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You are not logged in. #1 20130218 13:14:37
InductionHi, I think i have found a way to use induction for integer set too (don't know if it was discovered before this post) by extending the axiom of induction we get "if S is a subset of Z and (a) 0 belongs to S,(b) for n belonging to S (n+1) belongs to S,(c)for n belonging to S (n1) belongs to S ,then S=Z." if I'm wrong I hope others will correct me. There are 10 kinds of people in the world,people who understand binary and people who don't. #2 20130218 13:25:46
Re: InductionI think (c) is unneccessary there, and the rest just represents the axiom of induction. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #3 20130218 13:52:20
Re: InductionBy (c) i stated that 1,2,3,4,... are in S,the general axiom doesn't state that. There are 10 kinds of people in the world,people who understand binary and people who don't. #4 20130218 22:26:02
Re: InductionIt is necessary. This is the induction law for , not . Last edited by scientia (20130218 22:36:12) #5 20130218 23:33:22
Re: InductionSo,can this axiom be used to prove stuff for integers? There are 10 kinds of people in the world,people who understand binary and people who don't. #6 20130218 23:41:08
Re: Induction
It looks fine to me. However proving stuff about integers is usually much more straightforward: first prove that the property holds for all nonnegative integers, then show that it also holds for where is a nonnegative integer.
