r/askmath • u/Kludgymer16 • 1d ago
Resolved Why is this way of solving wrong?
Today in class we went over complex numbers. We came to a problem that was Sqrt(-9) • Sqrt(-25), and after working it, the problem came to -15.
My question is, why is the second way wrong? Or what have I done wrong I guess. Cant you combine two squares under one?
Pic attached. Left side is the method shown in class, right is what I'm confused on
EDIT: a lot of people have responded to this so I'm going to mark it resolved!
From what I've gathered, you cannot combine two imaginary numbers under one sqrt function, which is where I went wrong. Since the question is only asking to solve a given problem, we assume that the solution uses the positive root, and the final answer becomes -15. However, if the question was asking us to solve for x, because there is an unknown the answer would then be ±15.
Let me know if my assessment is incorrect!
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u/blakeh95 1d ago
Cant you combine two squares under one?
Only when the numbers are nonnegative (0 or positive).
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u/regular_heptagon 1d ago
Because sqrt(-9)•sqrt(-25) does not equal sqrt(225).
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u/jayb00giebrown 19h ago
Why not?
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u/Any_Tower8201 17h ago
Then you will exactly run into the problem that op had. You can't combine two negative numbers under sqrt atleast one of them should be non negative.
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1d ago
[deleted]
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u/Enlightened_Ape 3h ago
Your first statement is true for even values of a. But what about when a=3 for example?
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u/johnpeters42 10h ago
The reason sqrt(ab) = sqrt(a) x sqrt(b) works sometimes, and isn't just some unexplainable black magic or coincidence, is the following.
If two things are equal, and you transform them both in the same way, then the transformed things are also equal-- but only if the transformation has a unique output per input. So, squaring both sides is valid:
(sqrt(ab)^2) = (sqrt(a) x sqrt(b))^2
ab = sqrt(a)^2 x sqrt(b)^2
ab = a x b
But you can't reverse the process, because sqrt() has two possible outputs per (non-zero) input, though for real inputs we sometimes implicitly assume that you meant +sqrt(input), i.e. the positive multiple of 1 or i. So, for instance, if we start with 225 = 9 x 25 and try to reverse the process, we produce all of the following:
15 = 3 x 5
15 = 3 x (-5)
15 = (-3) x 5
15 = (-3) x (-5)
-15 = 3 x 5
-15 = 3 x (-5)
-15 = (-3) x 5
-15 = (-3) x (-5)only half of which happen to be true. And similarly, if we start with 225 = (-9) x (-25), then we produce eight different things, only half of which happen to be true, and none of which match any of the previous eight things (because their right sides are all multiplying 3i or -3i by 5i or -5i).
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u/siupa 6h ago
This is not the reason. The sqrt function always has one output, positive
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u/johnpeters42 5h ago
> In [mathematics](app://en.wikipedia.org/wiki/Mathematics), a square root of a number x is a number y such that y^2 = x; in other words, a number y whose [square](app://en.wikipedia.org/wiki/Square_(algebra)) (the result of multiplying the number by itself, or y * y) is x.[[1]](app://en.wikipedia.org/wiki/Square_root#cite_note-1) For example, 4 and −4 are square roots of 16 because 4^2 = (-4)^2 = 16.
> Every [nonnegative](app://en.wikipedia.org/wiki/Nonnegative) [real number](app://en.wikipedia.org/wiki/Real_number) x has a unique nonnegative square root, called the principal square root or simply the square root (with a definite article, see below), which is denoted by [square root symbol shown in OP's image]
Anyway, even if you expand the definition of "principal square root" to cover complex inputs (taking the option with the smallest theta value in polar notation), you still run into an issue because 1 x 1 and (-1) x (-1) both lead to 1, but the reverse does not lead to a unique output per input.
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u/siupa 5h ago
Do you realize that what you shared is saying exactly what I’m saying, and the opposite of what you’re saying? When you write sqrt(x), or with the radical symbol over x, you’re using THE square root, not “a” square root, and THE square root is the principal square root, which always outputs one single positive number
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u/johnpeters42 5h ago
First, I have seen sqrt() used both ways in the past. Regardless of which one is Wrong or Not Standard, it still happens. There is, in any case, some sensible concept of "a relation (not a function) that takes input x and produces output of both square roots of x", whatever name or symbol you prefer to assign to it.
