You brought me a couple of years back to my high school math studies ! I really hope you continue to inspire young people to understand, like and enjoy the beauty of mathematics !
Nice... we never got imaginary numbers in high school. I learned about them in college, but as an EE minor we called it j rather than i so as not to confuse it with the symbol for current.
Wow. Thank you so much for your videos! I have recently been watching them late at night for fun lol! Today I taught my daughter some of your methods, in preparation for the SAT. She completely understood it all, and then solved some problems correctly. Again thank you 🙏
you helo bringing my math confidence back after going from a straight A student to being bullied and sick I could no longer perform in school, thank you for helping me find my old self again❤ your videos are always very clear and informative, appreciate it
... Good day to you sir, When encountering for instance SQRT( - 9) in a calculation, to avoid possible problems in the Complex world, I like to apply i^2 = - 1 immediately, so SQRT( - 9) = SQRT(9 * i^2) = SQRT(3^2 * i^2) = 3 * i, personally my way of avoiding errors ... thanking you for as always clear and instructive presentations and last but not least math efforts ... best regards, Jan-W
@@atussentinelBy convention, we use +i to keep it simple, unless it is otherwise specified that we want all possible roots. There is a convention to select principal roots in the complex world, which is the following: #1: if there is a positive real root, the positive real root is the principal root #2: if there is no positive real root, but there is a negative real root, the negative real root is the principal root #3: if there are no real roots, the complex root that is closest to the positive real axis is considered the principal root #4: if #3 is inconclusive because the complex roots in question are a conjugate pair, the root with a positive imaginary component is considered the principal root. Examples: #1: principal square root of 25 is +5 #2: principal cube root of (-27) is -2 #3: principal square root of (-18*i), is 3 - 3*i #4: principal 4th root of (-64), is 2 + 2*i
@@atussentinel You can say something like, pick the root in the complex upper half plane (cutting out the negative real axis). Doesn’t always do what you want, but it’s often a decent default.
Without any rigor, I can say why this makes sense. We'd expect sqrt(-1)sqrt(-1) = -1, and we would want to extend that in general. We would want sqrt(-9)(sqrt(-9) = -9 (where 9 plays the role of 1), so whatever sqrt(-9) is, multiplying it by itself should give -9. The number 3sqrt(-1) seems like a good candidate for sqrt(-9). If we have two different numbers in the root, such as sqrt(-2)sqrt(-3), then again we want the answer to be the square root of the product of the magnitudes of the numbers under the root, multiplied by -1, or in this case sqrt(-2)sqrt(-3) = -sqrt(2*3) = -sqrt(6).
One simply gives the 6 answer and knows to "naively" add the negative sign anyway. But the step-by-step is to show the answer by the justification of the imaginary and to remind that it is not permitted to attach the negative sign except as "outside the sign". Given: It is not the case that negative roots have solutions AND roots have positive and only positive sign. QED: It is the case that negative roots have solutions with negative sign as demonstrated.
Thank you very much for your videos! As we get older including Math in our daily routine is great for our brain health. I will keep watching! Thanks again!:-)
If you ask me, this means when the rule sqrt(a) * sqrt(b) = sqrt(a * b) is taught, it should be taught that it is true unless both a and b are negative.
Wow that was a very complex explanation for something very simple bar the concept of imaginary numbers. Exponetial operations take priority over multiplication (weird American use of . for multiplication rather than x but I will let you of that). Square root of -9 is 3i, square root of -4 is 2i. 3i x 2i = 6 x i^2 = -6
When it comes to multiplication I've seen the "center dot" notation far more often than the (slightly smaller) x, both during lectures etc. in my studies (in Europe) and in text books. I imagine the reason is to avoid confusion with "x the variable".
√(-9) = 3i, and √(-4i) = 2i, where i = √(-1), then 3i * 2i = 6i² = -6.. Actually, if we plot 3i and 2i on an Argrand diagram, we see these numbers have a magnitude of 3 and 2 respectively and both make a 90° angle anti-clockwise on the x-axis (real number axis). So they are on the y-axis (imaginary axis). Now when we multiple these two complex numbers together on and Argrand diagram, then the resulting number has a magnitude 2*3 = 6 and the angle subtebded to the real number axis is 90° + 90° = 180° The complex number (6, 180°) on the Argrand diagram corresponds to -6.
Remember. Two minus values under square roots like√(−1)×√(−1) when multiply will result in minus. Fout times multiplication like √(−1)×√(−9)×√(−1)×√(−9) will result in positive.
The question is how do we explain that sqrt(1) * sqrt(1) = sqrt(1 * 1), but sqrt(-1) * sqrt(-1) != sqrt(-1 * -1). Is it always right that sqrt(m) * sqrt(n) = sqrt(m * n)?
The property that sqrt(m) * sqrt(n) = sqrt(m * n) holds only when at least one of m and n is non-negative. So, we have to be careful when it's possible for both numbers to be negative.
If we have two numbers x and y, such that y is the square root of x, then y = sqrt(x) y = sqrt(x * 1) y = sqrt(x * (-1 * -1)) y = sqrt(x) * sqrt(-1) * sqrt(-1) y = sqrt(x) * i * i y = sqrt(x) * i^2 y = sqrt(x) * (-1) y = -sqrt(x) y = -y ?
The domain of what? This is not a formula with unknown numbers. You want to define the domain of an equation with all numbers known? I'd love to see that.
@@DemanaJaire This is a formula containing functions. We can think of the square root as a function. For specific part of the domain it takes specific part of the possible values. My math professor used to say it doesn't matter that much if we know or don't know the numbers. But I forgot the exact context, it was a long time ago in another life.
Hi Mr. H, From your comment on the sq. root of 36 NOT being both +6 and -6, but just the former when we have a radical sign (@ c. 2:34). Okay. Could you clarify when we would have both a positive and negative answer to a square root? As I recall this seems to happen with quadratic equations esp. when using the quadratic formula (for example--and even if you have to check the correctness of both answers). So, I'm a bit confused on when we have just a positive answer to a square root vs. both positive and negative answers. Is there a rule here? Thanks!
@@akamajoseph9415 hey, hope i can help. in general √ refers to the positive square root of a number. eg. √4 = 2. when we solve the equation x² = 4, there are two numbers that solve the equation, the positive square root and the negative square root. the positive one we write as √4 as mentioned before, and the negative one is -√4 (negative of the positive root makes the negative root). we often write both of these two separate solutions together with one sign, putting ±√4 = ±2. so, √4 refers to the positive root of 4, and ±√4 refers to both the positive root of 4, AND the negative root of 4, which come from solving the equation x² = 4
in a nutshell, √x is the positive number that multiplies by itself to make x, but when solving equations, we don't only want the positive number, we also want the negative one, and hence we also have to consider -√x as a solution, giving our ±√x.
