@@JubeiKibagamiFezBut any number to the power of zero is one. An exponential function always crosses the y axis at 1, and it never crosses the x axis, meaning n^x can never be zero.
A) 1/(x+2)=0 => 1=0 Contradiction B) 3^x=0 can never be true, as an exponential is never equal to zero C) (1+x)/(2-x)=-1 (1+x)=-1(2-x) x+1=x-2 Contradiction D) cbrt(x)=-1 x=(-1)³=-1 cbrt(-1)=-1 So the answer is D
Your proof/reasoning for A is a bit incomplete. You are missing the step that in order for a fraction to equal zero, the numerator must be zero. From there, your 1=0 contradiction is apparent. For B, another way to express it is that log of zero (regardless of base) is undefined by definition. Proof for C is also, technically, incomplete. (x) can be cancelled on both sides, leaving you with 1=-2, which is indeed a contradiction.
@@ericgoldman7533 His A isn't incomplete at all, multiply both sides by (x+2) and you directly get 1=0. The fact that a fraction is only zero when the numerator is zero is a generalization of this fact, but it is by no means necessary for this to be "complete".
@@ericgoldman7533 His proof is not incomplete. "for a fraction to equal zero, the numerator must be zero" is not a step, it is a (generalized) result; one which must be proven using axioms and logical reasoning (you're back at square one now having to justify your result with the same type of proof). Even if it _were_ a fundamental axiom, though, you don't need to use any one specific (set of) axiom(s) in order to prove a result. If two entirely different (sets of) axioms can be used to prove the same result (via different paths), then neither proof is "incomplete", they just take different paths (they are equally valid, one may just be more "desirable"). That clarification is a basic one taught to students in any introductory formal proofs/rigorous math uni course, but I am sure you have experienced this yourself already: If you have ever been in a formal proofs course, or even a geometry, it is often, and even expected, for students to take different paths to arrive to the same result, yet simultaneously get full marks. This would not be possible if either proofs were incomplete.
@@ericgoldman7533 1=0 being a contradiction is already apparent, it does not need to follow your step. His proof was a proof by contradiction. It goes like this: "For the sake of a contradiction, assume there exists a number x such that 1/(x+2) = 0." "Then multiplying the equation by the non zero number (x+2) yields 1=0 (which is valid by the "multiplication property of equality", an axiom)." "1=0 is a contradiction, thus proving that 1/(x+2) = 0 has no solution. ∎" Of course, x=-2 is trivial because it's already not in the domain and yields the LHS undefined, which means it can't = 0 either. So no solution. And I left out some formal language for the sake of brevity, but the logical steps after the setup the problem are complete.
@@JubeiKibagamiFez well, what do you mean by a "positive square root"? A "positive square root" can't be equal to i, because that would require that i is positive, which it clearly isn't.
@@xinpingdonohoe3978 The square root of a positive number, therefor a real number greater than 0, could never be a negative number, therefor a real number less than 0. You're suddenly talking about the imaginary number 'i' even though that was not mentioned in the comment, you're saying that a "positive square root" can't be equal to i, which obviously isn't true, as i is defined as the square root of -1, -1 is not a positive number. Additionally, i is neither positive nor negative, and therefor, has no correlation to the original comment at all.
@@westy9447 the point is, in the video the same operator is being applied to get the answer of i, so we can't be using a "positive square root" operator. And there isn't one square root of -1, so it doesn't make sense to say that "i is the square root of -1", in definition of otherwise.
There is no positive and negative concept for complex numbers (at most one can speak about 4 different quadrants), so if you allow extension of the problem to complex plane, this all makes no sense.
This is the kind of problem you'll rarely see in real life, since it's really about the convention of "principal square root" rather than about mathematics.
There is no convention, function can return only one output and it is defined as |x|, if it returned 2 values, it is not a function. I really don't know where this nonsense is coming from. Perhaps study what Df and Hf of any function are, they are explicitly defined.
What I can never remember is whether the principle root has a range arg(√x) ∈ [0, π) or arg(√x) ∈ (-π/2, π/2]. I'm fairly certain it's one of those two, but I can never remember which.
It's a good point, but I think it needs to be said more clearly. This is a matter of definition: The square root *operator notation* specifically means only the positive square root. The number still has 2 square roots, but the notation is only referring to one of them (the positive one). Note that this is why the quadratic formula has the +/- in front of the square root. That wouldn't be necessary if the symbol meant both of the square roots.
That's an important clarification. Note also that i is not positive or negative, so defining square root to just be the positive root doesn't work for cases C and D: the more general definition is to define square root of x as "the complex number z with the smallest principle argument such that z^2=x", where principle argument is the (smallest non-negative) angle between that number and the positive real axis, if you interpret numbers as vectors in the complex plane.
@@rhysbyt And if we are working in the complex realm specifically, we always have to consider principal and secondary roots. That can be confusing to people, learning that the square root of a real number is only the positive option, but that the square root of 3 + 4i is both 2 + i and -2 - i, in that the notation suddenly changes meaning based on the number set you're in. It;s so confusing I still screw it up often, and I get it.
@toddblackmon what is the difference between the positive square root and the normal one? I mean just optical from the notation. How can I tell them both apart? I have never heard from a positive square root. I've always worked with the amount dashes instead
@@bprpmathbasics Thanks for the video! I think that the following answer to the 'question A' may confuse someone: SQRT(X)=-1 => SQRT(i^4) => i^2 = -1. So, x = i^4 (I know that this answer is incorrect).
Answer is D A) x has be to very very large, approaching (positive or negative) infinity. But no matter how large the number is, you *cannot get exactly 0* . B) Same as A. But x has to be negative. C) Simpilfication gives 1 = -2, which is not possible. D) x = -1
A) A fraction is zero when the numerator is zero. The numerator is not equal to zero here. B) x would be log₃(0). But logarithms of zero are undefined.
Can you tell me more about why x=1 isn't a solution to Equation (A)? Kinda seems like I can represent 1 as e^(i*2*pi), and then the square root can work when I substitute that and I can get -1 back as the unambiguous answer to that square root. Maybe my question is, what does this "functional positive square root" operator do when I plug in e^(i*2*pi)? Or, if that is somehow a nonsense question, why is e^(i*2*pi) not a valid Complex value for x when it comes to solving equations like these?
In short, the square root operation takes what is known as the principle square root, which basically is just the most obvious answer, in the case of a square root that means the positive, for a cube root it is a bit different, as the principle root isn't always positive. Why? This is so that you can graph the operation of sqrt(x), if the square root operation had 2 solutions, not just would that be a mess, it would also mean that there would be 2 y values for any x value, meaning f(x)=sqrt(x) would not be a function, it "has" to be a function, because all operation should be defined as such.
@@kentagent6343For things like that it is that many of the square root properties don't translate when you move out of its domain, because you could also say that 1=-1 since 1=√(1)=√(-1×-1)=√(-1)×√(-1)=-1
The radical symbol (√) refers only to the PRINCIPAL root. Positive square roots only apply when the value inside the radical is a positive real number, we are looking for complex solutions here so we cannot refer to this. By definition, the argument of a principal square root lies in the range (-π/2, π/2] so can never be negative. A similar argument applies to your concluding question as the argument of a principal cube root lies in the range (-π/3, π/3] so also cannot be negative. As such, no option has any solutions. Sure, we all learnt that ³√(-1) = -1 at school before we found out about complex numbers but this is technically incorrect, -1 is the real cube root of -1 but not the principal root as required by the radical symbol. The principal root is actually e^(iπ/3).
Whilst it is true that if x^2=1, then x can be either + or - 1, however, that is not quite the same as saying sqrt(1), sqrt(1) takes what is known as the principle root, which is +1, why is this different? So that it can be graphed. If you graph f(x)=x^2, then every y value, except the vertex and any point above or under (depending on which way the graph is flipped) has 2 x values, this in turn means there are 2 solutions for any y, however, if you graph f(x)=sqrt(x), then you might think it would look the same as f(x)=x^2 with x and y flipped, however, an important rule of functions is that it is not a function if there are 2 y values for any x value, this is why the graph actually looks like f(x)=x^2 rotated 90 degrees, and half cut off. In short, sqrt(x) takes the positive root, because f(x)=sqrt(x) has to be able to be expressed as a function.
Yes! All this f(x) = sqrt(x) with graphs only applies when x is a real number. f(z) = z^(1/2) always has 2 solutions in the complex numbers. Correct if I'm wrong, but I seem to recall that if you restrict theta to be in the interval -pi to pi, then f(z) is analytic except at (0,0) and the negative real-axis. Not sure.
The cube root of -1 is -1 so the answer to the question asked at the end is D. A leads to a contradiction by multiplying both sides by the denominator, for B log_3 (0) diverges, and C also leads to a contradiction by multiplying both sides by the denominator and then subtracting X from both sides
@@elladunham9232 Sure. But you'd be forcibly doing the equation too fast, without giving it a little time to grow. Do me a favor. Consider the Express Sqrt(x). Let X = U^4 SQRT(U^4) = U^2 Sqrt(X) = U^2 = -1. I appreciate you humoring this so far. Though I really wish the gentleman in the video would have addressed this part. Humor this last part. (A) If you wanted to algebraically solve for Sqrt(U^4) one would do, Sqrt (U^4) = U^(4/2) = U^2 (B) This Formula applies to all Real and all Complex Numbers. (C) Therefore what is the Sqrt (i^4)? (If you made it this far then ignore everything in the video.) (D) Since Sqrt (U^4) = U^2 for all U then Sqrt (i^4) = i^2. Do you agree or disagree with anything A through D?
