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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??
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??