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Math Help - Ordered Fields, 0<a<b

  1. #1
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    Ordered Fields, 0<a<b

    I just read a proof that says 0 < a ==> 0 < a^-1.

    How do I prove 0 < a < b ==> 0 < b^-1 < a^-1?

    Also, the text did not give a definition for 0 < a < b.

    Would it be correct to assume 0 < a and a < b <=> 0 < a < b, similarly for 0<= a and a<=b <=> 0 <= a <= b?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Noxide View Post
    I just read a proof that says 0 < a ==> 0 < a^-1.

    How do I prove 0 < a < b ==> 0 < b^-1 < a^-1?

    Also, the text did not give a definition for 0 < a < b.

    Would it be correct to assume 0 < a and a < b <=> 0 < a < b, similarly for 0<= a and a<=b <=> 0 <= a <= b?
    Of course this last definition is the correct one. For the second part recall (c.f. Rudin) that one of the ordered field axioms is that x<y and z>0 then xz<yz. Moreover, it's easy to show that x,y>0\implies xy>0. So you have that 0<a<b, a^{-1},b^{-1}>0 and thus from previous discussion a^{-1}b^{-1}>0 it follows that 0<(a^{-1}b^{-1}}a<(a^{-1}b^{-1})b\cdots
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  3. #3
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    My text only gave me axioms for <=
    Are < and <= equivalent ... I know they're not, but can one replace the other in the ordered field axioms?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Noxide View Post
    My text only gave me axioms for <=
    Are < and <= equivalent ... I know they're not, but can one replace the other in the ordered field axioms?
    In all cases that immediately occur to me, yes.
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