(correct) Solution: since √ of negative numbers is not defined, you have to split it up: √-9 * √-4 = √(-1 * 9) * √(-1 * 4) = √-1 * √9 * √-1 * √4 = (√-1)² * 3 * 2 Now, the special constant "i" is defined as i² = -1, so that i = √-1. Therefore: = i² * 6 = -1 * 6 = -6
Addition after the video: The error, that √36 = ±6 comes from the believe, that x² = 36 results in x = ±6, which is not directly true, because taking the √ or x² does not equal x, but |x| (meaning the absolute/positive value of x). Therefore: x² = 36 |√ |x| = 6 This now has to be evaluated for x ≥ 0 and x < 0. For x ≥ 0, it just becomes x = 6 For x < 0, it has to becomes -x = 6, because x is negative, but the result is positive. Therefore we have to negate x to make it correct. Now we just multiply both sides with -1 and get x = -6 As a result, we now have x = 6 and x = -6, which can be written as x = ±6. Most teacher don't really explain this process and just skip to "x² = 36, therefore x = ±6", which is technically correct, but confusing to the students. Because they skip this process, they also almost never teach the fact, that √x² ≠ x but instead it is √x² = |x|
the square root symbol only represents the principled result, same with x^0.5, its why the plus-minus sign is in the quadratic formula i don’t think there’s any good notation for exponentials that return all results unfortunately, if anyone does know then i’d like to know
@@Theooolone "i don’t think there’s any good notation for exponentials that return all results unfortunately, if anyone does know then i’d like to know" There is. As an example, to indicate that we want all 3 solutions to cbrt(8), we can write it as: e^((ln(8) + 2*pi*k)/3), where k is any integer With k=0, we get e^(ln(8)/3), which is the principal root +2 With k=1, we get -1 + sqrt(3)*i With k=2, we get -1 - sqrt(3)*i And when k passes 3, we continuously repeat this cycle, where k is reduced to its remainder when dividing by 3.
Before watching video, I keep getting the answer is plus or minus six depending on how tackle the problem, thefore, it not being sqrt(36) is correct. I'll now watch video. After watching I'm assuming we cannot do the sqrt(a)•sqrt(b)=sqrt(a•b) even after converting to sqrt(9i^2) and sqrt(4i^2) since by order of operations, we would normally have to do exponents first (sqrt is same as 1/2 power) and since that would still make 3i and 2i, we cannot get 6 as part if the final answer. It kind of makes sense but is kind of confusing.
The trouble with this explanation is that you rely on sqrt(ab)=sqrt(a)*sqrt(b) to show that sqrt(-9)=sqrt(-1)*sqrt(9) and everything that follows, but then if sqrt(a)*sqrt(b)=sqrt(ab) it also follows that sqrt(-9)*sqrt(-4)=sqrt(-9*-4)=sqrt(+36), which isn't true. It's no wonder students get confused.
@@mrhtutoring Yes, exactly. So you shouldn't be using sqrt(ab)=sqrt(a)*sqrt(b) to give sqrt(-9)=sqrt(-1)*sqrt(9) because that formula only applies when a and b are positive. Instead make clear that sqrt(-x)=sqrt(x)i.
is there a "principal root" for sqrt of negative numbers (or convention)? I'm not sure. If there is not, then the notation of sqrt(-9) is ambiguous. To me it's very weird to define a "principal root" for complex numbers, that is, sqrt(-9) is always +/-3i, then you can definitely get 6 as a second valid answer. I'll accept the answer is "undefined/not exist" as taking the sqrt of negative number is undefined in real domain. It's better than using the methods defined with real numbers and directly use it to solve a complex number problem.
The principal root of -9 does not exist. What exists is a change to the complex numbers and then calculation -9 as 9i² = 3²i² Also √(-9) is only 3i, not -3i Only in equations like x² = -9 √(x²) = √(-9) is "the result" x = ±3i - which comes from the x. The solution of √(x²) = √(-9) is |x| = 3i, and we don't want absolute values, so we write x = ±3i
The answer is yes. There is a convention for principal roots of negative numbers, and for complex numbers in general. Priority #1: positive real roots Priority #2: negative real roots Priority #3: complex root closest to the positive real axis Priority #4, if priority #3 produces a complex conjugate pair of roots, the one with a positive imaginary part is the one that is the principal root.
Very interesting and useful, but not entirely accurate. If we discuss roots of negative numbers, we must take into account ambiguity in the complex plane, so the answer is 6*EXP( i*m *π), where "m" - is any integer.
If you can't do sqrt(-4)*sqrt(-9)=sqrt[(-4)*(-9)] because those numbers do not exist, then why can you do sqrt[(-1)*9]=sqrt(-1)*sqrt(9) (so the reverse), since sqrt(-1) also does not exist?
@@alexbork4250Sure, fair enough, but in my mind the same applies though. If you can't multiply square roots of negative numbers according to what he says, why would the reciprocal be allowed (albeit one of them is indeed a real number)?
@@alexbarac The reason is that the square root of -1 is +i, by definition and by convention it is positive i. The square root sign in general, tells you to produce the principal root. There is a convention for the principal square root, where for positive real numbers, it is a positive real number, and for negative real numbers, it is the positive imaginary number. The principal root convention in general is: Priority #1: is there a positive real root? Then it's the positive real root. Priority #2: is there a negative real root? Then it's the negative real root. Priority #3: is there single root closest to the positive real axis? Then that's the principal root. Priority #4: is there complex conjugate pair of roots closest to the positive real axis? Then the one with a positive imaginary part is the principal root.
Consider the polar form. Where "i" represents a 90 degree anticlockwise phase shift. Multiplication of two 90 degree phase shifts results in a rotation of 180 degrees (-1). Multiply the real numbers 3 x 2=6. The combined result -1x6=-6. Without the imaginary number "i", (or "j" in engineering), there would not be algebraic solutions to many important mathematical and engineering issues.
sqrt(-9) is ambiguous. It's undefined in real numbers but in complex numbers it's always 3i/-3i. Throwing one away (like the thing did in the vid) looks weird to me.
@@atussentinel Yeap, to work with a complex function as if it were a real function, you'd first make it unambiguous. But it does not explain what you cannot do and which of 1) and 2) is wrong
@@YTRusViewer #1 is wrong You got 3/i by converting -9 to 9/-1, then attempting to take sqrt(9/-1), and then using properties to make that into sqrt(9)/sqrt(-1). But, the property that sqrt(a/b) = sqrt(a) / sqrt(b) doesn't hold when b is a negative number, because it results in the wrong (non-principal) root.
