Will the basis case always be the first value within the range given in proof by induction? I just started learning proofs by induction, and typically in the example problems the first value within the range of values of consideration satisfies the base case, but I was curious if there is ever an instance where the first value of the range given would not satisfy the induction proof?
Would this be possible given that questions typically ask for a proof by induction over a specific range? Would we not be skipping a value if the base case is not the first value from the range of values?
 A: An induction is meant to prove that a certain statement $A(n)$ is true for all integers $n$ equal or greater than a certain $n_0$. If $A$ fails fails at $n = n_0$, the proposition is false as you've got a counterexample.
Of course, what I've said is valid for "homework" questions, in which you are supposed to prove a proposition which explicitly gives you the range of the variable. In practice (and in more interesting homeworks) you often need to experience values until you find out what is the range. For example, to show that $n! > 3^{n+1}$ for $n$ large enough requires you to guess the minimal value for $n$ in which the inequality holds. Only after this you can prove by induction that your guess was indeed correct.
Well, I am not sure if that's exactly what you've asked, but I hope it helps.
A: If the base case doesn't work for the given range then the statment to prove is false.
For example if you are given "Show that for every $\mathbb{N}$ the inequality $0<x-1$ holds" as for $x=1$ the inequality fails then the initial steatment is false
It doesn't matter if it work from $2$ onwards, you just found a conterexample
A: Yes. If the proposition $P(n) $ holds true for $n \geq n_0$, and you want to prove it by induction, you could try to prove $P (n_0)$ is true, $P (n_1) $ is true, and so on. This is equivalent to prove the base case many times... but since you are going to need an inductive step anyway (cause you are not going to be around here long enough to prove infinitely many propositions), there's no need to prove more than one case to serve as base case, which will determine the first element $n_0$ for which $P (n)$ is true. Of course you could choose $n_1 > n_0$ for the base case, but then you will  be proving that $P (n) $ is true for $n \geq n_1$, and still need a separate proof for $P (n_0)$.
