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My apologies, I couldn't find a proper title for this question, if you do after reading question please edit it.

I read about mathematical induction theory but still I don't see how is it useful? Can anyone please list me some trivial practical usage of it rather than theoretical questions?

I was seeing this example: Prove that $4^n + 15n -1$ is divisible by $9$.

Statement $S(n)$ : $4^n + 15n -1$ is divisible by $9$.

  1. Of course, statement $S(1)$ is true (first case in proving is done)

  2. Now let's assume $S(r) = 4^r + 15r -1$ be divisible by $9$.
    (though I don't understand what is the rationale behind assuming that: just because theory says so?)

  3. We write $S(r^+)\;=\; 4^{r+1} + 15(r+1) -1$
    ... and the proof goes on

I had learned for natural numbers $x+1 = x^+$ , in above 2nd part of proof , does $S(r^+)$ = $S(r) +1$? or what does it mean?

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You are on the right track. FYI $S(r^+)=S(r+1)$. Once you have assumed that $S(r)$ is divisible by 9, show that $S(r+1)$ is also divisible by 9. – Daryl Sep 18 '12 at 9:00
If $x^+$ means $x+1$, then $S(r^*)$ means $S(r+1)$, which is not the same thing as $S(r)+1$. – Gerry Myerson Sep 18 '12 at 9:01
aah right, thanks daryl and gerry for correcting me but still my main question remains – Mr.Anubis Sep 18 '12 at 9:02
If you look at the list of related questions running down the right side of this page, you'll see dozens of other questions about induction. You can probably learn a lot about induction by reading those pages. – Gerry Myerson Sep 18 '12 at 9:03
Induction is "useful" to prove statements like your S(n), i.e. propositions that assert a property holds true for all naturals. How else do you propose to prove such universal statements if not by induction? – Bill Dubuque Sep 18 '12 at 14:38
up vote 1 down vote accepted

Following Gerry's advice, begin with these Arturo Magidin's and Pete Clark's posts.

Once you are comfortable with the logic of induction, try to come back at your problem and solve it yourself.
Calling $S(n)$ the expression $4^n+15n-1$, you already checked that S(1) is divisible by 9. Then, assuming that $S(k)$ is divisible by 9, you must show that this implies that $S(k+1)$ is also divisible by 9.

Hint: If you have to prove an implication, you must make use of the inductive hypothesis. So, when writing explicitly $S(k+1)$, try to reconstruct the expression of $S(k)$, about which you can already say something relevant (by hypothesis).

P.S.: As you see, there's no need to introduce further notation defining $x^+\equiv x+1$. If you want to do it, however, then $S(n^+)=S(n+1)=4^{n+1}+15(n+1)-1 \neq 4^n+15n = S(n)+1$.

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About the line in your hint, what if the hypothesis is wrong, s(r+1) won't reduce to expression having S(r) as it's sub-expression then? – Mr.Anubis Sep 18 '12 at 9:51
Well, the expression reduction does not care if either of the them is divisible by 9 or not. You'll note that $S(r+1)-S(r)$ can be shown divisible by 9 - thus only after that you'll conclude that $S(r+1)$ is divisible by 9 if $S(r)$ is. – Hagen von Eitzen Sep 18 '12 at 10:08
Hagen has already answered to the main issue: remember, in the inductive step you're not proving the validity of the statement, you are proving the validity of the implication $S(k) \Rightarrow S(k+1)$. The reduction of $S(k+1)$ to something "similar" in some way to $S(k)$ is just the condition you have to expect in order to exploit the inductive hypothesis. I'm being vague about the similarity because it can take different forms in different problems. What you'll find out in this one is quite typical of divisibility problems. That said, it can be very hard to actually make the connection. – Andrea Orta Sep 18 '12 at 10:13

This is how I would do it:

If $S(r) = 4^r + 15r -1$ is divisible by $9$,

then $4S(r) = 4^{r+1} + 60r -4 = S(r+1) +45 r- 18$ is divisible by $9$,

but $45 r- 18$ is divisible by $9$,

so $S(r+1)$ is divisible by $9$.

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