2 are solutions to x²=4.
If we wanted to indicate both possible roots, we'd need to explicitly write ±√(-4)⋅±√(-9). The √ notation alone has a well-defined meaning that gives us one specific result: (2i)(3i) = -6.
You're absolutely right that both roots exist mathematically! But the question is about what the specific notation √(-4)√(-9) evaluates to, given standard mathematical conventions.
The actual thing here is, that the task solving sqrt(-4) within complex numbers can easily be rewritten to sqrt((i^2)*4), which then translates to | i | * sqrt(4)
I was learned in school that i has definition i²=-1 and √(negative number) is not defined
The problem is one of definition, how do you choose which root is the principal root of a number? Easy to do for the positive reals, but why choose i as root(-1) and not -i? And what about root(i) or root(-i) or any other complex number, which of the two "roots" count as the principal one? Choosing arbitrarily is fine, of course, but there would be a continuity problem in the resulting function (from C to C). This is typically why I try to avoid using the square root symbol with negative or complex numbers, to avoid ambiguity.
Exactly why it is undefined, even given complex numbers
The radical symbol √ is defined as a positive number by convention. It may be an arbitrary convention, but it's still a convention.
@@thenonsequitur If you insist on that strict definition, then √(-1) simply does not exist.
@@TonyFisher-lo8hh I didn't give a strict definition. Why would you assume that I did? Obviously I was talking about an implied range of positive real numbers.
A more technical definition of √x is the principle square root of x.
When x is a positive real number, the principle square root of x is defined to be the positive square root.
When x is -1, the principle square root is defined to be the complex number "i". So √(-1) does exist.
@@thenonsequitur It's not a convention, because it's different here.
It may be the convention in the USA, but it's not so in the rest of the world.
I do love the argument about the domain... Don't limit the range of 'x' in f(x), but do limit the number of f(x) used to only the principal root function.
I come from a country where conventions were maybe a little different.
when writing the Radical sign ("Square root sign"), the convention is that you work in R.
If you want to work in C, we normally write ^(1/2)
And to avoid any confusion, always, always write the definition domain. I assume that if this is an interview, the very first thing to do is clarify the definition domain. The last thing you want is to be "right" but misunderstood by half of your co-workers, or by authorities.
I solved it as -6 but I have an argument for undefined which you didn’t mention. One can say it’s undefined as there was no domain of x mentioned so I can not be sure what was meant. The reason for this is that most of the time real numbers are used as domain and that is the reason for fair assumption that this is also the case.
Yes, but we can't exclude imaginary numbers from the domain of the answer because the question contains them (sqrt(-4) and sqrt(-9)).
No real solutions =/= undefined
Exactly. I stopped watching at the half-baked integer mathematics analogy. The domain is the real numbers unless otherwise specified. If this question came up in a textbook in a chapter about complex numbers I wouldn't expect them to state it again in each problem.. but in isolation it's intentionally ambiguous about what domain to use -- there is no right answer because you can't know which of the two right answers it wants.
@@seedmole I disagree, the domain is all numbers unless otherwise specified. Without that assumption the fundamental theorem of algebra breaks. Unless there is something that specifically prohibits them. The characteristic equations of second order differential equations wind up with complex roots even if you can use them to work your way to purely real valued functions.
That makes no sense. If you know about complex numbers then you know how to evaluate this expression. You do not need the author's permission to use that knowledge.
at 3:19 you state that the product of 2 roots only applies if the numbers under the root are positive. why?
because them being the root of a negative number changes how the maths works
it is more that the product of 2 roots is more a guide and a quick shortcut that can be used under some circumstances rather than a good rule
Pretty much just to avoid this specific problem having the answer 6. If you're using the version of square root with both branches present then the rule is fine, if you use the one with only one branch, sometimes the rule gives you the other branch. Since 6 would be the positive and thus correct branch for sqrt(36) but not for sqrt(-4)*sqrt(-9).
@@Stereomoo It would also end up working if there was only one negative number, TBF.
Like
radical(-9)*radical(4) = 3i*2 = 6i
And
radical(-36) = 6i
Same answer.
It's only deceptive when there are multiple negatives, because you'll end up on the non-principal branch.
Actually, going to more than 2 numbers...
