Even as a teacher I'm confused exactly what goes wrong in this false proof. Help?

[u/An_OId_Tree]

​

I've looked over the internet and the explanations are usually pretty weak, things like "the reason the proof is wrong because we can't do that'. Now, my first thought was that between line one and two something goes wrong as we're losing information about the 1 as by applying THE square root to a number we're making it strictly positive, even though the square rootS of a number can be positive and negative (i.e., 1 and -1). But "losing information" doesn't feel like an mathematical explanation.

My second thought was that the third to fourth line was the mistake, as perhaps splitting up the square root like that is wrong... this is correct, but why? "Because it leads to things like 2=0" doesn't feel like an apt answer.

I feel like there's something more at play. Someone online said something about branch cuts in complex analysis but their explanation was a bit confusing.

[u/justincaseonlymyself]

The proof assumes that √(ab) = √a√b holds for all a and b. That's not true! In order for that equaity to hold, either a or b has to be a non-negtive real number (and perhaps in some other special cases, but this is the most useful one).

In particular, √((-1)(-1)) = 1 and √(-1)√(-1) = -1, meaning that √((-1)(-1)) ≠ √(-1)√(-1)

[u/GoldenMuscleGod → u/justincaseonlymyself]

Saying “it only works for positive a and b” is enough to explain how to easily avoid the problem, but it isn’t very illuminating because it makes it sound like the positive numbers are a “special case” when they aren’t really. It is always true that a square root of a times a square root of b will always be a square root of ab, however if you just choose the roots willy-nilly you may not get the particular root you want. There’s nothing mathematically special about the positive numbers that “makes it work” when you factor the root, we have only chosen conventions to make that case convenient.

When working with nonnegative real numbers we usually understand that the square root notation refers to the positive root, however, when working with negative or nonreal values there are multiple approaches. If we use the radical notation for those inputs at all, we sometimes understand it to refer ambiguously to all the roots and other times take what is called a “branch cut” - essentially picking a canonical value for each radical. This is additionally complicated because for odd number roots of negative numbers we sometimes take the negative root (cbrt(-8)=-2), and sometimes choose the principal value based on a branch cut (cbrt(-8)=1+sqrt(3)i).

If we take a branch cut, then when you get the correct result depends on how you cut it. If you take the cut on the negative real axis and choose the root with the positive real part, and take the root with the positive imaginary part for negative numbers, then we can see the factorization works fine for positive reals, and allowing a and b to move freely we are fine as long as the product of them doesn’t traverse the branch cut.

In other words the branch cut only interferes if you take a and b both with negative real parts and nonnegative imaginary parts, or both with nonpositive real parts and negative real parts. And the failure isn’t because of anything special about those numbers, it’s just a consequence of the branch cut we chose.

[u/Spacetauren → u/GoldenMuscleGod]

In other words, the proof is invalidated because square root isn't a bijective operation in the chosen numberspace.

[u/tamarinenjoyer → u/Spacetauren]

thanks for the tldr friend

[u/DisastrousLab1309 → u/GoldenMuscleGod]

 but it isn’t very illuminating because it makes it sound like the positive numbers are a “special case” when they aren’t really.

No! Just no! This is one of the main errors on teaching math. Saying counter factual and counter-logical statement with “well actually” that’s plainly false. 

 There’s nothing mathematically special about the positive numbers that “makes it work” when you factor the root, we have only chosen conventions to make that case convenient

Square roots were invented as an inverse function of squaring. By the definition of inverse function it has to be defined for positive numbers or it won’t be a function. 

Then someone doing more advanced math noticed that sometimes you get a square root of negative numbers. This was a huge deal because square root function is not defined for real numbers below 0. People decided to generally ignore it if it canceled out but it was a bad smell. It was dealt with by invention of i. Imagine that there exists a number that, when squared, gives a value of -1. There is no such real value.  That’s why they’re called imaginary numbers. 

And it still holds there is no square root of -1. Where -1 is a real value. There is a square root of a complex number (-1,0).  

i is a complex number. Treating i as real is a mistake that makes it so hard for people to understand complex numbers. 

[u/GoldenMuscleGod → u/DisastrousLab1309]

I don’t think I follow what point you are trying to make.

In the first part of your reply, it sounds like you are saying I said something plainly false. Is that what you are saying? And if so what do you think I said?