My usual rule for understanding complex square root is that the magnitude is the square root or the original magnitude and the angle is half. But to get 180 degree angle, the original angle would have to be 360 degrees, but the angles only go up to less than 360 degrees. So you get 0 angle instead of 360 at 1 and half of 0 is still 0 not 180. This implies that we can't get any square roots who's answer is on the negative side of the complex plane. We can't have sqrt(x)=-i either.
I solved this graphically in my head. Since learning about complex numbers I think of negative numbers as positive numbers rotated in complex plane by 180. Rooting the number divides the angle this number is at. Division by n>1 will result in a smaller angle. In these examples 1 is not rotated, i and -i are 1 rotated by pi/4 and -pi/4 deg while -1 is 1 rotated by pi/2deg. There is no number that can have it's angle divided by 2 that will yield pi/2.
For the last part: A has no solution. 1/(x+2) tends to zero as x approaches infinity or negative infinity, but it never can equal zero. In fact, using horizontal asymptote rules, since the degree of the denominator is larger than the numerator, its horizontal asymptote is 0. B has no solution because an exponential function with base a, a =/= 0, can never output zero. There is no power you can put a number to in order to return zero. We can get close- as x approaches negative infinity- but a^x for any real number a, a =/= 0, can never equal zero. C has no solution because, using horizontal asymptote rules, since the degree of both the numerator and denominator is equal, its asymptote is equal to the ratio of the coefficients of the leading terms; so, 1/(-1) = -1. This means -1 is an asymptote so the function can never output -1. D is the only option which has a solution. The cube root of -1 is -1, so x = -1 is the solution. (-1)(-1)(-1) = -1.
3:55 -- the answer is D). x = -1 is a valid solution. As for the others: A) LHS is a 'unit fraction', i.e., its numerator is fixed to be 1, so that can never be 0. It can approach 0, but never equal 0. No solutions. B) This is a parent exponential function, and that also can't equal 0 but can approach it. No solutions. C) You could cross multiply here and you get 1 + x = x - 2. But that'd be saying 1 = -2...false statement, so no solutions.
But Wolfram Alpha said sqrt1 can be -1 (e^ipi), which is in complex domain. It is just not the principle solution and sqrt1 is very often to be considered an operation in real domain only.
Not exactly. If you ask Wolfram Alpha for sqrt(1), it tells you (correctly) that 1 and -1 are both 2nd roots of 1, but it does not say that both are equal to sqrt(1) (or √1).
WolframAlpha is designed to give all possible answers, which sometimes means the answer isn't exactly correct, e.g. it'll tell you the second roots of 1 is +1 and -1, because (-1)^2 is also 1, however, sqrt(1) is technically taking the principle root of 1, which I believe WolframAlpha will also tell you is only +1, or a complex expression of +1.
D has a solution of x = -1 A, B, and C have no solutions A fraction is only equal to 0 if and only if the numerator is 0 B the exponent operation just can't have zero as an output C the xs cancel leaving a contradiction
It's only because of the definition of the radical sign. Principal Square Root ( i.e. radical sign) definition requires a non-negative real input and outputs the same. In the complex numbers negative output is just fine. That's why you need to specify whether the input is taken to be real or complex. Radical sign only applies to non-negative real numbers. That's why, in complex numbers, we use the notation: (4)^(1/2) = +/-2 instead of sqrt(4) = 2.
For equation A, if x = 1 then √x ≠ -1, but x = (-1)² works. The problem is that the square root (as the inverse of the square function) is a multi-valued function. The square root of a positive number is positive by convention, but if it wasn't for that convention -2 and 2 would be both legitimate square roots of 4. And honestly if we allow a square root to be complex I don't see why we wouldn't allow it to be negative. It's purely conventional. What always bugged me out however is that i is the imaginary number such that, by definition, i² = -1. But as (-i)² also equals to -1, it means that i cannot be distinguished from its opposite by definition 😵💫
It's necessary to have a convention so that the function returns an unique value, but strictly speaking it's not necessary that the chosen value is positive. Even if of course it's more natural. With the opposite convention, sqrt(4) would be -2. But the solutions of the equation x² = 4 would still be sqrt(4) = -2 and -sqrt(4) = +2. Of course that would make you add minus signs in many formulas where only the positive value is intended (like when using Pythagora's theorem), which is very impractical, but the fundamental logic wouldn't change.
I was confused for a second because I thought the question was "Which eq has NO sol," but it was asking which one DOES have a solution. Answer is D. cubedroot(x)^3 = -1^3 x = -1 Plug it back in cubedroot(-1) = -1
Clearly not b or c. 1 has solution in the limit only. d concerns me, if the root symbol means positive root regardless of being cube or square, then there is no answer, but taken as what cubed is -1, then -1.
I thought taking a square root of something always gives two solutions, one positive and one negative. It is the inverse of squaring a number, and since both number and negative of the number squared gives the same outcome, by reversing that step via square root both positive and negative are a solution. 2² => 4 √4 => 2 (-2)² => 4 √4 => -2 At least that is what I remember having learnt in school, but maybe I just remember wrong...
I don't get it. If √4=+/-2, then √1= +/-1. So, if you let X=1 in the answer choice A, how is it wrong that one possible answer to √X is not -1? And, I also don't get how the answer to a negative square root can be a Real Number (√-X=1, answer choice B). Square roots of negative numbers have to be Imaginary (√-4=2i, or √-X=i, as in answer choice D).
I disagree. If you are working with complex numbers, it should be assumed that the roots are complex numbers. Then, sqrt(1+0i) should have two solutions, namely sqrt(1+0i) = exp(πi) = -1+0i and sqrt(1+0i) = exp(2πi) = +1+0i, with the former satisfing your equation.
Unfortunately math doesn't work like this. You gotta think from all branches. Take for instance 1/0. We can assume the answer to be 1 due to 0*1=0 However the reason why it doesn't work is there's technically infinite answers and no answers at the same time. That's why it's called undefined. The way you can tell is by looking at a graph of f(x)=1/x. It just doesn't work since x approaches negative infinity from the left and approaches positive infinity to zero from the right. It's the the same with this. There's not a definitive answer. That's why it's called undefined. You can see it how there's no solution from a complex graph. Even from your viewpoint if we interpret i as 0+i it doesn't work in those graphs. Trust me i wondered the same thing
@@trifortay 1/0 is undefined in the real line, but not on Projectively Extended Real Line, though the answer is not 1, but rather 1/0 = ∞. Likewise, sqrt(1) is not only defined on the complex plane but also has finitely many answers - two in fact - unlike what you claimed. Serch, for instance, "math.libretexts Roots of Complex Numbers" and see Corollary 6.3.1.
@@alanpommer infinity is not the correct answer due to how limits work... now if you asked for the limit of 1/x as x approaches 0 from the right the answer is positive infinity. But that is only if you ask for limits. Limits aren't answers. You learn this in calculus. Simple arithmatic doesn't tell you the whole story. This is why Calculus exists. What you are describing are limits which there are 2 when describing 1/x. The reason why 1/x doesn't have an answer is because the limit of 1/x at x=0 doesn't exist. But they do if you have x approach from either side giving different answers. Math is more complicated than just that. Also the roots of the complex numbers... yeah you do solve them through polar form. However again it's more complicated than that, we're assuming there is a solution to sqrt(x)=-1 is 1. Squaring both sides even in complex form doesn't give you a definitive answer. sure plugging in 1 does yield sqrt(1)=-1 being true. But going backwards it also doesn't work like that. That's why sometimes when solving roots answers sometimes you need to use absolute values since numbers can end up incorrect. It's the same here. It works but not all the way. It's also not specified what specific root we're solving for. It's hard to explain but in calculus it just clicks. Since calculus does make math much easier and a lot of things including roots just makes how they ACTUALLY work just click. Short answer is, it's complicated and it CAN work in the complex world but there is no definitive answer. Again that's why it's called "undefined". Also we can't just assume the answer is asking for a complex number as an answer.
i⁴ = 1 U could also use e^2πi instead of 1, and it would work as -1 too. However, the principal branch only takes the first solution, when k = 0 (the smallest argument) we assume it is the principal branch unless they specifically say ALL SOLUTIONS
@@MikehMike01 Incomplete! e^(2pi*i) = 1. By definition in the complex numbers there are TWO square roots of 1: +1 or -1. So the square roots of e^(2pi*i) are +1 or -1. DeMoirvre: e^(2pi*i) ^(1/2) = (1)^(1/2) = cos(k*pi) + isin(k*pi) for k = 0, k=1 k = 0 yields 1 k = 1 yields -1
positive numbers always have a positive and negative square root, so the operator notation was designed so you could specify which square roots you are referring to: no sign in front of the radical refers to the positive square root, negative sign in front of the radical refers to the negative square root, and +/- sign means both. so there is no solution because positive square roots can’t be negative
@TheFinalChapter - The last time I checked, your answer for: A) x = 1 is very wrong. How can sqrt(1) = -1 ? The Principal Square Root can *not* return a negative number. So, no ...
@@TheFinalChapters Everyone, except for you, knows that the Principal Square Root does *not* return a negative value. You should go back to school and learn about Square Roots before you reply. The symbol √ means Principal Square Root. Maybe, you are just too lazy to learn?
