# Thread: Cuts and Archimedean Property

1. ## 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: $-(x \cdot (-y))$?

2. function fields are not the only ones. the most important ones are the p-adic ones. and they have a immensely rich analysis.

3. Originally Posted by Sampras
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.

4. Originally Posted by HallsofIvy
?? 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).

5. Originally Posted by Sampras
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!