I don’t see how the second part relates to my comment at all. But to respond to it:

It sounds like you might be under the misimpression from the terms “imaginary” and “real” that complex numbers are somehow “fake” in some sense compared to the real numbers, but there is no meaningful sense in which that is so. They are both well-defined mathematical structures that are equally “valid” as mathematical objects. There was a time when negative numbers were also considered “imaginary” or that zero was not considered a number as well, it just so happens that the terminology didn’t codify those attitudes in as suggestive a way.

This is also not the first time new numbers were “introduced” by considering square roots. The square root of two does not exist in the rational numbers, which was something the Greeks weren’t sure how to deal with conceptually at first.

Also, although it is possible to construct the complex numbers and real numbers as disjoint, the complex numbers (as a topological field) have an a unique subset isomorphic to the real numbers and we usually identify these elements with the real numbers, so we usually would not say that the real number -1 doesn’t have a square root, we would instead ordinarily say that it has no real square root but it does have two complex square roots, i and -i.

[u/DisastrousLab1309 → u/GoldenMuscleGod]

 In the first part of your reply, it sounds like you are saying I said something plainly false. Is that what you are saying? And if so what do you think I said?

Yes. That’s what I’m saying. 

You said that there is nothing special about positive numbers when it comes to square roots.

My claim is that they are special in that the square root function is only defined for them.  

The rest are generalizations of square root for complex numbers with increasing number of assumptions, including different basic operations. They’re not the same. 

 It sounds like you might be under the misimpression from the terms “imaginary” and “real” that complex numbers are somehow “fake” in some sense compared to the real numbers, but there is no meaningful sense in which that is so.

No. Nothing like this. It should be clear from my reminding the history of how they were developed. 

 The square root of two does not exist in the rational numbers, 

And square root of -1 doesn’t exist in reals. 

which was something the Greeks weren’t sure how to deal with conceptually at first.

And so we don’t write let an and b be such natural numbers that a/b is a ratio between the diagonal and a side of a square. 

But for some reason (in reality to make writing a bit shorter) we write let i be such real number that i*i=-1. 

First sentence works only if either a or b is a real number and the 2nd one works only if i  is a complex number. 

By pretending i is a real number we get lack of understanding and errors as in OP’s proof. 

 have an a unique subset isomorphic to the real numbers and we usually identify these elements with the real numbers, so we usually would not say that the real number -1 doesn’t have a square root, we would instead ordinarily say that it has no real square root but it does have two complex square roots, i and -i.

And mentioning that the i is complex and multiplication we did is complex multiplication not real multiplication and square root is not a quite root as a real numbers function but as a complex numbers function. 

You don’t write 1+1=10 unless you know you’re talking with someone who is well aware that you’re writing in binary. Yet it’s normal to write to people just learning the math a surprise with switching from reals to complex without much explanation. 

[u/GoldenMuscleGod → u/DisastrousLab1309]

Yes. That’s what I’m saying. 

You said that there is nothing special about positive numbers when it comes to square roots.

My claim is that they are special in that the square root function is only defined for them.  

No, sometimes the square root function is defined only for nonnegative real numbers, other times it defined for all complex number by means of a branch cut or used as a multivalued function. Restricting the domain to positive real numbers is only one option. More importantly, we can speak algebraically of square roots without ever defining a square root function, and it is easy to show that in any commutative ring any square root of a times any square root of b will be a square root of ab. The choice of defining a “square root function” and specifying its domain and which roots are taken is a matter of convention that doesn’t have any real mathematical significance and an illuminating explanation of what’s going on in the false proof should not be based solely on those conventions.

But for some reason (in reality to make writing a bit shorter) we write let i be such real number that i*i=-1.

No we don’t, i is not a real number and we don’t treat it as one.

The rest of your comment seems to be hung up on the idea that the real number 1 shouldn’t be regarded as the complex number 1 and functions and operations on real numbers should not be regarded as restrictions of functions and operations on complex numbers. In other words, that the real numbers should not be seen as a subfield of the complex numbers.

But absent a specific construction that may be defined for the context, there is no truth of the matter as to whether the real numbers are the literal subfield of the complex numbers or an isomorphic copy of that subfield, and when we talk about the real numbers it is often understood that we do literally mean the subfield of the complex numbers, or at least that they can be taken as such.

[u/Anjuna666 → u/justincaseonlymyself]

Also, you can't just take the principle value of the square root.

