0

It's well known that if you want to prove a theorem that involves $n$ elements (terms,sets,variables,sequences,etc.) you should think about induction.

But is there another way to proof a theorem with that characteristics?

For instance:

What if I prove that the result holds for the base case $(n=1)$ and then for $n=2$ and $n=3.$ Could I conclude that the result holds for the $n$ case$?$

Martin Sleziak
  • 53,687
user441848
  • 1,658
  • 1
    You really need to clarify your question, because as it stands, I could give a statement like "all natural numbers are less than 4". It works for 1,2,3, but obviously not for all n. – Ted Aug 20 '17 at 01:00
  • 1
    In induction, you show that a statement $P(k)$ holds for $k=1$ (or some other base case), and then show that $P(k)$ implies $P(k+1)$. There is a variant called strong induction, in which you show $P(1)$, then show that for any $k$, $$\left[ , P(1) \text{ and } P(2) \text { and } P(3) \text{ and } \cdots \text{ and } P(k), \right] ,,\implies,, P(k+1)$$ – Zubin Mukerjee Aug 20 '17 at 01:01
  • @Ted haha I really like you counterexample.. – user441848 Aug 20 '17 at 01:19
  • I personally choose to believe that there are no integers greater than 3 (because I'm really bad at counting). For me, what you wrote definitely works. – Clement C. Aug 20 '17 at 01:51
  • :D hahaha! yea I'm starting to think that it's one of my best ideas that I've ever had @ClementC. – user441848 Aug 20 '17 at 01:58

2 Answers2

4

Nope. You have only proved it for $n=1,2,3$.

In general, you can only prove that kind of theorem by induction.

If you seem to avoid induction, you will have to use theorems that themselves are proved by induction.

For example, consider this proof that, if $s(n) =\sum_{k=1}^n k $ then $s(n) =\frac12 n(n+1) $.

Changing the order of summation, $s(n) =\sum_{k=1}^n (n+1-k) $. Therefore

$\begin{array}\\ 2s(n) &=\sum_{k=1}^n k+\sum_{k=1}^n (n+1-k)\\ &=\sum_{k=1}^n (k+n+1-k)\\ &=\sum_{k=1}^n (n+1)\\ &=n(n+1) \end{array} $

Look - no induction!

However, the definition of $\sum_{k=1}^n$ and the manipulation of the summation indices both depend on induction to prove that they can be used as stated.

Another way to see this is to note that the Peano axioms which define the integers only have one way to do the definition: induction.

marty cohen
  • 107,799
2

The thing is you can't guarantee it'll hold for the $n$ case. One example for instance are the Borwein integrals, who follow a pattern up to $n=14$ then fail at the next iteration. This is the worry with many conjectures out there like the Collatz conjecture that's been checked for $ 5×2^{60}$ iterations, but could technically fail at one point, until we prove it always holds (induction doesn't necessarily work for Collatz, but you get my drift.)

Induction doesn't "right off the bat" prove it works for all integers, you just show it works for a base case and induction guarantees it'll hold for the next integer -- and as a result, the one after, and one after, etc.

Andrew Tawfeek
  • 2,555