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# square root

The square root of a nonnegative real number $x$, written as $\sqrt{x}$, is the unique nonnegative real number $y$ such that $y^{2}=x$. Thus, $(\sqrt{x})^{2}\equiv x$. Or, $\sqrt{x}\times\sqrt{x}\equiv x$.

###### Example.

$\sqrt{9}=3$ because $3\geq 0$ and $3^{2}=3\times 3=9$.

###### Example.

$\sqrt{x^{2}+2x+1}=|x+1|$ (see absolute value and even-even-odd rule) because $(x+1)^{2}=(x+1)(x+1)=x^{2}+x+x+1=x^{2}+2x+1$.

In some situations it is better to allow two values for $\sqrt{x}$. For example, $\sqrt{4}=\pm 2$ because $2^{2}=4$ and $(-2)^{2}=4$.

Over nonnegative real numbers, the square root operation is left distributive over multiplication and division, but not over addition or subtraction. That is, if $x$ and $y$ are nonnegative real numbers, then $\sqrt{x\times y}=\sqrt{x}\times\sqrt{y}$ and $\displaystyle\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$.

###### Example.

$\sqrt{x^{{2}}y^{{2}}}=xy$ because $(xy)^{2}=xy\times xy=x\times x\times y\times y=x^{2}\times y^{2}=x^{{2}}y^{{2}}$.

###### Example.

$\displaystyle\sqrt{\frac{9}{25}}=\frac{3}{5}$ because $\displaystyle\left(\frac{3}{5}\right)^{2}=\frac{3^{2}}{5^{2}}=\frac{9}{25}$.

On the other hand, in general, $\sqrt{x+y}\neq\sqrt{x}+\sqrt{y}$ and $\sqrt{x-y}\neq\sqrt{x}-\sqrt{y}$. This error is an instance of the freshman’s dream error.

The square root notation is actually an alternative to exponentiation. That is, $\sqrt{x}\equiv x^{\frac{1}{2}}$. When it is defined, the square root operation is commutative with exponentiation. That is, $\sqrt{x^{a}}=x^{\frac{a}{2}}=(\sqrt{x})^{a}$ whenever both $x^{a}>0$ and $x>0$. The restrictions can be lifted if we extend the domain and codomain of the square root function to the complex numbers.

Negative real numbers do not have real square roots. For example, $\sqrt{-4}$ is not a real number. This fact can be proven by contradiction as follows: Suppose $\sqrt{-4}=x\in\mathbb{R}$. If $x$ is negative, then $x^{2}$ is positive, and if $x$ is positive, then $x^{2}$ is also positive. Therefore, $x$ cannot be positive or negative. Moreover, $x$ cannot be zero either, because $0^{2}=0$. Hence, $\sqrt{-4}\notin\mathbb{R}$.

For additional discussion of the square root and negative numbers, see the discussion of complex numbers.

The square root function generally maps rational numbers to algebraic numbers; $\sqrt{x}$ is rational if and only if $x$ is a rational number which, after cancelling, is a fraction of two perfect squares. In particular, $\sqrt{2}$ is irrational.

The function is continuous for all nonnegative $x$, and differentiable for all positive $x$ (it is not differentiable for $x=0$). Its derivative is given by:

$\frac{d}{dx}\big(\sqrt{x}\big)=\frac{1}{2\sqrt{x}}$ |

It is possible to consider square roots in rings other than the integers or the rationals. For any ring $R$, with $x,y\in R$, we say that $y$ is a square root of $x$ if $y^{2}=x$.

When working in the ring of integers modulo $n$, we give a special name to members of the ring that have a square root. We say $x$ is a quadratic residue modulo $n$ if there exists $y$ coprime to $x$ such that $y^{2}\equiv x\;\;(\mathop{{\rm mod}}n)$. Rabin’s cryptosystem is based on the difficulty of finding square roots modulo an integer $n$.

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## Attached Articles

square root of square root binomial by pahio

existence of square roots of non-negative real numbers by PrimeFan

square root of 5 by MathNerd

square root of 2 by MathNerd

square root of 3 by PrimeFan

table of continued fractions of $\sqrt{n}$ for $1 < n < 102$ by PrimeFan

squaring condition for square root inequality by pahio

square roots of rationals by pahio

Bombelli's method of computing square roots by pahio

## Comments

## I have *no idea* why this doesn't work

I'm new to this. Somebody tell me what I'm doing or

not doing to cause this entry to not show up.

I'm just writing plain text with latex stuff like $x$

embedded in the text. Am I supposed to use HTML tags

or something?

## okay, fixed now, but...

There's one thing I don't like that I don't know how to fix.

The exponentiated fraction in the middle, (3/5)^2, doesn't render well.

