Exercise 8.3

Question

Obtain the definitions of $a^n$ and $n!$ for $n \in {{\mathbb{Z}}_+}$ as special cases of \href{./exercise_8-2}{Exercise~8.2}.

Dan Whitman · Accepted Answer

ewcommand\seqinf[1]{\left(#1_1, #1_2, \ldotsight)}
\def{ho}

Regarding $a^n$, defined the sequence $\seqinf{b}$ by $b_i = a$ for every $i \in {{\mathbb{Z}}_+}$, which we could denote by $(a,a,\ldots)$.
Then define $a^n = \prod_{k=1}^{n} b_k$ as it is defined in \href{./exercise_8-2}{Exercise~8.2}, and we claim that this satisfies the inductive definition given in \href{./exercise_4-6}{Exercise~4.6} and Example~8.2.
\begin{proof}
First, we clearly have $a^1 = \prod_{k=1}^{1} b_k = b_1 = a$.
Next, for $n > 1$, we have
\begin{gather*}
*      a^n = \prod_{k=1}^{n} b_k = \left(\prod_{k=1}^{n-1} b_kight) \cdot b_n = a^{n-1} \cdot a \,,
\end{gather*}
which shows that the inductive definition is satisfied.
\end{proof}

Since it does not seem to be given in the book thus far, we reiterate the typical inductive definition for $n!$:
\begin{align*}
1! &= 1 \,, \
n! &= (n-1)! \cdot n \hspace{1cm} 	ext{for $n>1$.}
\end{align*}
Now, define the sequence $\seqinf{b}$ by $b_i = i$ for $i \in {{\mathbb{Z}}_+}$.
We then claim that defining $n! = \prod_{k=1}^{n} b_k$ as defined in \href{./exercise_8-2}{Exercise~8.2} satisfies this definition.
\begin{proof}
First, we have $1! = \prod_{k=1}^{1} b_k = b_1 = 1$.
Then we also have
\begin{gather*}
n! = \prod_{k=1}^{n} b_k = \left(\prod_{k=1}^{n-1} b_kight) \cdot b_n = (n-1)! \cdot n
\end{gather*}
for $n > 1$ so that the definition is clearly satisfied.
\end{proof}

Exercise 8.3

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Exercise 8.3

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