@@RealMesaMike so sqrt(-9)/sqrt(1) is correct while sqrt(9)/sqrt(-1) is not. Though the wrong root is in the numerator sqrt(-9), isn't it? :) I just want to understand in a simple manner where the error comes up
@@YTRusViewer sqrt(a/b) means that you have to resolve a/b first, then take the square root of the result. This means the convenient property sqrt(a/b) = sqrt(a) / sqrt(b) doesn't necessarily hold when negative numbers are involved in the denominator, because you get a different answer than if you resolved a/b first then took the square root of that result.
If a and b are real variables and from the formula sqrt(a) * sqrt(b) = sqrt(a*b) then if a = -4 and b = --9 then a*b = +36 and the sqrt(36) = 6. OK and the sqrt(-4) = sqrt(-1) * sqrt(4) which is 2i and sqrt(-1) * sqrt(9) is 3i and 2i * 3i = 6 * i^2 which equals 6 * -1 or -6. So why the two answers? What logic was violated in the first proof? If I say that a and b are real variables that means the they can be positive or negative. I think you are playing here with the order of precedence like in PEMDAS (please excuse my dear aunt sally) or parenthesis then exponents, multiplication, division then add and subtract. Where you are not clear here is that a square root symbol is really a to the 1/2 power exponent (a^1/2) and is an exponentiation and must be done first by convention from the order of operations. Yes we all know that -4 is really -1 * 4 and -9 is really -1 * 9. So by the order of operations you must exponentiate the product (a*b)^1/2 before you multiply.
So, you are saying in that in the algebraic expression x^2=36, the solution cannot be -6, it can only be +6? I'm sorry, but that is absurd, both +6 and -6 when squared is 36.
The question is just square root of 36 and that is not an equation, if it is an equation, then you can put -6 and 6 to satisfy the equation, but the above question is square root of 36, square root any number cannot be a negative answer .For instance, square root 4 is 2, square root -4 is 2i. No negative answer for square rooting a number.A lot of people are confused at this, spend some time to understand it.Besides, when it comes to square root, what is the length of a square with 16cm^2 area size? 4 only.Size can't be negative number.Despite that is not a good example,square root a number can't be negative.
@@biucmm7072 Firstly, the concept of i comes from trying to solve the equation x^2+1=0, which is an algebraic expression. Now, he is partially correct by saying that 6 is the primary solution, which he really means that is is the solution along the primary branch. Any square-root is a multivalued function which means that it has a brach point and requires a branch cut to distinguish solutions. Along the primary branch, 0 to 2 pi, the only solution is 6. However, nothing the problem is restricting to the primary branch, meaning that any branch is acceptable. As for writing it as -sqrt(36), which I do agree with, without restriction to the principle branch, then the correct answer is -(+/-6)=-/+6 or simply +/-6. Finally, for your example, if I told you that the charge squared of an object was 4 C^2, does that determine that the only possible amount of charge that the object has is 2 C, hence positively charged? Of course not since negative charges exist so it could easily have -2 C worth of charge. Even stating that the length of an object is negative does not mean that is wrong, since the negative (in a vectorial sense) can give a nothing of orientation of the object (hence direction it is pointing, for example). You are correct in the fact that certain scalar quantities are chosen to be positive and only positive quantities, only by convention, but that does not mean that any number cannot be negative. The point is, without being careful and restricting to the principle branch, his statement, and hence video, is in correct and can lead to misunderstanding.
Both +6 and -6 when squared is 36 is true but try actually calculating square roots and you will never get a -n result. There are ONLY positive square roots despite this video.
Complex numbers are used in electrical engineering, to expand the concept of resistance, to also cover capacitors and inductors with the more general concept of impedance. The real part of impedance is the resistance, indicating that energy leaves the electrical domain, while the imaginary part is called reactance, which refers to energy that is stored and released later in the cycle. This allows you to keep track of impedance for all three of these kinds of components, and go through the same setup as you would if they were all just resistors. When you use the impedance to solve for the current or voltage of interest, the magnitude corresponds to the amplitude of the wave, and the angle of the voltage or current phasor (i.e. phase vector) corresponds to the phase shift from the source's waveform.
For god’s sake, we are doing operations not equations Operations itself only have 1 value Also, many values of the operation can be used simultaneously in an equality if it were the case because there is no restriction on that, but when you use the same variable in an equality, you can’t use two values of the same variable simultaneously That is why, we only use the principal root which explains as to why sqrt(36)=6
Because the radical sign, √, is defined as the positive square root of a number. If you want the negative square root, you have to write −√. And if you want both square root, you have to write ±√. So, if x² = 36, then x = ±6. But if x = √36, then x = 6.
@@Nikioko to add on to that, if you were to do sqrt{(-6)^2} then cancelling the square and the square root would be what most people would do. But by the order of operations, you’re supposed to deal with (-6)^2 first and then take the square root of that If you think about it using logic, if you have to square (-6)^2 and then take the square root of it, then it can’t equal to -6 since you’re unable to directly cancel the square and the square root I also have another analogy to think Let’s take a simple equation, say 2+2=4 If you were to square root this on both sides, then it’s just sqrt(2+2) = sqrt(4) Then sqrt(2+2)=+/-2 But if you think carefully using logic, you’d think, “How can the square root of the sum of two positive integers even result in a negative number?” So I’d say sqrt(2+2) = +2
@@Brid727 Actually, by the order of operations, roots and exponentiations have the same priority. In fact, roots are just exponentiations with inverted values, just like divisions are multiplications with inverted values. And subtractions are additions of negative values.
Multiply square root of negative number, Don’t directly combine it, First u should take the comment factor (i) , let it become square root of positive numbers and then combine it.😅😅
Example: x^2 = 4 sqrt(x^2) = sqrt(2^2) |x| = 2 x = 2 or x = -2 because both 2^2 and (-2)^2 equal 4, and you want to find all possible solutions, not just the convenient one. In case of sqrt(36) you don't need it. Sure, you can do sqrt(36) = sqrt(6^2) = |6| = 6, but it's redundant, because |6| can only equal 6. There are no multiple solutions.
Yes, but taking the square root takes precedence over multiplication, so you shouldn't be multiplying those two numbers first (remember your PEMDAS obedience training?) With that in mind, you can see that the property sqrt(a) × sqrt(b) = sqrt(a×b) only holds if at least one of a or b is non-negative.
The radical sign specifies that we are only interested in the principal root, unless context specifies otherwise. For roots of positive numbers, the convention is that the positive root is the principal root. The +/- sign in front, indicates that we're interested in using both roots as possible contributions to the rest of the equation.
