Numbers: Rational numbers
Integer powers of fractions
Integer power of a real number
For each number \(g\) distinct from #0# and any positive integer \(n\)
\[ \begin{array}{rcl}
g^n & =& \underbrace{g \times g \times \cdots \times g}_{n\textrm{ times}} \\ g^0 & =& 1 \\ g^{-n} & = &\frac{1}{g^n}\end {array}\]
With this \(g^n\), the #n#-th power of #g#, for every integer \(n\) is determined. The number \(g\) is called the base and \(n\) is called the exponent.
If #n# is a natural number, then the definition of #g^n# is known from the theory the notion of integer. In this case the law #g^{m}\cdot g^n=g^{m+n}# applies, where #m# is a natural number.
The definition of #g^n# for negative #n# is chosen in such a way that this law remains valid: #g^{1}\cdot g^{-1}=g\cdot\frac{1}{g}=1#.
If the base is a fraction, then the exponentiation is simple:
Rules of calculation for exponents
For each number \(a\) and \(b\) unequal to \(0\) and each integer \(n\) , \[\left(\frac{a}{b}\right)^n=\dfrac{a^n}{b^n}\tiny.\] holds.
Furthermore, if \(n\lt0\) , then also \[\left(\frac{a}{b}\right)^n=\dfrac{b^{-n}}{a^{-n}}\tiny.\] holds.
If #n\ge0#, then following from the definition: \[\left(\frac{a}{b}\right)^n=\underbrace{\frac{a}{b} \times \frac{a}{b} \times \cdots \times \frac{a}{b}}_{n\textrm{ times}}=\dfrac{a^n}{b^n}\tiny.\] This proves the first statement in the case #n\ge0#.
If #n\lt0#, from this follows: \[ \begin{array}{rclcl}\left(\frac{a}{b}\right)^n&=&\frac{1}{\left(\dfrac{a}{b}\right)^{-n}}&\phantom{x}&\color{blue}{\text{definition integer power of fraction}}\\&=&\frac{1}{\ \dfrac{a^{-n}}{b^{-n}}\ }&&\color{blue}{\text{previous case for }-n}\\&=&\frac{b^{-n}}{a^{-n}}&&\color{blue}{\text{divided by a fraction because}}\\&&&&\color{blue}{\text{multiplied with inverse fraction}}\end{array}\] With this the second statement is proven.
We still have to check that the first statement is true for #n\lt0#. For this end, we continue calculating:\[ \begin{array}{rclcl}\left(\frac{a}{b}\right)^n&=&\frac{b^{-n}}{a^{-n}}&\phantom{x}&\color{blue}{\text{just reduce}}\\&=&{\ \dfrac{1}{b^n\ }}/{\ \dfrac{1}{a^n}\ }&\phantom{x}&\color{blue}{\text{definition integer power of fraction}}\\&=&{\frac{1}{b^n}}\cdot {\frac{a^n}{1}}&\phantom{x}&\color{blue}{\text{divided by a fraction because}}\\&&&&\color{blue}{\text{multiplied with inverse fraction}}\\&=&\frac{a^n}{b^n}&\phantom{x}&\color{blue}{\text{fractions multiplied}}\\ \end {array}\]
Because \(\left(\frac{7}{2}\right)^{2}=\frac{{7}^{2}}{{2}^{2}}={{49}\over{4}}\)
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