|
|
![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Message
|
|
|
| ``Re: Proof of Chebyshev's inequality by induction''
by TomLK on 2004-05-19 02:53:41 |
|
| Beg to disagree: there's only one way to prove that a certain set S of natural numbers is equal to the set N of all natural numbers, and that is to show that S is inductive: (i) 1 belongs to S; (ii) for any n, 'n belongs to S' implies 'n+1 belongs to S'.
The claim that 1+2+...+n=n(n+1)/2 for any n may seem utterly obvious, since it's high school derivation is independent of the 'value' of n; but ever since Euclid, the proof is still needed, and it is by induction. You have to play by the rules of an axiomatic theory.
Inductive conjectures are quite common, and sometimes not easily proved or disproved; e.g.,
(1) 2^{2^n}+1 is a prime for n=0,1,2,3,4; conjecture: it is a prime for any n (Fermat); conjecture is false: 2^{2^5}+1=6700417*641 (Euler).
(2) 991n^2+1 is not a square for none of n's you would ever possibly thing of; however (W. Sierpinski)
991*(12055735790331359447442538767)^2+1=(379516400906811930638014896080)^2.
|
| | [ reply | up | top ] | |
|
|
|
|
|
|
|
