Exercise 4.6

Question

Let $a \in {\mathbb{R}}$.
Define inductively
\begin{align*}
  a^1 &= a \,, \
  a^{n+1} &= a^n \cdot a
\end{align*}
for $n \in {{\mathbb{Z}}_+}$.
(See \S 7 for a discussion of the process of inductive definition.)
Show that for $n,m \in {{\mathbb{Z}}_+}$ and $a,b \in {\mathbb{R}}$,
\begin{align*}
  a^n a^m &= a^{n+m} \
  \left(a^night)^m &= a^{nm} \
  a^m b^m &= (ab)^m \,.
\end{align*}
These are called the 	extbf{\emph{laws of exponents}}.
[Hint: For fixed $n$, prove the formulas by induction on $m$.]

Dan Whitman · Accepted Answer

\def\multcom{since multiplication is commutative}
\def\multass{since multiplication is associative}
The following lemma is the familiar proof by induction, which is more straightforward than having to frame everything in terms of inductive sets.
Henceforth we use this whenever induction is required.
\begin{lemma}
  (Proof by Induction)
  Suppose that $P(x)$ is a statement with parameter $x$.
  Suppose also that $P(1)$ is true and that $P(x)$ implies $P(x+1)$.
  Then $P(n)$ is true for all $n \in {{\mathbb{Z}}_+}$.
\end{lemma}
\begin{proof}
  Define the set $X = \left\{x \in {\mathbb{R}} \mid P(x)ight\}$.
  We show that $X$ is inductive.
  Clearly, since $P(1)$ is true we have $1 \in X$.
  Now suppose that $x \in X$ so that $P(x)$ is true.
  Then $P(x+1)$ is also true so that $x+1 \in X$.
  This shows that $X$ is inductive so that ${{\mathbb{Z}}_+} \subset X$.
  So, for any positive integer $n$ we have that $n \in X$ since ${{\mathbb{Z}}_+} \subset X$.
  Therefore $P(n)$ is true.
  Since $n$ was arbitrary, this shows the desired result.
\end{proof}

extbf{Main Problem.}

In what follows, suppose that $a,b \in {\mathbb{R}}$.

First we show that $a^n a^m = a^{n+m}$ for all $n,m \in {{\mathbb{Z}}_+}$.
\begin{proof}
  Fix $n \in {{\mathbb{Z}}_+}$.
  We show the result by induction on $m$.
  First, we clearly have $a^n a^1 = a^n \cdot a = a^{n+1}$ by the inductive definition.
  Now suppose that $a^n a^m = a^{n+m}$.
  Then
  \begin{align*}
    a^n a^{m+1} &= a^n \cdot (a^m \cdot a) & 	ext{(by the inductive definition)} \
                &= (a^n a^m) \cdot a & 	ext{(\multass)} \
                &= a^{n+m} \cdot a & 	ext{(by the induction hypothesis)} \
                &= a^{(n+m)+1} & 	ext{(by the inductive definition)} \
                &= a^{n + (m+1)} \,, & 	ext{(since addition is associative)}
  \end{align*}
  which completes the induction step.
  Therefore the result holds for all $m \in {{\mathbb{Z}}_+}$ by induction.
\end{proof}

Next we show that $\left(a^night)^m = a^{nm}$ for all $n,m \in {{\mathbb{Z}}_+}$.
\begin{proof}
  We again fix $n \in {{\mathbb{Z}}_+}$ and use induction on $m$.
  First, we have $\left(a^night)^1 = a^n = a^{n \cdot 1}$ by the inductive definition.
  Supposing now that $\left(a^night)^m = a^{n \cdot m}$, we have
  \begin{align*}
    \left(a^night)^{m+1} &= \left(a^night)^m \cdot a^n & 	ext{(by the inductive definition)} \
                           &= a^{n \cdot m} a^n & 	ext{(by the induction hypothesis)} \
                           &= a^{n \cdot m + n} & 	ext{(by what was shown above)} \
                           &= a^{n \cdot m + n \cdot 1} \
                           &= a^{n \cdot (m+1)} \,. & 	ext{(by the distributive property)}
  \end{align*}
  This completes the induction so that the result holds for all $m \in {{\mathbb{Z}}_+}$.
\end{proof}

Lastly, we show that $a^m b^m = (ab)^m$ for all $m \in {{\mathbb{Z}}_+}$.
\begin{proof}
  We show this by induction on $m$.
  First, we have $a^1 b^1 = ab = (ab)^1$ by the inductive definition.
  Now suppose that $a^m b^m = (ab)^m$ so that
  \begin{align*}
    a^{m+1} b^{m+1} &= (a^m \cdot a)(b^m \cdot b) & 	ext{(by the inductive definition)} \
                    &= (a \cdot a^m)(b^m \cdot b) & 	ext{(\multcom)} \
                    &= ((a \cdot a^m) b^m) \cdot b & 	ext{(\multass)} \
                    &= (a \cdot (a^m b^m)) \cdot b & 	ext{(\multass)} \
                    &= (a (ab)^m) \cdot b & 	ext{(by the induction hypothesis)} \
                    &= ((ab)^m a) \cdot b & 	ext{(\multcom)} \
                    &= (ab)^m (ab) & 	ext{(\multass)} \
                    &= (ab)^{m+1} \,. & 	ext{(by the inductive definition)}
  \end{align*}
  This completes the induction.
\end{proof}

Exercise 4.6

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Exercise 4.6

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