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

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

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

Here lies the reader who will never open this book. He is forever dead.

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

By (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.

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**scientia****Member**- Registered: 2009-11-13
- Posts: 222

anonimnystefy wrote:

It is necessary. This is the induction law for , not .
I think (c) is unneccessary there, and the rest just represents the axiom of induction.

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

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

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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**scientia****Member**- Registered: 2009-11-13
- Posts: 222

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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