I'm curious. If 0.999... equals 1 then can we do something like 1-0.999...to divide by zero without actually dividing by zero in a limit sort of way. To me 0.999... is just an endless string of 9s followed by a 9. Could we have 0.000...1 with an endless string of zeros followed by a 1. It's just the remainder of 1- 0 999... which equals zero but is infinitesimal and this just seems like a notation issue. What am i missing?
@@MrSummitville yeah, you're right. Dividing by an infinitesimal is the same as multiplying by infinity. It's meaningless. I think limits cover it and the rest of the idea is just misguided and half thought. Sorry.
The radical square root symbol by definition means the _positive_ square root. Hence the quadratic formula having "+/- sqrt(...)" - the sqrt() is positive by definition, thus for both you need the +/-.
A) 1 / (x + 2) = 0: No solution. A fraction is zero when the numerator is zero. But that's not the case here. B) 3ˣ = 0: No solution. x = log₃(0) = lg(0) / lg(3). And logarithm of zero is undefined. C) (1 + x) / (2 − x) = −1. No solution. 1 + x = −(2 − x) = −2 + x. And 1 ≠ −2 D) ³√x = −1 ⇒ x = (−1)³ = −1.
So it looks like I stand to be corrected. That's actually a good thing - for me anyway. Seems that there is such a thing as the "Principal Square Root" of a Complex Number. Did not know that. So 5 + 12i has 2 square roots : 3 - 2i and -3 + 2i. Which one is Principal? and why? It's all in the way you define it. Think about it before using Google. Thanks rex
What would a^2 be then? I'm not saying that a smart mathematician won't be able to rigorously define such a number, but the genius of i is that it's defined as the solution to x^2=-1. The number you're proposing can't be defined in such a way.
@@caspianberggren4195 why not? We're getting in trouble because of a square root of i⁴ being both 1 and -1. We have a field with the defined number epsilon, where (epsilon)²=0. Therefore, nothing can stop us from defining "a" as a solution of √a=-1.
@@Bruh-bk6yoThen by squaring both sides you get a = 1, and you get a contradiction. Complex logarithms do not work the same as the natural one in the positive reals.
Well, for the first it depends on your definition of √. For fhe inverse of x², then all have solutions. As for the second: 1/(x+2)=0→x=1/0 -2, which fails 3^x=0→x=log3(0), which fails (1+x)/(2-x)=-1→3=0, which fails ³√x=-1→x=-1, which succeeds
@@bjorneriksson2404there are more than one definition for every object in Mathematics. The point is: are they standardized? Does anyone use them? Are they convenient?
The second one isn't a convincing argument. You just assert that the log isn't defined at 0, which is of course equivalent, but doesn't convince anybody that there aren't any complex solutions either unless they know all about the complex logarithm (many don't)
@@quantumbuddha777 Except that it's not. The square root function by definition provides the principal square root of its argument, which is always the positive root. sqrt(x^2) is NOT x, it is |x| That's what you are doing, you've taken -1, in the form of e^(pi*i), squared it to get 1 and are now trying to claim that the square root is -1, when the square root of 1 is always 1, no matter how you write it.
Convention. We define sqrt(y) to give the positive solution to y = x^2. If it's specified with a radical or "sqrt", it is positive by convention. If specified indirectly, such as all solutions for x of y=x^2, it could be both solutions.
@@carultch But mathematically it was right to square both sides... mh... So the mistake is the equation itself. If you have a false equation you can do the right maths but will not get the right result... I think I got it. Sorry for asking. Now I feel dumb
I mean I saw that it will be wrong but then I did not understand why his solution would show it is wrong because his maths was right but obviously I am just dumb hahaha
Overall, without regard to this situation, when yiu square the equation, you get equation of a higher order, which may have more solutions than the original one. For instance x=1 equation has, obviously, one solution, but squaring it, x^2=1 has two solutions, one of which does not satisfy the original equation. So in general yiu need to be careful. Here he is just playing on a definition of what he understands as square root.
It is because sqrt(x) needs to be able to be defined as a function, such that f(x)=sqrt(x) is a function, this means, that there can't be 2 roots of a given x, as that would result in there being 2 y values for an x value, which would mean it's not a function.
Also square root is not defined when performing on negative numbers! Only exponentiation such as to the power of 0.5 can be applied to negative numbers, in which case the many roots are considered. The correct answer to the first part is none but B!
@@kro_me "The correct answer to the first part is all but B" implies all of the equations, except B, has no solutions, which is clearly wrong as sqrt(-1)=i even if i is defined as i^2=-1.
for the question A, we have sqrt(x)=-1 sqrt(x)=e^(i*(pi+2*n*pi)) x=e^(i*(2pi+4*n*pi)) 2pi and aadding as much 4*pi as we want, we still getting 1. so, we het x=1 now, if we put 1 into the square root, we get : sqrt(e^(i*(2pi+4*n*pi))), and we just do the thing backward, so we get e^(i*(pi+2*n*pi)) wich is -1. so, what happened here ?
sqrt(1) = {1+0i; -1+0i} if we consider complex numbers, or so I was taught in school 1 is not positive if we consider complex numbers, nor is it negative, neither is -1, nor any other number All of the options would be wrong by my book, because neither is a set of two numbers
@@Mythical_Myths16 are you looking at the assignment at the end, where he asks which one is the only one that has a solution? If not, start looking there.
@@MrSummitville that is not entirely accurate as it could lead to other complications. Square root of x^4 = x^2 for all values of x real and imaginary. If we follow our logic then the quadratic formula should not yield the -ve value of the square root
@@siddharthchabra9022 We are *not* taking the Square Root of x^4. We are taking the the Square Root of X. There is no quadratic equation in this example. There is no x^4 in this example. You are creating a completely different equation. i^4 equals 1. So, no! You clearly do *not* understand the *difference* between a Square Root and roots of polynomial.
@@MrSummitville you misunderstood put x=i^4 then square root of x=i^2 which solves the first equation. since there were no limits placed on what values x can take. the reasoning that the square root of a number has to be positive is incomplete or incorrect. as just like in the quadratic formula we take the positive and negative value of the square root we dont ignore the negative value of the square root
The square root works differently for complex numbers : If z is complex : √z=√|z|.e^(2ikπ/2) √z=√|z|.e^(ikπ) for k in {0;1} Hence, for z=1 : √1=√|1|.e^(2ikπ/2) √1=e^(ikπ) for k in {0;1} =|1|.e^0=1 if k=0 =e^(iπ)=-1 if k=1 Hence, X=1 is a solution of the equation √X=-1
@@madarab tell me then, how do you define a "negative" number of the complex plane ? For example, is i negative ? Is -i ? I'm making a point here, negative/positive numbers make no sense when we're talking complex numbers, there's an infinite number of directions on the complex plane, not just left/right, and the square root function's definition changes as a consequence.
@@brocolive1950 you are not making any point nor any sense, since the function is not defined for for -1. End of story. You are not aware of the basics.
@@madarab the basics change once you consider complex numbers. You would be right if we were considering real numbers only, but it's not the case here. The same way that the exponential, the logarithm, the absolute value, and many other functions, have different definitions on the complex plane than on the real set, the square root's definition changes too. You can look it up on the internet : "square root of complex". The basics are what they are : basics. They're incomplete, and omit a part of the whole story. Complex numbers go just slightly beyond what you can learn from the basics.
@@brocolive1950 the basics don't change at all and you cannot define a new solution via complex numbers if the function is not defined for the number, you do not understand the basics and spitting nonsense. There is a reason you lecture a math teacher on TH-cam, because you know nothing. It is not defined for D(f) and H(f) and no amount of your nonsense will change that, there is no solution. And you have no idea how complex numbers work either and what comple number is. A complex number is expressed by formula a+bi and simply choosing a SINGLE REAL NUMBER from the fomula, in your case pick 1 IS NOT A SOLUTION. Seriously dude, back to reddit.
You are assuming that sqrt(a²) = a, but this is not necessarily true. In particular, if a is negative (as is the case here), then a² is positive and sqrt(a²) is still positive, so it does not equal a.
i^4 is equal to 1, sqrt(1) is only +1. You are going by the assumption that sqrt(i^4) = i^2 = -1, however, that is not entirely true as sqrt takes the principle root, which in case of a square square root, is the positive root.
@@westy9447 i^4 has 2 square roots in the complex numbers. We all realize that i^4 is real, but it is also a complex number and hence has 2 square roots in the complex numbers. Principal square root does not apply here, even though i^4 is also real. We are taking it to be a complex number. Why is this so difficult?
as always, let's forget that you can square both sides of equation A to get x= (-1)^2 wich simplifies to x=1 I know it's hard to believe but the square root of a number has TWO solutions. one positive and one negative. another proof of that hard to believe idea is that ∜1 has FOUR different solutions. (1,-1, i and -i) that's because we can rewrite it as √(√1) or √(±1) which clearly solves as two positive numbers (1 and i) a root can have multiple results. even multiple positive solutions. like it or not. by the way, there are no equations without solution. there may be equations we can't calculate the values of all the variables since we lack enough information (like having a single equation with more than one variable) but all equations have solution. I hoped to find here proof of an unsolvable equation but all I found was somebody that refuses common sense because it doesn't fit what he believes even when given proof of otherwise.
@@WilliamWizer-x3m The answer is still *NO* . The Principal Square Root does *NOT* have two answers - *only* a non-negative answer. It is not hard to believe, because it is a *FACT* that you want to ignore. And *NO* you cannot "Square both sides" , which then creates 2 answers, out of thin air.
