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After asking a question on the square root yesterday (On the real square root and branches of the complex square root.) I saw a lot of arguments of the form,

If we solve the equation $x^{2}=4$ then the solutions are $\pm 2$.

and

If we want to define the square root as a function then we have to pick a branch. And then we have $\sqrt{4}=2$ and only that.

I suppose that when we solve and equation we can apply a function to both sides and get the answer on sets where the function is well defined. Hence we get $2$ as an answer to the above equation if we pick the principal branch of the root.

Following this logic the above equation have unique solution in a universe where we picked a branch.Hence using a function arising in this way i.e resticting a multivalued function to a branch we "loose" or drop som solutions.

So my question is weather this is complete nonsense or if the multivaluedsness of the root and the multiple solutions to above equation is related.

  • It might be better to think of $x^2=4$ as having multiple solutions rather than $\sqrt{\cdot}$ being multivalued. Generally when we take roots we have in mind a certain subset of solutions (a branch) within which the root is single valued. In the case of non negative real numbers, when we write $\sqrt{c}$ we mean, unambiguously, the non negative solution to $x^2 = c$. When we wish to denote the other solution we write $-\sqrt{c}$. – copper.hat Dec 30 '17 at 08:05
  • @copper.hat I think I kind of resolved it. The $\sqrt{}$ is not a proper inverse unless we pick a branch and hence the $^{2}$ dosnt get canceled. We end up with the absolute value instead which give us a reasonable answer. –  Dec 30 '17 at 08:09
  • @copper.hat By the way in general, what is best is not always right :) –  Dec 30 '17 at 08:10
  • ...and what is right is not always best :-). – copper.hat Dec 30 '17 at 08:13

4 Answers4

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There are two square roots of $4$, and we choose one to be "the" square root of $4$, namely $2$, and we say that $\sqrt4 = 2$.

There are still two solutions to $x^2=4$, because it turns out that the two square roots are opposite of each other, i.e. if $a$ is a square root of $b$, then $-a$ is also a square root of $b$.

Therefore, the other square root of $4$ can be expressed as $-\sqrt4$, i.e. $-2$.

Kenny Lau
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Let we be in the region of real numbers and we say about equations with one variable.

A root of the equation it's a number such that if we'll substitute this number instead variable in the equation then we'll get a true statement.

To solve an equation it says to find all set of roots of the equation.

An arithmetic square root of a non-negative number $a$ (we'll write it $\sqrt{a}$) it's a non-negative number $b$ for which $b^2=a$.

Now, if we want to solve an equation $x^2=4$ then we get $\{2,-2\}$ by the definition.

By the way, $\sqrt4=2$ by the definition again.

Maybe do you mean to use $\sqrt{x^2}=\sqrt{4}$?

If so, then we obtain $|x|=2$ and $\{2,-2\}$ again.

Michael Rozenberg
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  • What do you think about my argument in the above comment? i.e The $\sqrt{}$ is not a proper inverse unless we pick a branch and hence the $^{2}$ dosnt get canceled. We end up with the absolute value instead which give us a reasonable answer. I cant help to end up with the $2=-2$ following your arugment but maybe you have an explanation for that? –  Dec 30 '17 at 08:16
  • $\sqrt{a}$ is not "multivaluedsness" function. $\sqrt{a}$ it's always one number in $\mathbb R$. See please my definitions. – Michael Rozenberg Dec 30 '17 at 08:20
  • Yes, if we are thinking of it as an element of an algebraic structure. But surely this must be compatible with functions on it. –  Dec 30 '17 at 08:23
  • Maybe do you mean $\sqrt{x^2}=\sqrt{4}$? If so, then we obtain $|x|=2$ and ${2,-2}$ again. – Michael Rozenberg Dec 30 '17 at 08:27
  • yes, and I also claim that if we pick the princiapal branch then the $\sqrt{*}$ is a proper inverse and we end up with 2. So your claim $\sqrt{4}=2$ is based on picking a branch. If we think like this then I think it makes sense. Dont you? –  Dec 30 '17 at 08:29
  • But the function $f(x)=\sqrt{x^2}$ has no an inverse function. You can not rape nature and demand that this function will have an inverse function. – Michael Rozenberg Dec 30 '17 at 08:31
  • unless we pick a branch. Which I think you have to do for the $\sqrt{4}=2$ aswell. However picking a branch makes us "loose" solutions –  Dec 30 '17 at 08:37
  • $\sqrt4=2$ by definition and you don't need to choose any branches. If you want to get all roots then you obtain: $x_1=\sqrt4=2$, $x_2=-\sqrt4=-2.$ – Michael Rozenberg Dec 30 '17 at 08:39
  • I got it from here, https://math.stackexchange.com/questions/2583970/on-the-real-square-root-and-branches-of-the-complex-square-root. But maybe I missunderstood him –  Dec 30 '17 at 08:45
  • In $\mathbb C$ we have $\sqrt4={-2,2}$ of course. If you want to solve $x^2=4$ in $\mathbb C$ you don't get a contradiction. The definition of square root in $\mathbb C$ and the definition of square root in $\mathbb R$ they are different. – Michael Rozenberg Dec 30 '17 at 09:00
  • According to two different peopole, the definitions are not different. It is in the comments of the answer –  Dec 30 '17 at 09:37
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The word "root" means solution. The square roots of $n>0$ are the solutions to the equation $x^2=n.$ Since $x^2 = 4$ has two solutions, it is therefore rediculous to call $2$ the square root. It makes perfect sense, however, to call $\sqrt 4$ the positive square root of $4$.