Like, 4 negatives and 5 negatives both work. i^4 = 1, i^5 = i, both on the principal path if you multiplied everything together.
0,1 work; 2,3 don't; 4,5 work; 6,7 don't; etc.
@@durandle9226 in my limited knowledge of pure math, i consider the whole explanation unnecessarily complicated, because sqrt(-4)*sqrt(-9) could be simplified to sqrt ((-4)*(-9)) which simplifies to sqrt (36) = 6
Once you leave the number line and venture into the complex plane, the world changes.
So if we were given the sqrt(8 - 6i) instead of sqrt(-4) then we would get the 2 values: 3 - i and -3 + i. Which one should we pick?
According to the definition of principal square root of a complex number (en.m.wikipedia.org/wiki/Square_root) the answer is 3-i
Basically it's an agreement that the square root of a number is _always_ positive; for the x² = c, the answer is plus-minus root c, which means that root c must have a fixed sign.
My initial thought was -6, however after hearing you explain why +/-6 is not correct, I am now of the opinion that +/-6 is the correct answer.
Imaginary numbers are a real (mathematical) thing, they are not real numbers.
I won't lie, at first I thought maybe 6, but realized you can't square root negative numbers. It must be imaginary!
That's what I was thinking
Both terms were imaginary, but if you multiply two imaginary numbers you get a real.
@@captsorghum Ah, you're right!
I think Presh's observation that the (±2i)(±3i) camp would be the hardest to convince is exceptionally astute, as I understand but continue to disagree with his belief that the principal square root bears on the evaluation of this expression. Given the PEMDAS stipulation that exponentiation supersedes multiplication, the ± has to be produced twice and therefore winds up in the final answer, as well. Denying half the complete answer in order to solve an irrelevant manufactured issue (the not-a-function issue) is logically fallacious. The expression has two valid values, not one. // Do I sound terribly opinionated? Well, I probably am -- except for the *terribly*. Cheers. :)
@@jim55price the expression has one value. ALL arithmetic expressions involving addition, subtraction, multiplication, division and powers (including the √ symbol) ALWAYS produce a single value. Even in the context of complex arithmetic no simple expressions ever has more than a single value. There is no equasion here. There are no functions. √-4*√-9 is a simple arithmatic expression with a SINGLE value of -6.
@@nbooth Its not just arithmetic operators, _all functions_ should return a single value, since by definition a function is a one-to-one (or many-to-one, but never one-to-many) mapping between two sets
@@nbooth Your comment is preposterous, given that far simpler expressions, e.g. ±6, ±x, ±i, have two values. I've no idea what you're going on about with your claims of "ALL" and "ALWAYS". All it takes to illustrate your error is a single "±".
@jim55price yes the ± symbol allows you to write two values with a single expression. Congratulations. {-1, 1} is also an expression with more than a single value and [0, 1) is uncountably many.
Those aren't arithmetic operations however. They're ways of writing sets, which is what the ± symbol does as well.
Raising a number to a power is an arithmatic operation that produces a SINGLE value. That's why we need the ± symbol in the first place.
@@jim55price you didn't read what I wrote. I said all expressions involving the following operations (+, -, /, *, ^) yield a single value and you're claiming some other operation is a counterexample to that.
Regardless of what the correct answer is for this question, it is important to remember that the exponent x^a y^a = (xy)^a only applies for positive real numbers *x* and *y* and real exponents *a* where here, a=1/2.
Also, sqrt has a meaning only for pos real numbers but in marvel universe everything is possible...
@@pelasgeuspelasgeus4634 Usually the square root of any complex number is defined as x^(1/2) = e^(1/2 * log(x)) but the important part is choosing which value of log to use.
I was of the opinion that it could be +/-6, but having looked up mathematical notation, the √ sign implicitly means that one should take only the principal root. It's a matter of convention. I don't think the explanation involving functions was convincing to me (or others in the comments I have noticed), it was probably better to just say that the notation implies we should discard the non-principal roots.
Trick questions are designed to be traps, which is why they aren't really useful as means to evaluate or teach anything. Education requires honesty.
• When dealing with square roots of negative numbers in the complex number system, each square root operation returns a single principal value.
• The principal square roots of -4 and -9 are 2i and 3i, respectively.