@@MrSummitville so ∜1≠i but √-1 =i? makes no sense. but, of course, spitting nonsense like "principal square root" does wonders to defend your point of view. just out of curiosity. is there a principal cube root too? because, for example. ∛8 has 3 answers (even if two of them are complex): 2, −1+√3i and −1-√3i. interesting. cubic roots can have multiple positive results.and, strangely enough, if you include complex solutions like you did in the video, every single cubic root has 3 possible results. and it's damn easy to prove using polar coordinates. by the way, just in case you didn't notice, any equality remains an equality if you aply the same operation on both sides. so yes, I can square both sides and it will still be true.
@@WilliamWizer-x3m Everybody, except you, know that √ means Principal Square Root. If you are too lazy to learn that, then I can't help you. i^4 equals 1. The √1 = 1. End of discussion. You are trying to *pretend* (like a child) the fact that ( i^2 ) equals -1, so that you can *magically* get a negative answer for √x . The answer is still .. Hell No! You childish "word games" don't work with math.
A nice manipulation! But a square root is defined as a non-negarive value for REAL numbers ONLY! And when speaking about COMPLEX number, this isn't applicable, as complex numbers cannot be negative nor positive, at all. By the way, in your examples C and D the sq.roots are complex (imaginary) - neither negative nor positive. So, x=1=1+0*i=exp(i*(2П+4Пk)) is indeed the solution of the example A. And x=1=1+0*i=exp(i*4Пk) is not...
No, the square root function is.. a function. And a function only has one outcome. You are confusing calculation square roots with solving quadratic equations.
No, the square root function is… a function. And a function only has one outcome. You are confusing calculation square roots with solving quadratic equations.
For equation A this is what I got: since sqrt(x) = x ^(0.5), and since e^(i pi) = -1, then sqrt(x) = -1 can be written as: x ^ 0.5 = e ^ i pi 0.5 lnx = i pi ln x = 2i pi x = (e^2ipi) = (e^ipi)^2 = (-1)^2 = 1 which would mean that sqrt(1) = -1. While it may not be the principle root, that would be sqrt (1) = 1, it still seems like a valid solution to me.
About C: A fine example of sloppy math. Since by definition i² = -1, it should be x² = -1 → x² = i² → √x² = √i² → so x = i. Please include the required middle steps.
Where did you get x^2 = i^2 or x^2 = -1 from? Both of those equations are wrong because x = -1. So x^2 would be 1. Also if we plug in your final solution we get that sqrt(i) is i which is wrong
@@NLGeebee i is a number. It is a constant whose value squared is equal to -1. In addition the principal square root of -1 is i. Idk where you heard that i is not a number, but every source i can find online says it’s a number.
You will have to define the principal branch of sqrt root before you proceed.Also if you are working with complex numbers,please use z instead. This notation confuses students. You have a large channel and you should follow the standard notation. Also remember how exponent work in the complex set.You can not simply square when you want and what you want.For instance students are accepeting that 1=-1 because we can start with -1=-1 =(-1)^1=(-1)^2/2=((-1)^2)^1/2=1^1/2=1.This is wrong of course.But students are squaring and taking square root everywhere with no rules.
This video is just wrong though. The expression "i = sqrt(-1)" is a convenient intuition, but not really true. If you're allowing negative numbers under a square root sign, then you're implicitly choosing a branch of the complex logarithm to define the square root, since a^b is defined to be e^(b*log(a)) when a is not positive. If youre allowing that, then sqrt(x)=-1 is absolutely possible for a given branch of the logarithm.
But it's not wrong. A radical results in a positive result. That's just the standard. Yes, you can defy the standard and say it results in a negative number, but if you're not gonna follow the standard that's just a whole different story.
3:55
I guess D as cbrt (-1) =-1
Could U give pf of the identity of sin(x/3) or cos of it.
*If it exists*
It's D. ()^3 to both sides. The cube root and the ^3 will cancel leaving just x. On the right side, (-1)^3 is just -1.
B has the solution. 3⁰=0
@@JubeiKibagamiFezBut any number to the power of zero is one. An exponential function always crosses the y axis at 1, and it never crosses the x axis, meaning n^x can never be zero.
A) 1/(x+2)=0 => 1=0
Contradiction
B) 3^x=0 can never be true, as an exponential is never equal to zero
C) (1+x)/(2-x)=-1
(1+x)=-1(2-x)
x+1=x-2
Contradiction
D) cbrt(x)=-1
x=(-1)³=-1
cbrt(-1)=-1
So the answer is D
Your proof/reasoning for A is a bit incomplete. You are missing the step that in order for a fraction to equal zero, the numerator must be zero. From there, your 1=0 contradiction is apparent.
For B, another way to express it is that log of zero (regardless of base) is undefined by definition.
Proof for C is also, technically, incomplete. (x) can be cancelled on both sides, leaving you with 1=-2, which is indeed a contradiction.
@@ericgoldman7533 His A isn't incomplete at all, multiply both sides by (x+2) and you directly get 1=0. The fact that a fraction is only zero when the numerator is zero is a generalization of this fact, but it is by no means necessary for this to be "complete".
@@ericgoldman7533 His proof is not incomplete. "for a fraction to equal zero, the numerator must be zero" is not a step, it is a (generalized) result; one which must be proven using axioms and logical reasoning (you're back at square one now having to justify your result with the same type of proof).
Even if it _were_ a fundamental axiom, though, you don't need to use any one specific (set of) axiom(s) in order to prove a result. If two entirely different (sets of) axioms can be used to prove the same result (via different paths), then neither proof is "incomplete", they just take different paths (they are equally valid, one may just be more "desirable"). That clarification is a basic one taught to students in any introductory formal proofs/rigorous math uni course, but I am sure you have experienced this yourself already: If you have ever been in a formal proofs course, or even a geometry, it is often, and even expected, for students to take different paths to arrive to the same result, yet simultaneously get full marks. This would not be possible if either proofs were incomplete.
@@ericgoldman7533 1=0 being a contradiction is already apparent, it does not need to follow your step. His proof was a proof by contradiction. It goes like this:
"For the sake of a contradiction, assume there exists a number x such that 1/(x+2) = 0."
"Then multiplying the equation by the non zero number (x+2) yields 1=0 (which is valid by the "multiplication property of equality", an axiom)."
"1=0 is a contradiction, thus proving that 1/(x+2) = 0 has no solution. ∎"
Of course, x=-2 is trivial because it's already not in the domain and yields the LHS undefined, which means it can't = 0 either. So no solution. And I left out some formal language for the sake of brevity, but the logical steps after the setup the problem are complete.
And the two complex solutions for D: 1/2 + ((√3)/2) i and 1/2 - ((√3)/2) i
I never would have known that a positive square root can never equal a negative number, but the second you pointed it out, it just clicked.
@@JubeiKibagamiFez well, what do you mean by a "positive square root"? A "positive square root" can't be equal to i, because that would require that i is positive, which it clearly isn't.
@@xinpingdonohoe3978 The square root of a positive number, therefor a real number greater than 0, could never be a negative number, therefor a real number less than 0.
You're suddenly talking about the imaginary number 'i' even though that was not mentioned in the comment, you're saying that a "positive square root" can't be equal to i, which obviously isn't true, as i is defined as the square root of -1, -1 is not a positive number.
Additionally, i is neither positive nor negative, and therefor, has no correlation to the original comment at all.
just like me ❤
@@westy9447 the point is, in the video the same operator is being applied to get the answer of i, so we can't be using a "positive square root" operator.
And there isn't one square root of -1, so it doesn't make sense to say that "i is the square root of -1", in definition of otherwise.
There is no positive and negative concept for complex numbers (at most one can speak about 4 different quadrants), so if you allow extension of the problem to complex plane, this all makes no sense.
This is the kind of problem you'll rarely see in real life, since it's really about the convention of "principal square root" rather than about mathematics.
There is no convention, function can return only one output and it is defined as |x|, if it returned 2 values, it is not a function. I really don't know where this nonsense is coming from. Perhaps study what Df and Hf of any function are, they are explicitly defined.
@@madarab The principal square root of a positive number could as easily be chosen to be always negative.
@@MattMcIrvinSo if question asks what satisfies x - 5 = 0 I can just change it to x - pi = 0 and get pi
What I can never remember is whether the principle root has a range arg(√x) ∈ [0, π) or arg(√x) ∈ (-π/2, π/2].
I'm fairly certain it's one of those two, but I can never remember which.
It's a good point, but I think it needs to be said more clearly. This is a matter of definition: The square root *operator notation* specifically means only the positive square root. The number still has 2 square roots, but the notation is only referring to one of them (the positive one).
Note that this is why the quadratic formula has the +/- in front of the square root. That wouldn't be necessary if the symbol meant both of the square roots.
That's an important clarification. Note also that i is not positive or negative, so defining square root to just be the positive root doesn't work for cases C and D: the more general definition is to define square root of x as "the complex number z with the smallest principle argument such that z^2=x", where principle argument is the (smallest non-negative) angle between that number and the positive real axis, if you interpret numbers as vectors in the complex plane.
@@rhysbyt And if we are working in the complex realm specifically, we always have to consider principal and secondary roots. That can be confusing to people, learning that the square root of a real number is only the positive option, but that the square root of 3 + 4i is both 2 + i and -2 - i, in that the notation suddenly changes meaning based on the number set you're in. It;s so confusing I still screw it up often, and I get it.
No it doesn't stop with this nonsense. It can have only one root otherwise it is not a function.
