TITLE: Proof writing: how to write a clear induction proof?
QUESTION [24 upvotes]: What is an effective way to write induction proofs? 
Essentially, are there any good examples or templates of induction proofs that may be helpful (for beginners, non-English-native students, etc.)? 
To guide readers, please state whether your answer handles:

Case 1: a simple induction $(P_n \implies P_{n+1}$), or
Case 2: a strong induction ($P_1,\ldots,P_n \implies P_{n+1}$), or
Case 3: a more exotic induction (e.g. over $\Bbb Q$ on $|p|+q$).

PS: I have seen many induction related questions, and very often the problem lies with the OP's lack of a proper methodology (or style) in writing the proof whereas the answers focus on the particular case of the OP's question. The matter of style is obviously subjective, but it seems to me that the "craft" of writing good proofs is almost as important as understanding underlying concepts, so advice on proper proof-writing practices should fall within the scope of MSE (under the proof-writing tag).

REPLY [3 votes]: This is an answer covering a somewhat more exotic proof. The claim is one I used in my answer at prove or disprove $H$ is a subgroup but I will expand on the details here and write it as nicely as possible.
For more about when induction can be done on these more exotic examples, see my blog post http://math.blogoverflow.com/2015/03/10/when-can-we-do-induction/.
Claim: Let $A\subseteq \mathbb{Z}$ be the smallest subset containing $1$ and closed under the operation $(n,m)\mapsto -(n+m)$.
Then $A = \{ n\in \mathbb{Z}\mid n\equiv 1\pmod{3}\}$.
Proof of claim: Let $B = \{ n\in \mathbb{Z}\mid n\equiv 1\pmod{3}\}$. First, we show that $A\subseteq B$ (this part does not use induction, but I will include it for completeness), which by definition means we need to show that $1\in B$ and that for any $n,m\in B$ we have $-(n+m)\in B$. The first part is trivially true, and for the second, we note that if $n\equiv m\equiv 1\pmod{3}$ then $n+m\equiv 2\pmod{3}$ and thus $-(m+n)\equiv -2 \equiv 1\pmod{3}$ which was the second part.
To show that $B\subseteq A$ we would like to proceed by induction, using the well-founded ordering "is closer to $0$" (more precisely, we say that $x\leq y$ if either $x = y$ or $|x| < |y|$). That this order is in fact well-founded I will leave as an exercise, since it is the actual induction proof that is important here.
For $n\in \mathbb{Z}$ we define the statement $P_n =$ "$n\in A$", and we would like to show that $P_n$ is true for all $n\in B$. Note that the assumption on $A$ now translates to $P_1$ and $P_m\wedge P_n\implies P_{-(m+n)}$.
Let $n\in B$ and assume that $P_m$ is true for all $m < n$ (in the order defined above).
We will first need to deal with the $n=1$ case, since there are no smaller elements in $B$. But $P_1$ is true by assumption, so this is clear.
We split into two cases, depending on whether $n$ is positive or negative.
If $n$ is negative, then $-n - 1 < n$ (in the order we are using). So by the induction hypothesis, $-n-1\in A$ since $-n-1\in B$. But $n = -((-n - 1) + 1)$, so since both $P_{-n-1}$ and $P_1$ are true, we also have $P_n$ (by the definition of $A$).
Now, if $n$ is positive, then $-n + 2 < n$, so by the induction hypothesis we have $-n + 2 \in A$. But we saw above that $-2\in A$ since $-2$ is negative and in $B$. So we have both $P_{-n+2}$ and $P_2$ and since $n = -((-n+2) + -2)$ this shows that we have $P_n$, which finishes the proof.