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Basic Real Analysis Question

  1. Oct 31, 2007 #1
    I have just started in a Real Analysis textbook. It starts "In this chapter we construct the real numbers. We assume that the rational numbers and their arithemtic and order properties are known."

    What exactly does this assumption mean?

    Here is an example of where I get caught up. One of the first exersizes is this:
    "Prove that for all intervals I, 0 is conatined in I - I."

    My proof would be something like this,

    Let I = [r, s]
    by definition I - I = [r - s, s - r]
    since s >= r, subtracting s from both sides yields 0 >= r - s
    also since s>= r, subtracting r from both sides yields s - r >= 0
    we now see that r - s <= 0 <= s - r, which by definition implies 0 is contained in [r-s , s-r] = I - I, as required.

    I am guessing that this is good, if properties of inequalities are covered in the assumed "arithmetic and order" properties??
  2. jcsd
  3. Oct 31, 2007 #2


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    The proof looks good, the "order properties" refer to the relations "<", "<=", etc.
  4. Nov 1, 2007 #3

    Gib Z

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    It seems quite trivial that certain order properties should hold, and we can use inequality relations to describe certain numbers. That is until you realise that the complex numbers in general can not be ordered :( So sometimes we establish these relations for the reals, though as it did in that book, it is usually left out.
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