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Difference between revisions of "Square root"

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==Notation==
 
==Notation==
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The square root (or the principle square root) of a number <math>x</math> is denoted <math>\sqrt x</math>.  For instance, <math>\sqrt 4 = 2</math>.  When we consider only [[positive number|positive]] [[real number|reals]], the square root [[function]] is the [[Function/Introduction#The_Inverse_of_a_Function|inverse]] of the squaring function.  
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The square root (or the principal square root) of a number <math>x</math> is denoted <math>\sqrt x</math>.  For instance, <math>\sqrt 4 = 2</math>.  When we consider only [[positive number|positive]] [[real number|reals]], the square root [[function]] is the [[Function/Introduction#The_Inverse_of_a_Function|inverse]] of the squaring function.  
  
 
==Exponential notation==
 
==Exponential notation==

Latest revision as of 10:14, 18 July 2024

A square root of a number $x$ is a number $y$ such that $y^2 = x$. Generally, the square root only takes the positive value of $y$. This can be altered by placing a $\pm$ before the root. Thus $y$ is a square root of $x$ if $x$ is the square of $y$.

Notation

The square root (or the principal square root) of a number $x$ is denoted $\sqrt x$. For instance, $\sqrt 4 = 2$. When we consider only positive reals, the square root function is the inverse of the squaring function.

Exponential notation

Square roots can also be written in exponential notation, so that $x^{\frac 12}$ is equal to the square root of $x$. Note that this agrees with all the laws of exponentiation, properly interpreted. For example, $\left(x^{\frac12}\right)^2 = x^{\frac12 \cdot 2} = x^1 = x$, which is exactly what we would have expected. This notion can also be extended to more general rational, real or complex powers, but some caution is warranted because these do not give functions. In particular, if we require that $x^{\frac 12}$ always gives the positive square root of a positive real number, then the equation $\left(x^2\right)^{\frac 12} = x$ does not hold. For example, replacing $x$ with $-2$ gives $2$ on the left but gives $-2$ on the right.

See also

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