How do you get the fraction to appear in parenthesis?

## Re: okay, fixed now, but...

use \frac{numerator}{denominator}

-apk

## Re: okay, fixed now, but...

I was already doing that. The like with (2/5)^3, you can't really tell whether the

exponent is on the 2 or the whole fraction. Maybe there's no way to do it...

## Re: okay, fixed now, but...

Try this:

$$\left ( \frac{2}{5} \right )^3$$

In LaTeX, you use \left and \right to specify that a pair of braces should be resized to fit what's inside them.

vampyr

## "Clearly, there is only one square root of 2"

>> There is no "the" square root of a positive real number, but rather >>two possible square roots (+ or -).

>>

>> Alvaro

>Sure there is. X squared equals 2; what's X? "Plus-minus square root >of 2". Clearly there is only one "square root of 2", and both it and >its negative, when squared, produce 2.

Absolutely not, there are TWO distinct reals numbers such that X^2=2. Why would we give a special relevance to the positive solution?

>Besides, I wrote this with a sixth grade or younger reader in mind; >everyone else knows perfectly well what a square root is. The whole >plus-minus thing is confusing; I thought it best to use only positive >numbers at first and work up to the rest.

Even in that case, there is no loss being precise. You may write "a square root of 2 is a number x such that x^2=2" instead of "the square root". A sixth grader will understand, and mathematicians will not flinch.

Alvaro

## Re: "Clearly, there is only one square root of 2"

In any case, you don't even specify in the definition that the square root should be the positive one, so even if you're thinking of 6th graders the definition is confuse.

## Re: "Clearly, there is only one square root of 2"

> Why would we give a special relevance to the positive solution?

Because we commonly write the negative solution as -sqrt(2), or the full solution as +-sqrt(2), as if to imply in the notation that sqrt(2) must be a positive number.

But your alternate phrasing sounds good to me.

## Re: "Clearly, there is only one square root of 2"

Hi there,

Same discussion about

\sqrt{-1}\sqrt{-1}=i^2=-1,

or

\sqrt{-1}\sqrt{-1}=sqrt{(-1)(-1)}=sqrt{1}=1

## Re: "Clearly, there is only one square root of 2"

The square root sign, and the word is very often used for the positive square root without specifying. You will find most analysis will for example use notation $\sqrt{u^+v^2}$ for the modulus of $u+iv$, and obviously they mean the positive part. Similarly all places I've seen that list the quadratic formula use the plus minus notation and that pretty much means that $\sqrt{x}$ is meant for the positive square root

(regardless of sign then). Perhaps the entry should be more clear in this case, and while not specifying the square root as a unique number (which it is clearly not), it should note that very commonly $\sqrt{x}$ is used to mean the positive root.

## Re: "Clearly, there is only one square root of 2"

> Absolutely not, there are TWO distinct reals numbers such

> that X^2=2. Why would we give a special relevance to the

> positive solution?

Because we wish the radical sign to denote a function.

## Re: "Clearly, there is only one square root of 2"

Well, maybe it is not a good idea to have \sqrt(x) to denote a function in the real case, since \sqrt(x) is really a "multivalued" function in the more general case where x is a complex number. It should at least be emphasized that there is a choice involved. We *chose* that the radical sign denote the positive solution when the index of the root is even.

## Re: "Clearly, there is only one square root of 2"

> Well, maybe it is not a good idea to have \sqrt(x) to denote

> a function in the real case, since \sqrt(x) is really a

> "multivalued" function in the more general case where x is a

> complex number. It should at least be emphasized that there

> is a choice involved. We *chose* that the radical sign

> denote the positive solution when the index of the root is

> even.

Given the PM definition of "function", the notion of a multi-valued function is self-contradictory. In fact, "multi-valued function" is a confusing anachronism that ought to be snuffed.

Instead of generalizing the base from real to complex, generalize the exponent from rational to real: sqrt(x) = exp(0.5*ln(x)). The illusion of choice is dispelled.

## Re: "Clearly, there is only one square root of 2"

Well, that's clear that \sqrt{2} is the positive root.

## Re: "Clearly, there is only one square root of 2"

>

> Given the PM definition of "function", the notion of a

> multi-valued function is self-contradictory.

Well, sometimes a "red herring" is neither red nor a herring. A multivalued function is by definition a (continuous) function from a covering space. So that for example, the square root is really a function from the double covering of the Riemann sphere branched over 0 and oo (this covering happens to be conformally isomorphic to the sphere).

I think that this point of view is a very useful and fruitful one.

> In fact,

> "multi-valued function" is a confusing anachronism that

> ought to be snuffed.

