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#1 2013-02-17 14:14:37

{7/3}
Member
Registered: 2013-02-11
Posts: 210

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.


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#2 2013-02-17 14:25:46

anonimnystefy
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Registered: 2011-05-23
Posts: 15,399

Re: Induction

I think (c) is unneccessary there, and the rest just represents the axiom of induction.


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#3 2013-02-17 14:52:20

{7/3}
Member
Registered: 2013-02-11
Posts: 210

Re: Induction

By (c) i stated that -1,-2,-3,-4,... are in S,the general axiom doesn't state that.


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#4 2013-02-17 23:26:02

scientia
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Registered: 2009-11-13
Posts: 222

Re: Induction

anonimnystefy wrote:

I think (c) is unneccessary there, and the rest just represents the axiom of induction.

It is necessary. This is the induction law for
, not
.

Last edited by scientia (2013-02-17 23:36:12)

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#5 2013-02-18 00:33:22

{7/3}
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Registered: 2013-02-11
Posts: 210

Re: Induction

So,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.

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#6 2013-02-18 00:41:08

scientia
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Registered: 2009-11-13
Posts: 222

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.

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