Can you use proof by contradiction inside simple induction?

Question

During a proof using simple induction, I assumed P(k) is true. Now in order to show P(k+1) is true using P(k), can I do a proof by contradiction on P(k+1) and say P(k) would be wrong if P(k+1) is wrong, therefore P(k+1) is true as well?

Answer 1

Surely reaching a contradiction is acceptable, though it is usually not the most convenient path to follow. Anyway, here's a futile example.

We have $1^2=0^2+2\cdot0+1$. Assume $(k+1)^2=k^2+2k+1$ but $(k+2)^2\ne k^2+4k+4.$ Then, $$\require{cancel} (k+2)^2-(k+1)^2\ne2k+3\\(\cancel{k}+2-\cancel{k}-1)(k+k+2+1)\ne2k+3\\ 2k+3\ne2k+3.$$ Hence, $(n+1)^2=n^2+2n+1$ for all $n$.

Answer 2

Yes. This proof technique is used to prove the Recursion Theorem in Nathan Jacobson's Basic Algebra I, for example. Although $P(k+1)$ doesn't have to contradict only $P(k)$. $P(k+1)$ can contradict some other assumption, assuming $P(k)$