@toddblackmon what is the difference between the positive square root and the normal one? I mean just optical from the notation. How can I tell them both apart? I have never heard from a positive square root. I've always worked with the amount dashes instead
@@paulhein9815 The difference between the positive square root and the negative square root is twice the positive square root. :D
Just want to say respects man, you’re doing a great job helping math students ✌️
Thank you!
@@bprpmathbasics Thanks for the video!
I think that the following answer to the 'question A' may confuse someone:
SQRT(X)=-1 => SQRT(i^4) => i^2 = -1. So, x = i^4 (I know that this answer is incorrect).
Answer is D
A) x has be to very very large, approaching (positive or negative) infinity. But no matter how large the number is, you *cannot get exactly 0* .
B) Same as A. But x has to be negative.
C) Simpilfication gives 1 = -2, which is not possible.
D) x = -1
whAT HAPPEND we aplied the answer to x
@@DragonReallynothing changed
@@DragonReally that's would be infinite possibility
A) A fraction is zero when the numerator is zero. The numerator is not equal to zero here.
B) x would be log₃(0). But logarithms of zero are undefined.
@@Nikioko I was about to make this exact comment
The problem with problems like this is that it doesn't actually say anything mathematically. It's purely an exercise in definition and convention.
D: x^(1/3) = -1
x = -1 (+ 2 complex roots)
Is the answer to the question D?
Can you tell me more about why x=1 isn't a solution to Equation (A)? Kinda seems like I can represent 1 as e^(i*2*pi), and then the square root can work when I substitute that and I can get -1 back as the unambiguous answer to that square root. Maybe my question is, what does this "functional positive square root" operator do when I plug in e^(i*2*pi)? Or, if that is somehow a nonsense question, why is e^(i*2*pi) not a valid Complex value for x when it comes to solving equations like these?
the square root of e^(2iπ) is 1
In short, the square root operation takes what is known as the principle square root, which basically is just the most obvious answer, in the case of a square root that means the positive, for a cube root it is a bit different, as the principle root isn't always positive.
Why? This is so that you can graph the operation of sqrt(x), if the square root operation had 2 solutions, not just would that be a mess, it would also mean that there would be 2 y values for any x value, meaning f(x)=sqrt(x) would not be a function, it "has" to be a function, because all operation should be defined as such.
@@westy9447How about i^4? Shouldnt the sqrt(i^4) = i^2 = -1?
@@kentagent6343For things like that it is that many of the square root properties don't translate when you move out of its domain, because you could also say that 1=-1 since 1=√(1)=√(-1×-1)=√(-1)×√(-1)=-1
@@MikehMike01 Sorry to be blunt but, wrong. e^(2pi*i) = 1 and the square roots of e^(2pi*i) are 1 or -1.
The radical symbol (√) refers only to the PRINCIPAL root. Positive square roots only apply when the value inside the radical is a positive real number, we are looking for complex solutions here so we cannot refer to this.
By definition, the argument of a principal square root lies in the range (-π/2, π/2] so can never be negative.
A similar argument applies to your concluding question as the argument of a principal cube root lies in the range (-π/3, π/3] so also cannot be negative. As such, no option has any solutions.
Sure, we all learnt that ³√(-1) = -1 at school before we found out about complex numbers but this is technically incorrect, -1 is the real cube root of -1 but not the principal root as required by the radical symbol. The principal root is actually e^(iπ/3).
but isn't a complex number has n n-th roots, so 2 square roots of 1 are 1 and -1?
Whilst it is true that if x^2=1, then x can be either + or - 1, however, that is not quite the same as saying sqrt(1), sqrt(1) takes what is known as the principle root, which is +1, why is this different? So that it can be graphed.
If you graph f(x)=x^2, then every y value, except the vertex and any point above or under (depending on which way the graph is flipped) has 2 x values, this in turn means there are 2 solutions for any y, however, if you graph f(x)=sqrt(x), then you might think it would look the same as f(x)=x^2 with x and y flipped, however, an important rule of functions is that it is not a function if there are 2 y values for any x value, this is why the graph actually looks like f(x)=x^2 rotated 90 degrees, and half cut off.
In short, sqrt(x) takes the positive root, because f(x)=sqrt(x) has to be able to be expressed as a function.
Yes! All this f(x) = sqrt(x) with graphs only applies when x is a real number. f(z) = z^(1/2) always has 2 solutions in the complex numbers. Correct if I'm wrong, but I seem to recall that if you restrict theta to be in the interval -pi to pi, then f(z) is analytic except at (0,0) and the negative real-axis. Not sure.
The cube root of -1 is -1 so the answer to the question asked at the end is D. A leads to a contradiction by multiplying both sides by the denominator, for B log_3 (0) diverges, and C also leads to a contradiction by multiplying both sides by the denominator and then subtracting X from both sides
Can you explain why for A the answer is not X = i^4?
@@Quadratic4mula i^4 is 1. If we plug that into the equation we get 1/(1 +2) = 0 which is false
@@elladunham9232 for the first half of the video. I mean. Not the second half.
@@Quadratic4mula i^4 is 1, the principal square root of 1 is 1, therefore the equation has no solution
@@elladunham9232 Sure. But you'd be forcibly doing the equation too fast, without giving it a little time to grow.
Do me a favor.
Consider the Express Sqrt(x).
Let X = U^4
SQRT(U^4) = U^2
Sqrt(X) = U^2 = -1.
I appreciate you humoring this so far.
Though I really wish the gentleman in the video would have addressed this part.
Humor this last part.
(A) If you wanted to algebraically solve for Sqrt(U^4) one would do, Sqrt (U^4) = U^(4/2) = U^2
(B) This Formula applies to all Real and all Complex Numbers.
(C) Therefore what is the Sqrt (i^4)? (If you made it this far then ignore everything in the video.)
(D) Since Sqrt (U^4) = U^2 for all U then Sqrt (i^4) = i^2.
Do you agree or disagree with anything A through D?
My usual rule for understanding complex square root is that the magnitude is the square root or the original magnitude and the angle is half. But to get 180 degree angle, the original angle would have to be 360 degrees, but the angles only go up to less than 360 degrees. So you get 0 angle instead of 360 at 1 and half of 0 is still 0 not 180. This implies that we can't get any square roots who's answer is on the negative side of the complex plane. We can't have sqrt(x)=-i either.
Sqrt(i^4) is a sol for A no?
No because i⁴ is just 1
@kornelviktor6985 ok but complex square roots have infinite values therefore -1 would be a valid solution since i^(4/2)=i^2=-1
@@redroach401No! Because by *definition* the Principal Square Root can *not* return a negative number. ( i^2 ) = -1. So, no ...
I solved this graphically in my head. Since learning about complex numbers I think of negative numbers as positive numbers rotated in complex plane by 180. Rooting the number divides the angle this number is at. Division by n>1 will result in a smaller angle. In these examples 1 is not rotated, i and -i are 1 rotated by pi/4 and -pi/4 deg while -1 is 1 rotated by pi/2deg. There is no number that can have it's angle divided by 2 that will yield pi/2.
For the initial question: Even the first equasion can have a solution if you imagine a third dimension.
It's i⁴, but you must not equal it to 1.
For the last part:
A has no solution. 1/(x+2) tends to zero as x approaches infinity or negative infinity, but it never can equal zero. In fact, using horizontal asymptote rules, since the degree of the denominator is larger than the numerator, its horizontal asymptote is 0.
B has no solution because an exponential function with base a, a =/= 0, can never output zero. There is no power you can put a number to in order to return zero. We can get close- as x approaches negative infinity- but a^x for any real number a, a =/= 0, can never equal zero.
C has no solution because, using horizontal asymptote rules, since the degree of both the numerator and denominator is equal, its asymptote is equal to the ratio of the coefficients of the leading terms; so, 1/(-1) = -1. This means -1 is an asymptote so the function can never output -1.
D is the only option which has a solution. The cube root of -1 is -1, so x = -1 is the solution. (-1)(-1)(-1) = -1.
I guess D as cbrt (-1) =-1
why can we not write 1 as i^4? or -1 as i^2?
3:55 -- the answer is D). x = -1 is a valid solution. As for the others:
A) LHS is a 'unit fraction', i.e., its numerator is fixed to be 1, so that can never be 0. It can approach 0, but never equal 0. No solutions.
B) This is a parent exponential function, and that also can't equal 0 but can approach it. No solutions.
C) You could cross multiply here and you get 1 + x = x - 2. But that'd be saying 1 = -2...false statement, so no solutions.
Sqrt (x)=-1=i^2 >> (Sqrt (x))^2=(i^2)^2 >> x=i^4
Check: Sqrt (i^4)=i^2=-1
About A: what if x = i⁴? √i⁴ = i² which is -1.
No! Because by *definition* the Principal Square Root can *not* return a negative number. ( i^2 ) = -1. So, no ...
Could it be e^(2*pi*i*n), only for n=odd integer ? That is, excluding 1 and others from the domain.
√x=-1
The square root function's range is y>0
But Wolfram Alpha said sqrt1 can be -1 (e^ipi), which is in complex domain. It is just not the principle solution and sqrt1 is very often to be considered an operation in real domain only.
when the square root sign appears in an equation it is always the principle value
Not exactly. If you ask Wolfram Alpha for sqrt(1), it tells you (correctly) that 1 and -1 are both 2nd roots of 1, but it does not say that both are equal to sqrt(1) (or √1).
@@Steve_Stowers Yes, but that is an error in Wolfram Alpha.
