Mathematical induction and shades

Question

I do not understand the following from my lower grades. No one cleared my doubt properly with good examples. Why we need to learn Mathematical induction is I know very well. But, in which cases we treat or need to learn weak and strong induction. I really dont know. Why to learn weak and strong induction. Where it is really applicable. Please discuss with good example.

Answer 1

The answer is simply that you use one that works. You don’t choose ahead of time which form to use; you use the one that gives you the strength of hypothesis needed to make your proof work.

Sometimes the hypothesis $P(n)$ simply isn’t strong enough to let us derive $P(n+1)$, but we can derive $P(n+1)$ if we assume $P(n)$ and $P(n-1)$. Sometimes we have to assume $P(k)$ for all $k$ such that $n_0\le k\le n$ in order to be able to infer $P(n+1)$. (Here $n_0$ is the initial value for the induction.) In practice you might as well simply assume that $P(k)$ holds for $k=n_0,\dots,n$ when trying to prove $P(n+1)$; if it turns out that you don’t actually need that strong a hypothesis, no harm has been done. In other words, when attacking a new proof, always remember that you can use the full strength of strong induction, though in many cases you won’t need to do so.

It’s unfortunate that so-called strong and weak induction are so often taught as different things, when in fact they are just very slightly different special cases of a considerably more general concept that covers transfinite induction and structural induction as well. Roughly speaking, it’s a method that applies whenever the setting is such that it’s meaningful to talk about a minimal counterexample to the theorem that you’re trying to prove. In the case of induction over the integers, a minimal counterexample is simply the smallest $n$ for which $P(n)$ is false. You can think of a proof by induction as a proof that no such minimal counterexample can exist. You suppose that $n$ is a minimal counterexample, and you get a contradiction. Sometimes the contradiction can be obtained just from the hypothesis that $P(n-1)$ is true; sometimes you find that you need a bit more $-$ the truth of both $P(n-1)$ and $P(n-2)$, for instance, or even of all $P(k)$ for $n_0\le k<n$. Since you’re assuming that $n$ is a minimal counterexample, however, you are assuming that $P(k)$ is true for $n_0\le k<n$, so you can use as much of that assumption as you need in order to get your contradiction.