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Math Help - Axiomatic System Problem: Checking Models and Isomorphism

  1. #1
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    Axiomatic System Problem: Checking Models and Isomorphism

    Consider an infinite set of undefined elements S and the undefined relation R that satisfies the following axioms:

    Axiom 1. If a, b \in S and aRb, then a does not equal b.
    Axiom 2. If a, b, c \in S, aRb, and bRc, then aRc.

    (i) Show that an interpretation with S as the set of integers and aRb as "a is less than b" is a model for the system.
    (ii) Would S as as the set of integers and aRb interpreted as "a is greater than b" also be a model?
    (iii) Are the models in parts (i) and (ii) isomorphic?
    (iv) Would S as the set of real numbers and aRb interpreted as "a is less than b" be another model?
    (v) Is the model in Part(iv) isomorphic to the model in Part (i)?

    Part (i) and (ii) are models for the system because the axiom statements are correct.
    Part (iii) The models are isomorphic because they are the same except the notation changes.
    Part (iv) This interpretation is also a model.
    Part (v) The model is not isomorphic because Part i's S is set of integers while Part iv's S was set of real numbers.

    I was wondering if someone could check and make sure everything is okay.
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  2. #2
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    Your answer for (iii) is a little lacking. The difference between (i) and (ii) is more than a notation change. What you need to exhibit is an isomorphism mapping S to S that "converts" the first relation to the second (and its inverse converts the second back to the first).

    Your answer for (v) is true, but lacking in detail. For the two models to be isomorphic, you're going to have to produce an isomorphism that retains the relation. Why is this not possible between the integers and the reals?
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  3. #3
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    Ackbeet, thanks for the feedback. I was a little lazy and summarized my answer for Part iii. Here is the detailed answer:

    The models are isomorphic because the only thing that changes is the notation.

    a \leftrightarrow a
    b \leftrightarrow b
    c \leftrightarrow c

    a<b \leftrightarrow a>b
    b<c \leftrightarrow b>c
    a<c \leftrightarrow a>c

    Part V:
    Let me think this over some more.
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  4. #4
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    I disagree that the only thing that changes for (iii) is the notation. If you think about using the relation \le instead of <, only the notation changes, but axiom 1 is not satisfied. I think you have to produce the isomorphism and show that it works in order to prove the isomorphic property. That's the usual procedure, in any case. Make sense?
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  5. #5
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    You've got me confused. I understand that axiom 1 would not be satisfied if \leq is used. How would I produce the isomorphism? My textbook example for this problem was done similarly to what I did for part iii.

    Thanks for the help again, I am taking geometry class as an independent study and my textbook was rather confusing.
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  6. #6
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    Well, what is an isomorphism? Can you think of an isomorphism that changes < to > (and, by definition of isomorphism, it'll have an inverse that changes > to <)?
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  7. #7
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    Ackbeet thanks for the help. I have made a tutor appointment at my school.
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  8. #8
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    Ok. Good luck!
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