• Multiplying these roots results in 6i^2, which simplifies to -6. Therefore, in the complex number system, the answer to -4 x -9 is -6, not ±6.
There should be no ambiguity here.
who decided complex numbers must be principal roots? If so, would (-1)^(1/2) be i, or +-i? I really don't think we should always ONLY talk about principal roots. Now if the square root *symbol* is about principle, then that should have been mentioned.
P.S. not to mention -i^2 is in fact -1
@entityredstoneonyt to express the fact that the principal square root of 9 is 3, we write √9=3. X²=9 is a different thing. The same applies to negative numbers. That's what I've learned in school, and that's what wikipedia says.
@@entityredstoneonytthat’s like saying “who decided that 5^-1 = 1/5?”
We did, as a collective over thousands of years.
@@alinzmeul Maybe that's what you do. It's not what we do.
For us √9 = ±3
Seems there is a difference in definitions between countries.
I feel like there should be a new mathematical function where it returns not just the principal root but all the roots:
(I'm not a mathematician so the following might sound a bit wierd)
let root(a,b) be all solutions to x^b = a
For example: root(25, 2) = {5, -5} because 5^2 = (-5)^2 = 25.
Now we can find all "solutions" to (sqrt(-4))(sqrt(-9)).
root(-4, 2) = {2i, (-2)i}
root(-9, 2) = {3i, (-3)i}
Multiply them together means take all the possible combinations and multiply them: {-6, 6, 6, -6}.
Remove the duplicates, we get: {-6, 6}
Yeah, that's correct. That's the way it is normally done in algebra. You simply cannot choose a good way to pick just one of nth complex roots.
On the other hand, if √4̅ = ±2 = {2; −2}, then for √4̅ + √4̅ we have four options:
• 2 + 2 = 4
• 2 + (−2) = 0
• (−2) + 2 = 0
• (−2) + (−2) = −4
So it turns out that there are three different possible outcomes for a simple sum, and also √4̅ + √4̅ is no longer equal to 2√4̅ = ±4 ¯\_(ツ)_/¯
@@cyberaguaoh yikes that’s something I’d never considered. If you expand the definition of √x to be the set {+√x, -√x}, then yeah √x + √x ≠ 2√x. That’s bound to mess some stuff up…
@@cyberagua I guess it only makes sense if you multiply or (?)divide.
@@121DSpCe-Tile In fact, it makes sense when solving irrational equations over the field of complex numbers ℂ - there you have to take into account all possible combinations of the root values.
The problem is that you considered i as sqrt(-1), it ain't. i is the number such as i² is -1, while the answer for sqrt(-1) is i and -i , both branches should always be considered, unless the question amde has a context and only one branch makes sense.
is nice that you talked about functions and all, but this isn't a question about functions, is a question about an equation. I don't know how you gringos learn equations, but when i did in primary school, if we had to answer a problem of the kind: "what are the roots of 3x² + 4 - 2" we would solve 3x² + 4 - 2 = 0 and we knew we could put that on the in the square equation formula (A.K.A. Bhaskara) and we could find up to two roots, unless they coincided or one or both couldn't be calculated in real numbers (it was primary school after all).
So, it is + or - 6.
when we looks at sqrt(-4)*sqrt(-9) we immediately know that we could have four answers because each square root can have two answers and we know that two of hose answers would coincide in the same number, and the other two as well therefore we can have up to two unique answers.
then when we expand that to sqrt(-1)*sqrt(+4)*sqrt(-1)*sqrt(+9) we know this expression, written as it is can give us eight answers and we know we will have, at most, two unique answers because we know that (+or-)i*(+or-)2*(+or-)i*(+or-)9 have eight anwers.
Here, have a *±* to cut and paste.
@@noname_atall there is no equasion here! "√-9*√-4" is a simple arithmatic expression. There are no equasions or functions involved. Even in the realm of complex numbers, simple arithmatic expressions have SINGLE values.
But the "√" is _not_ defined as the solution to a quadratic equation. It's defined as the principle square root. This symbol always denotes a single positive number.
Recall the quadratic formula that you referenced. Note that it includes the construction "±√...". If "√" could be either a positive or negative number, why would the quadratic formula have to include the ± here? If it was already implicit in the √ symbol, it wouldn't be a necessary to include it in the formula.
