Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Exponentiation
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Rational exponents== [[File:Mplwp roots 01.svg|thumb|From top to bottom: {{math|''x''<sup>1/8</sup>}}, {{math|''x''<sup>1/4</sup>}}, {{math|''x''<sup>1/2</sup>}}, {{math|''x''<sup>1</sup>}}, {{math|''x''<sup>2</sup>}}, {{math|''x''<sup>4</sup>}}, {{math|''x''<sup>8</sup>}}.]] If {{mvar|x}} is a nonnegative [[real number]], and {{mvar|n}} is a positive integer, <math>x^{1/n}</math> or <math>\sqrt[n]x</math> denotes the unique nonnegative real [[nth root|{{mvar|n}}th root]] of {{mvar|x}}, that is, the unique nonnegative real number {{mvar|y}} such that <math>y^n=x.</math> If {{mvar|x}} is a positive real number, and <math>\frac pq</math> is a [[rational number]], with {{mvar|p}} and {{mvar|q > 0}} integers, then <math display="inline">x^{p/q}</math> is defined as :<math>x^\frac pq= \left(x^p\right)^\frac 1q=(x^\frac 1q)^p.</math> The equality on the right may be derived by setting <math>y=x^\frac 1q,</math> and writing <math>(x^\frac 1q)^p=y^p=\left((y^p)^q\right)^\frac 1q=\left((y^q)^p\right)^\frac 1q=(x^p)^\frac 1q.</math> If {{mvar|r}} is a positive rational number, {{math|1=0<sup>''r''</sup> = 0}}, by definition. All these definitions are required for extending the identity <math>(x^r)^s = x^{rs}</math> to rational exponents. On the other hand, there are problems with the extension of these definitions to bases that are not positive real numbers. For example, a negative real number has a real {{mvar|n}}th root, which is negative, if {{mvar|n}} is [[odd number|odd]], and no real root if {{mvar|n}} is even. In the latter case, whichever complex {{mvar|n}}th root one chooses for <math>x^\frac 1n,</math> the identity <math>(x^a)^b=x^{ab}</math> cannot be satisfied. For example, :<math>\left((-1)^2\right)^\frac 12 = 1^\frac 12= 1\neq (-1)^{2\cdot\frac 12} =(-1)^1=-1.</math> See ''{{slink||Real exponents}}'' and ''{{slink||Non-integer powers of complex numbers}}'' for details on the way these problems may be handled.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)