The question asks me to state True or False.
Answer: I prefer True.
I'm I correct?
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2Who knows? Some situations start with $n=0$, others with $n=3$. – almagest May 22 '16 at 05:49
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1I’m afraid not. A proof by induction can start at any integer. – Brian M. Scott May 22 '16 at 05:50
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@BrianM.Scott More generally, at any integer or ordinal :-). – YoTengoUnLCD May 22 '16 at 05:54
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2@YoTengoUnLCD: Or the base of a well-founded partial order, but it seemed a bit counterproductive to head off in that direction! (Actually, it's quite likely that induction as the OP has seen it defined so far does have to be over a tail of the integers.) – Brian M. Scott May 22 '16 at 06:03
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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.
Ittay Weiss
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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.
gnasher729
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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 $1$.