Why do we assume principal root for the notation $\sqrt{}$

Question

I'm wondering why when $n$ is even we always assume the positive root for $\sqrt[n]{}$.

For example, if we have $x = \sqrt{4}$, we always assume $x = 2$. But if we have $x^2 = 4$, we do $x = \pm\sqrt{4} \Longrightarrow x = -2, 2$.

The problem is that if I take the 1st equation, and square both sides, I get $$ x = \sqrt{4} \Longrightarrow x^2 = 4 \Longrightarrow x = \pm\sqrt{4} \Longrightarrow x = -2, 2 $$ My teacher says that if you introduce the radical sign, use $\pm$, but those two equations are the same under the rules about radicals she taught us, so in reality, (I think) "taking the positive root" rule has no meaning.

I get that people want to make the radical mean something without ambiguity, but making arbitrary rules (at least I thing they're arbitrary), like taking the positive answer, seems to confuse things and leads to inconsistency, like what is shown above.

There are other examples, such as solving for $x$, then substituting the original equation with the value of $x$, and not getting it to work because you can only take the positive root.

For things like the Pythagorean theorem, people like to say it's obvious to take the positive one only, but there are ways to represent the Pythagorean theorem without relying on human judgement to decipher the final answer. Say you have a right triangle with legs $a=3$, $b=4$, and you want to find c. So you do: $3^2 + 4^2 = c^2$, $c > 0$. and solve the systems of equations, just like any other: $c = \pm\sqrt{25} \Longrightarrow c = -5, 5$. $c = -5, 5$ intersects $c > 0$ at $c = 5$, so the answer is $c = 5$. I don't see why you would need to redefine square root: $\pm\sqrt{}$ to mean principal square root: $\sqrt{}$ to find the correct answer.

Does it have something to do with imaginary numbers, or am I missing something? Thanks for any help.

Edit: I am not asking if $\sqrt{}$ means positive, I'm asking why.

Answer 1

Here is a short answer to the question in the title of OP:

Well, if we don't do so, what could a better alternative be?

What is the notation $\sqrt{}$?

The confusion seems to be from understanding of the notation $\sqrt{}$. When writing, for instance $\sqrt{16}$, one pronounces it as "square root of $16$". However, what one really means is "the principal square root of $16$".

Let's go back to the definitions. A square root of a real number $a$ is a number $y$ such that $y^2 = a$; in other words, a number $y$ whose square is $a$. For example, $4$ and $−4$ are square roots of $16$ because $4^2=(-4)^2=16$. Note carefully that the notation $\sqrt{}$ is not involved in this definition at all.

Now, for every given positive real number, say $16$ again, there are two "square roots" (note carefully again that we don't write $\sqrt{x}$ for "square roots of $x$" yet) of it. What if one wants specifically to refer to the positive one? Instead of explicitly saying "I'm refering to the positive square root of $16$", one uses the notation $\sqrt{}$ to define $\sqrt{16}$ as the positive square root of $16$. Here comes the notation $\sqrt{}$. Of course you are losing "information" when you write $\sqrt{16}$ to mean "the positive square root of $16$". Because it is by definition so. What does one do for the "lost information"? One naturally has $-\sqrt{16}$ as the negative square root of $16$.

One can put two definitions together to see what is really going on:

• A "square root" of a real number $a$ is a number $y$ such that $y^2=a$;

• Given a positive real number $x$, the notation $\sqrt{x}$ is defined as a positive real number $y$ such that $y^2=x$. And in this case, we write $y=\sqrt{x}$.

Why is $\sqrt{}$ defined in the way above?

If one does not define $\sqrt{a}$ as the positive square root of $a$ and instead as the "square roots of $a$", then one would have $\sqrt{16}=\pm 4$. Now how would you write the answer to the following question?

What is the positive real number $x$ such that $x^2=\pi$?

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[Added: ]Compare the following two possible definitions for the notation $\sqrt{}$:

• I. For any positive real number $a$, define $\sqrt{a}$ as the square roots of $a$;
• II. For any positive real number $a$, define $\sqrt{a}$ as the positive square root of $a$;

Now, if one uses definition I, then $\sqrt{16}=\pm4$. With this definition, you have perfectly what you might want: $$ x^2=16\Rightarrow x=\pm 4;\quad\text{and }x=\sqrt{16}=\pm4. $$

If one uses definition II instead, on the other hand, one would have $\sqrt{16}=4$.

You might be happier with definition I and ask why on earth one prefers II. Here is "why". Suppose you are asked to solve the following problem.

Find the solution to the equation $x^2-\pi=0$ such that $x>0$.

If one uses definition II, then one immediately has $x=\sqrt{\pi}$.