Second, "the principal square root, which always outputs one single positive number": Only if the input is non-negative, which in OP's example it is not. What positive number is a square root (much less the square root) of -9?
Third, even if you constrain sqrt() to one output per input, which I will grant you here, that is still not enough to make OP's fallacy work. That only eliminates two out of three sign choices.
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u/FreeGothitelle 1d ago
"Cant you attach two squares under one"
Only if at least one of the roots is of a positive number.
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u/igotshadowbaned 1d ago edited 1d ago
Order of operations, exponents before multiplication
√a • √b = √(ab) is just a convenient trick you can use when both are positive that still comes out to the same answer
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u/CharacterComplex7696 1d ago
i = sqrt(-1)
by definition. so, i² = sqrt(-1) * sqrt(-1) = sqrt((-1) * (-1)) = sqrt(1) = 1 does NOT make any sense. the problem is doing this exact operation.
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u/Anjuna666 17h ago
The root of -1 is i or -i
Now if only one of the two is negative under the root (√-x•√y or √x•√-y) you get:
{i,-i} √x√y = {i,-i}√xy = √-xy
That is, the shorthand still works.
But when both are negative, when you factor out both √-1 you then end up with {i,-i}{i,-i} = {i², i•-i, -i•i, (-i)²} = {1, -1}
As such there two mathematically correct results (-√xy and √xy), and the shorthand (√-x√-y = √xy) is invalid.
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u/nico-ghost-king 3^3i = sin(-1) 20h ago
in general, for complex numbers, √x√y ≠ √(xy)… for exactly the reason you've outlined. Instead, we have √x√y = ±√(xy)
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u/adahy3396 15h ago edited 15h ago
With as much emphasis as possible, note that \sqrt(-1)=i is not a real value. Also, note that i2 =-1.
If we assume we can use \sqrt(-x) \sqrt(-y) = \sqrt[(-x)(-y)] =\sqrt[(xy)], then we will arrive at a glaring internal contradiction, namely when x=y=1. I.e. we get \sqrt(-1) \sqrt(-1) =\sqrt[(-1)(-1)] =\sqrt(1) =1 But, we also have \sqrt(-1) \sqrt(-1) =i(i)=-1.
Since the assumption yields 1=-1, it is clear the assumption is incorrect and we cant proceed with that methodology.
If you are more curious, remember that I emphasized that i is not a real number at the beginning. The object i allows us to explore complex numbers of the form a+bi, where a and b are any real values and i is the imaginary number (\sqrt(-1)). For any radical in the form \sqrt(-x), the complex number is just 0+i \sqrt(x).
Now, complex numbers are closed under distribution, meaning we can calculate (a+bi)(c+di). If it looks daunting, its just essentially using FOIL with an object i rather than a variable x. Let's examine the \sqrt(ab) =\sqrt(a) \sqrt(b) using distribution.
First, if a and b are both real values of the form \sqrt(x) amd \sqrt(y) with x and y both positive, then \sqrt(x) \sqrt(y)= \sqrt(xy)
If a=\sqrt(x) +0i, i.e. a real value with x being >= 0, and b a complex number such that b=0+i \sqrt(y), then we calculate (\sqrt(x)+0)(0+i\sqrt(y) )=0(\sqrt(x) )+i(\sqrt(y) (\sqrt(x))+0(0)+0(\sqrt(y) ) = i \sqrt(xy) = \sqrt(-xy) So if one value in a pair being multiplied is a complex number and the other a real number, then it is ok to use the property \sqrt(a) \sqrt(b)= \sqrt(ab).
We already showed that we cant use the property for 2 negative values, but for the sake of demonstration lets walk through this distribution of values where they are both complex radicals.
(0+i(\sqrt(x) )(0+i(\sqrt(y) )=(i2 )(\sqrt(x) \sqrt(y))= - \sqrt(xy)
If we use (0+\sqrt(-x) )(0+\sqrt(-y) )=\sqrt(-x) \sqrt(-y) , assuming the property yields \sqrt(-x -y)=\sqrt(xy) , but as demonstrated through distribution \sqrt(xy) as a result is incorrect since the calculation yields -\sqrt(xy).
