Find the flaw with the following "proof" that $a^n = 1$ for all nonnegative integers $n$, whenever $a$ is a nonzero real number.
Basis Step: $a^0 = 1$ is true by definitino of $a^0$.
Inductive Step: Assume that $a^j = 1$ for all nonnegative integers $j$ with $j \leq k$. Then note that
$$
a^{k+1} = \frac{a^k \cdot a^k}{a^{k-1}} = \frac{1 \cdot 1}{1} = 1
$$
Where is the flaw in the above proof? My answer is that the basis step needs to iterate nonnegative integers beyond $0$, like $1$, $2$, etc.
Any thoughts?
Best Answer
$a^1=\cfrac {a^0\cdot a^0}{a^{-1}}$ and your hypothesis doesn't cover the case of $a^{-1}$.