WolframAlpha is designed to give all possible answers, which sometimes means the answer isn't exactly correct, e.g. it'll tell you the second roots of 1 is +1 and -1, because (-1)^2 is also 1, however, sqrt(1) is technically taking the principle root of 1, which I believe WolframAlpha will also tell you is only +1, or a complex expression of +1.
@@Emilia333g In such a case the variable must be a non-negative real number and hence the Principal Value applies.
So why isn't x= i⁴ a solution for A? i^(4/2) = i² = -1 or what did i miss?
sir why you can't multiply both sides of an equation by zero?
would you be able to explain this? thank you!!
Nothing is preventing you from doing that. It's just that changing an equation to 0 = 0 is rarely useful.
@@Llortnerof in theory it allows you to change a false equation into a true one?
@@CalebSu-pv1bj
0*a = 0*b => 0 = 0 doesn't imply that a = b, since you would divide by zero to make that conclusion.
@@CalebSu-pv1bj It's really closer to just erasing the original equation. That doesn't make it true, just no longer present.
What about i^4 ?
1/(x+2) = 0 => 1 = 0. False.
3^x = 0 => x = log_3(0). log_3(0) is undefined
(1+x)/(2-x) = -1 => 1+x = x-2 => 1 = -2. False.
³sqrt(x) = -1 => x = (-1)³ = -1
This is Number Theory. If x isna complex number then a) can eg a x:=exp(ipi) be correct?
isnt x in a in the first equations is i?
What about sqrt(x)=-i?
What about polar notation?
The cube root one, too obvious as I don't even need to try to solve it, just knowledge that all cube roots have solutions
Multiple options can be correct. Not the case here, but can be some time.
@@daakudaddy5453 No, this one has only one answer, all others are "can approach but not exactly equal"
Non of them has a solution
Why? @@Ahmed-kg2gf
@@Ahmed-kg2gfD has a solution
If one considers sqrt sign as a operator for positive reals, then only B has a solution, namely x = -1.
*Positive* Reals only? There is no such restriction. You made that up! B, C & D all have real solutions.
D has a solution of x = -1
A, B, and C have no solutions
A fraction is only equal to 0 if and only if the numerator is 0
B the exponent operation just can't have zero as an output
C the xs cancel leaving a contradiction
Why disallow the square root to output negative numbers but not nonreal numbers (i, -i)? I’m not sure about the logic
It's only because of the definition of the radical sign. Principal Square Root ( i.e. radical sign) definition requires a non-negative real input and outputs the same. In the complex numbers negative output is just fine. That's why you need to specify whether the input is taken to be real or complex. Radical sign only applies to non-negative real numbers. That's why, in complex numbers, we use the notation: (4)^(1/2) = +/-2 instead of sqrt(4) = 2.
For equation A, if x = 1 then √x ≠ -1, but x = (-1)² works. The problem is that the square root (as the inverse of the square function) is a multi-valued function. The square root of a positive number is positive by convention, but if it wasn't for that convention -2 and 2 would be both legitimate square roots of 4. And honestly if we allow a square root to be complex I don't see why we wouldn't allow it to be negative. It's purely conventional.
What always bugged me out however is that i is the imaginary number such that, by definition, i² = -1. But as (-i)² also equals to -1, it means that i cannot be distinguished from its opposite by definition 😵💫
"positive by convention"
It's positive by necessity. If it weren't, then it'd cease to be a function *at all*
It's necessary to have a convention so that the function returns an unique value, but strictly speaking it's not necessary that the chosen value is positive. Even if of course it's more natural. With the opposite convention, sqrt(4) would be -2. But the solutions of the equation x² = 4 would still be sqrt(4) = -2 and -sqrt(4) = +2. Of course that would make you add minus signs in many formulas where only the positive value is intended (like when using Pythagora's theorem), which is very impractical, but the fundamental logic wouldn't change.
I was confused for a second because I thought the question was "Which eq has NO sol," but it was asking which one DOES have a solution. Answer is D.
cubedroot(x)^3 = -1^3
x = -1
Plug it back in
cubedroot(-1) = -1
No, only "A" has *no* solution.
Clearly not b or c. 1 has solution in the limit only. d concerns me, if the root symbol means positive root regardless of being cube or square, then there is no answer, but taken as what cubed is -1, then -1.
I thought taking a square root of something always gives two solutions, one positive and one negative.
It is the inverse of squaring a number, and since both number and negative of the number squared gives the same outcome, by reversing that step via square root both positive and negative are a solution.
2² => 4 √4 => 2
(-2)² => 4 √4 => -2
At least that is what I remember having learnt in school, but maybe I just remember wrong...
Why i^4 is not a solution for A ?
No! Because by *definition* the Principal Square Root can *not* return a negative number. ( i^2 ) = -1. So, no ...
I don't get it. If √4=+/-2, then √1= +/-1. So, if you let X=1 in the answer choice A, how is it wrong that one possible answer to √X is not -1? And, I also don't get how the answer to a negative square root can be a Real Number (√-X=1, answer choice B). Square roots of negative numbers have to be Imaginary (√-4=2i, or √-X=i, as in answer choice D).
Your *IF* statement is *FALSE* . √4 = +2 !!!
A) x= i^4; √x = i^(4/2) = i^2 = -1
³√x = -1
X = -1 ( inside cuberoot can be a negative number)
It has to be A because that is the only equation in which the value on the right can't be a principal square root.
I disagree. If you are working with complex numbers, it should be assumed that the roots are complex numbers. Then, sqrt(1+0i) should have two solutions, namely sqrt(1+0i) = exp(πi) = -1+0i and sqrt(1+0i) = exp(2πi) = +1+0i, with the former satisfing your equation.
Unfortunately math doesn't work like this. You gotta think from all branches. Take for instance 1/0. We can assume the answer to be 1 due to 0*1=0 However the reason why it doesn't work is there's technically infinite answers and no answers at the same time. That's why it's called undefined. The way you can tell is by looking at a graph of f(x)=1/x. It just doesn't work since x approaches negative infinity from the left and approaches positive infinity to zero from the right. It's the the same with this. There's not a definitive answer. That's why it's called undefined. You can see it how there's no solution from a complex graph. Even from your viewpoint if we interpret i as 0+i it doesn't work in those graphs. Trust me i wondered the same thing
@@trifortay 1/0 is undefined in the real line, but not on Projectively Extended Real Line, though the answer is not 1, but rather 1/0 = ∞.
Likewise, sqrt(1) is not only defined on the complex plane but also has finitely many answers - two in fact - unlike what you claimed. Serch, for instance, "math.libretexts Roots of Complex Numbers" and see Corollary 6.3.1.
@@alanpommer infinity is not the correct answer due to how limits work... now if you asked for the limit of 1/x as x approaches 0 from the right the answer is positive infinity. But that is only if you ask for limits. Limits aren't answers. You learn this in calculus.
Simple arithmatic doesn't tell you the whole story. This is why Calculus exists. What you are describing are limits which there are 2 when describing 1/x. The reason why 1/x doesn't have an answer is because the limit of 1/x at x=0 doesn't exist. But they do if you have x approach from either side giving different answers.
Math is more complicated than just that. Also the roots of the complex numbers... yeah you do solve them through polar form. However again it's more complicated than that, we're assuming there is a solution to sqrt(x)=-1 is 1. Squaring both sides even in complex form doesn't give you a definitive answer. sure plugging in 1 does yield sqrt(1)=-1 being true. But going backwards it also doesn't work like that. That's why sometimes when solving roots answers sometimes you need to use absolute values since numbers can end up incorrect. It's the same here. It works but not all the way. It's also not specified what specific root we're solving for. It's hard to explain but in calculus it just clicks. Since calculus does make math much easier and a lot of things including roots just makes how they ACTUALLY work just click.
Short answer is, it's complicated and it CAN work in the complex world but there is no definitive answer. Again that's why it's called "undefined". Also we can't just assume the answer is asking for a complex number as an answer.
Wait, don’t some square roots have two solutions? Like sqrt(4) can be +- 2, so why can't sqrt(1) be +- 1?
No! Because by *definition* the Principal Square Root can *not* return a negative number. So, no ...
X=i^4 would get you -1 for problem A. Why would that not work?
i⁴ = 1
U could also use e^2πi instead of 1, and it would work as -1 too. However, the principal branch only takes the first solution, when k = 0 (the smallest argument)
we assume it is the principal branch unless they specifically say ALL SOLUTIONS
D has a solution and the answer is -1.
What is the square root of i^4?
No! Because by *definition* the Principal Square Root can *not* return a negative number. ( i^2 ) = -1. So, no ...
Given e^(pi*i) = -1, sqrt(x) = -1 => x = e^(2*pi*i)
the square root of e^(2iπ) is 1
sqrt(x) takes the principle root, in case of square roots, that is always positive.
@@MikehMike01 Incomplete! e^(2pi*i) = 1. By definition in the complex numbers there are TWO square roots of 1: +1 or -1. So the square roots of e^(2pi*i) are +1 or -1.
DeMoirvre: e^(2pi*i) ^(1/2) = (1)^(1/2) = cos(k*pi) + isin(k*pi) for k = 0, k=1
k = 0 yields 1
k = 1 yields -1
@@ianfowler9340the principal branch root (normal sqrt symbol) takes k=0 only
Last time I checked, -1 squared was 1. As for the rest:
(A) x = 1
(B) x = -1
(C) x = -1
(D) x = 1
positive numbers always have a positive and negative square root, so the operator notation was designed so you could specify which square roots you are referring to: no sign in front of the radical refers to the positive square root, negative sign in front of the radical refers to the negative square root, and +/- sign means both. so there is no solution because positive square roots can’t be negative
@TheFinalChapter - The last time I checked, your answer for: A) x = 1 is very wrong. How can sqrt(1) = -1 ? The Principal Square Root can *not* return a negative number. So, no ...