@@mrhtutoring Ahhh! Okay. This seems, I think, to answer the question I just posted (even if the correct answers to the posted problem should be x= +2 or x = -2 . . . ). From my question, when using the quadratic formula you are solving for x. Here in the above example you are not solving for x.
I think the answer is wrong, and that is basically because you used bad notation. We usually don't write it like that sqrt(-9), but rather -9^(1/2). if z = -9^(1/2) then z2 = -9 then z = +- 3i Also -4^(1/2) = +- 2i (+- 3i) * (+- 2i) = +- (6 * i^2) /* we take both signs here because signs in the left hand side dont have to be synchonized */ = +- 6
@@mrhtutoring it is impossible to properly define principle root for negative numbers, because i is defined as i^2 = -1, but (-i)^1 also equals -1. In other words there are 2 numbers such as i^2 = -1.
@@mrhtutoring well may be But isn't it a strange definition? It makes sence for rational numbers because they are ordered, so you've got negatives, positives and zero in between. It is easy to define a principal root. It also has lots of usages esp in geometry where negative length wouldn't make any sense. Complex numbers are unordered. Definition that I googled is quite long. And is this notation usable?
@@mrhtutoring I'm just curious, what is the "standard" form of √(i) in the US then? or (4)√(-9) (the fourth root of -9)? If you say "ahh those don't have standard form" then comes my following question: why sometimes the "standard" form exists, but sometimes doesn't? √-9 = 3i is just weird. this notation will only appear in informal math afaic, where "informal" means one can define anything just based on their taste. those things will probably break math at some point.
@@husseinabdulkadir6707 No. The definiton is as I said, for the reason I mentioned: (√−1)² = √−1 ⋅ √−1 = √[(−1) ⋅ (−1)] = √1 = 1. In complex numbers, radical sign and square do not simply cancel out.
Only in Excel, or other programs (none that I'm aware of) that give priority to the unary negative operator. In standard math, the convention is that a minus needs to be snared, in order to be squared. (-6)^2 = 36, but -6^2 = -36.
![Why you didn't learn tetration in school[Tetration]](http://i.ytimg.com/vi/Ea1_1aVfwl4/mqdefault.jpg)
Excellent, sir! I find your videos mentally stimulating and fascinating. Please, continue. Thank you.
You brought me a couple of years back to my high school math studies !
I really hope you continue to inspire young people to understand, like and enjoy the beauty of mathematics !
Nice... we never got imaginary numbers in high school. I learned about them in college, but as an EE minor we called it j rather than i so as not to confuse it with the symbol for current.
More of a convention trick, less math.
@@kaasmeester5903lol I am doing that now
Note: i^2 isnt equal to +1 because squaring a root just gives you the number without the radical
of course i^2 isn't equal to +1, it is equal to -1.
@@warblerab2955 yeah i just wrote it for me, feel free to ignore it
Wow. Thank you so much for your videos! I have recently been watching them late at night for fun lol! Today I taught my daughter some of your methods, in preparation for the SAT. She completely understood it all, and then solved some problems correctly. Again thank you 🙏
you helo bringing my math confidence back after going from a straight A student to being bullied and sick I could no longer perform in school, thank you for helping me find my old self again❤ your videos are always very clear and informative, appreciate it
Thank you. you are a star. Please keep on I enjoy watching your videos 😊
... Good day to you sir, When encountering for instance SQRT( - 9) in a calculation, to avoid possible problems in the Complex world, I like to apply i^2 = - 1 immediately, so SQRT( - 9) = SQRT(9 * i^2) = SQRT(3^2 * i^2) = 3 * i, personally my way of avoiding errors ... thanking you for as always clear and instructive presentations and last but not least math efforts ... best regards, Jan-W
the problem is (-i)^2=-1 as well, why not use -i?
there is no i > (-i) or i > 0 such thing to help you define a "principle root" in the complex domain
Dat is hetzelfde als wat hij doet, duh.
@@atussentinelBy convention, we use +i to keep it simple, unless it is otherwise specified that we want all possible roots. There is a convention to select principal roots in the complex world, which is the following:
#1: if there is a positive real root, the positive real root is the principal root
#2: if there is no positive real root, but there is a negative real root, the negative real root is the principal root
#3: if there are no real roots, the complex root that is closest to the positive real axis is considered the principal root
#4: if #3 is inconclusive because the complex roots in question are a conjugate pair, the root with a positive imaginary component is considered the principal root.
Examples:
#1: principal square root of 25 is +5
#2: principal cube root of (-27) is -2
#3: principal square root of (-18*i), is 3 - 3*i
#4: principal 4th root of (-64), is 2 + 2*i
@@atussentinel You can say something like, pick the root in the complex upper half plane (cutting out the negative real axis).
Doesn’t always do what you want, but it’s often a decent default.
Appreciate all your videos. I enjoy solving mental math Problems.
best tutor ever thank you
Glad you think so!
LOL. I love it! You are the man.
Thanks sir
i liked your comment
Without any rigor, I can say why this makes sense. We'd expect sqrt(-1)sqrt(-1) = -1, and we would want to extend that in general. We would want sqrt(-9)(sqrt(-9) = -9 (where 9 plays the role of 1), so whatever sqrt(-9) is, multiplying it by itself should give -9. The number 3sqrt(-1) seems like a good candidate for sqrt(-9). If we have two different numbers in the root, such as sqrt(-2)sqrt(-3), then again we want the answer to be the square root of the product of the magnitudes of the numbers under the root, multiplied by -1, or in this case sqrt(-2)sqrt(-3) = -sqrt(2*3) = -sqrt(6).
Thanks. Very nice work.
One simply gives the 6 answer and knows to "naively" add the negative sign anyway.
But the step-by-step is to show the answer by the justification of the imaginary and to remind that it is not permitted to attach the negative sign except as "outside the sign".
Given: It is not the case that negative roots have solutions AND roots have positive and only positive sign.
QED: It is the case that negative roots have solutions with negative sign as demonstrated.
Thank you very much for your videos! As we get older including Math in our daily routine is great for our brain health. I will keep watching! Thanks again!:-)
Thank you for watching.
relearning basic- math rules I learned a long time ago is soo cool
If you ask me, this means when the rule sqrt(a) * sqrt(b) = sqrt(a * b) is taught, it should be taught that it is true unless both a and b are negative.
Obrigado pelos esclarecimentos de multiplicação de números negativos, na radiciação !
The BEST, You're not Mr H : Your title should be :Dr H Tutoring.
