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Math Help - Cuts and Archimedean Property

  1. #1
    Senior Member Sampras's Avatar
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    Cuts and Archimedean Property

    So we know that not all fields obey the Archimedean property. For instance, the field  \mathbb{R}(x) of all rational functions with real coefficients does not. But is there any utility in doing analysis in these fields?

    Also suppose we have a positive cut  x and a negative cut  y . Then we should (and do) get a negative cut. But why do we write it as follows:  -(x \cdot (-y)) ?
    Last edited by Sampras; August 28th 2009 at 12:32 PM.
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  2. #2
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    function fields are not the only ones. the most important ones are the p-adic ones. and they have a immensely rich analysis.
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    Quote Originally Posted by Sampras View Post
    Also suppose we have a positive cut  x and a negative cut  y . Then we should (and do) get a negative cut. But why do we write it as follows:  -(x \cdot (-y)) ?
    ?? Just having a positive cut and a negative cut doesn't necessarily give us anything! If we have a positive cut, x, and a negative cut, y, then adding the two, which is what you appear to be doing, does not necessarily give a negative cut. But if we have a positive cut, x, and negative cut, y, and |y|> x, then x+ y is a negative cut. As for writing it as -(x)(-y), I suspect that was because they define the sum of two positive cuts first and wanted to write the general case in terms of that.
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    Senior Member Sampras's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    ?? Just having a positive cut and a negative cut doesn't necessarily give us anything! If we have a positive cut, x, and a negative cut, y, then adding the two, which is what you appear to be doing, does not necessarily give a negative cut. But if we have a positive cut, x, and negative cut, y, and |y|> x, then x+ y is a negative cut. As for writing it as -(x)(-y), I suspect that was because they define the sum of two positive cuts first and wanted to write the general case in terms of that.
    I am not adding them. I am multiplying them. If  x = A|B and  y= C|D are non-negative cuts, then their product is  x \cdot y = E|F where  E = \{ r \in \mathbb{Q}: r \leq 0 \ \text{or} \ \exists a \in A, \ \text{such that} \ a>0, c>0, \ \text{and} \ r= ac \} and  F = \{\text{rest of} \ \mathbb{Q} \} . If we multiply a positive cut by a negative cut, we get a negative cut. But why is this represented as  -(x \cdot (-y)) ? Because this "looks" positive (e.g. a negative multiplied by a negative is a positive).
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    Quote Originally Posted by Sampras View Post
    If we multiply a positive cut by a negative cut, we get a negative cut. But why is this represented as  -(x \cdot (-y)) ? Because this "looks" positive (e.g. a negative multiplied by a negative is a positive).
    Don't forget that y is a negative cut, so y is a positive cut, as is x.(y). Then (x.(y)) is negative. If you think of y as having a minus sign built into it, we have altogether three minuses, which make a minus!
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