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Topic: Peano's Integers Modulus n.
Replies: 12   Last Post: Mar 23, 2014 4:44 AM

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LudovicoVan

Posts: 3,201
From: London
Registered: 2/8/08
Re: Peano's Integers Modulus n.
Posted: Mar 22, 2014 12:32 PM
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"Julio Di Egidio" <julio@diegidio.name> wrote in message
news:lgkdrn$mu$1@dont-email.me...
> "William Elliot" <marsh@panix.com> wrote in message
> news:Pine.NEB.4.64.1403220146430.27476@panix3.panix.com...

>> Let N be a set and S:N -> N an injection and 0 & z be two elements of N.
>>
>> Assume Sz = 0 and the schemata,
>> if A subset N, 0 in A and for all n in A, Sn in A,
>> then A = N.
>>
>> Are the integers modulus z a model for these axioms?

>
> Yes, but I don't know how to prove it: how do you prove that something (a
> structure?) is a model for some axioms?
>

>> Is omega_0 + 1 with z = omega_0 another model?
>
> No, in omega_0 + 1, the successor of omega_0 is omega_0, not zero.
>

>> Does the inductive definitions:
>> for all n,m, 0 + n = n, Sm + n = m + Sn
>> 0 * n = 0, Sm * n = m*n + m
>> give arithmetic modulus z?

>
> Addition works, but multiplication does not. For instance:
>
> 3 * 2 = 2 (mod 4)
> 2*2 + 3 = 3 (mod 4)


On the other hand, if you rewrite that as:

0 * n = 0, Sm * n = m*n + n

...

Julio

> Finally, does the model, omega_0 + 1
>> with those axioms yield Peano's arithmetic?
>
> No: as noticed above, firstly, this is not a model for your axioms,
> secondly, in Peano's arithmetic omega_0 isn't a natural number at all.
>
> Please correct me where I am wrong.







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