I Agree
Wow that was a very complex explanation for something very simple bar the concept of imaginary numbers. Exponetial operations take priority over multiplication (weird American use of . for multiplication rather than x but I will let you of that).
Square root of -9 is 3i, square root of -4 is 2i. 3i x 2i = 6 x i^2 = -6
When it comes to multiplication I've seen the "center dot" notation far more often than the (slightly smaller) x, both during lectures etc. in my studies (in Europe) and in text books. I imagine the reason is to avoid confusion with "x the variable".
True it can avoid confusion between x's (tend to use italic to avoid this) but it causes confusion with the decimal point@@joeltimonen8268
I'm European, and we've never used × for multiplication. We've always been using •
√(-9) = 3i, and √(-4i) = 2i, where i = √(-1), then 3i * 2i = 6i² = -6..
Actually, if we plot 3i and 2i on an Argrand diagram, we see these numbers have a magnitude of 3 and 2 respectively and both make a 90° angle anti-clockwise on the x-axis (real number axis). So they are on the y-axis (imaginary axis).
Now when we multiple these two complex numbers together on and Argrand diagram, then the resulting number has a magnitude 2*3 = 6 and the angle subtebded to the real number axis is 90° + 90° = 180°
The complex number (6, 180°) on the Argrand diagram corresponds to -6.
Actually, √(-9) = 3i *OR* √(-9) = -3i. You can work out the rest of the thought process.
That’s how I did it in my head too
Beancounters everywhere...
Sir your biceps 💪
i liked ur comment
Wow.
I wish you were my math teacher!
I like the way you use sqr(ab)=sqr(a)sqr(b) to show that sqr(-9)sqr(-4) does not equal sqr(-9x-4)
thanks. I'm mastering it.
Geometric Algebra FTW. + And - are just vector directions.
Remember. Two minus values under square roots like√(−1)×√(−1) when multiply will result in minus. Fout times multiplication like √(−1)×√(−9)×√(−1)×√(−9) will result in positive.
Superb sir 🎉🎉🎉
The question is how do we explain that sqrt(1) * sqrt(1) = sqrt(1 * 1), but sqrt(-1) * sqrt(-1) != sqrt(-1 * -1). Is it always right that sqrt(m) * sqrt(n) = sqrt(m * n)?
The property that sqrt(m) * sqrt(n) = sqrt(m * n) holds only when at least one of m and n is non-negative. So, we have to be careful when it's possible for both numbers to be negative.
Only those who love numbers
idk if this works but what I do is stack the negatives
sqrt(-9) * sqrt(-4) = sqrt(-^2 36) = -6
I whent “BUT i^2 IS 1”
Sqrt(-1)= +/-i ( nonzero complex numbers always have two complex square roots).
But when we use the √ symbol, we only use the principal root
If we have two numbers x and y, such that y is the square root of x, then
y = sqrt(x)
y = sqrt(x * 1)
y = sqrt(x * (-1 * -1))
y = sqrt(x) * sqrt(-1) * sqrt(-1)
y = sqrt(x) * i * i
y = sqrt(x) * i^2
y = sqrt(x) * (-1)
y = -sqrt(x)
y = -y
?
First things first - look at the domain. We can't ignore the complex domain here. If we don't - then we immediately see that -6 != 6.
The domain of what? This is not a formula with unknown numbers. You want to define the domain of an equation with all numbers known? I'd love to see that.
@@DemanaJaire This is a formula containing functions. We can think of the square root as a function. For specific part of the domain it takes specific part of the possible values. My math professor used to say it doesn't matter that much if we know or don't know the numbers. But I forgot the exact context, it was a long time ago in another life.
Hi Mr. H,
From your comment on the sq. root of 36 NOT being both +6 and -6, but just the former when we have a radical sign (@ c. 2:34). Okay. Could you clarify when we would have both a positive and negative answer to a square root? As I recall this seems to happen with quadratic equations esp. when using the quadratic formula (for example--and even if you have to check the correctness of both answers).
So, I'm a bit confused on when we have just a positive answer to a square root vs. both positive and negative answers. Is there a rule here? Thanks!
When you take the square root of both sides of an equation, you add the ±.
For example, x²=4
x=±2.
@@mrhtutoring
This helps.
Thank you! Thank you!
I'm not yet convinced.
@@akamajoseph9415 hey, hope i can help. in general √ refers to the positive square root of a number. eg. √4 = 2. when we solve the equation x² = 4, there are two numbers that solve the equation, the positive square root and the negative square root. the positive one we write as √4 as mentioned before, and the negative one is -√4 (negative of the positive root makes the negative root). we often write both of these two separate solutions together with one sign, putting ±√4 = ±2. so, √4 refers to the positive root of 4, and ±√4 refers to both the positive root of 4, AND the negative root of 4, which come from solving the equation x² = 4
in a nutshell, √x is the positive number that multiplies by itself to make x, but when solving equations, we don't only want the positive number, we also want the negative one, and hence we also have to consider -√x as a solution, giving our ±√x.
(correct) Solution:
since √ of negative numbers is not defined, you have to split it up:
√-9 * √-4
= √(-1 * 9) * √(-1 * 4)
= √-1 * √9 * √-1 * √4
= (√-1)² * 3 * 2
Now, the special constant "i" is defined as i² = -1, so that i = √-1. Therefore:
= i² * 6
= -1 * 6
= -6
Addition after the video:
The error, that √36 = ±6 comes from the believe, that x² = 36 results in x = ±6, which is not directly true, because taking the √ or x² does not equal x, but |x| (meaning the absolute/positive value of x).
Therefore:
x² = 36 |√
|x| = 6
This now has to be evaluated for x ≥ 0 and x < 0.
For x ≥ 0, it just becomes x = 6
For x < 0, it has to becomes -x = 6, because x is negative, but the result is positive. Therefore we have to negate x to make it correct. Now we just multiply both sides with -1 and get x = -6
As a result, we now have x = 6 and x = -6, which can be written as x = ±6.
Most teacher don't really explain this process and just skip to "x² = 36, therefore x = ±6", which is technically correct, but confusing to the students.
Because they skip this process, they also almost never teach the fact, that √x² ≠ x but instead it is √x² = |x|
I don't understand how you can go into the complex plane and then say the answer isn't both 6 and negative 6.
Why? He is right.