Now if one uses definition I, $x=\sqrt{\pi}$ would be the WRONG answer.

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One more lesson from Terry Tao:

It’s worth bearing in mind that notation is ultimately an artificial human invention, rather than an innate feature of the mathematics one is working on; sometimes, two writers happen to use the same symbol to denote two rather different concepts, but this does not necessarily mean that these concepts have any deeper connection to them.

Answer 2

What happens is that when you squared you got an equation that seems to be equivalent to the original, but in reality it is not. The original implicitly has the restriction that the $x$ must be a non-negative number, but the second does not.

As the one you are solving is the first and not the second, the restriction must remain present, and if it cannot be implicitly, you must write it explicitly:

$$ x^2 = 4, x\geq0 $$

And so nothing has changed. When we solve an equation we must identify firstly what the restrictions are and keep them present, because sometimes algebraic manipulations change the set of solutions.

Answer 3

Instead of looking at a specific equation, like $x^3 = 8$, you need to look at the bigger problem $x^3 = y$. What you want is, given $y$, to find what $x$ is. In other words, you want a function $f$ such that $x = f(y)$. In this example, $x = \sqrt[3] y$ and you problem is solved. This point of view works great when the exponent is an odd number.

Now consider the equation $x^2 = y$. We have a problem because, for example, $3^2 = 9$ and $(-3)^2 = 9$. A function $x = f(y)$ can only return one value for each $y$. So either $3 = f(9)$ or $-3 = f(9)$ but we can't have both if we want $f$ to be a function. So, if we want $f(y) =\sqrt y$ to be a function, then we have to choose. The choice was $f(y) = \sqrt y$ is the positive square root of $y$.

So, when you see an equation like $x^2 = 25$. Then $x = \sqrt{25}$ gives you a solution $x=5$. If you want both solutions, then you have to write $x = \pm \sqrt{25}$.

Answer 4

Mathematicians use the principal square root more often.

We would rather use $ \pm \sqrt x $ in a few cases then have to write $ |\sqrt x|$ all the time.

That's pretty much the only reason why.

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$ \sqrt \cdot $ is usually how we notate a principal square root. It is not an inverse operation.

$ \pm \sqrt \cdot $ is usually how we notate a (true) square root. Is is the inverse operation of $ \cdot^2 $.

Answer 5

The problem is that it shouldn't be presented as a "rule". The reason we don't apply the plus and minus when solving even radical equations is because the domain for these functions is intentionally restricted. It must be restricted to values of x for which the radicand is positive. In the "Real Number System" we cannot take even roots of negative numbers. This would require complex numbers. Therefore we take only the "Principal Root" because under the restricted domain, there is no alternative.

Answer 6

It depends entirely on the context in which the square-root is being used.

$\pm$ is used when finding solutions to a condition to accurately describe the domain in which the condition is true. For instance, take $x^2 + 5 = 14$. Isolate $x^2$: $x^2 = 9$. This condition is satisfied such that $x = \pm 3$; thus, to encompass the full domain in which the condition is true, we use the $\pm$ operator.

$\pm$ is not used for algebraically manipulated functions that happen to involve radicals, such as $f(x) = \sqrt{x}$. Functions are defined as input-output systems such that each input has one output (i.e. only one branch). $f(x) = \pm \sqrt{x}$ contradicts this definition because it outputs two quantities for one input. Now, that doesn't necessarily mean that you cannot declare $f(x)$ to be $\pm \sqrt{x}$, but $f(x)$ will not be an I/O function and will lose certain properties that comes with the consequence of having more than one branch when dealing with applied mathematics. Multi-valued I/O systems are almost exclusively used for solving a condition (e.g. the Lambert W Function $W(x)$).

Answer 7

The problem is that I get $$x = 2 \color{red}\implies x^2 = 4 \implies x = \pm\sqrt{4} \implies x = \pm2\tag1$$

1. This statement is really not a problem: it can be summarised as $$\text{if $x$ equals $2,\,$ then $x$ equals either $-2$ or $2$},$$ which is a perfectly reasonable claim. Analogously, $$\text{if today is Monday, then today is a weekday}.$$

2. To be clear: statement $(1)$ claims neither that $$x = \pm2\implies x = 2\\\text{(if today is a weekday, then today is Monday)}$$ nor that $$x = 2 \color{red}\iff x = \pm2\\\text{(today is a weekday means that today is Monday)}.$$

3. Let $A(x)$ and $B(x)$ be equations.

   The statement $$A(x)\color{red}{\implies}B(x)$$ means that every solution of $A(x)$ is a solution of $B(x);$ on the other hand, the statement $$A(x)\color{red}{\iff}B(x)$$ means that $A(x)$ and $B(x)$ have the same solutions.