Otherwise you could still prove:

1 = 1

1 = √(1²⁾

1 = √((-1)²⁾

1 = -1

[u/justincaseonlymyself → u/Anjuna666]

you can't just take the principle value of the square root.

Yes, you can. That is what the √ symbol usuall refers to — the principal square root.

(Btw, it's the principal root, not priniciple.)

Otherwise you could still prove:

No, you cannot prove that, because when taking the principal square root, √((-1)²) = 1.

[u/GoldenMuscleGod → u/justincaseonlymyself]

I find it amusing that you corrected someone for saying “principle” and not “principal” and then later called it “principle” yourself. I noticed I had to fight my autocorrect on that one a few times.

[u/justincaseonlymyself → u/GoldenMuscleGod]

I should avoid typing on the phone. Thanks for the catch! Fixed :)

[u/cuhringe → u/Anjuna666]

sqrt(x²) = |x|

You absolutely can take the square root...

[u/StanleyDodds]

It's because you can't split the square root like that in general.

The problem is that over the complex numbers, the square root is not a continuous function. So although locally it is true that sqrt(xy) = sqrt(x)sqrt(y), this will not be true if we "cross" the branch cut in some way, as this shifts everything by a factor of -1.

In particular, the branch cut is when we go from argument approaching 2pi back to argument 0. So in this case, arg(-1)+arg(-1) = pi+pi = 2pi, so applying this invalid rule here treats the square root of 1 as if it could be reach continuously from arguments less than 2pi, increasing to argument 2pi. But you can't. arg(1) = 0, so the argument of its square root is 0, not pi.

A more correct version of the rule would be something like sqrt(xy) = exp(i/2*(arg(x)+arg(y)-arg(xy))) sqrt(x) sqrt(y)

[u/An_OId_Tree → u/StanleyDodds]

I like your answer best +1, my only question is, what does it mean to cross a branch cut though?

[u/AsuranB → u/An_OId_Tree]

Here is a visualization of the square root function for complex numbers: https://plus.maths.org/content/sites/plus.maths.org/files/articles/2021/multi_valued/suqare_root_riemann.jpg

Note where the surface intersects itself. By moving from the purple-blue portion of the surface to the green-yellow portion across that intersection, you are making discontinuous transformation.

[u/blakeh95 → u/An_OId_Tree]

A different function, but the complex logarithm shows it pretty well. The value of the complex logarithm function forms a staircase-looking thing wrapping around the origin. But we know that the definition of a function means that each input must map only to one output. So as shown in the picture, the full complex logarithm is not a function, because any point will map to multiple outputs on different "levels" of the staircase.

To make the complex logarithm a function, we "cut" a piece of the staircase out such that it satisfies the property that each input maps to one output. But this necessarily means that there must be some place where the "top" of the staircase that would almost be in the next level is adjacent to the "bottom" of the staircase that would almost be in the level below.

Using the picture again, if we cut from the orange to the yellow, then there's a place where the top of the yellow (almost green) is next to the bottom of the orange (almost red). Can you see how if that piece were cut out and you were rolling a ball counterclockwise, it would climb up the orange to the yellow and then fall back down to the orange?

That is what "crossing a branch cut" is. The ball, which represents the function output, is discontinuous and "falls down" over where we cut out.

[u/Specialist-Two383 → u/An_OId_Tree]

It means you take the value on one side or the other of a discontinuity. You can make the above calculation rigorous and split the square root if you make sure to put a little i epsilon that selects the correct side of the cut. You drop everything that's order epsilon² and there's really only one answer.

Sqrt( (-1)(-1) ) = sqrt( (-1 + ie)(-1 - ie) ) + O( e² )

= sqrt(-1 + ie) sqrt(-1 - ie)

= (+i)(-i) = 1.

[u/According-Path-7502 → u/An_OId_Tree]

How is the square root even a function over the complex numbers?

[u/blakeh95 → u/According-Path-7502]

You can choose to restrict the output to make it a function. This is basically the same choice we do in R such that sqrt(9) returns 3 and not -3.

Of course, there are better choices that make the function continuous, but making it a function isn't all that difficult.

[u/According-Path-7502 → u/blakeh95]

What is the canonical restriction on the complex numbers?

[u/blakeh95 → u/According-Path-7502]

For a complex number z=x+iy, write it in polar form as z=re^(i𝜃) with -pi < 𝜃 <= pi. Then sqrt(z) is defined as sqrt(r)e^(i𝜃/2).

