Basis Step: a^0 = 1 is true by the definition of a^0.
Inductive Step: Assume that a^j = 1 for all non negative integers j with j <= k. Then note that
a^(k+1) = (a^k*a^k)/(a^k-1) = 1*1/1 = 1
2. Find the flaw with the following "proof" that every postage of three cents or more can be formed using just three-cent and four-cent stamps.
Basis Step: We can form postage of three cents with a single three-cent stamp and we can form postage of four cents using a single four-cent stamp.
Inductive Step: Assume that we can form postage of j cents for all non negative integers j with j <= k using just three-cent and four-cent stamps. We can then form postage of k + 1 cents by replacing one three-cent stamp with a four-cent stamp or by replacing two four-cent stamps by three-cent stamps.
The inductive step is correctly done, but on basis step we need to check not only for n=0 but for n=1 and sometimes for n=2.
While for n=0 is true that a^0 = ...
Flaws and inductive proofs for two mathematical induction problems are provided in the solution.