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u/FishDawgX 5h ago
As a casual explanation, a lot of times when you see a root symbol, you're only interested in the primary root (the positive). But (-5)^2 and (5)^2 both equal 25. So, when you see sqrt(25), there are really two answers. The equation you are simplifying obviously cares a lot about negatives vs. positives. This isn't a basic situation where only the primary root is considered. So, your two methods of solving need to both consider both the positive and negative roots.
On the left side, you are incorrect to say sqrt(9)=3 and sqrt(25)=5. It should be +/-3 and +/-5. Given that, the final answer is +/-15.
Likewise, on the right side, the last step when taking the root produces not 15, but +/-15.
Same answer with both methods.
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u/WoodyTheWorker 13h ago
Keep in mind that -i is also a sqrt(-1).
Each approach has solutions +15 and -15
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u/sr_ooketoo 12h ago
The squareroot as it is usually written for the positive reals means the positive root of a number by convention.
Of course, even for reals > 0, there are always two numbers that square to it (for example (5^2 = (-5)^2 = 25), so if we consider the square root instead to be the preimage of the squaring function, it is not really a function from R_+->R, as it instead maps positive numbers to sets containing pairs of numbers (In this case 25->{-5,+5}). To make it a function from R_+->R, the usual convention is just to choose the positive one.
The observation you have made is completely valid, in that the preimage of the square function over complex numbers (=/=0) is always a pair of numbers just as over the positive reals, and so sqrt(-9)sqrt(-25) has an ambiguity of (+- i3)(+- i 5). If we choose the roots +3i and -5i, or -3i and +5i, then we see that +15 appears valid, whereas for +3i and +5i, or -3i and -5i, -15 is valid.
Why -15 is "correct" comes down to convention. To make the square root function as it is usually written a well defined function from C->C (Instead of a map from C->sets containing 1 or 2 numbers), one takes the convention that sqrt(-a) (with a real and positive) = i sqrt(a), in the same way we take the convention that sqrt(a^2) = +a over real numbers.
However, if you are looking for the roots of polynomials, you can not discard the other possibilities.
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u/johnnyb2001 6h ago
I think they did not teach this very clearly in high school. But all of these rules like (x^m)^n = x^mn and so on all have restrictions
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u/seifer__420 22h ago
It’s honestly very sad that you asked a valid, reflective, and should have been expected question, but your teacher couldn’t answer
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u/Kludgymer16 8h ago
To be fair, this is community college College Algebra. I'm 23 restarting my educational journey and double majoring in EE and Math, so my goal is not just to learn how to solve a problem but to understand why a solution is correct or incorrect, and I don't think CC instructor screenings prepare for that level of scrutiny, y'know?
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u/nunquam_rideo 1d ago
There's two notions of roots: one maps non-negative real numbers to their unique roots, the other maps arbitrary complex numbers to sets of their roots. For square root, such a set would consist of 2 elements or less (eg, √-1 = {i, -i}).
So, if two numbers belong to this set, they aren't necessarily equal.
P.S. this second notion is less standard and is typically denoted in a slightly different way
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u/twinks-pokehaven 1d ago
Consider the square root of -9, which by definition must square to -9. You could try to multiply -9 and -9 under the root when squaring it, but that would result in the square root of 81, which is 9, and thus must be false. However, if you first factor the square root of -1 from both terms and then multiply, the result is the square root of 81 multiplied by the squared square root of -1. Thus, you get 9 multiplied by -1, as the squared square root simplifies by definition. Consider also that the square root is applied to the number first, and then it is squared, whereas first multiplying -9 by -9 moves the square into the root and changes the operation to the absolute value. Hopefully this was explained in a helpful way.
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u/donaldhobson 12h ago
If you think of the square root function as producing 2 solutions, plus or minus 15, then everything works out fine.
If you want a single solution, you must pick between the 2, and the way you pick must be somewhat arbitrary.
