S. D. V. Stephens

notes / Integrated Mathematics / chapter 17: The Real Numbers

17.1 Archimedes’ Method and the Axiomatic Approach

Archimedes’ method finds the area of a parabolic segment by trapping it between sums of rectangles, the lower and upper sums and of the exercises at the end of this page. This page writes the result in integral notation, asks what such an argument rests on, and sets out the approach taken for the rest of this topic: the real numbers are pinned down by axioms.

Archimedes’ result in integral notation

The post-Archimedean show of would be Let us examine its anatomy:

  • is the integrand;
  • and are the limits of integration;
  • the process is called integration.

Deductive systems

Definition 17.1.1. Deductive systems contain “undefined” concepts (axioms or postulates) and define all other matters in terms of these axioms. Statements deduced from these axioms are called theorems.

Example 17.1.2 (Deductive system (Aristotelian)).

  • Premise 1: All men are potatoes.
  • Premise 2: I am a man.
  • Conclusion: I am a potato.

The real numbers as an axiomatic system

We will let the real numbers themselves () be an axiom, and also some of their properties. (TF does this mean?? Is that to say that the class of the real numbers is , where is the irrational numbers?)

Remark (Correction). The real numbers are not themselves an axiom. We take the real numbers as undefined objects, as in a deductive system, and take some of their properties as axioms (the field, order and completeness axioms on the following pages). It is true that , but is not introduced that way here: the irrationals are defined later, as the real numbers that are not rational.

There are groups of axioms: the field axioms, the order axioms, and the completeness axiom.

What is the “least-upper-bound” axiom / completeness / continuity axiom? It literally just means there are no gaps in ’s number line.

Exercises

Exercise 17.1.3. Modify the region of the parabolic segment by taking the ordinate at each to be (a) , (b) , (c) , (d) , (e) instead of , and see what effect this has on the calculation of the area.

Show solution

(a)

Graph of f(x) = 2x squared rising from the origin, with dotted vertical segments at x and at b marking the ordinates 2x squared and 2b squared.

Each rectangle has base and the th one has height . So the th rectangle has area : Thus, just like when produced and the summation was we now get and since we know that , we can substitute and multiply by . We get and simplifying, and following the same steps for the lower sum,

(b) Now we repeat with :

(c) And with :

(d) : the th rectangle has area so summing gives .

(e) :

Exercise 17.1.4. Take the ordinate at each to be instead of .

Show solution

(a)

Graph of f(x) = x cubed rising from the origin, with dotted vertical segments at x and at b marking the ordinates x cubed and b cubed.

With base and height ,

(b) which gives the area .

(c) : .

Exercise 17.1.5. Take the ordinate at each to be .

Show solution

With and , and