I have a hard time understanding your second sentence : i is the number such as i×i = -1. So by your definition i = -i ?
I was looking for this answer. This is the reason that the answer should be 'undefined'.
The principal value of the root can be expressed and dissuaded from debate by noting that we start with the root already taken. The plus/minus is only necessary to solve when applying the root as a solution to the quadratic equation.
Here, a product of squares is NOT equal to the square of the product. The answer is -6 as you need to use complex numbers.
well, we actually also needed to only think about principle sqrt to arrive at negative 6. Just plain complex numbers also give +-6.
There is no such limit to negative numbers under the root. We all have happily used this with variables, not even wasting a single thought about signs.
The undefined solution is largely given to students learning square roots initially. In order to not overwhelm students, the teachers simply want students to say it is undefined. But, once students learn of complex numbers, negative square roots can be taught in their entirety and should be viewed as the actual correct answer.
many people might see 2 negetives in multiplication and get the answer as +6 but it actually goes like this:
since i=sqrt(-1) we get,
-> [{i*sqrt(4)}{i*sqrt(9)}]
-> (i^2)*(2)*(3)
and since {sqrt(-1)}^2=-1
the answer is -6
-2i squared equals -4, so I can understand why they might say the original problem's answer is plus or minus 6. I confirmed this approach by trying it with apples, and ended up with plus or minus 6 apples, which did kind of mess up my bookkeeping, to say nothing of determining how much space is left in my refrigerator!
Even if we extend the principal branch of the square root to complex numbers, it is usually not defined on a ray. That ray is typically negative real numbers. If we extend it continuously to negative numbers (and remove another ray instead), a choice must be made as to whether sqrt(-1) is i or -i.
Fortunately, for either choice, sqrt(-4) * sqrt(-9) evaluates to -6.
All you need is the definition of the radical symbol. From Wikipedia - The two square roots of a negative number are both imaginary numbers, and the square root symbol refers to the principal square root, the one with a positive imaginary part. Additionally, I don’t think the tree pruning for functions explanation is relevant. There’s nothing in the original question that requires a function. We’re simply asked to evaluate an expression. -6 😎
This is a much better explanation that the one given in the video
I’m a bit closer to agree with the Principal Root argument. However, as a counter example in the quadratic formula +/- is part of it even when the radicand is negative. There is no enough information in the problem to exclude the positive answer. What am I missing?
It’s just the definition of the radical symbol. The two square roots of a negative number are both imaginary numbers, and the square root symbol refers to the principal square root, the one with a positive imaginary part. So when evaluating the expression, we use the positive imaginary part for both radicals and the answer is -6. The answer is very straightforward. The quadratic formula is a different situation. In this case we’re solving an equation (not simply evaluating an expression) and it will have two answers (unless the radicand is 0). Here we want to include both roots, positive and negative, no matter if the radicand is real or imaginary. Since the radical symbol only refers to the positive square root, we must add the plus minus sign to specify both roots and find both solutions.
@@johnhoslett6732 Interesting. So you are saying "Simplify the following term" and "x equals the following term, solve for x" are two different things. But who says that I am not allowed to pick the second option to solve this problem?
@@manfredlemke4671they are different things. If you take a square root of something (call it a) you get two possibilities: ±√a. If you're given a square root √a, that *always* refers to just one of them.
The fact that the square root symbol refers to only the principal square root is similar to the fact that 1 is not a prime number. It might not fit with your world view, but mathematicians have adopted is as a standard because it makes mathematics easier, and you should too.
The problem has nothing to do with the principle value.
Neglecting all but the principle root is a good way to pass the vertical line test. Just like kicking the ball into the hole is a good way to make par. If you were given more than just an equation, some real world situation, for which negative values could not be valid, then you would be justified in ignoring the negative roots. But, this problem did not give details of any particular situation, we don't know that x represents a real value. Since the notation given does not indicate a function ( it is ?, not f(x)) and doesn't say to graph the result, the vertical line test for a function is also irrelevant. So it all comes down to your assumptions.
If you assume a real value function, then you would have undefined, because you would not have i to deal with the negative square. If you assume complex numbers, then it makes sense to use the complex plane for a graph if you want to graph it, in the complex plane using both positive and negative roots is just fine, then you get ±6. You only have the answer -6 if you assume complex numbers but choose to graph the answer in rectangular coordinates as a function of x.