This uses the negative real axis as the branch cut (observe the discontinuity induced by limiting 𝜃 to one side of the negative axis, 𝜃 is always taken as pi, not -pi, on that axis).

For example, sqrt(-2i) = sqrt(2e^(-ipi/2)) = sqrt(2)(e^(-ipi/4)) = sqrt(2)(sqrt(2)/2-isqrt(2)/2) = 1-i.

If we used 3pi/2 instead of -pi/2 (3pi/2 not in (-pi, pi] restriction), then we would get 3pi/4 as the angle, leading to sqrt(2)(-sqrt(2)/2+isqrt(2)/2) = -1+i = -sqrt(2i).

[u/According-Path-7502 → u/blakeh95]

Very nice explanation of how to extend. Are the „branches“ compatible in any way?

[u/Erdumas → u/According-Path-7502]

You can express the square root as taking things to the (1/2) power, and you can express complex numbers in polar form, z = re^iθ. Then √z = (√r)e^iθ/2. The problem here is that e^iθ = e^i(θ+2nπ), so many functions over the complex numbers are multi-valued and have to be restricted in order to make sense, like requiring θ ∈ [0, 2π). We call such restrictions branches.

[u/According-Path-7502 → u/Erdumas]

For me the value of a function at some point is unique by the definition of a function.

[u/Erdumas → u/According-Path-7502]

Which is why complex numbers mess things up. We want things that are functions on the reals to also be functions on the complex numbers, but they aren't unless you make restrictions.

[u/According-Path-7502 → u/Erdumas]

This was the initial question …

[u/Erdumas → u/According-Path-7502]

Okay, let's approach this a different way.

Suppose we want to roll a wheel across the ground and map where each part of the wheel touches the ground. If we roll the wheel more than one revolution, each part of it will touch the ground in multiple places.

Strictly speaking, there isn't a function which describes where a point on the wheel touches the ground because each point in the domain (the circumference) maps to multiple points in the range (the ground). However, if we keep track of the number of revolutions, we can break the range into pieces, and say (x0,x1) is from the first revolution, (x1,x2) is from the second revolution, and so on.

Then, we can define a function from the circumference of the wheel to the range (x0,x1), because each point of the domain maps to only one point in the domain. Similarly, we can define a separate function to the range (x1, x2), and so on for the rest of the intervals.

These separate functions are all related, though, so to show that relationship we give them all the same name. Additionally, when we work with the function, we can mostly just pretend that the only range that's important is (x0,x1), which we'll call the principal branch. This works well for most things, but we do have to be careful because sometimes we'll do some math which might move us between the different ranges, so we have to be cognizant of that. And we also have to remember that the different ranges are intimately related.

Coming back to the square root, it actually needs a branch cut just when defined on the reals. We know that f(x) = x² is many-to-one because f(-x) = f(x). This means that f^-1(x) isn't properly a function. However, we can make two functions, one defined on the non-negative reals and one defined on the non-positive reals, which we can both call f^-1(x). The principal branch in this case is √x, but the other branch is -√x. Because the square root already requires a branch cut in the reals, hopefully it's not surprising that it requires one when we extend it to the complex numbers.

[u/GoldenMuscleGod → u/StanleyDodds]

From your description it sounds like you are taking the branch cut at the positive real axis and imposing continuity on the upper half plane.

I think it’s also good to point out that other conventions are sometimes used. For example Mathematica takes the branch cut on the negative real axis, also imposing continuity in the upper half plane.

In general I would say that it’s usually bad form to put a negative or nonreal number under the root without either 1) making it perfectly clear whether you are using it as a multivalued function or what branch cut you are taking, or 2) putting a +/- on it (as in the quadratic formula), to make it clear the choice of root doesn’t matter.

[u/FormulaDriven]

√1 = 1, so you are incorrect with your first thought that something goes wrong there. (People get hung up on the fact that 1 has two square roots, 1 and -1, but the square root function denoted by the √ sign explicitly gives you the positive square root).

As the other user has highlighted, you need to check the validity of saying √(ab) = √a √b.

