Let \(a \in \mathbb{R}\) . Define inductively
\[ \begin{array}{c} {a ^ {1} = a,} \\ {a ^ {n + 1} = a ^ {n} \cdot a} \end{array} \]
for \(n \in \mathbb{Z}_{+}\) . (See §7 for a discussion of the process of inductive definition.) Show that for \(n, m \in \mathbb{Z}_{+}\) and \(a, b \in \mathbb{R}\),
\[ \begin{array}{l} a ^ {n} a ^ {m} = a ^ {n + m}, \\ (a ^ {n}) ^ {m} = a ^ {n m}, \\ a ^ {m} b ^ {m} = (a b) ^ {m}. \end{array} \]
These are called the laws of exponents. [Hint: For fixed \(n\), prove the formulas by induction on \(m\) .]
Solution
We prove the three identities by induction.
First fix \(n\in\mathbb Z_+\). For \(m=1\), \[ a^na=a^{n+1} \] by definition. If \(a^na^m=a^{n+m}\), then \[ a^na^{m+1}=a^n(a^ma)=(a^na^m)a=a^{n+m}a=a^{n+m+1}. \] Thus \[ \boxed{a^na^m=a^{n+m}}. \]
Next fix \(n\). For \(m=1\), \((a^n)^1=a^n\). If \((a^n)^m=a^{nm}\), then using the first identity, \[ (a^n)^{m+1}=(a^n)^ma^n=a^{nm}a^n=a^{nm+n}=a^{n(m+1)}. \] Hence \[ \boxed{(a^n)^m=a^{nm}}. \]
Finally, for \(m=1\), \(a^1b^1=ab=(ab)^1\). If \(a^mb^m=(ab)^m\), then commutativity and associativity give \[ a^{m+1}b^{m+1}=a^mab^mb=(a^mb^m)(ab)=(ab)^m(ab)=(ab)^{m+1}. \] Therefore \[ \boxed{a^mb^m=(ab)^m}. \]