I always used to tell my students that they could use reasonable assumptions, but they must identify the assumptions they make.
In all honesty, ±6 is still incorrect, all complex functions have infinitely many values depending on what branch you evaluate the function on. In the complex plane there is simply isn't really an answer, so the assumption about where branch cuts are and the which branch is being chosen for your function is explicitly stated
I think the assumption of a principal square root is the best assumption to make since that's literally what the radical sign means. Yes, numbers have two square roots, but as soon as you use "√", you're automatically referring to the positive one. It's just what the notation means.
@@ric6611 Unfortunately the radical symbol is in and of itself, ambiguous. As the same radical symbol itself is used for the square-root function over the positive reals as well as the square-root over the entire complex plane.
i.e. f: ℝ⁺ → ℝ⁺ is f(x) = √x, and pertains to the positive square root of a non-negative real number.
while, f: ℂ → ℂ is f(z) = √z, and may pertain to any of the square roots of any complex number, on any branch of the function.
I think personally, the original square root function itself, f: ℝ⁺ → ℝ⁺ is f(x) = √x does not apply here, as the second you put a negative real number in there, we have to expand our definition of f, to that of f: ℂ → ℂ is f(z) = √z, wherein we must define which branch of the square function we are working on, that usually being the principal branch. I think that √-3 itself implies we are talking f(z) here, since if we used f(x), -3 is simply not in the domain of the function.
An important distinction he made is that although a number has more than one square root, the square root function has only one answer which is why we are careful to write +/- in front of the square root symbol in the quadratic formula for example.
They way I like to think about it is the sqrt(x) is really a shorthand for (x)^(1/2). That square root symbol implies powers and parentheses, so you have to do them before you perform any multiplication. Unfortunately, many people are taught in school that sqrt is "fundamental".
Totally agree. It is glad that more than half of the people get it right.
Since square root is multivalue function and we have 2 roots in the original question, we have 4 ways of choosing branches (sqrt(-4) = 2i or = -2i). So my answer was undef. until branches are chosen (+-6 is not proper way to write uncertainty in my opinion)
Upd: Also, your argument about +-6 is ruined by your previous one about undef. Why you has limited the domain of the function to C, why not to choose the Rieaman surface for square root, so both branches could exist simultaneously.
The Square Root function is *NOT* a multi-value function.
@@bigpushing7167 The Square Root function returns a Single Value for this math problem of ... √(-4) . You will learn this.
@@MrSummitvillewdym "for this problem". Math isn't working like this. We need domain to consider whether function is multivalued or not. If it is, we need either specified the leaf on which we are, or use structure called Rieman surface to make a well-defined term "square root".
you don't know the complex analysis doesn't mean complex analysis disappears...
@@MercuriusCh You can do whatever analysts you want. But for this problem ... √-4 = 2i and √-9 = 3i. You can pretend, that you have other answers.
@@MrSummitvilleDifferent areas of mathematics may use diffrent notation. Different countries and mathematical schools may use different notation. Diffrent contexts require different approaches.
If you assume that the radical sign denotes the principal square root of complex number, this does not imply that this is the convention that is always used by everyone else. In complex analysis it is convenient to view the square root as a multifalued function and so we usually use the radical sign to denote the *set* of all square roots of a complex number. As it was mentioned by @MercuriusCh, you could even consider the square root as a nice and well-defined function on the Riemann surface if that works for a particular problem. It is all the matter of what is convenient in any given situation
for the +/- argument
it can't be the case because we have the square root of a negative number
it is only valid when taking the square root of a positive number (or more broadly the even root of a positive number)
because the number was defined as negative we know that were are evaluating a product of positive number in the sub step of "i root 4 and i root 9"
What is the value of (-2i)*(-2i)?
If you get -4, then -2i is a square root of -4.
PEMDAS (or BODMAS for folks in the UK, Australia, etc.) applies. A square-root is part of the "E" (or "O") portion of those rules.