Because you still take the principal square root
the square root symbol only represents the principled result, same with x^0.5, its why the plus-minus sign is in the quadratic formula
i don’t think there’s any good notation for exponentials that return all results unfortunately, if anyone does know then i’d like to know
@@Theooolone
"i don’t think there’s any good notation for exponentials that return all results unfortunately, if anyone does know then i’d like to know"
There is. As an example, to indicate that we want all 3 solutions to cbrt(8), we can write it as:
e^((ln(8) + 2*pi*k)/3), where k is any integer
With k=0, we get e^(ln(8)/3), which is the principal root +2
With k=1, we get -1 + sqrt(3)*i
With k=2, we get -1 - sqrt(3)*i
And when k passes 3, we continuously repeat this cycle, where k is reduced to its remainder when dividing by 3.
@@carultch That's so awesome! Of course it would be something with e and ln, thanks for letting me know!
Before watching video, I keep getting the answer is plus or minus six depending on how tackle the problem, thefore, it not being sqrt(36) is correct.
I'll now watch video.
After watching I'm assuming we cannot do the sqrt(a)•sqrt(b)=sqrt(a•b) even after converting to sqrt(9i^2) and sqrt(4i^2) since by order of operations, we would normally have to do exponents first (sqrt is same as 1/2 power) and since that would still make 3i and 2i, we cannot get 6 as part if the final answer. It kind of makes sense but is kind of confusing.
I agree!
Thank you for handling the radical properly.
Sorry, that was posted to the wrong video lesson by a different person
I wish i could edit or delete. The person writing on black background is indeed. The one on a green background routinely mishandles the radical.
The trouble with this explanation is that you rely on sqrt(ab)=sqrt(a)*sqrt(b) to show that sqrt(-9)=sqrt(-1)*sqrt(9) and everything that follows, but then if sqrt(a)*sqrt(b)=sqrt(ab) it also follows that sqrt(-9)*sqrt(-4)=sqrt(-9*-4)=sqrt(+36), which isn't true. It's no wonder students get confused.
Square root of negative numbers have different properties as they're imaginary.
Hence, different rules
@@mrhtutoring Yes, exactly. So you shouldn't be using sqrt(ab)=sqrt(a)*sqrt(b) to give sqrt(-9)=sqrt(-1)*sqrt(9) because that formula only applies when a and b are positive. Instead make clear that sqrt(-x)=sqrt(x)i.
👏👏👏👏
My math experience stop at square(x) with x>=0
I didn't do all the steps and Mr. H did, but I arrived at the correct answer, -6, knowing about imaginary numbers. Thank you, Mr H!
Nice. 👍🏻
I would say the imaginary number "i" was discovered rather than invented. An important distinction.
Imaginary number is a concept, and concepts are inventions, not discoveries.
You saved my life. My exam is tomorrow lol.
Good luck in your exam. 👍
Invented by Leonhard Euler, Swiss Mathematician
The direct answer will be -6 because the I
“Square roots of negatives do not exist”
Yes they do
Complex number theory
Root -1 = i
You are correct and complex so-called numbers are a fraudulent abomination.
Thanks!
Wow. Thank you for your support!
I appreciate it very much. 🙏
why we can't use sqrt(a)*sqrt(b) = sqrt(a*b) generally? While the converse, sqrt(a*b) = sqrt(a)*sqrt(b) is true?
√negative is an imaginary number.
Imaginary numbers work differently.
is there a "principal root" for sqrt of negative numbers (or convention)? I'm not sure. If there is not, then the notation of sqrt(-9) is ambiguous. To me it's very weird to define a "principal root" for complex numbers, that is, sqrt(-9) is always +/-3i, then you can definitely get 6 as a second valid answer.
I'll accept the answer is "undefined/not exist" as taking the sqrt of negative number is undefined in real domain. It's better than using the methods defined with real numbers and directly use it to solve a complex number problem.
The principal root of -9 does not exist. What exists is a change to the complex numbers and then calculation -9 as 9i² = 3²i²
Also √(-9) is only 3i, not -3i
Only in equations like x² = -9 √(x²) = √(-9) is "the result" x = ±3i - which comes from the x. The solution of √(x²) = √(-9) is |x| = 3i, and we don't want absolute values, so we write x = ±3i
The answer is yes. There is a convention for principal roots of negative numbers, and for complex numbers in general.
Priority #1: positive real roots
Priority #2: negative real roots
Priority #3: complex root closest to the positive real axis
Priority #4, if priority #3 produces a complex conjugate pair of roots, the one with a positive imaginary part is the one that is the principal root.
You just made that up ! ha
I hope you're kidding.
Yes sir ! :) @@mrhtutoring
😆
sqrt("-"9) and sqrt("-"4) are mean 3i and 2i so solution is "-6" Not 6
Very interesting and useful, but not entirely accurate. If we discuss roots of negative numbers, we must take into account ambiguity in the complex plane, so the answer is 6*EXP( i*m *π), where "m" - is any integer.
If you can't do sqrt(-4)*sqrt(-9)=sqrt[(-4)*(-9)] because those numbers do not exist, then why can you do sqrt[(-1)*9]=sqrt(-1)*sqrt(9) (so the reverse), since sqrt(-1) also does not exist?
when he says "doesn't exist" he just means "it isn't real number"
@@alexbork4250Sure, fair enough, but in my mind the same applies though. If you can't multiply square roots of negative numbers according to what he says, why would the reciprocal be allowed (albeit one of them is indeed a real number)?
@@alexbarac The reason is that the square root of -1 is +i, by definition and by convention it is positive i. The square root sign in general, tells you to produce the principal root. There is a convention for the principal square root, where for positive real numbers, it is a positive real number, and for negative real numbers, it is the positive imaginary number.
The principal root convention in general is:
Priority #1: is there a positive real root? Then it's the positive real root.
Priority #2: is there a negative real root? Then it's the negative real root.
Priority #3: is there single root closest to the positive real axis? Then that's the principal root.
Priority #4: is there complex conjugate pair of roots closest to the positive real axis? Then the one with a positive imaginary part is the principal root.
Consider the polar form. Where "i" represents a 90 degree anticlockwise phase shift. Multiplication of two 90 degree phase shifts results in a rotation of 180 degrees (-1). Multiply the real numbers 3 x 2=6. The combined result -1x6=-6.
Without the imaginary number "i", (or "j" in engineering), there would not be algebraic solutions to many important mathematical and engineering issues.
√a×√b = √(ab), with a,b ≥ 0
you can't totally ignor the negative root in both navigation and elctronic engineering direction maters not always but in many cases
ขอบคุณครับ
It violates the order of operations. sqrt(9)*sqrt(4)=sqrt(36) does as well, but the positives make it behave nicely.
SQRT(-1) times SQRT(-1) equal to -1 straightaway. There is no need to change it to i and then i times i equal to -1.
I see. How about sqrt(2) times sqrt(-1)?
