In the book How to Prove It, they say that strong induction requires no base case. My professor's notes also say this. However, while I understand weak and strong induction, as well as the difference and reasoning as to why we would use one versus the other, something about "not having a base case" really bothers me.
The argument made by all those texts which say strong induction requires no base case go as follows. Strong induction requires: if we have that $[\forall k \in N, k<n, P(k)] \rightarrow P(n)$ holds true, then $P(n)$ is true for all $n\in N$.
The argument for $P(0)$ holds true vacuously since $[\forall k \in N, k<0, P(k)]$ is always false (i.e. there are no natural numbers that are less than zero) so we have that $[\forall k \in N, k<0, P(k)] \rightarrow P(0)$. Therefore, $P(0)$ is proven as the base case while proving the inductive hypothesis.
However this is obviously not true. The statement $a \rightarrow b$ doesn't prove $b$ unless $a$ is true. If $a$ is always false, we don't know anything about $b$; we just know that the statement if $a$ then $b$ is correct. So my issue is that we know nothing about $P(0)$. Here I can use "strong induction" to prove something that is false if I don't have to prove a base case:
All natural numbers are even. No base case required. Well, suppose that $[\forall k \in N, k<n, P(k)]$ this is true. Then it's clearly true for $n$, since $n=(n-2)+2$. Since $n-2$ is even by the hypothesis, and $n-2+2$ is just adding 2 to an even number, we still have an even number.
If you required base cases, this would obviously fail. Am I missing something? Why do these books still say that no base case is required? The Wikipedia page for induction still says a base case is needed, and many similar questions on StackExchange regarding this subject seem to be in the middle of a debate around this.