TITLE: Why is the even root of a number always positive?
QUESTION [14 upvotes]: 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?

REPLY [2 votes]: 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.