@@mrhtutoring Since sqrt(2) is not the same as sqrt(-1), then sqrt(-1) will have to change to i. Final answer is i(sqrt(2))
👍
@@mrhtutoring it will be sqrt(-1)*sqrt(2); often written shorthand as sqrt(-2). When we write a+ib, a and b are implied real but i is not.
In your video ( 10 = -10 ) you tell the logic which is totally inverse of this 🤔🤔🤔
Can I use the same logic as follows?
1) sqrt(-9) = sqrt(9/-1) = 3 / i = -3i
2) sqrt(-9) = sqrt(-1*9) = +3i
Nice proof that 0=1 :)
sqrt(-9) is ambiguous. It's undefined in real numbers but in complex numbers it's always 3i/-3i. Throwing one away (like the thing did in the vid) looks weird to me.
@@atussentinel Yeap, to work with a complex function as if it were a real function, you'd first make it unambiguous. But it does not explain what you cannot do and which of 1) and 2) is wrong
@@YTRusViewer #1 is wrong
You got 3/i by converting -9 to 9/-1, then attempting to take sqrt(9/-1), and then using properties to make that into sqrt(9)/sqrt(-1).
But, the property that sqrt(a/b) = sqrt(a) / sqrt(b) doesn't hold when b is a negative number, because it results in the wrong (non-principal) root.
@@RealMesaMike so sqrt(-9)/sqrt(1) is correct while sqrt(9)/sqrt(-1) is not. Though the wrong root is in the numerator sqrt(-9), isn't it? :) I just want to understand in a simple manner where the error comes up
@@YTRusViewer
sqrt(a/b) means that you have to resolve a/b first, then take the square root of the result.
This means the convenient property sqrt(a/b) = sqrt(a) / sqrt(b) doesn't necessarily hold when negative numbers are involved in the denominator, because you get a different answer than if you resolved a/b first then took the square root of that result.
Why can't you raise both sides to the second power to get rid of the square root?
The given problem isn't an equation.
We are simplifying the expression on the left.
@@mrhtutoring I see, thanks
If a and b are real variables and from the formula sqrt(a) * sqrt(b) = sqrt(a*b) then if a = -4 and b = --9 then a*b = +36 and the sqrt(36) = 6. OK and the sqrt(-4) = sqrt(-1) * sqrt(4) which is 2i and sqrt(-1) * sqrt(9) is 3i and 2i * 3i = 6 * i^2 which equals 6 * -1 or -6. So why the two answers? What logic was violated in the first proof? If I say that a and b are real variables that means the they can be positive or negative. I think you are playing here with the order of precedence like in PEMDAS (please excuse my dear aunt sally) or parenthesis then exponents, multiplication, division then add and subtract. Where you are not clear here is that a square root symbol is really a to the 1/2 power exponent (a^1/2) and is an exponentiation and must be done first by convention from the order of operations. Yes we all know that -4 is really -1 * 4 and -9 is really -1 * 9. So by the order of operations you must exponentiate the product (a*b)^1/2 before you multiply.
Why coulnd you write sqrt(-9) * sqrt(-4) as sqrt(-4*-9) = sqrt(9*4) = sqrt(36)?
Different rule is applied when you deal with sqrt(negative number)
what reasoning is behind therule that sqrt(36) is only +6 ?
It's a convention that the mathematicians agreed upon.
If we use both the positive and the negative roots, it would be a mess.
Geometry precedes algebra. Lol.
I hate how we teach mathematics to our youth.
So, you are saying in that in the algebraic expression x^2=36, the solution cannot be -6, it can only be +6? I'm sorry, but that is absurd, both +6 and -6 when squared is 36.
The question is just square root of 36 and that is not an equation, if it is an equation, then you can put -6 and 6 to satisfy the equation, but the above question is square root of 36, square root any number cannot be a negative answer .For instance, square root 4 is 2, square root -4 is 2i. No negative answer for square rooting a number.A lot of people are confused at this, spend some time to understand it.Besides, when it comes to square root, what is the length of a square with 16cm^2 area size? 4 only.Size can't be negative number.Despite that is not a good example,square root a number can't be negative.
@@biucmm7072 Firstly, the concept of i comes from trying to solve the equation x^2+1=0, which is an algebraic expression. Now, he is partially correct by saying that 6 is the primary solution, which he really means that is is the solution along the primary branch. Any square-root is a multivalued function which means that it has a brach point and requires a branch cut to distinguish solutions. Along the primary branch, 0 to 2 pi, the only solution is 6. However, nothing the problem is restricting to the primary branch, meaning that any branch is acceptable. As for writing it as -sqrt(36), which I do agree with, without restriction to the principle branch, then the correct answer is -(+/-6)=-/+6 or simply +/-6. Finally, for your example, if I told you that the charge squared of an object was 4 C^2, does that determine that the only possible amount of charge that the object has is 2 C, hence positively charged? Of course not since negative charges exist so it could easily have -2 C worth of charge. Even stating that the length of an object is negative does not mean that is wrong, since the negative (in a vectorial sense) can give a nothing of orientation of the object (hence direction it is pointing, for example). You are correct in the fact that certain scalar quantities are chosen to be positive and only positive quantities, only by convention, but that does not mean that any number cannot be negative. The point is, without being careful and restricting to the principle branch, his statement, and hence video, is in correct and can lead to misunderstanding.
Both +6 and -6 when squared is 36 is true but try actually calculating square roots and you will never get a -n result.
There are ONLY positive square roots despite this video.
Where in life do we apply
this?
You'd need to learn this to go to college.
Complex numbers are used in electrical engineering, to expand the concept of resistance, to also cover capacitors and inductors with the more general concept of impedance. The real part of impedance is the resistance, indicating that energy leaves the electrical domain, while the imaginary part is called reactance, which refers to energy that is stored and released later in the cycle.
This allows you to keep track of impedance for all three of these kinds of components, and go through the same setup as you would if they were all just resistors. When you use the impedance to solve for the current or voltage of interest, the magnitude corresponds to the amplitude of the wave, and the angle of the voltage or current phasor (i.e. phase vector) corresponds to the phase shift from the source's waveform.
√−9 ⋅ √−4 = −√36
If you multiply a number by 1 you don’t change the value but if you multiply a number by -1 you change the value
technically you are just rotating it, not changing its value.
What a complex video...
You should explain why √36 does not equal +/- 6, rather the answer is only +6. WHY?
For god’s sake, we are doing operations not equations
Operations itself only have 1 value
Also, many values of the operation can be used simultaneously in an equality if it were the case because there is no restriction on that, but when you use the same variable in an equality, you can’t use two values of the same variable simultaneously
That is why, we only use the principal root which explains as to why sqrt(36)=6
Because the radical sign, √, is defined as the positive square root of a number.
