Results 1 to 2 of 2

Thread: axiomatization of real numbers

  1. #1
    Newbie
    Joined
    Jul 2011
    Posts
    10

    axiomatization of real numbers

    Hi guys,

    Suppose following axiomatization of real numbers:
    ($\displaystyle \exists !$ is syntactical sugar for uniqueness quantifier)

    (+ 1) $\displaystyle \forall a \forall b~a+b = b+a$
    (+ 2) $\displaystyle \forall a \forall b\forall c~ a+(b+c) = (a+b)+c$
    (+ 3) $\displaystyle \forall a \forall b \exists ! x~a+x = b$

    (. 1) $\displaystyle \forall a \forall b~ab = ba$
    (. 2) $\displaystyle \forall a \forall b \forall c~a(bc) = (ab)c$
    (. 3) $\displaystyle \forall a \forall b~(a \neq 0 \rightarrow \exists ! x~ax = b)$

    (+ .) $\displaystyle \forall a \forall b \forall c~(a+b)c = ac + bc$


    (< 1) forall $\displaystyle a$ forall $\displaystyle b$, one of the following holds: $\displaystyle a < b$, $\displaystyle a = b$, $\displaystyle b < a$
    (< 2) $\displaystyle \forall a \forall b \forall c~a<b \wedge b < c \rightarrow a < c $
    (+ <) $\displaystyle \forall a \forall b \forall c~a<b \rightarrow a+c < b+c$
    (. <) $\displaystyle \forall a \forall b~0 < a \wedge 0 < b \rightarrow 0 <ab$

    (sup) Let $\displaystyle A$ be non-empty set, s.t. it has upper bound. Then $\displaystyle \exists \alpha (\forall x~x \leq \alpha~\wedge~\forall \beta ~ \beta < \alpha \rightarrow ~ \exists a~ \beta < a)$

    Now following theorem holds: There is precisely one element (which will be denoted $\displaystyle 0$), which is solution of equation $\displaystyle \forall a~ a+x = a$.
    Proof: Assume some fixed $\displaystyle b$. Let the only solution of $\displaystyle b+x = b$ be denoted by symbol $\displaystyle 0$. So $\displaystyle b+0=b$ holds.

    Now proof is straightforward and I will not finish it. My problem is that $\displaystyle 0$ is formal symbol of our first-order theory (i.e. it is non-logical constant) and no axiom define any property that should $\displaystyle 0$ obbey. However the first step of the theorem says something like $\displaystyle \exists b~b+0 =b$. My question is why is this correct? I don't see any inference rule which interlinks $\displaystyle 0$ and axiom (+ 3).

    any ideas why is that correct? thanks
    Last edited by jozou; Sep 5th 2011 at 05:07 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790

    Re: axiomatization of real numbers

    I am not sure I fully understand the question, but there is a theorem allowing the introduction of new functional symbols. Namely, let T be the original theory. We proved $\displaystyle \exists! x\forall a\,a+x=a$. Let T' be T with a new constant 0 and a new axiom $\displaystyle \forall a\,a+0 = a$. We can define a natural translation from formulas of T' to formulas of T: e.g.,

    $\displaystyle \forall b\exists c\,b + c = 0$

    is translated into

    $\displaystyle \forall b\exists c\,(\forall x\,((\forall a\,a+x=a)\to b+c=x))$

    If $\displaystyle A$ is a formula in T', let $\displaystyle A^*$ denote the translation of A, which is a formula in T. Then $\displaystyle T'\vdash A$ iff $\displaystyle T\vdash A^*$, and for all formulas $\displaystyle A$ in the language of T, $\displaystyle T'\vdash A$ iff $\displaystyle T\vdash A$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Sep 27th 2010, 03:14 PM
  2. Real Numbers
    Posted in the Algebra Forum
    Replies: 6
    Last Post: Feb 21st 2010, 02:31 AM
  3. Real Numbers
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jan 20th 2009, 02:12 PM
  4. Real Numbers - Real Anaylsis
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Sep 3rd 2008, 10:54 AM
  5. should I use real numbers ??
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: Sep 17th 2006, 11:16 AM

Search Tags


/mathhelpforum @mathhelpforum