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Whenever I use the inductive method to prove some questions, I usually start from the $n=1$ case and assume it holds for all $n$.

However, is the reason why we do not consider the $n=0$ case because induction is defined on natural numbers, or is it too trivial that we do not check this case?

Peter Phipps
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jessie
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  • You can start in fact at any integer $a$ (even negative) for the base case. Check $P(a)$ is true, then show for $n \ge a$ that $P(n) \implies P(n+1).$ This is appropriate if your goal is to show $P(n)$ is true for all $n \ge a$ – coffeemath Nov 24 '15 at 20:24
  • If you want to prove the hypothesis is true for all non-negative integers $n$ then you have to start at $n \le 0$. You do not then "assume" it holds for all $n$, you show that if it is true for $n=k$ then it is true for $n=k+1$ and then use the axiom of induction. – Henry Nov 24 '15 at 20:24
  • Induction can be done on every infinite set,this might be helpful http://math.stackexchange.com/questions/450652/how-can-i-expand-mathematical-induction-to-rational-numbers mostly the answer by zarathustra – kingW3 Nov 24 '15 at 20:28
  • @kingW3 Although technically you are correct, this is rarely useful outside set theory if e.g. you want to prove some property $P$ holds for all real numbers $r$, because the well-order on $\Bbb R$, even if it exists, is very complicated, and in no way resembles standard ordering of real numbers. – Wojowu Nov 24 '15 at 20:31

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(for the purpose of this answer I assume $0\in\Bbb N$)

This really depends on what exactly you are trying to prove. If you want that property $P$ holds for all $n\in\Bbb N$, then it is necessary to either start the induction with base case $n=0$, or to start it with base case $n=1$ and then verify $P$ for $0$ separately (which might not be at all trivial!).

If you want to prove it only for positive $n\in\Bbb N$, then using base case $n=1$ is perfectly fine.

In greater generality, the following principle is valid (which coffeemath mentions in their comment): If $a$ is any integer (possibly negative!) and you want to prove $P$ holds for all integers $n\geq a$, then it's enough to prove that $P(a)$ holds (base case), and that $P(n)\Rightarrow P(n+1)$ (inductive step), when $P(x)$ is shorthand for "$P$ holds for $x$".

Wojowu
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Mathematical Induction is true for every well ordered set. $\mathbb{N}$ is a natural example of well ordered set, but we have induction in other sets like: $$W=\left \{ \alpha |\alpha <\epsilon_0 \right \}$$ and $\epsilon_0=\sup{\left \{\omega,\omega^{\omega},\omega^{\omega^{\omega},...} \right \}}$ is an ordinal number. See Transfinite induction for more details.

Erfan Khaniki
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