Theorem B.27.
For all \(m,n,p\in\nonnegints\) with \(m\neq0\text{,}\) \(m^{n+p}=m^n\,m^p\text{.}\)
Section B.9 Exponentiation
We now define a binary operation called exponentiation which is defined only on those ordered pairs \((m,n)\) of natural numbers where not both are zero. The notation for exponentiation is non-standard. In books, it is written \(m^n\) while the notations \(m**n\text{,}\) \(m\wedge n\) and \(\exp(m,n)\) are used in-line. We will use the \(m^n\) notation for the most part.
When \(m=0\text{,}\) we set \(0^n=0\) for all \(n\in\nonnegints\) with \(n\neq0\text{.}\) Now let \(m\neq0\text{.}\) We define \(m^n\) by (i) \(m^0=1\) and (ii) \(m^{k+1}=mm^k\text{.}\)
Proof.
Let \(m,n\in\nonnegints\) with \(m\neq0\text{.}\) Then \(m^{n+0}=m^n=m^n\,1=m^n\,m^0\text{.}\) Now suppose that \(m^{n+k}=m^n\,m^k\text{.}\) Then
\begin{equation*}
m^{n+(k+1)}=m^{(n+k)+1}=m\,m^{n+k}
= m(m^n\,m^k)=m^n(m\,m^k)=m^n\,m^{k+1}.
\end{equation*}
Theorem B.28.
For all \(m,n,p\in\nonnegints\) with \(m\neq0\text{,}\) \((m^n)^p=m^{np}\text{.}\)
Proof.
Let \(m,n\in\nonnegints\) with \(m\neq0\text{.}\) Then \((m^n)^0=1=m^0=m^{n0}\text{.}\) Now suppose that \((m^n)^k=m^{nk}\text{.}\) Then
\begin{equation*}
(m^n)^{k+1}=m^n(m^n)^k=m^n(m^{nk})=m^{n+nk}=m^{n(k+1)}.
\end{equation*}
You have attempted of activities on this page.