If you want the negative square root, you have to write −√. And if you want both square root, you have to write ±√.
So, if x² = 36, then x = ±6. But if x = √36, then x = 6.
@@Nikioko to add on to that, if you were to do sqrt{(-6)^2} then cancelling the square and the square root would be what most people would do. But by the order of operations, you’re supposed to deal with (-6)^2 first and then take the square root of that
If you think about it using logic, if you have to square (-6)^2 and then take the square root of it, then it can’t equal to -6 since you’re unable to directly cancel the square and the square root
I also have another analogy to think
Let’s take a simple equation, say 2+2=4
If you were to square root this on both sides, then it’s just sqrt(2+2) = sqrt(4)
Then sqrt(2+2)=+/-2
But if you think carefully using logic, you’d think, “How can the square root of the sum of two positive integers even result in a negative number?”
So I’d say sqrt(2+2) = +2
@@Brid727 Actually, by the order of operations, roots and exponentiations have the same priority. In fact, roots are just exponentiations with inverted values, just like divisions are multiplications with inverted values. And subtractions are additions of negative values.
@@Nikioko ok got it
Multiply square root of
negative number,
Don’t directly combine it,
First u should take the
comment factor (i) , let
it become square root of
positive numbers and
then combine it.😅😅
I hope you make a decent living with this.
Haha, me too. Unfortunately, no....
@@mrhtutoring Sorry. I appreciate the videos.
Video is 3:14 long. Coincidence? I think not.
Negative numbers don't a root. Forget all that imaginary nonsense.
Tell that to the electrical engineers. 😁
You’re wrong
👍
✓(-9) x ✓(-4) = 3i x 2i = -6 = -✓36
Could you clarify (please!) why we have the +- in front of the discriminant part of the quadratic formula ?
When solving an equation, we add +-.
In that case we do expect two (possibly identical) solutions.
Example:
x^2 = 4
sqrt(x^2) = sqrt(2^2)
|x| = 2
x = 2 or x = -2
because both 2^2 and (-2)^2 equal 4, and you want to find all possible solutions, not just the convenient one.
In case of sqrt(36) you don't need it. Sure, you can do sqrt(36) = sqrt(6^2) = |6| = 6, but it's redundant, because |6| can only equal 6. There are no multiple solutions.
May I know which chalk piece holder are you using ?
I use the Hagoromo chalk.
@@mrhtutoringthank you
When multiplying two negatives it becomes a positive.
Yes, but taking the square root takes precedence over multiplication, so you shouldn't be multiplying those two numbers first (remember your PEMDAS obedience training?)
With that in mind, you can see that the property sqrt(a) × sqrt(b) = sqrt(a×b) only holds if at least one of a or b is non-negative.
6
i think it is 2i x 3i = 6i^2 = 6 x (-1) = -6?
edit: I make this comment before watching
👍
I did it in my head in just a few seconds, and took a much simpler route: “3i x 2i = 6 x i^2 = 6 x -1 = -6”
I did this too, but I tried justifying 6 also being an anwser and failed to do so, since it's not Q_Q
Sir why sometime root of number is ± and sometime just + whyy
The radical sign specifies that we are only interested in the principal root, unless context specifies otherwise. For roots of positive numbers, the convention is that the positive root is the principal root. The +/- sign in front, indicates that we're interested in using both roots as possible contributions to the rest of the equation.
@@carultch i get it now thanks!
So we only use + or - when we are to solve for x right?
Like x² = 4,
x = +4 or x = -4
Absolutely correct 👍
@@mrhtutoring
Ahhh! Okay. This seems, I think, to answer the question I just posted (even if the correct answers to the posted problem should be x= +2 or x = -2 . . . ). From my question, when using the quadratic formula you are solving for x. Here in the above example you are not solving for x.
why is that so?
@@mrhtutoring
That's the convention.
@@mrhtutoring u mean it's just a thing mathematicians have agreed upon to reduce confusion? or what?
Who ever said it equals to the square root of 36. Before clicking on the solution I got -6.
Now I understand how they calculated Big Bang. Math is flexible and you cannot rely on it solely when doing scientific research.
I think the answer is wrong, and that is basically because you used bad notation. We usually don't write it like that sqrt(-9), but rather -9^(1/2).
if z = -9^(1/2) then z2 = -9 then z = +- 3i
Also -4^(1/2) = +- 2i
(+- 3i) * (+- 2i) = +- (6 * i^2) /* we take both signs here because signs in the left hand side dont have to be synchonized */ = +- 6
Have you heard of principal root?
@@mrhtutoring it is impossible to properly define principle root for negative numbers, because i is defined as i^2 = -1, but (-i)^1 also equals -1. In other words there are 2 numbers such as i^2 = -1.
I don't know about how it's done in your country, but in the USA, √-9 = 3i is standard notation
@@mrhtutoring well may be
But isn't it a strange definition? It makes sence for rational numbers because they are ordered, so you've got negatives, positives and zero in between. It is easy to define a principal root. It also has lots of usages esp in geometry where negative length wouldn't make any sense. Complex numbers are unordered. Definition that I googled is quite long. And is this notation usable?
@@mrhtutoring I'm just curious, what is the "standard" form of √(i) in the US then? or (4)√(-9) (the fourth root of -9)? If you say "ahh those don't have standard form" then comes my following question: why sometimes the "standard" form exists, but sometimes doesn't?
√-9 = 3i is just weird.
this notation will only appear in informal math afaic, where "informal" means one can define anything just based on their taste. those things will probably break math at some point.
-6
1:22: Actually, the definition of i is i² = −1.
The simple reason is to avoid exactly that if i = √−1, then i² = √−1 ⋅ √−1 = √(−1)² = √1 = 1.
i stands for √−1 hence (√−1)(√−1)=i square means (√−1)square, therefore cancel the square with root sign, the final is -1
@@husseinabdulkadir6707 No. The definiton is as I said, for the reason I mentioned:
(√−1)² = √−1 ⋅ √−1 = √[(−1) ⋅ (−1)] = √1 = 1.
In complex numbers, radical sign and square do not simply cancel out.
@@Nikioko
The property that sqrt(a) × sqrt(b) = sqrt(a×b) only holds if at least one of a or b is non-negative.
= -6
Imaginary numbers are 'real'
Of course, -6^2=36. A distinction without a difference.
Only in Excel, or other programs (none that I'm aware of) that give priority to the unary negative operator. In standard math, the convention is that a minus needs to be snared, in order to be squared. (-6)^2 = 36, but -6^2 = -36.