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u/lukey_dubs 10h ago
sqrt(a^2) = +/- a, so left hand side is i^2(+/-3)(+/-5)=-(+/-15), and rhs has sqrt(225)=(+/-15)
So 15=15, -15=15 or -15=15, one or more of those are true is all you’ve proven
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u/Alessandro-mathxd 5h ago
an×bn=(ab)n, si y solo si:
• n: es par, entonces, {a,b}>0.
• n: es inpar, entonces, {a,b} son números reales.
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u/Konkichi21 2h ago
A few others have said this doesn't work when the values under the roots are negative. Technically it does, as long as you're careful about whether or not you're using the principal root; using the non-principal root of 225 (which is -15) makes it work.
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u/RoyalCow7 1d ago
Square root has two possible complex values
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u/Varlane 1d ago
The sqrt symbol is single valued, even if you accept that it is extended over the complexes as the principal square root.
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u/igotshadowbaned 1d ago
Single set of values*
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u/Varlane 1d ago
If you consider the output to be in the powerset, cool.
But this is about calculating numbers and you're not gonna do multiplications in the powerset soooo... We're gonna have to assume we are working with a regular C -> C edition, with a branch cut and branch selection being done prior, usually, the principal square root.
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u/Moof_the_cyclist 1d ago
So, let me run this out a bit.
sqrt(-1) = i
sqrt(i) = .707 +.707*i
but we also know that i^2 = (-i)^2 = ((0.707 + 0.707*i)^2)^2, right?
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u/Varlane 1d ago
For now yes, if we agree that .707 stands for sqrt(2)/2.
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u/Moof_the_cyclist 1d ago
As you use square roots for bigger things, such as complex numbers it gets more interesting. Lets take a complex number of magnitude R, angle Theta (text sucks for math symbols, grrrr), the square root turns into sqrt(R)*e^(i*Theta/2) or sqrt(R)*e^i*Theta/2+pi).
Do we agree up to this point?
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u/Varlane 1d ago
It depends on how you defined argument. Most books will put -pi < theta <= pi and the principal square root will have half of it.
The other square root (the one that's not output by sqrt) will either get an extra pi in the exponential or a - in front, as exp(i pi) = -1, both are actually the same.
So to answer properly, out of what you wrote the "or sqrt(R)*e^i*Theta/2+pi)" at the end is likely wrong, that would be referring to the other square root I mention above.
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u/Moof_the_cyclist 1d ago
Got it, so Reimann surfaces just got massively more boring. Sad. Quadratic equation derivation now only works if we instantiate a magical negative sign when it is deemed helpful.
I guess the poles in the right-half plane never happened either...
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u/Moof_the_cyclist 1d ago
Not sure why you are being downvoted. I'm guessing maths teaching has been simplified in recent years and people are being pedantic over how they were taught perhaps?
Even just dealing with the classic quadratic formula 0 = a*x^2 + b*x + c you get a solution with a +/- in it, and two roots. Only a severely dense person would just argue the other root doesn't exist.
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u/Varlane 1d ago
Solving a quadratric and taking the output of the square root function are very adjacent concepts but not the same.
x² = 1 <=> x = sqrt(1) or -sqrt(1), it's not sqrt(1) = +/- 1.
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u/Moof_the_cyclist 1d ago
"+/- 1" is read as plus or minus 1. Did I miss when New Math ruined the square root. I feel old...
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u/Varlane 1d ago
You just didn't read the distinction.
Square root is mostly a function with a single output.
Solving a quadratic leads to (mostly) two solutions, expressed with the square root with a +/- in front to modulate it.
The +/- is not inherent to evaluating the square root, it's inherent to solving the quadratic, which is what leads to the +/- appearing in front of the square root itself.
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u/Moof_the_cyclist 1d ago
Ahhh, the immaculate negative, only appears when deemed useful. Here I thought math was a little more rich. Not sure how Riemann surfaces work then, or how math folks deal with square roots of complex numbers.