@@MrSummitville The square root of 1 is both -1 and 1. No one cares about the "principal" square root.
@@TheFinalChapters Everyone, except for you, knows that the Principal Square Root does *not* return a negative value. You should go back to school and learn about Square Roots before you reply. The symbol √ means Principal Square Root. Maybe, you are just too lazy to learn?
I'm curious. If 0.999... equals 1 then can we do something like 1-0.999...to divide by zero without actually dividing by zero in a limit sort of way.
To me 0.999... is just an endless string of 9s followed by a 9. Could we have 0.000...1 with an endless string of zeros followed by a 1. It's just the remainder of 1- 0 999... which equals zero but is infinitesimal and this just seems like a notation issue.
What am i missing?
No!
@@MrSummitville yeah, you're right. Dividing by an infinitesimal is the same as multiplying by infinity. It's meaningless.
I think limits cover it and the rest of the idea is just misguided and half thought.
Sorry.
A.
√x > or = 0
A, because if C has a solution than D also has a solution while B just works with x = -1
I should have been taking notes
All my life I've been told that a square root of a positive number has both the positive and negative solutions 😮
Yes. But that's not the question. √x is non-negative, and -√x is non-positive. So, √x < 0 has no solution, as well as -√x > 0.
The radical square root symbol by definition means the _positive_ square root.
Hence the quadratic formula having "+/- sqrt(...)" - the sqrt() is positive by definition, thus for both you need the +/-.
@@cigmorfil4101 Exactly. If y = x², then x₁ = √y and x₂ = −√y. With √y ≥ 0 and −√y ≤ 0, respectively.
That’s completely wrong, whoever told you that is incorrect
@@MikehMike01 Unfortunately, that is indeed the way many schools introduce the square root.
A) 1 / (x + 2) = 0: No solution. A fraction is zero when the numerator is zero. But that's not the case here.
B) 3ˣ = 0: No solution. x = log₃(0) = lg(0) / lg(3). And logarithm of zero is undefined.
C) (1 + x) / (2 − x) = −1. No solution. 1 + x = −(2 − x) = −2 + x. And 1 ≠ −2
D) ³√x = −1 ⇒ x = (−1)³ = −1.
So it looks like I stand to be corrected. That's actually a good thing - for me anyway. Seems that there is such a thing as the "Principal Square Root" of a Complex Number. Did not know that. So 5 + 12i has 2 square roots : 3 - 2i and -3 + 2i. Which one is Principal? and why? It's all in the way you define it. Think about it before using Google. Thanks rex
Ok, but WHAT if there is SUCH number a, so √a=-1?
What would a^2 be then?
I'm not saying that a smart mathematician won't be able to rigorously define such a number, but the genius of i is that it's defined as the solution to x^2=-1. The number you're proposing can't be defined in such a way.
@@caspianberggren4195 why not? We're getting in trouble because of a square root of i⁴ being both 1 and -1. We have a field with the defined number epsilon, where (epsilon)²=0. Therefore, nothing can stop us from defining "a" as a solution of √a=-1.
@@Bruh-bk6yoThen by squaring both sides you get a = 1, and you get a contradiction. Complex logarithms do not work the same as the natural one in the positive reals.
@@giovanni1946 i⁴=1
Find a square root of i⁴ and you'll get another example of contradiction.
@@giovanni1946 these "imaginary" numbers are tended to contradict the operations based on real numbers, so this doesn't matter.
Well, for the first it depends on your definition of √. For fhe inverse of x², then all have solutions.
As for the second:
1/(x+2)=0→x=1/0 -2, which fails
3^x=0→x=log3(0), which fails
(1+x)/(2-x)=-1→3=0, which fails
³√x=-1→x=-1, which succeeds
"...it depends..." That would imply that there is more than one definition of the radical sign, which I believe isn't the case.
@@bjorneriksson2404there are more than one definition for every object in Mathematics. The point is: are they standardized? Does anyone use them? Are they convenient?
@megachonker4173 Yes, sure. I was just too lazy to write out the rest... 🙂 (accepted throughout the math community, standardized, used, etc)
@@bjorneriksson2404the radical sign literally does have more than one meaning... and the context here is poorly specified if at all
The second one isn't a convincing argument. You just assert that the log isn't defined at 0, which is of course equivalent, but doesn't convince anybody that there aren't any complex solutions either unless they know all about the complex logarithm (many don't)
@0:33. Why not x = i^4
because i^4 = 1
@TazPessle - No! Because by *definition* the Principal Square Root can *not* return a negative number. ( i^2 ) = -1. So, no ...
What about x=e^(2pi*i) as a solution to a?
e^(2pi*i) is equal to +1, the square root of it is also +1.
@@westy9447 I agree that e^(2pi*i)=1, however, its square root equals e^(pi*i), which is -1
@@quantumbuddha777 Except that it's not. The square root function by definition provides the principal square root of its argument, which is always the positive root.
sqrt(x^2) is NOT x, it is |x|
That's what you are doing, you've taken -1, in the form of e^(pi*i), squared it to get 1 and are now trying to claim that the square root is -1, when the square root of 1 is always 1, no matter how you write it.
@@phiefer3 agree to disagree
@@quantumbuddha777 What are you disagreeing with?
are you actually claiming that -1=1?
Why does the first one not work... I mean I saw it that it doesnt but which rule was broken if you understand what I mean?
Convention. We define sqrt(y) to give the positive solution to y = x^2. If it's specified with a radical or "sqrt", it is positive by convention. If specified indirectly, such as all solutions for x of y=x^2, it could be both solutions.
@@carultch But mathematically it was right to square both sides... mh... So the mistake is the equation itself. If you have a false equation you can do the right maths but will not get the right result... I think I got it. Sorry for asking. Now I feel dumb
I mean I saw that it will be wrong but then I did not understand why his solution would show it is wrong because his maths was right but obviously I am just dumb hahaha
Overall, without regard to this situation, when yiu square the equation, you get equation of a higher order, which may have more solutions than the original one. For instance x=1 equation has, obviously, one solution, but squaring it, x^2=1 has two solutions, one of which does not satisfy the original equation. So in general yiu need to be careful. Here he is just playing on a definition of what he understands as square root.
It is because sqrt(x) needs to be able to be defined as a function, such that f(x)=sqrt(x) is a function, this means, that there can't be 2 roots of a given x, as that would result in there being 2 y values for an x value, which would mean it's not a function.
i is defined as i^2=-1 not i=sqrt(-1) !! It is important to make this distinction!
Also square root is not defined when performing on negative numbers! Only exponentiation such as to the power of 0.5 can be applied to negative numbers, in which case the many roots are considered. The correct answer to the first part is none but B!
@@kro_me "The correct answer to the first part is all but B" implies all of the equations, except B, has no solutions, which is clearly wrong as sqrt(-1)=i even if i is defined as i^2=-1.
@@westy9447 sorry yeah i meant none but B
@@forbidden-cyrillic-handle yeah that is why i and -i are indistinguishable
we say that both are correct solutions, where if you label one as i the other would be -i
for the question A, we have
sqrt(x)=-1
sqrt(x)=e^(i*(pi+2*n*pi))
x=e^(i*(2pi+4*n*pi))
2pi and aadding as much 4*pi as we want, we still getting 1.
so, we het x=1
now, if we put 1 into the square root, we get :
sqrt(e^(i*(2pi+4*n*pi))), and we just do the thing backward, so we get e^(i*(pi+2*n*pi)) wich is -1.
so, what happened here ?
sqrt(1) = {1+0i; -1+0i} if we consider complex numbers, or so I was taught in school
1 is not positive if we consider complex numbers, nor is it negative, neither is -1, nor any other number
All of the options would be wrong by my book, because neither is a set of two numbers
the only one that has a solution is letter D (sol: -1)
x=1
@@Mythical_Myths16 the cube roots of 1 are 1, e^(2πi/3) and e^(-2πi/3). None of those are -1, so x≠1.
@@xinpingdonohoe3978 no idea what that means, just said that x=1 for D, cuz sqrt(-1)=i
@@Mythical_Myths16It isn't sqrt
@@Mythical_Myths16 are you looking at the assignment at the end, where he asks which one is the only one that has a solution? If not, start looking there.
A has no solution. A principle root cannot be negative.
B: x = -1
C: x = -1
D: x = 1
* *principal root*
It’s D, and x = -1
what about x = i^4
No! Because by *definition* the Principal Square Root can *not* return a negative number. ( i^2 ) = -1. So, no ...
@@MrSummitville that is not entirely accurate as it could lead to other complications.
Square root of x^4 = x^2 for all values of x real and imaginary.
If we follow our logic then the quadratic formula should not yield the -ve value of the square root
@@siddharthchabra9022 We are *not* taking the Square Root of x^4. We are taking the the Square Root of X. There is no quadratic equation in this example. There is no x^4 in this example. You are creating a completely different equation. i^4 equals 1. So, no! You clearly do *not* understand the *difference* between a Square Root and roots of polynomial.