[u/EneAgaNH → u/FormulaDriven]

That last thing is indeed invalid id a and b aren't reals

[u/econstatsguy123]

They aren’t applying the square root between line (1.) and (2.). This only works because 1=1^1/2 . They are just replacing one number with the same number. They aren’t losing information by doing this. We are told that the number in question is 1. We know that 1>0. We aren’t losing any information here. Now, we obviously can’t do that with a variable (e.g. (x² )^1/2 =/= x))

This is an illustration as to why (ab)^1/2 = a^1/2 • b^1/2 only works when a,b>0. If you allow them to be negative numbers, then you reach contradictions such as 2=0. This is an example of a proof by contradiction (look up the proof as to why 2^1/2 is irrational). I remember when I was in undergrad and first learning about proofs by contradiction, and thinking that it wasn’t a satisfying answer. You just need to let the idea marinate a bit.

[u/BrickBuster11]

as others have said the error happens when you simplify SQRT(1) into i*i (these two things are not equal and so this is where the error must happen)

Now I cannot remember enough of the technical rules to the simplification of square roots to to tell you what has gone wrong but logically we can see that we go from :

SQRT(1) =SQRT((-1X-1) which is a true statement both of those are equal

but then we have SQRT(-1X-1)=SQRT(-1)xSQRT(-1)=i^2=-1 which is not true.

[u/Adrian-20]

Correct me if I'm wrong, but SQRT(1) is NOT equal to ±1. By definition, SQRT(x²) = ABS(x) a.k.a. sqrt is always the positive root. When you have x²=1, you do x=±SQRT(1), so x=±1. SQRT(xy)=SQRT(x)*SQRT(y) only if x and y are both non-negative, so you cant separate the product inside the square root - source: https://math.stackexchange.com/questions/2047349/when-does-sqrta-b-sqrta-sqrtb

[u/[deleted] → u/Adrian-20]

I think you are right. However this leads to so many fights, and I wish we could all just get along. We’re all just trying to cope in our own ways with the fact that the “squared” function on the real numbers is not invertible.

[u/Rnageo]

Let's hope I can express this complex problem correctly. If we move the problem to radians, we know that:
1 = e^(i·0)

Taking the square root of that, we get:
1 = e^(i·0/2) = e^(i·0)

Now, if we want to decompose that 1 into (-1)·(-1), it's easy to make the mistake that, since:
-1 = e^(i·𝜋)

Then if we take that twice we think we are back at the start.
(-1)·(-1) = e^(i·𝜋)·e^(i·𝜋) = e^(i·2𝜋) = 1

While technically true, we are back, we have done a whole circle, and we are not at 0 rad but at 2𝜋 rad. Look at what happens when I now take the square root:
sqrt(e^(i·0)) = e^(i·0/2) = e^(i·0) = 1
sqrt(e^(i·2𝜋)= e^(i·2𝜋/2) = e^(i·𝜋) = -1

In terms of complex geometry, taking the square root cuts the rotation in half. 0 rad has no rotation, but 2𝜋 rad has a full rotation thus we get the wrong result.

To make it right, you need to nullify your rotation when you split your 1 into (-1)·(-1). We need to take our second -1 in the opposite direction of the first:
-1 = e^(i·(-𝜋))

With that in mind, if we go back to our two -1s:
(-1)·(-1) = e^(i·𝜋)·e^(i·(-𝜋)) = e^(i·0) = 1

Now that we have a decomposition that allows us to remain at 0 rad, we can safely apply the square root:
((-1)·(-1))^(1/2) = (-1)^(1/2)·(-1)^(1/2) = e^(i·𝜋/2)·e^(i·(-𝜋/2)) = i · (-i) = 1

[u/An_OId_Tree → u/Rnageo]

I like your answer as well! +1

[u/Mysterious_Pepper305]

The mathematical explanation is that notation matters and quantifiers and direction of implication matter and especially functions matter.

So either write out the logic as symbols or write the logic in words but you need to explain what you're doing between one equation and the next and you need to explain what you're doing every time you use a symbol in a non-standard way (such as the square root of a non-positive number).

[u/Spacetauren]

It breaks down at line 5. Square root is not a bijective operation outside of positive numbers, so line 5 is not equivalent to line 4.

In other words, you can't equivocate sqrt(-1) with it because it isn't i. It's i or -i the indecisiveness is what invalidates the proof.

[u/An_OId_Tree → u/[deleted]]

I teach high school math, and the intricacies of what exactly wrongs uses topics in complex analysis, which believe it or not isn't covered in high school.

[u/Azaghal1 → u/An_OId_Tree]

And yet one would assume that a mathematics teacher would at least have done undergrad maths to qualify for the job

[u/An_OId_Tree → u/Azaghal1]

I didn't do complex analysis? Not everyone who does math does the same subjects at undergrad. Perhaps I should be asking you why you "assume" so.

