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#1 2013-02-18 13:14:37
- {7/3}
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Induction
Hi, 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 (n-1) belongs to S ,then S=Z." if I'm wrong I hope others will correct me.
#2 2013-02-18 13:25:46
- anonimnystefy
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Re: Induction
I 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.
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#3 2013-02-18 13:52:20
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Re: Induction
By (c) i stated that -1,-2,-3,-4,... are in S,the general axiom doesn't state that.
#4 2013-02-18 22:26:02
- scientia
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Re: Induction
It is necessary. This is the induction law for , not .anonimnystefy wrote:
I think (c) is unneccessary there, and the rest just represents the axiom of induction.
Last edited by scientia (2013-02-18 22:36:12)
#6 2013-02-18 23:41:08
- scientia
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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 non-negative integers, then show that it also holds for where is a non-negative integer.