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?

Solution 1:

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.

Solution 2:

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,

Solution 3:

I have two questions and two answers that should clear up your confusion.

So here's the first question. Why are we only allowed to choose one value for a square root?

The answer is this: Because any numerical value must only have one real answer. Otherwise, what you're essentially saying is $\sqrt{1}=1=-1$. Well no, see, right there, there's a contradiction between your statement; $1 \neq -1$. Therefore we can only have one answer; the positive root or the negative root.

$2$nd question: Why do we take the positive value for the square root?

The best answer I can come up with is this. Let's take a number like $256$. Okay, now let's say the square root of $256$ is $16$ (Oh wait, it's also equal to $-16$) Now let's take the square root of $16$. The answer to that is $4$ (Oh wait, it's also equal to $-4$) Lastly, let's take the square root of $4$. The answer is $2$, but at the same time it's $-2$.

Now, the $2$nd root of $256$ is $16$, the $4$th root of $256$ is $4$, and the $8$th root of $256$ is $2$. However, $256$ would never have a fourth or an eighth root if you were to use the negative square root. Otherwise, then the square root of $256$ is $-16$, and the square root of $-16$ would not exist.

I hope this helped answer your question.