Exponentiation
lastmod 2018-08-17created 2015-12-232 edits
Exponentiation is a binary operation meaning the operation of multiplying a number by itself several times. It is written with the symbol \(a^n\), where \(a\) is called the base and \(n\) the exponent.
Definition
For a natural number \(n\), the power \(a^n\) is defined as follows.
\[ {{a^n = } \atop {\ }} {{\underbrace{a \times \cdots \times a}} \atop n} \] This is similar to how multiplication repeats addition. Also, by definition, the following hold.
- \(a^1 = a.\)
- \({a^b}{a^c} = a^{b+c}.\)
- \((a^n)^m = a^{nm}.\)
- \((a^b)^c = (a^c)^b\)
- \({a^c}{b^c} = (ab)^c.\)
- \(a^m \div a^n = a^{m-n}\)
The following inductive definition is also possible:
- \(a^1 = a.\)
- \(a^{n+1} = a \times a^n,\ n=1, 2, 3, \cdots.\)
Roots
Just as, for a real number \(a\), \(a^2,a^3,a^4,a^5 ... a^n ...\) (= the square of a, the cube of a, the fourth power of a, the fifth power of a ... the nth power of a ...) are collectively called the powers of a,
the square root of a, the cube root of a, the fourth root of a, the fifth root of a, ... the nth root of a ... are collectively called the roots of a.
Here, the positive square root of a is written with the symbol
- \(\sqrt[2]{a}\) or \(\sqrt{a}\)
the positive cube root of a with the symbol
- \(\sqrt[3]{a}\)
the positive fourth root of a with the symbol
- \(\sqrt[4]{a}\)
and the positive nth root of a with the symbol
- \(\sqrt[n]{a}\)
as shown.
That is, a root is an nth root of some real number, such as \(\sqrt{a},\sqrt[3]{a}, \sqrt[4]{a}, ... , \sqrt[n]{a} ...\).
Extending the range of the exponent
Besides the definition of multiplying a number several times, exponentiation can be defined in various ways. Depending on the definition of exponentiation, the range of the exponent n in \(a^n\) can then be taken to be wider than the positive integers (natural numbers).
That is, from here on, we will think of exponentiation in terms of the meaning of exponent 1 as how many times a number is multiplied.
Integers
When n is a negative integer, it is defined as follows.
\[ a^{-n} = \frac{1}{a^n} \] And for \(a^n \times a^m = a^{n+m}\) to hold in this case too, \(a^n \times a^{-n} = a^{n+(-n)}\) must hold, so \(a^0\) is defined as follows. (when a≠0[^0])
\[ a^0 = 1\, \]