Problem 23

The following is a recursive defintion of $a^n$: \[ \begin{align*} a^1 &= a \\ a^{n + 1} &= a^n \cdot a \\ \end{align*} \] Prove, by induction, that: \[ \begin{align*} a^{n + m} &= a^n \cdot a^m \\ (a^n)^m &= a^{nm} \\ \end{align*} \] We start by proving the first proposition for $n = 1$: \[ \begin{align*} a^1 \cdot a^m &= a^m \cdot a \\ &= a^{1 + m} \\ \end{align*} \] Now, follows the induction step frpm $n = k$ to $n = k + 1$: \[ \begin{align*} a^1 \cdot a^{k + m} &= a^1 \cdot a^x \\ &= a^{1 + x} \\ &= a^{(k + 1) + m} \\ \end{align*} \] Next, we prove the second proposition and start again witg $m = 1$: \[ \begin{align*} (a^n)^1 &= a^n \\ &= a^{n \cdot 1} \\ \end{align*} \] Now, follows the induction step frpm $m = k$ to $m = k + 1$: \[ \begin{align*} (a^n)^{m + 1} &= (a^n)^m \cdot a^n \\ &= [a^n \cdot ... \cdot a^n] \cdot a^n \\ &= a^{[n + ... + n]} \cdot a^n \\ &= a^{nm} \cdot a^n \\ &= a^{nm + n} \\ &= a^{n(m + 1)} \\ \end{align*} \]

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