This idea carries over into other problems. The fifth roots of $32$ are the five complex numbers, $z$, satisfying the equation $z^5 = 32$. Since there is only one real-valued solution, $z=2$, we say that the fifth root of $32$ is $2$ and no one will complain about us ignoring the other four solutions. Yet, to be correct, we must say that the positive sixth root of $64$ is $2$.

The point is that, given one solution, it is often much easier to find the other solutions. In trigonometry, for example, the equation $\sin x = \frac 12$ has an infinite number of solutions. Yet, on most scientific calculators, pressing $``\mathbf{inv \phantom{0} sin \phantom{0} 0.5 \phantom{0} ="} $ will return $30$ if the calculator is set for degrees. Knowing this, we can find all of the other solutions.

Steven Alexis Gregory
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The word function is important. A function can has one or less values for every input. When we say $\sqrt x$ we define it to be the positive answer to the equation $b^2=x$.

Unlike functions equations can have more than one solutions, so we don't need to choose, the equation $b^2=x$ has 2 solutions, $\pm\sqrt x$, but $\sqrt x$ is defined to be positive.

When we apply the function to an equation we do have 2 values: apply $\sqrt \cdot$ to $b^2=x$ to get $|b|=\sqrt x$, this has one value, this also answer the definition of $\sqrt x$ because it is always positive, but we want to solve for $b$ not for $x$, so let's remember what $|\cdot|$ is:$$|b|=\begin{cases}b&b\ge0\\-b&b<0\end{cases}$$hence the equation $|b|$ has to solutions($\pm |b|$) lets apply this to the result we got($|b|=\sqrt x$) and we get $\sqrt x=\begin{cases}b&b\ge0\\-b&b<0\end{cases}$ we can now find that $b=\pm|b|=\pm\sqrt x$.

So even with the definition of $\sqrt x$ to be positive the answer of the equation is $\pm\sqrt x$, two solutions

ℋolo
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  • I think we have to pick a branch of the square root to use it. Even when its real –  Dec 31 '17 at 09:27
  • @l33tg33k no, we don't. we don't need to pick because someone already did, the notation $\sqrt \cdot$. it is called the principal square root and it is defined to be the positive value. when we are talking about complex numbers we will define $\sqrt {re^{i\theta}}=\sqrt re^{i\frac\theta2}$ where $r$ and $\theta$ are real. this is how we define it, you are right that someone somewhere somewhen had to pick, but we do not need to do it – ℋolo Dec 31 '17 at 09:41
  • but every real number is complex so why dont we think of it as a restriction? –  Dec 31 '17 at 10:52
  • @l33tg33k because there is no reason for restrictions. What do you want to restrict? – ℋolo Dec 31 '17 at 10:53
  • To apply a function it must be a function and can only have one value. What operation do you apply to your equation? –  Dec 31 '17 at 11:10
  • I am not sure what you are trying to ask, if you are asking how could we get 2 answers even though the function gives one answer then square root gives us 2 solutions, we can't work with this so we defined the principal square root($\sqrt\cdot$) where the set ${-\sqrt\cdot,\sqrt\cdot}$ is the possible outcome of the square root. so in the equation $x^2=4$ we apply the principal square root to get $x=2$ and from this i get the set ${-2,2}$, this is the set of the action square root. so the answer is ${-2,2}=\pm2$ – ℋolo Dec 31 '17 at 11:52
  • I am asking if us getting two solutions is a consquense of the square root having 2 values. Regarding your explanation, if we apply the principal root to get $x=2$ then $x=2$ and not $-2$. I think we get $-2$ if we use the other branch. –  Dec 31 '17 at 12:02
  • if we are talking about square root then you are right but if you are talking principal square root then no. I know it probably weird that i am repeating this but it is very important because i am almost sure that when one study square root one study the principal square root without knowing it is something different. but square root does really have 2 branches and we need both of them to get all of the possible answers – ℋolo Dec 31 '17 at 12:13
  • I dont see how they are different. They are both the inverse of $x^2$. How do you define it as a fuction on $\mathbb{R}$? (when it is not principal). Two guys on my previous question seem to agree aswell. –  Dec 31 '17 at 18:06
  • defining what function is is not a simple thing, i'll try to explain it: imagine 2 sets: $D={1,2,3},C={a,b,c}$($D$=domain, $C$=codomain). a function is a way to pair every value in $D$ to single value in $C$, where $1,2,3,a,b,c$ can be any type of variable($\Bbb R,\Bbb R^n,\Bbb C\cdots$) now the square root of $x$ is not a function because it is not a single value! the principal square root is defined to be positive to make it a function. now, $x^2$ has no inverse, but $f:\Bbb R\to\Bbb R^+,x^2$ has a inverse, the principal square root. – ℋolo Dec 31 '17 at 18:21
  • Right, I was refering to the square root (not principal) in my comment. Ofc if we pick a branch and get the pricipal sqrt we only obtain one solution. I see your point in not mixing these up. –  Dec 31 '17 at 18:27