The core of the discussion between -6 and +-6 is the square root. The trouble is the numeric answer, and that is a "bad manner" in math. Just write -sqrt(36). If the root was sqrt(2) nobody would calculate the numeric answer and would let the reader to interpret the meaning depending on the context. Only 10-14 years olds care about giving a precise numeric anwer with 987 decimals. Adult, math-educated, people are not afraid of amswering with the simplest roots or fractions. It is the same if the answer was 22/7, only children would calculate the numeric answer, normal and functional people would let the answer be 22/7 since it retains all the information (all the decimals with 0 error). So the answer should be -sqrt(36) and it would not delete any information.
A LOT of people are confused about this. Every number has (except zero) two square roots but only ONE of them is ever denoted with the √ symbol.
When you evaluate an expression, you take all possible results. The correct answer should be +/-6 as it takes all possible outcomes into picture.
I like your logic of principal square roots but that falls short of a complete answer.
I agree.
The symbol √ has a meaning, it will represent one number. For example
√ 2 is the exact value of 1.414….. It is a number, not several numbers.
If √ 2 could mean both 1.414… and -1.414… (± √ 2) then how could you possibly write just one of them. You wouldn’t be able to.
X=√ 2 is not the same as x^2=2
I disagree, we are simply performing a binary operation on two complex numbers, which by definition has one output. We define the square root on positive reals as the principal square root so there is no reason to assume otherwise for the complex numbers
@@bobh6728 is -1.414... not a square root of 2?
@@bobh6728when you do calculations on physical objects, value or principal roots are considered. However, when it is asked as a general question to evaluate, all possible results are to be used.
It isn't correct. There's no principal square root in complex world. In this case you must consider it as a power 1/2. Power has multiple values in complex world.
There is.
Source: wikipedia
this is my understanding
Put a complex number in polar form: r * e^(i * θ)
With a restriction on θ: -π < θ ≤ π
Now, the principal square root would be defined as:
sqrt(r * e^(i * θ)) = sqrt(r) * e^(i * θ/2)
With this definition
sqrt(-1) = 1 * e^(i * π/2) =
= i
How stupidly inconsistent. It makes sense if the radical symbol to be the positive square root only for all the reasons they give then the exact same reasoning must apply to the imaginary verson.
The thing is, ±6 comes with the assumption that we can take both square root values both times. That's (usually) a bigger assumption than "there's a square root that is explicitly and obviously that of a negative number, we can use i", which makes ±6 feel still correct, but not quite as correct as just -6. Especially when you can tell that the i's are gonna "cancel each other out" into -1, it feels much neater to remove that uncertainty even if it takes two weird steps, than if you were to take two more uncommon weird steps and still have this "uncertain" result in the end.
It would only be the negative solution since those were principal square roots.
Limiting to the principal root is an assumption you are making.
Also, we're not physicists. "Wolfram Alpha said so" isn't a valid argument here.
Not an assumption: the radical symbol is defined to represent the principal value of the square root of a number.
it is not correct to use an ambiguity like that in the first place solving the sub step of " i root 4" is not something to consider the 4 might be the product of 2 negative numbers
we KNOW it is a product of positive numbers in that state because we started with the square root of a negative number
Limiting to the principal root is the definition, not the assumption.
Also, "wolfram alpha said so" is a valid argument. WolframAlpha just follows the definitions we gave it.
1+1=2 is defined. We follow that definition. So do we follow the definition of a square root.
not an assumption lmfao
We can solve this without touching complex no.
Solution:
√-4×√-9= √(4×(-1)) × √(9×(-1))
= 2 × √(-1) × 3 × √(-1)
=6 × √(-1) × √(-1)
From, a^m × a^n = a^(m+n)
= 6 × (-1)^(1/2 × 1/2)
= 6 × (-1)^(1)
= -6
This the answer
1/2 * 1/2 = 1/4, what
nvm i'm a dumbass
By writing sqrt(-1) you are implicitly using complex numbers, you can't just use it without even defining it.
@@hayatara.no, you’re right. He made a mistake by putting 1/2 x 1/2 as the power of -1. It should say 1/2 + 1/2 instead.
@@alanharper23 fair
Its odd to me that the segment that went off on people who only used Real numbers is the followed by an assumption that we should take the positive values for both roots.
sqrt(4) =±2 is not a function, true, but sqrt(4) = -2 is.