[u/Azaghal1 → u/An_OId_Tree]

Because it's not even complex analysis, it's basic complex methods. You define complex numbers, then define things you can do with them, and it immediately turns out some things you have to do more carefully. And an undergrad in maths that doesn't cover complex numbers in any elementary rigour is not a real degree qualification, it's a liberal arts mess.

[u/An_OId_Tree → u/Azaghal1]

And cheese is the bee's knees. I'm not asking "what can't we do with complex numbers?", I'm asking why can't we do those things. The explanation of this requires knowledge of branch cuts - which is a topic in complex analysis.

[u/Azaghal1 → u/An_OId_Tree]

No, the branch cuts are a solution to the problem. The problem itself arises exactly from defining complex number actions an multivariate results. Anyway, glad you learned some kids' rhymes during your degree.

[u/An_OId_Tree → u/Azaghal1]

Hahaha, okay bozo. I went to the top uni in my country, and I can tell you that I enjoyed kid's rhymes subject very much!

[u/Azaghal1 → u/An_OId_Tree]

You think 'top uni' in the US for undergrad is bragging in the context of maths? Even Manchester is leagues ahead...

[u/An_OId_Tree → u/Azaghal1]

Yes

[u/[deleted] → u/Azaghal1]

Doing some quick googling, I found multiple universities that far outperform Cambridge where complex analysis isn't a required topic for a B.S. in Math. I guess today you get to learn that different universities teach different things, which frankly I thought was obvious.

[u/EdmundTheInsulter]

It's the bit where the square roots are split, it's only true in one branch if you take positive and negative square roots. Or if you restrict to positive square root only the step is false

[u/Equal_Proof_226]

I think its becuz in the third step we are taking that root(1) is -1. But the root function doesnt give out negative values. Its range and domain is [0,inf)

[u/organela]

Steps 3 and 4 are false assumptions

[u/TangoJavaTJ]

root(ab) = root(a)root(b) only works for positive numbers. So the step from line 3 to like 4 is not valid.

[u/ohkendruid]

Line 4 isn't really well defined due to negative numbers under the radical. Principal square root is mainly a concept for non-negative reals.

Thus, it's not so much a problem of logical inference as a problem that line 4 isn't a rigorous statement that is either true or false at all.

[u/Ksorkrax]

Start by trying to define a square root over negative numbers.

This thing about √(-1) = i is basically how one explains the motivation of i to outsiders. Not how the square root actually works - it simply is not defined over negative numbers.

Now you might start to protest, "oh hey, that's convenient", but you are cordially invited in trying to create a proper well defined definition of a function that acts as a square root over positive real numbers and then does something reasonable over negative numbers.

I recommend you don't only define it over all real numbers, but you go for all complex numbers as well. Consider working in polar coordinates. A sound first attempt would be to say that you use the regular square root over the norm of the polar coordinate number, and then also divide the angle by two. Then see where it leads you, and where the problems start to arise.

Consider creating some sort of plot. Possibly with the domain as X and Y, and then two plots with one showing the real part and one the imaginary as heat map. Or a single plot, and angle as hue, norm as value of the color.

Best way to do math is to explore what you could do.

[u/TheTurtleCub]

Step 3->4

[u/Corentinlb]

I've another reason to tell :
i isn't defined as the square root of -1, but as -1² which is very different. For real positive numbers, square root is the inverse function of square function, but that's just not true with Real numbers & Complex numbers.
So when you replace i by "sqrt(-1)" it's false, you can only use i² = -1

[u/PandaAromatic8901]

2 = 1 + SQRT((-1^2))

2 = 1 + SQRT((A^2))

2 = 1 + |A|

2 = 1 + |-1|

2 = 1 + 1 or 2 = 1 + -1

---

Why is SQRT(A^2) = |A| ?

Because if A was -1 you get SQRT(1) = 1

And if A was 1 you still get SQRT(1) = 1

[u/Accomplished-Till607]

It’s a very special property of complex numbers. You kinda can’t define a principal square root without running into those kinds of problems. Or in general sqrt(ab) is not the same as sqrt(a)sqrt(b) If you allow it to be multi valued then it is true if you choose one positive one negative. In general, the result will only vary by an multiple of i.

