# Why is the even root of a number always positive?

Let $n \in \mathbb N$ be a natural number and $a \in \mathbb R$ be a real number. The $n$-th root of the number $a$ is defined as follows:

Case I: $n$ is an odd number. In this case the $n^{\text{th}}$ root of $a$ is defined to be that number $b \in \mathbb R$ such that $b^n = a$.

Case II: $n$ is an even number. In this case the $n^{\text{th}}$ root of $a$ is defined to be that number $b \geq 0$ such that $b^n = a$.

Why is it that when $n$ is even, we only consider $b \geq 0$. For example, both $+2$ and $-2$ squared equal $4$, but when we say the square root of $4$ is $2$. Is there a reason for this?

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Nothing more than convention... –  Fabian May 28 '11 at 22:34
@Fabian: No, there is a lot more to it then convention! –  Eric Naslund May 28 '11 at 22:47
Actually Fabian is right. Even if you work with complex numbers and chose a particular branch, that is just a convention; you could pick another one. And while it becomes pretty clear that that particular branch is the "natural" choice, I don't see why this would be more natural than chosing the positive over the negative as the root of a positive number. –  N. S. May 28 '11 at 22:52
@user9176: A few things: The branch you pick can depend on what you are doing, and what you want your function to look like, so no that is not just convention. When you have functions with multiple branches, different choices can have cancellation in different ways. Most important, even when there is a convention, it is in no way helpful to say that it is "Nothing more then convention." There are always reasons to why we choose a particular convention, and understanding those reasons is very important. –  Eric Naslund May 28 '11 at 22:59
That's precisely why I asked this question! Frankly, as I went through high school a lot of these questions were answered as "because that's the way it is..". But I know that there is always a reason (in Mathematics at least), so I decided to ask here. –  neesh May 28 '11 at 23:21

There is a difference between "solutions to $x^n = a$" and "the $n$th root of $a$".

Basically, if you want the square root to be a (single valued) function, then you should get one and only one answer for any valid input. That means that you cannot simply say "the square root of $4$ is a number whose square is $4$", because then you can get different answers depending on who you ask, or when you ask; you always want to get the same answer. Which means you need to pick one of the numbers whose square is $4$ to be the square root of $4$.

This is done by convention (agreement). In principle, there is no reason to prefer the nonnegative solution to the nonpositive; in practice, you want to either always pick the nonnegative ones, or always pick the nonpositive ones (that makes the function "square root" a "nice" function, where nice has to do with properties of functions like continuity). And because people understood real positive numbers for a much longer time than they understood negative ones (even negative integers), the nonnegative choice is the one we all agree to use.

That's why $\sqrt{4}$ is $2$, and not $-2$, and not $\pm 2$. We want the square root to be a function, so we want a single answer, and we agree to give as an answer the nonnegative solution to $x^2=4$.

The same is true for other even powers: there are two possible answers for the equation, but we want the function to have a single answer, so we agree that it will be the nonnegative one.

This problem does not arise with cubic, fifth, seventh, odd-th roots, because then you don't have two possible answers for the equation, so there is no need to choose for the function.

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Actually, isn't there some reason to prefer the nonnegatives? e.g. $\sqrt{a}\sqrt{b}=\sqrt{ab}$ wouldn't hold if the convention were to take the nonpositive solution. –  trutheality Jun 12 '11 at 0:57
@trutheality: There are many reasons to prefer nonnegatives (e.g., historical; real numbers arose much earlier than negatives, after all), but any other choice simply means different conventions. If you instead decide you want your square roots to be nonpositive, then the equation would be $\sqrt{ab}=-\sqrt{a}\sqrt{b}$; no harder to remember than $\sqrt{x^2}=|x|$ is with the usual convention. –  Arturo Magidin Jun 12 '11 at 2:11

To make complete sense of this, we need to look at things in in the complex plane. When we choose to take the positive square root, we are just choosing a particular branch cut of the complex logarithm. In general, given a number $x\neq 0$, $x$ will have $n$ distinct $n^{th}$ roots.

You noticed that $-2$ and $2$ are both square roots of $4$. What about the cube roots of $2^{\frac{3}{2}}$? Well we have $\sqrt{2}$, but is that the only one? No, we also have $-1+i$ and $-1-i$. Similarly for the fourth root of $16$. We have the roots $2$ and $-2$, but we also have $2i,-2i$.

In general, when we take the $n^{th}$ root of a real number, we get $n$ different possibilities in the complex plane. In other words, taking $n^{th}$ roots is a multivalued function, so we have to make a choice, and this corresponds to choosing a branch cut for the logarithm function.

For most situations we choose the branch which sends the positive real line to the positive real line, but there are sometimes where we need to choose a different branch. (For example, evaluating certain integrals by complex methods.)

Hope that helps,

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