Its is completely possible to get
sqrt(4)i * sqrt(9)i
-2i * 3i
-6i^2
6
There are 4 potential routes for the equation. 2 give you -6 and 2 give you 6. Therefore the general solution is ±6. The answer is only -6 if you add a qualifier of using principle roots.
The question is not to find solutions for a quadratic function, where considering the negative root part is required, but it is about calculating the root, which is defined to be positive only.
Mathematician have agreed that whenever your are talking about "the square root" your are talking about "the principle square root."
And especially in this case, as when that sign (radical sign/radix) is used, it specifically denotes the principle square root.
@@arejaybee there's no equasion here. We just have arithmetic operations on complex numbers. The expression has a single value: -6. (Yes the √ does mean principal square root).
@@nbooth The question for the second poll was "What is the product of the square root of -4 and the square root of -9 equal to" and at 1:34 (where I'm currently paused at) it is even shown with an equal sign between the formula and the possible solutions.
So there is an equasion of root(-4)root(-9)=? implied.
@@TyroRNG but the notation was not in the original question (for the second poll, 1:34), so that argument does not apply. And based on the phrasing of the poll, it is root(-4)×root(-9)=?
So it comes down to finding ALL solutions for ?, which would mean, unless extracting root(-1)=i and then squaring root(4)i×root(9)i is forbidden by some rule ?=6 needs to be considered
Either we operate in the real domain, in which case it's undefined, or in the complex domain, where each non-zero number has exactly two square roots (yes, even if it is a positive real number). In the latter case, the expression has two values: 6 and -6. But this is why the square root symbol is better avoided in complex arithmetic, the usual assumption that an expression evaluates to one value breaks down and one has to be careful about the consequences. In typical uses it's better to start by saying that -1 has two square roots, we call one of them i, and then express everything in terms of i rather than using the tricky square root symbol.
I believe the people answering + or -6 do the same as the people answering 6, but and + or - to keep all answers available.
If the rule √x•√y = √(x•y) only applies to x,y≥0 then you can't split √-4 to i•√4
Once we start working with complex numbers we need to remember that roots have two results. So, the result of the product now depends on which value you get for √-4 and √-9. It turns out that all the combinations ends up giving you 6 or -6.
The result is ±6
You literally just talked about writing √(-4) as i√4
i√4 × i√9 is -6 and only -6.
"Most people think that 45% is 'most'"
55% get it right, you said.
45% might be a plurality in a three way split but it is never a majority.
And while I'm at it:
A consensus is 100%.
Agreement OR Acquiescence.
In the survey run by Brilliant, less than 50% got it right, so "most people get it wrong" is technically correct.
When the video maker re-ran the survey, 55% got it right. Presumably, this says something about the math knowledge of folks who watch this channel vs. folks who follow Brilliant.
@jmr5125,
and John.
Generally speaking while something over 50% is a majority, that does not mean that it is most as most requires nearly all. I'd be looking in the 80s or 90s for most.
However, if we are talking comparatives, many, more, most and comparing groups containing different numbers then many (of a large group) might require only 20%, with a group containing 35% being more and a remaining group containing 45% of the total might indeed represent the most.
Ain't words fun too.
The concept of the principal value is a standard that we invented and decide to use to make maths make more sense for us. I think the fact that when we don't use the principal value standard we get the same answer whether we take the square root of the numbers then multiply as we do when we multiply the numbers and then take the square root means that ditching the principal value standard would be a more consistent model.
Taking the principal value standard is like saying the answer is undefined because we're restricting ourselves to the real number line. It's a standard we can impose, not the actual answer.
Yes, when using only the principal values, we cannot solve irrational equations like √x̅ = -2 or ³√x̅ = -2 over the field of complex numbers ℂ, but if we let our complex nth roots be multivalued, then these equations have solutions.
The problem has nothing to do with the principle value.
If the problem does not state otherwise, the domain should be assumed to be complex.
At a guess, I think the reason why someone might go for +6 (which was my first instinct, before I thought that it could also be -6) is that when they're taught that √x√y = √(xy), complex numbers haven't been introduced yet, so the teacher doesn't need to specify that x and y have to be ≥ 0 for the rule to work - and then by the time complex numbers *are* introduced, this particular subtlety gets swept under the rug because we've got a lot of material to get through on these things! So possibly people might think that the rule still holds for negative numbers.