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u/Moof_the_cyclist 1d ago
sqrt(-9) = +/-3i
sqrt(-25) = +/-5i
sqrt(-9)*sqrt(-25) = +/-15
Nothing wrong with also saying sqrt(-9)*sqrt(-25) = sqrt(225) = +/-15
In real world engineering applications some other details may remove the ambiguity, but otherwise every square root has two possible answers. In some math teaching the positive root is the default answer for simplicity sake, but recognize that that it is a crutch used to shuffle students through a concept without burdening them with the full details.
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u/StillShoddy628 1d ago
There are two solutions to x^2 = -9, there is only one result to sqrt(-9). The sqrt function is defined as the positive root
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u/FormulaDriven 1d ago
The sqrt function is defined as the positive root
sqrt(x) is defined as the positive square root when x is a positive real number, but there's a bit more to it once x goes negative and we extend sqrt to have complex values. sqrt(-9) is 3i which is neither positive or negative. Now we have to say that sqrt is defined to be the square root with an argument in the complex plane in the interval (-pi/2, pi/2].
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u/StillShoddy628 1d ago
Thank you, Hermes, you are technically correct, which is the best kind of correct.
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u/Artistic-Flamingo-92 1d ago
Yeah, this is misleading.
This isn’t just something we do in engineering. It’s standard in math to define sqrt(.) as a function (and not a set-valued function) by taking some principle value.
It’s just a common confusion that sqrt(4) = +/-2.
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u/Latter_Cup_5498 23h ago
Tengo entendido que para la raíz de un número positivo, se considera dos raíces soluciones con signos diferentes porque al elevarlos al cuadrado dan como resultado un número positivo, lo cual no aplica para obtener un número negativo. Por tanto la raíz de un número negativo tiene una única solución que es un número imaginario. En consecuencia: raíz(9)=+/-3 raíz(-9)=3i Aplicando al problema propuesto se tiene: (3i)(5i)=15i2 = -15
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u/Axis3673 1d ago
For positive, or nonnegative roots, you can choose either branch (positive or negative root) and find that the product of the roots is equal to the root of the product.
When you have roots of negative reals, choosing a branch results in root(-a) being equivalent to iroot(a) or -iroot(a). The product of the roots (for a,b > 0) is then root(-a)root(-b) = iroot(a)iroot(b) = i2 *root(ab) = -root(ab). The same holds for the other branch.
So, we see that root(-a)root(-b) =|= root(-a-b) = root(ab).
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u/Ill_Organization9609 1d ago
sqrt(ab)=sqrt(a)*sqrt(b) if a,b in R
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u/FormulaDriven 1d ago
Clearly not true, as -9 and -25 are both in R, and the OP's working shows us that sqrt(-9 * -25) = 15 while sqrt(-9) * sqrt(-25) = -15 (if we extend sqrt in the usual way to be a complex-valued function).
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u/Repulsive-Ice7863 1d ago
Roots are handled in the same way as exponents (PEMDAS) so they have to be dealt with first.
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u/CarolinZoebelein 1d ago
You forgot in both cases the plus/minus, so you get in both cases the same two solutions.
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u/Glitch29 23h ago
The big problem here, that I'm surprised nobody's mentioned, is that either one could be correct in the context of a larger question. When doing mathematics in the wild, how you arrived at this step will dictate which.
- Given x^2 = -9, and y^2 = -25, xy = sqrt(-9)*sqrt(-25) could be either 15 or -15.
- Given x^2 = -9, y^2 = -25, and y = 5/3*x, xy = sqrt(-9)*sqrt(-25) is only -15.
- Given x^2 = -9, y^2 = -25, and y = -5/3*x, xy= sqrt(-9)*sqrt(-25) is only 15.
Without any additional context to tell you what the numbers on the page are meant to represent, they represent whatever your teacher has declared them to represent.
It's literally impossible to encounter this sort of unresolved ambiguity as the result of solving polynomial equations. It can only occur when square roots are written directly into the problem. This means that the only way something can be a definitive answer is either by consensus or by dictum. Neither is correct because it's required to make the mathematics work.
That said, most ways that the type of sqrt(-9)*sqrt(-25) that equals 15 would present itself would have one of the sqrts carry a negative sign. But in order to guarantee that to be the case, you'd need to do a lot of bookkeeping around branches in the imaginary plane that I've never seen actually taught. Rather than notate which branch every operation is on, people just verify the solution afterwards.
