The proof of the base case in proof by induction must always be to verify the claim is true for the number 1.

Question

The question asks me to state True or False.

Answer: I prefer True.

I'm I correct?

Answer 1

Think of the following: Suppose you want to prove for appropriate $n$ that $2^n>n^2$. Would the induction base be to verify for $n=1$? Try that and see what happens.

Answer 2

Absolutely not.

You prove that a(n) implies a(n+1) for n >= N. Then you prove a(n) for 1 <= n <= N.

N can be large. I think I heard of practical cases where N >= 10^16.

Answer 3

No. Suppose some proposition $P(n)$ is true for all integer $n \ge 1$.

Take any integer $a$. Then the proposition $Q(n) := P(n-a)$ is true for all integer $n \ge a + 1$.

A proof by induction for $Q(n)$ would begin with base case $n = a + 1$, which is not necessarily $1$.