>

When something confuses you it is maybe a good idea to try to understand it and clear the confusion. Of course, dismissing it as an anachronism may be an easier solution.

> Instead of generalizing the base from real to complex,

> generalize the exponent from rational to real: sqrt(x) =

> exp(0.5*ln(x)). The illusion of choice is dispelled.

>

Right. I don't disagree that you can consistently define the square root of x to be a positive number or that you can write fancy formulas in the process of doing so. The question was whether it is a good idea to do so without even mentioning that there is a choice involved, especially since one is forced to deal with the multivaluedness of the square root in the complex case.

## Re: "Clearly, there is only one square root of 2"

> Given the PM definition of "function", the notion of a

> multi-valued function is self-contradictory. In fact,

> "multi-valued function" is a confusing anachronism that

> ought to be snuffed.

A way to cope with the phrase "multi-valued function" rigorouly is as follows. Define a multi-valued function from a set D to a set R to be an ordinary (single-valued) function from D to P(R), the power set of R. Foir instance, sqrt(1) = {-1,+1}. A single valued function is then the special case where card (f(x)) = 1 for every x in D.

-RSP

## Re: "Clearly, there is only one square root of 2"

> Instead of generalizing the base from real to complex,

> generalize the exponent from rational to real: sqrt(x) =

> exp(0.5*ln(x)). The illusion of choice is dispelled.

So which logarithm do you choose?

In effect, you have replaced the choice of x>0 by the choice of k even in exp(2*pi*i*k)=1.

By the way (and quoted out of order):

> In fact,

> "multi-valued function" is a confusing anachronism that

> ought to be snuffed.

I'd like to think your overly generalizing to make your point here.

Multi-valued functions serve their purpose elsewhere. (Though the definition I have in mind differs from rspuzio's suggestion.)

Unless you demand abstention from using quotients with respect to group operations there are perfectly reasonable multi-valued functions.

Regards

T.

## Re: "Clearly, there is only one square root of 2"

> Unless you demand abstention from using quotients with

> respect to group operations there are perfectly reasonable

> multi-valued functions.

Please clarify this. By "quotients with respect to group operations" do you mean products of group elements of the form ab^{-1}? Perhaps you intended to refer to quotient groups, but bungled it? Or do you have some more abstruse relation between operations on distinct groups in mind? In any case, how do so called "multi-valued functions" enter into this picture?

## Re: "Clearly, there is only one square root of 2"

I apologize for the snide tone of my previous post. However, I would like to have the connection you allude to spelled out explicitly.

Thanks,

ratboy

## Re: "Clearly, there is only one square root of 2"

I mean taking orbits of an operation of a group on a set (the domain of single valued functions). The multi-valued functions the great-grandparent post wanted to "snuff" are nothing more than ordinary functions into this quotient space.

In the case at hand you have G=Z/2Z rotating the complex plane C, so if you could define a square root function C->C/G if you wish. Of course, the point 0 is somewhat singular. This leads to the same multi-valued function discussed, but gives it a nicer foundation than using some "ad hoc" definition of multi-valued functions.

I've seen multi-valued functions mostly in a different context:

When you want to look at the union of graphs of functions

f_i :Omega->R^n as a geometric object where you don't want to distinguish between different representations of the same object.

As you naturally get a metric space you can do some useful analysis such as defining Lipschitz continuity and differentials almost everywhere by considering the quotient of M^m by permutations of the entries of m-tuples, where M=R^n.

For example if Omega=R and n=1 and the function pairs

f_1(x)=x, f_2(x)=-x

and

f_1(x)=|x|, f_2(x)=-|x|

give the same union of graphs and are projected onto the same corresponding multi-valued function. Here the multi-valued function has Lipschitz constant 1 if you consider the quotient metric (and take max(|a_1|, |a_2|) as norm of the pair). Differentials are more tricky as they require some basic measure theory. Note that here, it makes sense intuitively to speak of 0 as a point of density 2, a notion that can be formalized.

Anyhow. I just thought the total rejection of the idea of multi-valued functions could be overly broad.

My apologies for my first post not being clear enough and this one being overly long.

Cheers

T.

## ad different error in sqrt definition definition

I think everyone is so busy arguing over this that they missed an even more blatant mistake in the given definition: This definition says "The square root operation is distributive for multiplication and division...". The square root operator is not "distributive" in any real mathematical sense of the word, and furthermore, it does not even meet the criteria for the mathplanet "distributive" definition linked right on the very word: The square root operator is NOT a binary operation!!! I think a better choice of words (without having to mention linearity or homs) would be to say that the square operator "preserves division and multiplication..."

## Re: ad different error in sqrt definition definition

Similarly, the word "associative" is also used incorrectly.