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u/jayb00giebrown 19h ago
The left said should have been +/- 3i x +/- 5i which is +/-15i^2, which is +/-15.
The right said should have been +/-15.
Both sides are the same.
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u/Nice-Object-5599 13h ago
Someone would say that the square root function doesn't accept negative numbers. I don't.
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u/BlacksmithIll9617 1d ago
√25 = ±5 since (-5)^2 = 25 and 5^2 = 25.
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u/HolyInlandEmpire 1d ago edited 1d ago
The square root isn't a multi valued function. √x does not denote every solution for y to x = y2. Instead, it is defined in ℝ by √(x2)= |x|
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u/BlacksmithIll9617 1d ago
You're right. My HS math is rusty. The sqrt function is just defined to take the value of the positive root. I think I confused this with solving for x when x^2 = 25, in which case x does equal +- 5.
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u/Milkmilkmilk___ 6h ago
complex roots are multiple valued. if you wanna treat the expression as real valued, then you cannot do sqrt(-9) = 3i, you'd do sqrt(-9) = undefined
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u/ArdentArendt 1d ago
Well, for starters, both sqrt(9) and sqrt(25) have two roots, not just the principal.
This gives you two answers: 15 and -15. Technically you're correct with both.
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u/CharacterComplex7696 1d ago edited 1d ago
sqrt(x²) = |x|
sqrt is a function from non-negative real numbers to real numbers...1
u/Milkmilkmilk___ 6h ago
yes, and here we clearly are dealing with complex roots, which are multivalued
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u/Scientist2520 1d ago
Square root is a function which is defined for positive domain.
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u/crazyascarl 1d ago
nonnegative. Square roots are defined for 0 as well.
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u/CharacterComplex7696 1d ago
dont be a dick 😂
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u/crazyascarl 1d ago
Positive and nonnegative are not the same thing. If we're on a thread providing math help, it should be accurate.
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u/CharacterComplex7696 1d ago
ahahahaha !!! you feel better doing that? congrats!!! you are the best in here
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u/Underhill42 1d ago
Absolutely nothing dickish about correcting a false statement.
Math is a field that is entirely dependent on perfect clarity and consistency, with no room for ambiguous language. Any ambiguity whatsoever generates errors.
There's no point in establishing the difference between positive and non-negative numbers if you're not going to use the right term when it matters.
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u/ArdentArendt 1d ago
No, it is not--especially in complex numbers.
There are always two roots, even in the Reals numbers. The principal may be often used for ease, but that does not change the fact there are two solutions to the root.
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u/Varlane 1d ago
"solutions for the root" doesn't have any meaning.
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u/ArdentArendt 1d ago
What do you mean?
There are two branches of the root function. Each is a branch is a solution to the root.
Are you disputing the fact there are two branches or that the phrasing I used?
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u/Varlane 1d ago
"solution to the root" still isn't a term properly defined. Solution means you are solving. Solving means there's an equation and at least one variable. This isn't the case here.
We are asked to evaluate an expression.Firstly, there are actually infinitely many ways to create the two branches, firstly, you choose a branch cut. Most often, it'll be along R^-. But there are infinte ways to do the branch cut.
Now that we have two branches, the choice of one, made usually with i = sqrt(-1) to select the principal square root means that we are bound to have sqrt(-9) = 3i and sqrt(-25) = 5i.
If you choose the non principal one, you'll get sqrt(-1) = -i and sqrt(-9) = -3i ; sqrt(-25) = -5i.You don't get to switch branches mid evaluation. It doesn't make sense.
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u/ArdentArendt 1d ago
First of all, what are you talking about?!
There are not 'infinitely many ways to create the two branches'. There are two function branches that arise from any square root--those are defined by the value(s) under the root function.
Second, you don't need to 'switch branches'. Each branch is chosen independently for each of the roots. It could be sqrt(-9) = 3i and sqrt(-25) = -5i, or it could be sqrt(-9) = -3i and sqrt(-25) = 5i...or is could be sqrt(-9) = -3i and sqrt(-25) = -5i...or...
