# Square Root as a multi-valued function

I was scrolling through this article on Wikipedia, and I was stumped when I came across this line:

Every real number greater than $$0$$ has two real square roots, so that the square root may be considered a multi-valued function. For exmaple, we may write $$\sqrt{4}=\pm 2=\{2,-2\}$$; although $$0$$ has only one square root, $$\sqrt{0}=\{0\}.$$

I very strongly believe that there is a conceptual gap that I need to bridge. I have always used the fact that the square root of a number is always positive (as seen by the graph of $$y=\sqrt{x}$$).

What particular fact am I overlooking?

As far as I understand upon reading the article again to look for clues, it is mentioned that the domain could be extended. So what I infer is that we can map $$4$$ under the (now multi valued) function $$\sqrt{.}\;$$to $$-2$$ and $$2$$ . Would I be right in saying so?

• Did you mean $\{2,\color{red}-2\}$? Usually $\sqrt\cdot$ means the positive square root Commented Jun 19, 2020 at 20:35
• The solution of $x^2=4$ , for example , can be written as $x=\pm 2$. But $\sqrt{x}$ is because of a clear convention the non-negative square root. For the other branch , we use $-\sqrt{x}$. So, $\sqrt{4}=\pm2$ is against a widely accepted convention. Commented Jun 19, 2020 at 20:43
• The article is about multi-valued functions, which are different from usual ones. It provides as an example an alternative definition of the square root, using the "features" of MVF which we don't have in normal functions. Since it's an alternative definition, it's not surprising that it is different from the one you are used to (which is using normal functions) Commented Jun 19, 2020 at 20:48
• @Peter, I'll edit the question to include the entire line Commented Jun 19, 2020 at 20:48
• The convention was made to make this choice unique, so I disagree that we have two square-roots. We have two branches instead as I mentioned above. Commented Jun 19, 2020 at 20:55

It is a standard convention to take $$\sqrt{x}$$ where $$x\in \Bbb{R}_{\ge0}$$ to be the non-negative solution of the equation $$y^2 = x$$, so that $$x \mapsto \sqrt{x}$$ is a well-defined function on $$\Bbb{R}_{\ge0}$$. We could equally well have chosen to take $$\sqrt{x}$$ to be the non-positive solution.

The Wikipedia article you cite is concerned with what happens when we let $$x$$ range over the complex numbers $$\Bbb{C}$$. In $$\Bbb{C}$$, every number has a square root, but as in $$\Bbb{R}$$, if $$x^2 = y$$, then $$(-x)^2 = y$$, so the square root function has two possible values for every non-zero $$y$$. Resolving this to choose a best possible single-valued approximation to $$x \mapsto \sqrt{x}$$ involves the notion of branch points discussed in the Wikipedia article.

NIST's Handbook of Mathematical Tables by Abramowitz and Stegun will doubtless give the definitive statement about how to read $$\sqrt{\cdot}$$ according to US standards. My copy of A&S is in the loft somewhere, but I'd bet my last but one dollar that it takes $$\sqrt{\cdot}$$ to have the negative real axis as its branch cut.

• Could you please give any reference to read about branch cuts? (Does reading a wiki article on it suffice for beginners?) Commented Jun 20, 2020 at 11:44
• Is it possible to define complex function f (on whole plane) which is continuous .and if f(z)=t then $t \times t=z$ ? Commented Jan 29, 2023 at 15:45

This was going to be a comment, but it got kind of long so I'm putting it here:

A big "gotcha' worth acknowledging is $$\sqrt{\cdot}$$ has no "positive" option once we allow complex inputs. The ability to take the positive root is an artifact of the real numbers. We obviously want $$\sqrt{\cdot}$$ to be continuous, so you might try to just extend the positive branch of $$\sqrt{\cdot}$$ to the complex numbers.

The unfortunate fact is there is no way to do this. The $$\sqrt{\cdot}$$ function is (in a way that complex analysis makes precise) inherently discontinuous. However, we can repair the situation in a number of ways:

1. Restricting the domain of $$\sqrt{\cdot}$$. This is basically the same as saying "don't take square roots of negative numbers". There are lots of possible "branch cuts" that you can take, but no matter what you do you'll be left with a discontinuity, which we typically treat as an undefined region.

2. Changing the domain to a riemann surface. This is really the "best" solution for a number of reasons, though it is conceptually difficult.

3. Allowing multi-valued functions. The issue with making $$\sqrt{\cdot}$$ continuous in $$\mathbb C$$ is this: Start at $$z$$, and remember $$\sqrt{z}$$. Then make a lap around the origin, taking note of $$\sqrt{\cdot}$$ as you go (this should be continuously, of course). When you get back to $$z$$, you'll find $$-\sqrt{z}$$ instead of what you started with! So we can "solve this problem" by just asserting $$\pm \sqrt{z}$$ is the output. Then when we make a lap around the origin, we start at $$\pm \sqrt{z}$$ and we end at $$\mp \sqrt{z}$$, which is the same! Our problem has "gone away", but at the cost of changing what it means to be a function.

This problem really doesn't come to light until you start playing with Complex Analysis, but $$\sqrt{\cdot}$$ and $$\log(\cdot)$$ are the two most famous examples of this phenomenon. Congratulations on running into it! This exact "issue" with complex numbers is responsible for tons of recent mathematics, with really beautiful geometry to show for it.

I hope this helps ^_^

• (+1) A good point ! I am not sure whether there is also a convention for $\sqrt{z}$, I guess there is one. Commented Jun 19, 2020 at 20:59
• This answer, I believe, would make much much more sense once I start learning complex analysis (I'm still in high school). However, I was able to grasp the basic point of how we widely accept the positive solutions to a square root. +1 and thanks for taking your time to help me out! Commented Jun 19, 2020 at 21:05
• How can we prove that it is not possible to define complex function f (on whole plane) which is continuous .and if f(z)=t then $t \times t=z$ ? Commented Jan 29, 2023 at 15:47
• @MeetPatel -- Good question! See here for an existing discussion. Commented Jan 30, 2023 at 0:41

First, I have to mention that the $$\sqrt(.)$$ function you mentioned and the multivalued function from wikipedia page are two different mathematical objects, which, unfortunately, happened to use the same notation.

Second, $$\sqrt a$$ for any positive real $$a$$ usually refer to its positive square root. However, in case of complex there might not be a generally accepted preferred square root due to which $$\sqrt(.)$$ might refer to the multivalued function from wikipedia.

Thus, in my opinion, the confusion arises because, unfortunately, the same notation happens to be used for two different things.It is not a matter of which is right, it is just different notations being used.

• would I be right in saying that the square root function is multi valued for complex numbers? Commented Jun 19, 2020 at 21:14
• No: it would be right to say that the square root function is neither defined for complex nor for negative reals. Then you could say that there is another, multi-valued function which is somewhat similar to the sqrt function which is defined for complex numbers. And then you can say that due to laziness we also call this second function sqrt function even though it different from the first function, and not even gives the same results for positive reals. Commented Jun 19, 2020 at 21:28
• @sai-kartik Ah, that question seems to be now answered by Engineer trying math above. Hope it is clear for you now. Commented Jun 20, 2020 at 5:49
• Yes, Deepak. Thanks again @Engineer ! Commented Jun 20, 2020 at 6:43