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Math Help - Question about ordered field...PLEASE HELP!!!

  1. #1
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    Talking Question about ordered field...PLEASE HELP!!!

    Assume { F, +,*,, <} is an ordered field. If a<b in F and c=(a+b)/2, prove
    a) a<c<b
    b) b - c = c - a = (b-a)/2. Thus the distance from c to either a or b is 1/2 the distance from a to b. C is called the mid point of [a,b].
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  2. #2
    Member
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    Aug 2009
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    Hi,
    I don't know what your favourite definition of an ordered field is so i'll use this one: A field is an ordered field if its nonzero elements can be split into two sets, P and N, such that P contains x iff N contains -x, and P is closed under addition and multiplication.
    The ordering < can be then defined as follows: x < y \mbox{ iff } y-x \in P.

    1 \neq 0 so either 1\in P or -1\in P, but if -1\in P then 1=(-1)(-1)\in P (since P is closed under multiplication) which is a contradiction, so -1\in N and 1\in P.

    Since P is closed under addition, 2=1+1\in P.
    Furthermore, \frac{1}{2}\in P, because if \frac{1}{2}\in N then -\frac{1}{2}\in P and since 2\in P we'd get -1 =2\left(-\frac{1}{2}\right) \in P (since P is closed under multiplication), a contradiction.

    Now we're ready to prove a), b).

    c-a=\frac{a+b}{2}-a=\frac{1}{2}(a+b)-\frac{1}{2}2a=\frac{1}{2}(a+b-2a)=\frac{1}{2}(b-a).

    Similarly, b-c=b- \frac{a+b}{2}=\frac{1}{2}2b-\frac{1}{2}(a+b)=\frac{1}{2}(2b-a-b)=\frac{1}{2}(b-a).

    Since  b-a\in P was given and we know \frac{1}{2} \in P, we get \frac{1}{2}(b-a)\in P. Thus a<c and c<b.
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