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Decidability of -1<1/0<1 using the ordered field axioms and first order logic

  1. Aug 31, 2008 #1
    Is the statement -1<1/0<1 decidable using the ordered field/real number axioms and first order logic? I have tried to prove that the statement is either true or false but have had no success since the axioms and theorems only make statements about objects that exist and do not give any clear way to treat those that do not. So I would like to know if it is even possible to prove that the statement is true or false or if it is fundamentally undecidable in the given system. Or alternatively is it possible to derive a contradiction regardless of whether the statement is true or false a la the liar paradox.
  2. jcsd
  3. Sep 1, 2008 #2
    If you assume that 1/0 acts just like the inverse of any other number, then you can show that it cannot exist in any field. (Note that then -1 < 1/0 < 1 is false because the number you called 1/0 doesn't exist)

    Assume that 0 * x = 1 (i.e., that x = 1/0).

    Now we also have that for any number in our field r: r * 0 * x = 0 * x = 1. Also, for all r (even when r is 0), we have (r / r) * 0 * x = 0 * x = 1/r. Which gives us that 1 = 1/r for all r. Since every number has an inverse, including 1/0, we have 1 = 1/(1/0) = 0. Therefore, we have only 1 number in our algebra. Other than not being a field, then is fairly uninteresting. Also, the statement x < y is always false since for any x and y in our algebra, x = y.
  4. Sep 1, 2008 #3
    Looking back on my first post I now see that I did not properly articulate what it was that I was trying to inquire about. I will be more carful next time.

    I already understand why 1/0 cannot exist in a field, I simply chose it as an example of a nonexistent object when what I am really interested in is the more general case of whether all statements about nonexistent objects are inherently false. While this seems obvious intuitively, I am not sure how derive it rigorously from the given axioms. I am wondering if there is such an axiom in first order logic that simply asserts this, but I am having trouble understanding the Wikipedia article on the topic. I believe that the following formula might shed some light on this issue and am asking for confirmation.

    [tex]Z(t)\rightarrow(\exists xZ(x))[/tex]
    where the expression Z(x) stands for any well-formed forumula with the additional convention that Z(t) stands for the result of substitution of the term t for x in Z(x).
  5. Sep 2, 2008 #4
    If t does not exists, then depending on your interpretation of Z(t), it is either not considered a statement, or it is interpreted as [tex]\exists t Z(t)[/tex], which is false.

    Try reading this Wikipedia article: http://en.wikipedia.org/wiki/Definite_description
  6. Sep 2, 2008 #5


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    If it helps, the following statement is provable from the ordered ring axioms:
    For all x: 0x = 1 implies -1 < x < 1.​
  7. Sep 3, 2008 #6
    What I am interested in knowing is which of these interpretations is part of first-order logic. Or are you saying that first-order logic does not itself specify which of these two interpretations to use and that that decision is left up to a particular model.

    What about the particular case of mathematical analysis which I believe is based on second-order logic? In order for the definition of a limit to mean what we want it to mean, statements about nonexistent objects must be false. (I think) So where does this rule derive from? Does second-order logic make a judgment on the matter of which interpretation to use or does it come from some other foundation of mathematical analysis?
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