## Loser66 one year ago 1)How to show an axiomatic system is a model? Please, help 2) How to show 2 axiomatic systems are isomorphic? Please, help

1. Loser66

I have an axiomatic system : Axiom 1: $$if ~~a, b\in \mathbb Z ~~and~~ a < b, then~~ a\neq b$$ Axiom 2: $$if~~ a, b, c\in \mathbb Z, ~~a< b~~and~~b<c, ~~then~~a<c$$ How to show it is a model?

2. Loser66

For part 2) I have 2 axiomatic system: A) $$P\leftrightarrow Bob$$ $$Q\leftrightarrow Ted$$ $$R\leftrightarrow Carol$$ $$\{P, Q\}\leftrightarrow Entertainment ~~committee$$ $$\{P, R\}\leftrightarrow Refreshment ~~committee$$ $$\{Q, R\}\leftrightarrow Finance ~~committee$$

3. Loser66

B) $$\{Bob, Ted\}\leftrightarrow P$$ $$\{ Ted, Carol\}\leftrightarrow Q$$ $$\{Bob, Carol \}\leftrightarrow R$$ $$\{P,Q\}\leftrightarrow Bob$$ $$\{P, Q\}\leftrightarrow Ted$$ $$\{Q,R\}\leftrightarrow Carol$$

4. Loser66

I don't know how to show that they are isomorphic.

5. anonymous

a model is a realization of an axiomatic system; all you've given in 1 is the axioms, not any possible model to verify

6. Loser66

By definition of a model, if all axioms are correct, then it is a model. But it means I just say that sentence as a proof?

7. anonymous

8. anonymous

"if all axioms are correct" doesn't make sense

9. Loser66

That is original one. ok, let me take a pic

10. Loser66

11. Loser66

Problem 1.2.16

12. Loser66

This is the definition of models

13. anonymous

okay, so we have a system speaking about $$S,R$$ with axioms: $$\forall a,b\in S\ (aRb\implies a\ne b)\\\forall a,b,c\in S\ (aRb\land bRc\implies aRc)$$ then they're asking to verify that the above axioms are valid in the model $$(S,R)=(\mathbb Z,<)$$, i.e. confirm: $$a<b\implies a\ne b\\a<b\land b<c\implies a<c$$ and likewise for $$(\mathbb Z,>)$$

14. Loser66

Got you, how about part 2)?

15. Loser66

How to show they are isomorphic?

16. anonymous

if the models are isomorphic then that means there is a one to one mapping between $$a<b$$ i.e. $$(a,b)\in\,<$$ as a relation and $$(a,b)\in\,>$$

17. anonymous

and this is true -- we can take $$(a,b)\in\, <$$ and assign it to $$(b,a)\, >$$ so $$a<b\Leftrightarrow b>a$$ and then our axioms clearly still hold: $$a<b\land b<c\implies a<c\\b>a\land c>b\Leftrightarrow c>b\land b>a\implies c>a$$

18. anonymous

and similarly for the first axiom $$a<b\implies a\ne b\\b>a\implies b\ne a\Leftrightarrow a\ne b$$

19. Loser66

So, we just show that they are " work" on the same way, right?

20. Loser66

@oldrin.bataku Please, explain me part d) I was doubt myself when I think they are isomorphic. I don't know why, just intuitively feel that. I thought in Real, like fraction, we have 1/2 = 3/6 . It is not like what the axiom1 say but don't know how to argue.

21. anonymous

in part (d) it is a valid model, but it is not isomorphic because there are uncountably many reals