The principle of mathematical induction works basically because of the following:
If we have a predicate $P(n)$, then if we have:
P(0) is true, and
P(n) $\implies$ P(n+1) for all nonnegative integers n,
P(m) is true for all nonnegative integers, m.
However, I was concerned what if $P(n)$ is false or what if the base cases were false, what that would mean about $P(n+1)$.
I think I know the answer to my question but I wanted to check the answer with the community.
Say that we know $P(n_0)$ is false. Thus, the implication $P(n_0) \implies P(n_0+1)$ is true vacuously (from the definition of implies). However, our goal in induction is to establish the truth about some implication $P(n)$ for as many n as we can. However, since $P(n_0)$ is false, then from the truth table of an implication $P(n_0) \implies P(n_0 + 1)$ remains true regardless of the truth value of $P(n_0 + 1)$. Thus, even though the implication is true, we are unable to conclude anything about $P(n_0+1)$ (or any other value of n > n_0) because $P(n_0 + 1)$ might be either true or false. The goal was to establish the truth about $P(n_0+1)$, but since we can't establish which fact is true it (about $P(n_0+1)$), then we cannot continue reasoning about values of $n \geq n_0 + 1$.
Is this correct? Is this why if we have a false statement we cannot have a proof by induction that proceeds correctly? Related to this question, if we find a case were $P(n_0) = False$ but then $P(n_1) = True$ for $n_1 > n_0$, can we still hope to show $P(n_0)$ for $n > n_0$? What about the other way round, if $P(n_0) = True$ but then $P(n_1) = False$, $n_1 > n_0$? Can we still hope induction to help us deduce anything meaningful?