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## #1 2013-02-18 13:14:37

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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.

There are 10 kinds of people in the world,people who understand binary and people who don't.

## #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.
“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 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.

There are 10 kinds of people in the world,people who understand binary and people who don't.

## #4 2013-02-18 22:26:02

scientia
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### 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 (2013-02-18 22:36:12)

## #5 2013-02-18 23:33:22

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### 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.

## #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.