@@MrSummitville you misunderstood put x=i^4 then square root of x=i^2 which solves the first equation. since there were no limits placed on what values x can take. the reasoning that the square root of a number has to be positive is incomplete or incorrect. as just like in the quadratic formula we take the positive and negative value of the square root we dont ignore the negative value of the square root
A. Final answer.
The square root works differently for complex numbers :
If z is complex :
√z=√|z|.e^(2ikπ/2)
√z=√|z|.e^(ikπ) for k in {0;1}
Hence, for z=1 :
√1=√|1|.e^(2ikπ/2)
√1=e^(ikπ) for k in {0;1}
=|1|.e^0=1 if k=0
=e^(iπ)=-1 if k=1
Hence, X=1 is a solution of the equation √X=-1
Jesus Christ, both D(f) and H(f) of the square root are non negative numbers. It cannot be any negative number by definition. Back to reddit.
@@madarab tell me then, how do you define a "negative" number of the complex plane ? For example, is i negative ? Is -i ? I'm making a point here, negative/positive numbers make no sense when we're talking complex numbers, there's an infinite number of directions on the complex plane, not just left/right, and the square root function's definition changes as a consequence.
@@brocolive1950 you are not making any point nor any sense, since the function is not defined for for -1. End of story. You are not aware of the basics.
@@madarab the basics change once you consider complex numbers. You would be right if we were considering real numbers only, but it's not the case here. The same way that the exponential, the logarithm, the absolute value, and many other functions, have different definitions on the complex plane than on the real set, the square root's definition changes too. You can look it up on the internet : "square root of complex". The basics are what they are : basics. They're incomplete, and omit a part of the whole story. Complex numbers go just slightly beyond what you can learn from the basics.
@@brocolive1950 the basics don't change at all and you cannot define a new solution via complex numbers if the function is not defined for the number, you do not understand the basics and spitting nonsense. There is a reason you lecture a math teacher on TH-cam, because you know nothing. It is not defined for D(f) and H(f) and no amount of your nonsense will change that, there is no solution. And you have no idea how complex numbers work either and what comple number is. A complex number is expressed by formula a+bi and simply choosing a SINGLE REAL NUMBER from the fomula, in your case pick 1 IS NOT A SOLUTION. Seriously dude, back to reddit.
POV
-1^2 =x
In A world option answer
lol
Terrance Howard would like a word with you...
Why is i^4 not a soln to A) ?
If x = i^4 then sqrt(x) = sqrt(i^4) = i^2 = -1.
Thus x = i^4 is a soln. Where am I wrong?
You are assuming that sqrt(a²) = a, but this is not necessarily true. In particular, if a is negative (as is the case here), then a² is positive and sqrt(a²) is still positive, so it does not equal a.
i^4 is equal to 1, sqrt(1) is only +1.
You are going by the assumption that sqrt(i^4) = i^2 = -1, however, that is not entirely true as sqrt takes the principle root, which in case of a square square root, is the positive root.
@@westy9447 i^4 has 2 square roots in the complex numbers. We all realize that i^4 is real, but it is also a complex number and hence has 2 square roots in the complex numbers. Principal square root does not apply here, even though i^4 is also real. We are taking it to be a complex number. Why is this so difficult?
as always, let's forget that you can square both sides of equation A to get x= (-1)^2 wich simplifies to x=1
I know it's hard to believe but the square root of a number has TWO solutions. one positive and one negative.
another proof of that hard to believe idea is that ∜1 has FOUR different solutions. (1,-1, i and -i) that's because we can rewrite it as √(√1) or √(±1) which clearly solves as two positive numbers (1 and i)
a root can have multiple results. even multiple positive solutions. like it or not.
by the way, there are no equations without solution. there may be equations we can't calculate the values of all the variables since we lack enough information (like having a single equation with more than one variable) but all equations have solution.
I hoped to find here proof of an unsolvable equation but all I found was somebody that refuses common sense because it doesn't fit what he believes even when given proof of otherwise.
Hard to believe but you are *wrong* . Because by *definition* the Principal Square Root can *not* return a negative number. So, no ...
@@MrSummitville yep. hard to believe that i is a positive number.
@@WilliamWizer-x3m The answer is still *NO* . The Principal Square Root does *NOT* have two answers - *only* a non-negative answer. It is not hard to believe, because it is a *FACT* that you want to ignore. And *NO* you cannot "Square both sides" , which then creates 2 answers, out of thin air.
@@MrSummitville so ∜1≠i but √-1 =i?
makes no sense.
but, of course, spitting nonsense like "principal square root" does wonders to defend your point of view.
just out of curiosity. is there a principal cube root too?
because, for example. ∛8 has 3 answers (even if two of them are complex): 2, −1+√3i and −1-√3i.
interesting. cubic roots can have multiple positive results.and, strangely enough, if you include complex solutions like you did in the video, every single cubic root has 3 possible results. and it's damn easy to prove using polar coordinates.
by the way, just in case you didn't notice, any equality remains an equality if you aply the same operation on both sides. so yes, I can square both sides and it will still be true.
@@WilliamWizer-x3m Everybody, except you, know that √ means Principal Square Root. If you are too lazy to learn that, then I can't help you. i^4 equals 1. The √1 = 1. End of discussion. You are trying to *pretend* (like a child) the fact that ( i^2 ) equals -1, so that you can *magically* get a negative answer for √x . The answer is still .. Hell No! You childish "word games" don't work with math.
D
A: unsolvable
B: x = -1
C: x = -1
D: x = 1
The solution for A, B and C are all easy and I have a wonderful solution for them, but I cannot fit the solutions here, so you'll have to trust me. 👋😁
Does sqrt(x)=-I has a solution?
A nice manipulation! But a square root is defined as a non-negarive value for REAL numbers ONLY! And when speaking about COMPLEX number, this isn't applicable, as complex numbers cannot be negative nor positive, at all. By the way, in your examples C and D the sq.roots are complex (imaginary) - neither negative nor positive.
So, x=1=1+0*i=exp(i*(2П+4Пk)) is indeed the solution of the example A. And x=1=1+0*i=exp(i*4Пk) is not...
Hmm, one of the square roots of 1 is -1 though...
No, the square root function is.. a function. And a function only has one outcome.
You are confusing calculation square roots with solving quadratic equations.
No way! the sq root of 1 is +/- 1
No, the square root function is… a function. And a function only has one outcome.
You are confusing calculation square roots with solving quadratic equations.
@@NLGeebee he isnt taking about the function
@@lisandro73 By *definition* the Principal Square Root can *not* return a negative number. Therefore, "A" has no solution.
For equation A this is what I got:
since sqrt(x) = x ^(0.5), and since e^(i pi) = -1,
then sqrt(x) = -1 can be written as:
x ^ 0.5 = e ^ i pi
0.5 lnx = i pi
ln x = 2i pi
x = (e^2ipi) = (e^ipi)^2 = (-1)^2 = 1
which would mean that sqrt(1) = -1. While it may not be the principle root, that would be sqrt (1) = 1, it still seems like a valid solution to me.
You can't do that. The codomain of sqrt is [0,infinity)
A square root cannot be a negative real number.
About C: A fine example of sloppy math. Since by definition i² = -1, it should be x² = -1 → x² = i² → √x² = √i² → so x = i. Please include the required middle steps.
Where did you get x^2 = i^2 or x^2 = -1 from? Both of those equations are wrong because x = -1. So x^2 would be 1. Also if we plug in your final solution we get that sqrt(i) is i which is wrong
@@elladunham9232 let me clarify that:
The sloppyness is in the explanation afterwards.
√-1 is not i, so the step via i² = -1 cannot be skipped.
@@NLGeebeehe says in the video during the explanation that i^ 2 is by definition -1 when he squares both sides. Also why would the sqrt(-1) not be i?
@@elladunham9232 because that is not the definition. i is not a variable and not a number.
@@NLGeebee i is a number. It is a constant whose value squared is equal to -1. In addition the principal square root of -1 is i. Idk where you heard that i is not a number, but every source i can find online says it’s a number.
You will have to define the principal branch of sqrt root before you proceed.Also if you are working with complex numbers,please use z instead. This notation confuses students. You have a large channel and you should follow the standard notation.
Also remember how exponent work in the complex set.You can not simply square when you want and what you want.For instance students are accepeting that 1=-1 because we can start with -1=-1 =(-1)^1=(-1)^2/2=((-1)^2)^1/2=1^1/2=1.This is wrong of course.But students are squaring and taking square root everywhere with no rules.
None of them has a solution , trick question
Only "A" ...
By the complex interpretation of square root, that'll give back all values that squared give 1. That's is 1 and -1, so that is correct in some sense
-1^1/3=-1
False. That equals -1/3 by the Order of Operations.
You know very well there's a completely valid and sound argument to say that x=1 is a valid solution.
This video is just wrong though. The expression "i = sqrt(-1)" is a convenient intuition, but not really true. If you're allowing negative numbers under a square root sign, then you're implicitly choosing a branch of the complex logarithm to define the square root, since a^b is defined to be e^(b*log(a)) when a is not positive. If youre allowing that, then sqrt(x)=-1 is absolutely possible for a given branch of the logarithm.
True, that 😅
But it's not wrong. A radical results in a positive result. That's just the standard. Yes, you can defy the standard and say it results in a negative number, but if you're not gonna follow the standard that's just a whole different story.
:D
This is just semantics.
Rewrite 1 = i⁴ gives us i^(4/2) = i² =-1 no? :D
No! The Principal Square Root can *not* return a negative answer. ( i^2 ) is a negative number. So no.
i might be dumb but since i² is -1 cant x be i⁴ so sqrt(i⁴) be come i² so -1 ?
D
A
√y
And Y = or > 0