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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

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  1. Loser66
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    I don't know how to show that they are isomorphic.

  5. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    please post the original questions

  8. anonymous
    • one year ago
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    "if all axioms are correct" doesn't make sense

  9. Loser66
    • one year ago
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    That is original one. ok, let me take a pic

  10. Loser66
    • one year ago
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  11. Loser66
    • one year ago
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    Problem 1.2.16

  12. Loser66
    • one year ago
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    This is the definition of models

  13. anonymous
    • one year ago
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    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
    • one year ago
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    Got you, how about part 2)?

  15. Loser66
    • one year ago
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    How to show they are isomorphic?

  16. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    and similarly for the first axiom $$a<b\implies a\ne b\\b>a\implies b\ne a\Leftrightarrow a\ne b$$

  19. Loser66
    • one year ago
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    So, we just show that they are " work" on the same way, right?

  20. Loser66
    • one year ago
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    @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
    • one year ago
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    in part (d) it is a valid model, but it is not isomorphic because there are uncountably many reals

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