HW4 Showing the rationals are dense in the reals

 

The following is the basis for a homework problem AND a different perspective on the issue of how to prove the rationals are dense in the real numbers. The text begins by proving some facts about integers without it being clear to the reader why it is being done. We alternatively start off with the goal of proving the rationals are dense and see what kinds of facts we need to establish that.

 

Assume we have the completeness axiom (p. 8), but we have not proved any results. Here is how we might proceed through a succession of simpler or more specific reformulations, simpler in that they involve integers rather than rationals.

 

The statement

1. The rationals are dense in the reals

may be restated

 

2. For each interval (a, b), where a and b are real numbers, there is a rational number m/n in (a, b).

 

It turns out that proving the rational number exists can be made constructive.

 

Problem 4.1 Follow the construction below for the interval (1/4, 1/3) and find the rational number that is produced. (Just finding any rational number in the interval is not acceptable.)

 

Choose n so that 1/n < b-a. The Archimedean property (p. 13) guarantees n exists.

Choose n as small as possible.

We can now restate 2 in an equivalent form:

 

3. For each real number b and each natural number n, there is a rational number in the interval [b – 1/n, b). If 2 is true, so is 1.

 

#3 can be strengthened so that if #4 below is true, then so is #3:

 

4. For each real number b and each natural number n, there is a unique integer m such that m/n is in the interval [b – 1/n, b).

 

For such an m/n, we would have b – 1/n ≤ m/n < b

 

Or     nb - 1 ≤ m < nb.

 

So #4 may be restated

 

4. For any real number c ( = nb -1 above) there is exactly one integer m in [c, c +1).

 

To prove this statement, use the completeness axiom by saying that since c is a lower bound for those integers that are ≥ c, there is a greatest lower bound G. Note G ≥ c. We claim that the proof of #4 reduces to 2 steps:

 

5a. The greatest lower bound G is an integer, and

 

(5b)      G < c+1.

 

If we establish that, then we have proved #4 where m = G, which implies #1 and we are done.

 

You can now review in Fitzpatrick’s Advanced Calculus how 5a and 5b are established.

 

Problem 5.1  A real number of the form m / 2^n, where m is an integer and n is a natural number, is called a dyadic rational.

 (a) Adapt the above procedure to show the so-called “dyadic” rationals are dense in the reals. Problem 27 on p. 22 is similar.

 (b) Do problem 4.1 for the dyadic rationals.