Constructing a non-standard model

[MLP]

I am trying to construct a non-standard model < A,0,S,+,*,E,< > that has as its domain the natural numbers plus the letter a such that the model
1. Makes all of the axioms of number theory true (Say, Mendelson's S)
2. Makes a < a.
So in this model the domain has already been specified. We take < to be just like < in the standard model except that we add the couple <a,a> to it. We take 0 to be the same as in the standard model. We take S to be just like S in the standard model with the addition that that S(a)=a. We take + to be just like the standard model except that
a + a = a
0 + a = a
for n > 0, a + n = n
We take E to be the same as in the standard model with the addition that aE0 = S(0).
Does that do the trick? I can't see where I've made any of Mendelson's axioms false.

 

[Office_Shredder]

I'm not familiar with Mendelson's axioms of number theory, but from googling I think I found the list. If so, then your definitions fail the requirement that 0 is not a successor of any element

 

[MLP]

I'm not familiar with Mendelson's axioms of number theory, but from googling I think I found the list. If so, then your definitions fail the requirement that 0 is not a successor of any element

Right. So let me amend.
The function S is just like that of standard number theory except that
S(a) = a
$\neg$S(0) = a

The relation < is just like that of standard number theory except that
a < a
0 < a
for any n > 0, S$\cdots_{}$S(0) < a where S$\cdots_{}$S are n occurrences of S.

Perhaps this will fare better.

 

[Citan Uzuki]

Either I've completely misunderstood what you are trying to do, or what you are trying to do is obviously impossible. The statement $\forall x, \neg (x<x)$ is provable from the axioms of number theory, and obviously would not be satisfied by a structure in which a<a.

 

[MLP]

Either I've completely misunderstood what you are trying to do, or what you are trying to do is obviously impossible. The statement $\forall x, \neg (x<x)$ is provable from the axioms of number theory, and obviously would not be satisfied by a structure in which a<a.

I would challenge you to prove $\forall x, \neg (x<x)$ from the axioms of number theory.

 

[AKG]

I would challenge you to prove ∀x,¬(x<x) from the axioms of number theory.

http://en.wikipedia.org/wiki/Peano_axioms#Equivalent_axiomatizations

See Axiom 9.

 

[MLP]

http://en.wikipedia.org/wiki/Peano_axioms#Equivalent_axiomatizations

See Axiom 9.

Ah, my bad. So the axiomatization I am using is the system N from Chris Leary's book A Friendly Introduction to Mathematical Logic and he produces a proof that the formula does not follow from those axioms. I got a little over-zealous there. Sorry

 

[Citan Uzuki]

Okay, I've looked up the axiomatization you're working with. You're right, this one doesn't prove $\forall x, \neg (x<x)$. However, your attempt at defining a nonstandard model possesses several shortcomings:

1: You have not specified n+a for n≠0 or a
2: You have not specified multiplication at all
3: You have not specified AEn for any n other than zero, nor have you specified nEa
4: You have not defined whether a<n or n<a for $n\in\mathbb{N}$ Perhaps you meant for a to satisfy neither of these relations, but if so, that is in contradiction to axiom 11.
5: Your definitions are inconsistent with axioms 3 and 4. Specifically you have for n>0, a+n = n. In particular, this would require that a+S0=S0. But axioms 3 and 4, together with your choice of Sa=a, require that a+S0=S(a+0)=Sa=a≠S0.

So I'm afraid your model will need some work. You have the right idea putting Sa=a -- in fact with only one additional element, you really have no choice but to do that. In fact, adding exactly one new element leaves you with very little choice in how to extend your model. For instance, you must set a+x=x+a=a (which means that to show the commutativity of addition doesn't follow from N, you'll need more than one new element). Similarly, you must set a·0=0 and x·a=a·x=a for x≠0. You do however have the choice of setting 0·a to either a or 0. I'll leave you to figure out how many ways you can extend exponentiation. Finally, when you get to the ordering, the only ordering that will extend the ordering on $\mathbb{N}$ and still get you a<a is to let x<a for all x.

 

[MLP]

Okay, I've looked up the axiomatization you're working with. You're right, this one doesn't prove $\forall x, \neg (x<x)$. However, your attempt at defining a nonstandard model possesses several shortcomings:

That sir, was excellent criticism. Thank you so much for taking the time to do that. Let me try again.

Successor is just like the standard successor except that
S(a) = a
¬S(0) = a
a + a = a

Multiplication is just like standard multiplication except that
a $\cdot$ 0 = 0
for any n > 0, a $\cdot$ n = a
a $\cdot$ a = a

Exponentiation is just like standard exponentiation except that
aEa = a
aE0 = S(0)
for any n > 0, nEa = a

The relation < is just like that of standard number theory except that
a < a
0 < a
for any n > 0, S⋯S(0) < a where S⋯S are n occurrences of S.

I hope I have repaired the shortcomings.

 

[Citan Uzuki]

Closer, but you need to specify +, *, and E when either side is a, not just when the left side is a.