Loser66
  • Loser66
1)How to show an axiomatic system is a model? Please, help 2) How to show 2 axiomatic systems are isomorphic? Please, help
Mathematics
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SOLVED
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chestercat
  • chestercat
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Loser66
  • 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
Loser66
  • Loser66
For part 2) I have 2 axiomatic system: A) \(P\leftrightarrow Bob\) \(Q\leftrightarrow Ted\) \(R\leftrightarrow Carol\) \(\{P, Q\}\leftrightarrow Entertainment ~~committee\) \(\{P, R\}\leftrightarrow Refreshment ~~committee\) \(\{Q, R\}\leftrightarrow Finance ~~committee\)
Loser66
  • Loser66
B) \(\{Bob, Ted\}\leftrightarrow P\) \(\{ Ted, Carol\}\leftrightarrow Q\) \(\{Bob, Carol \}\leftrightarrow R\) \(\{P,Q\}\leftrightarrow Bob\) \(\{P, Q\}\leftrightarrow Ted\) \(\{Q,R\}\leftrightarrow Carol\)

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More answers

Loser66
  • Loser66
I don't know how to show that they are isomorphic.
anonymous
  • 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
Loser66
  • 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?
anonymous
  • anonymous
please post the original questions
anonymous
  • anonymous
"if all axioms are correct" doesn't make sense
Loser66
  • Loser66
That is original one. ok, let me take a pic
Loser66
  • Loser66
Loser66
  • Loser66
Problem 1.2.16
Loser66
  • Loser66
This is the definition of models
anonymous
  • 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)\)
Loser66
  • Loser66
Got you, how about part 2)?
Loser66
  • Loser66
How to show they are isomorphic?
anonymous
  • anonymous
if the models are isomorphic then that means there is a one to one mapping between \(a\)
anonymous
  • anonymous
and this is true -- we can take \((a,b)\in\, <\) and assign it to \((b,a)\, >\) so \(aa\) and then our axioms clearly still hold: $$aa\land c>b\Leftrightarrow c>b\land b>a\implies c>a$$
anonymous
  • anonymous
and similarly for the first axiom $$aa\implies b\ne a\Leftrightarrow a\ne b$$
Loser66
  • Loser66
So, we just show that they are " work" on the same way, right?
Loser66
  • 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.
anonymous
  • anonymous
in part (d) it is a valid model, but it is not isomorphic because there are uncountably many reals

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