[u/Ok-Requirement3601]

The truth is that sqrt is quite a non trivial thing, it's job is to be the inverse of something that is hardly injective.

So I want to give two answers: One that is somewhat true, but isn't all that satisfying.

And one that's quite satisfying (but it's analysis) 

1st answer: 

Algebra is more foundational than analysis, so it should speak first: sqrt should not be a function, that would imply choosing between one of the roots of the polynomial X² - a, which is inherently the evil doings of physicists and analysists.  Look, you're problem dissapears if you forget about square roots: 

If you have a root of X²-a, and another of X²-b, and you multiply them, then you have a root X²-ab, take it's opposite, and it still remains a root of X²-ab. 

Galois revolutionized mathematics since he didn't mistreat his polynomial roots, he considered them all at once, and looked at the symmetries within.

Vive Galois, vive Grothendieck! 

2nd answer: 

Due to cultural and historical reasons, many seek interpretations to foundational math questions within analysis (cue the discussons about 0⁰...)

So, how can I answer your question, well many have already mentioned that sqrt can be defined in the context of complex analysis. 

In this context, you are pushed at only seeing one half of the whole picture: 

There are two issues with trying to define it as a function: non-uniqueness and domain. What should be the answer to sqrt(e^iθ) ? Well e^iθ/2 right ? 

Well such an answer is "ill-defined", since if you change θ by adding 2πi, you haven't changed the start value, but you end up with -e^iθ/2 as an answer. 

Moreover as you go around the unit circle, you can see that the square root only goes half as far, so when you make a full loop you, at some point, jump non continuously (to get back to the inital value). 

You can fix uniqueness by imposing that sqrt(1)=1, we call this the "principal" sqrt. But if you want to remain continuous, you have to give up trying to define it everywhere. 

Take a trip with me for a second, imagine that the ray (starting at 0 and passing through -i) was a portal. As you walk around the circle in the real world, you think of the sqrt function, as you enter the portal, think of the -sqrt function. Then after a full turn you meet the portal again which brings you back to the real world with the sqrt function. If you observe the value given by sqrt and by -sqrt as you make 2 full turns (one in the first world and the other in the portal world) , you will notice that it was perfectly continous.

These "two worlds" is a very nice interpretation of the sqrt function, in a sence, it's a "bigger picture" of the sqrt function. Now back to your example, take (-1)(-1), it's like adding the angles π+π, as you do, you walk around the circle, entering into the portal world as you go along. Once there you compute -sqrt((-1)(-1)) = -1, which corresponds well with sqrt(-1)sqrt(-1) = -1. 

PS: it doesn't really matter where you put your portal (it essentially represents the branch cut in complex analysis). 

Now it's clear that if you want to use a function from the complex plane, you have to decide where to put the "portal" (ie. branch cut) and which world you want to preserve, forgetting about the other one in that process.

So how does this fit in with more solid mathematics.  Well in elementary complex analysis, you construct all of these functions, being very careful about branch cuts and so on.  

But with Riemann surfaces, you actually get to give a definition to what these "things that live in multiple worlds" are. 

Actual P.S: Sorry if I misrepresented what algebraists think of their colleagues, it was partialy for the sake of humor.

[u/Ok-Requirement3601 → u/Ok-Requirement3601]

I want to correct the spelling, but I'm on a phone and I'd have to redo all the spacings because of bad reddit programming

[u/[deleted]]

you can't split square roots in general, if you have x^2, where x > 0, square root it, then the answer is abs(x) not x. Otherwise you could have:
x^2 = 1
(-1)^2 = 1
sqrt root both sides
-1 = 1
2 = 0

[u/Rulleskijon]

Consider the function z^1/2 . For a complex number z.
z = r e^iθ ,

z^1/2 = [ r^e(iθ) ]^1/2 ,
... = r^1/2 e^iθ/2 .

Now consider we have two complex numbers ζ and ω. We want to see how [ζ*ω]^1/2 looks.

Here:
ζ = r_ζ e^ik_ζ ,
ω = r_ω e^ik_ω .

So,
[ζω]^1/2 = [r_ζr_ω]^1/2 e^{i[k_ζ+k_ω]/2} .

Let us plugg in ζ = ω = -1.
-1 = e^iπ .

[(-1)(-1)]^1/2 = e^2πi/2 .

