Archimedean Property and Distribution of Q in R

1.2 Archimedean Property and Distribution of Q in R

Definition: We say the set S is dense in R if for every a < b the intersection
S ∩ (a, b) ≠ Ø. (In other words, there is an element of S between every two elements of
R.)

• The main goal of Section 1.2 is to show that the rationals are dense in the reals. To achieve
this goal, the book first proves The Archimedean Property.

Theorem 1.5 The Archimedean Property: For any positive number c there is a natural number n in N such that n > c. (Equivalently, for any positive number ε, there is a natural number n such that 1/n < ε .)

• Proof: That the two versions are equivalent is as simple as noting that when ε = 1/c then
the same n works for both. However, we must now prove one of the two versions of the claim.
Note that the first version is the same as saying N is not bounded. We will assume that it has
an upper-bound and reach a contradiction. If it is bounded above, then the completeness
axiom tells us that b = sup  is a number in R with the property that n < b for every
natural number n. But then, since n + 1 is also a natural number, we know n + 1 < b.
This is the same as n < b − 1 for every n, which means that b − 1 is an upper-bound for
N, which contradicts the minimality of b.

• The next thing the book proves is a fact that may seem completely obvious to us: “Theorem
1.8 For any number c, there is exactly one integer k in the interval [c, c + 1).” Note that it
takes about two pages to prove this result! Merely for the sake of time, we will accept this
fact here without discussing the proof, but you should look at the proof if only to see that it
can be proved using the axioms explicitly.

• Now, we will use the previous two results to prove the denseness of the rationals:

Theorem 1.9: If a and b are real numbers such that a < b then Q ∩ (a, b) ≠ Ø.

• Proof: We need to find m, n ∈ Z such that . We use Theorem 1.5 to get n
with . By theorem 1.8 there is an integer m in [nb − 1, nb). Thus:

nb − 1 ≤ m <nb
b − 1/n ≤m/n < b (dividing by n)
a <m/n< b because a = b − (b − a) < b − 1/n.

• Similarly, we can show that the irrationals are dense in R. This takes two steps...can we do
them together?

Show that the product of an irrational number and a rational number is irrational.
Show that there is an irrational number in (a, b) by using the fact that there is a rational
number in

1.3 Inequalities and Identities

• This section contains a number of formulas that will prove useful. Most important for us will
be:

Theorem 1.11 Triangle Inequality: where

• Proof: Since |c| ≤ d is equivalent to −d ≤ c ≤ d, we can add together the formulas
−|a| ≤ a ≤ |a| and −|b| ≤ b ≤ |b| to get this.

• Why is it called the “triangle” inequality?

• Other formulas and definitions here should be equally familiar to you from other classes, such
as:

and

Homework: 1.2 # 1 / 1.3 # 7