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Subsection 0.1.6 Exponentiation
We define exponentiation as repeated multiplication:
Definition 0.1.6.1.
For a real number \(b\) and a natural number \(n\) the \(n\)-th power of \(b\) is
\begin{equation*}
b^n=\underbrace{b\cdot b\cdot\ldots \cdot b}_{n \text{ copies of }b}.
\end{equation*}
We call \(b\) the base of \(b^n\) and \(n\) its exponent.
Theorem 0.1.6.2.
For all real numbers \(a\) and \(b\) and natural numbers \(m\) and \(n\) we have:
-
\(\displaystyle (b^m)\cdot (b^n)=b^{(m+n)}\)
-
\(\displaystyle (a^n)\cdot (b^n)=(ab)^{n}\)
-
\(\displaystyle {(b^m)}^n=b^{(m\cdot n)}\)
Proof.
-
\begin{align*}
(b^m)\cdot( b^n) \amp=
\underbrace{b\cdot b\cdot\ldots \cdot b}_{m \text{ copies of }} \cdot \underbrace{b\cdot b\cdot\ldots \cdot b}_{n \text{ copies of }b} \\
\amp=
\underbrace{b\cdot b\cdot\ldots\cdot b}_{(m + n) \text{ copies of } b} \\
\amp = b^{(m+n)}.
\end{align*}
-
\begin{align*}
a^n \cdot b^n
\amp = \underbrace{a\cdot a\cdot \ldots \cdot a}_{n \text{ copies of }a} \cdot \underbrace{b\cdot b\cdot \ldots \cdot b}_{n \text{ copies of }b }\\
\amp
=\underbrace{(a\cdot b)\cdot (a\cdot b)\cdot \ldots \cdot (a\cdot b)}_{n \text{ copies of }a\cdot b } \\
\amp= (a\cdot b)^n.
\end{align*}
-
\begin{align*}
{(b^m)}^n \amp= (\underbrace{b\cdot b\cdot \ldots \cdot b}_{m\text{ copies of }b})^n\\
\amp =
\underbrace{\underbrace{b\cdot b\cdot \ldots \cdot b}_{{m \text{ copies of } b}}\cdot \ldots \cdot \underbrace{b\cdot b\cdot \ldots \cdot b}_{m \text{ copies of }b}}_{n \text{ copies of } \underbrace{b\cdot b\cdot \ldots \cdot b}_{m \text{ copies of } b}} \\
\amp =
\underbrace{b\cdot b\cdot \ldots \cdot b}_{(m \cdot n) \text{ copies of } b}\\
\amp= b^{(m\cdot n)}.
\end{align*}
So far we only have considered exponents that are natural numbers. In a generalization we would like
Theorem 0.1.6.2 to still hold.
\begin{equation*}
b^0=1.
\end{equation*}
\begin{equation*}
b^{(-m)}=\frac{1}{b^m}.
\end{equation*}
Definition 0.1.6.3.
Let
\(a\in\RR\text{.}\) The square root of
\(a\) is the positive solution of
\(x^2=a\text{.}\)