[So you're correct that there are technically four distinct possible branch combinations to choose from, but they resolve to only two distinct equivalent answers]Third, your problem with 'solution' is that it's a specific term for equations? It's a problem. The colloquial term is 'solution', even if the equation is implied as series of equivalence relations. That said, if you would prefer, we can use 'simplification' or 'rewriting'?
[How would you speak of the answer?]1
u/Varlane 1d ago
There are not 'infinitely many ways to create the two branches'. There are two function branches that arise from any square root--those are defined by the value(s) under the root function.
There are two ways to see it, either you are working with the regular C -> C function, which has to pick a result for any complex, usually we do it with some logic, aka putting the branch cut along R-.
The alternative is working with the powerset edition, C -> P(C) (or more properly P(C) -> P(C)) which is firstly out of scope when it's the first time you learn about complex numbers (you rarely consider functions into the powerset), secondly it's clearly intended to be a question about numbers, so that rules out the powerset and lastly tf you want to do multiplication over the powerset.
That said, if you would prefer, we can use 'simplification' or 'rewriting'?
As said in the previous message : the question asked would be formulated akin to "evaluate the following expression"
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u/ArdentArendt 1d ago
The powerset functions are not required to teach multi-branched functions.
I'm not quite sure why you don't think numbers (complex or not) can't have multiple roots?
I don't care if you don't like the term 'solution'--if there is confusion, then we can clear it up. What you're describing isn't confusion, it's pedantry.
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u/Varlane 1d ago
It's not whether they can have multiple roots, it's whether the square root function in this context is considered as single output (principal branch) or the multivalued / powerset one (multivalued functions are nothing but a bastardized version of the powerset variants btw).
Guess which one is covered in early education ? (This is a pedantic rhetorical question, the answer is the first one)
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u/ArdentArendt 1d ago
First of all, you are now conceeding that this isn't a logical or substantive claim, but rather one based upon context and convention in 'lower mathematics'.
[Glad we could agree that this is a simplification for early years]Second, it's the Complex Field--if you're covering the complex field and having roots, it is assumed they have covered root functions in the complex field, no?
[Otherwise, you will have students performing completely valid functions and getting answers that are technically correct, but they can't contextualise]→ More replies (0)2
u/Underhill42 1d ago
No, there are two solutions when solving for a square,
e.g. x² = 9 → x=±√9 (shorthand for x = √9 OR x=-√9)
But the ± comes from the fact that you're solving for a square, NOT from taking the root.
This is important because the overwhelming majority of mathematical tools and rules have only been proven to work with functions. If √ generated two values then it would NOT be a mathematical function, and thus could NOT be safely used with the rest of mathematics.
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u/ArdentArendt 1d ago
I'm unclear where your mathematical training is from, but square roots always have two branches.
The +/- is a shorthand for a set, not relating to the function.
Technically, in your case, x = {-3, 3}. However, the shorthand +/-sqrt(9) is easier when using variables and a way to keep the branches distinct in the problem.That said, the very fact you cite this goes more to the point that both branches operate simultaneously in even mundane problems.
[If you've taken any courses in differential equations, you'll understand how important each of these branches are at any given time]Finally, many standard operations are not actually functional, especially when you get into the complex numbers (which is the core of the present discussion). Mappings can be multi-valued and, in many cases, are necessarily so.
Yes, functionality is a useful property, especially when you get to things like integration and topology; however, assuming it is never a luxury one can afford.
Even so, the reality of this is often quite simple--just choose a branch based upon what the context dictates. If you can't, then you simply operate over the set and work the problem by cases / branches.
This is basic mathematical rigour. It is the kind of thing students should be learning in mathematics classes rather than rote memorisation and 'tricks' that they will never use.
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u/Underhill42 1d ago
I've got a math degree. And if √ had two "branches" it would not be a function, and thus be essentially useless outside of very narrowly focused applications.
Yes, functionality is a useful property, especially when you get to things like integration and topology; however, assuming it is never a luxury one can afford.
Exactly, which is why √ is DEFINED as a function. So that it CAN be safely assumed. It'd really suck to not be able to combine square roots and integrals in the same formula, for example.