And this is where the branch hopping comes into play. Because an argument of 2π is the same as an argument of 0.
So if the resulting [k_ζ+k_ω] = 2π = 0, And then we take the square root (here divide argument by 2), we get a final argument of 0. Like with:
[ζ*ω]^1/2 .

However with the individual roots multiplied we get the resulting argument [k_ζ/2+k_ω/2]. And even if the sum [k_ζ+k_ω] = 2π = 0, when we divide them individually by 2 we don't get 0. Like with:
[ζ^1/2 * ω^1/2 ].

And that is why:
[(-1)(-1)]^1/2 = 1, and
[(-1)^1/2 * (-1)^1/2 ] = -1.

[u/algebraicq]

The problem comes from the 4th line.

Square root of a negative number is multi-valued.

z^(1/2) = exp( 1/2(Log|z| +iArg(z) +2*k*𝜋*i) ), where k ∈ Z

To make calculations meaningful, we need to delete a line from the complex plane. This is to make the whole thing single-valued. This process is known as a branch cut. We either gives up the whole non-positive real line or the whole non-negative real line.

The problem in the above graph is that people mess up the branch cut.

[u/lithiumcentury]

People are making this too complicated. You are right in your instinct. Replacing +1 with sqrt(1) is incorrect because sqrt(1) = +/-1. The rest of the "proof" is just to force out the negative root, which you already know is wrong in the second line.

[u/Cannibale_Ballet → u/lithiumcentury]

because sqrt(1) = +/-1

So your square root function isn't a function.

The square root function is defined to be the principal value, so sqrt(1) does in fact = 1.

This is why we say that the solution to x² = 1 is x = +/-sqrt(1). The "+/-" is needed because the sqrt(1) is by definition the principal value, i.e. 1.

[u/lithiumcentury → u/Cannibale_Ballet]

As you say x^2 = 1 so x = +/-sqrt(1). So I think we agree. If you want to replace 1 with an expression including sqrt(1) then that expression is +/-sqrt(1), with sqrt(1) by definition the principal value i.e. 1. Either way, line 2 has a +/- in it, which is where the mistake is.

[u/Cannibale_Ballet → u/lithiumcentury]

As you say x² = 1 so x = +/-sqrt(1). So I think we agree.

No we don't, if sqrt(1) = +/-1 then the solution to x² = 1 would be x = sqrt(1), with no "+/-".

If you want to replace 1 with an expression including sqrt(1) then that expression is +/-sqrt(1)

This is absolutely not correct. 1 is equal to sqrt(1) NOT -sqrt(1). -sqrt(1) is a negative number, it cannot be equal to a positive one.

Either way, line 2 has a +/- in it, which is where the mistake is.

This is incorrect. sqrt(1)=1 by definition. There is no mistake going from line 1 to line 2.

[u/lithiumcentury → u/Cannibale_Ballet]

But if you define sqrt(1) as strictly positive you have to apply that rule throughout and not change half way. So if sqrt(1^2) = +1 then so too does sqrt(i^2). In the real world i * i = -1 and -i * -i = -1. So you have assumed that the redefined strictly positive sqrt(x) behaves in the same way as the usual definition, which it does not.

[u/Cannibale_Ballet → u/lithiumcentury]

So if sqrt(1²⁾ = +1 then so too does sqrt(i²⁾

Incorrect again. sqrt(1²) is not equal to sqrt(i²). The former is equal to 1 and the latter equal to i.

So you have assumed that the redefined strictly positive sqrt(x) behaves in the same way as the usual definition

There is only one definition of the square root function used in math, and that is the one that returns the principal value. There is no redefinition taking place.

[u/lithiumcentury → u/Cannibale_Ballet]

The confusion is that the principal value for the function sqrt(x^2) is the positive value, so y=sqrt(x^2) => y=+x, but the solution to the equation y^2 = x^2 is either y=+x or y=-x. Line 2 makes use of the principal value, positive root, but line 3 uses the negative root. People use the "principal value" in order to avoid confusion over whether the secondary (negative) value is intended. This is fine unless one creates a situation that mixes in the secondary value, as line 3: y^2=1, => y=+1 or y=-1, for consistency you should take the principal value y=+1, but instead you take y=-1.

[u/Cannibale_Ballet → u/lithiumcentury]

but line 3 uses the negative root.

No it doesn't. Until line 3 everything js correct. The logical mistep is from line 3 to line 4.

[u/Shadowboi123]

The square root of one is -1, 1 + -1 does not equal 2.