Question

: Find the flaw with the following “proof” that${\mathit{a}}^{\mathbf{n}}\mathbf{=}\mathbf{1}$ for all nonnegative integersn , whenevera is a nonzero real number.

Basis Step: ${\mathit{a}}^{\mathbf{0}}\mathbf{=}\mathbf{1}$is true by the definition of ${\mathit{a}}^{\mathbf{0}}$.

Inductive Step: Assume that ${\mathit{a}}^{\mathbf{j}}\mathbf{=}\mathbf{1}$for all nonnegative integers jwith $\mathit{j}\mathbf{\le }\mathit{k}$. Then note that

${\mathit{a}}^{\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{=}\frac{{\mathbf{a}}^{\mathbf{k}}\mathbf{\cdot }{\mathbf{a}}^{\mathbf{k}}}{{\mathbf{a}}^{\mathbf{k}\mathbf{-}\mathbf{1}}}\mathbf{=}\frac{\mathbf{1}\mathbf{\cdot }\mathbf{1}}{\mathbf{1}}\mathbf{=}\mathbf{1}$

Short answer

The flaw is that$a^1=1$ is needed to the proved in the basis step also.

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The value of the nonzero real number isa .
• The nonnegative integer is termed asn .

Significance of the induction

The induction is referred to as the process that is useful for proving a particular theorem or statement. The induction is also comprised of two types such as simple and strong induction.

Determination of the flaw

In the inductive step, it has been identified that$a^{k-1}$ and$a^k$ is mainly equals to the number 1 . Moreover, in the basis step, the basis only accounts for $a^0$. If the result is needed to prove in this particular way, then it is not needed to be shown that the base case holds true for both$a^1$ and $a^0$. Hence, the inductive step can be applied on$a^{k-1}$ and$a^k$ .

Thus, the flaw is that$a^1=1$ is needed to the proved in the basis step also.