# 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 $3^{2}=3\times 3=9$.

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

###### 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   (http://planetmath.org/ProofByContradiction) 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 squares. In particular, $\sqrt{2}$ is irrational.

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 $n$ (http://planetmath.org/MathbbZ_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 $y^{2}\equiv x\pmod{n}$. Rabin’s cryptosystem is based on the difficulty of finding square roots modulo an integer $n$.

 Title square root Canonical name SquareRoot Date of creation 2013-03-22 11:57:19 Last modified on 2013-03-22 11:57:19 Owner Wkbj79 (1863) Last modified by Wkbj79 (1863) Numerical id 27 Author Wkbj79 (1863) Entry type Definition Classification msc 11A25 Related topic CubeRoot Related topic NthRoot Related topic RationalNumber Related topic IrrationalNumber Related topic RealNumber Related topic ComplexNumber Related topic Complex Related topic DerivativeOfInverseFunction Related topic EvenEvenOddRule