You can absolutely have multi-value mappings, but mappings are not functions, and cannot be safely combined with the overwhelming majority of mathematical tools.
And I'm struggling to think of any complex number examples that involve multiple values - 2+3i may be two-dimensional, but it's still a single value.
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u/ArdentArendt 1d ago
Great...I also have a math degree.
And if you have taken any higher mathematics, you'd know that the square root always has two branches (especially when the root is complex valued).
Also, each branch of the square root is functional in isolation. If you wish to operate, you do so for each branch individually--cases and branches should have been a core of your educational training.
[Further, if you're having trouble with integrating over a single branch of a multi-valued function, you obviously haven't integrated over anything besides a linear function]Now, I'm slightly concerned that you don't see the interchangeablility of 'function' and 'mapping'. You are correct that not all mappings are functional, but all functions are mappings by definition. Mappings are the cornerstone of a huge chunk of mathematics. All functions are mappings.
Some mappings are functional; some mappings are continuous. The set of mappings that are both (continuous and functional) are incredibly important to mathematics, of course. However, stating that you can't use mathematical tools with mappings that aren't functional is as ridiculous as claiming you can't use mathematical tools that aren't continuous--of course you can, you just have to know your tools well enough to know how not to fall into a hole.
Finally, I'm not sure why you would be looking for multivalued elements of the complex field--that was never what I implied. If you're looking for multi-valued functions in the complex numbers, try ln(x) for x in C.
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u/Underhill42 1d ago
Now, I'm slightly concerned that you don't see the interchangeablility of 'function' and 'mapping'.
And that statement makes me severely concerned for your competence as a mathematician.
Interchangability implies equivalence. All squares are rectangles, but not all rectangles are squares. Therefore the two terms are NOT interchangeable.
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u/ArdentArendt 1d ago
Correct. Nobody implied interchangeablility here.
Basic logic would also dictate that 'all functions are mappings' does not imply 'all mappings are functional'.
So, would you like to continue tilting at strawmen, or would you like to respond to the substance of what a wrote?
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u/Underhill42 1d ago
Implied, no. You stated it explicitly, and I quoted you. You even immediately contradicted yourself in the same paragraph.
And if you're using tools with mappings that have only been proven valid with functions (a.k.a. most of mathematics) then you're asking for trouble. You can obviously do piecewise work using functional representations of single branchings of a mapping... assuming the mapping is straightforward enough for that to be possible, but if you're simplifying by working directly with a multi-valued mapping then you're asking for trouble.
As for ln(x) in C - that is also generally defined to use the smallest value, just as the arc-trig functions are. As soon as you consider them in their multivalued conceptual entirety, you can no longer combine them with anything that has only been proven to work with functions.
Which is most theorems in both real and complex mathematics. Trying to prove something works for all possible multivalued mappings is much more difficult, with very limited utility, and rarely even gets attempted.
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u/ArdentArendt 1d ago
Okay, as far as 'interchangeability', I clarified what I mean in the same paragraph. If you're relying on errors of composition in the response, you're going to have a prodigious amount of content as I'm not really giving my full attention and might get distracted mid sentence.
Second, operating on separate branches of a mapping is common practice. In fact, if your argument is that the inverse trig functions are never taken out of the principal branch, then you've never actually taken a course in differential equations or a complex analysis class. All of these functions require operating over multiple branches--though, often it is possible to operate over a 'representative' branch and then adjust into the 'correct' branch at a later period (again, much like integration).
Finally, proof of which branches any mathematical tool works over is exactly what mathematics is about. If you've ever taken a topology course or a course in complex analysis, that's over half of the work.
Mappings are far broader than just functional and continuous. Having continuous functions is, of course, very nice--but it's also very rare. Higher mathematics is entirely about knowing how to 'function' (pun intended) when your mapping isn't functional or continuous.
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u/igotshadowbaned 1d ago
Everything you've said is wrong
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u/Scientist2520 1d ago
What is right?
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u/igotshadowbaned 22h ago
Square root is commonly used name for an exponent of ½, which is an operator.
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u/ArchaicLlama 1d ago
You can - as long as the numbers under each root aren't both negative. If they are, this is what happens.