**Theorem. (The Archimedean Property of R)** The set N
of natural numbers is unbounded

above in R.

Note: We will use the completeness axiom to prove this theorem. Although the
Archimedean

property of R is a consequence of the completeness axiom, it is weaker than
completeness.

Notice that N is also unbounded above in Q, even though Q is not complete. We
also have an

example of an ordered field for which the Archmidean property does not hold! N
is bounded

above in F, the field of rational polynomials!

**Proof by contradiction.** If N were bounded above in R, then by___

___N would have a___ . I.e., there exists m ∈

___such that m = ____. Since m is the ____,

___is not an upper bound for N. Thus there exists an
such that

___. But then ___ , and since
, this contradicts___

___.

The Archimedean property is equivalent to many other statements about R and N.

**12.10 Theorem.** Each of the following is equivalent to the Archimedean
property.

(a) For every z ∈ R, there exists an n ∈ N such that n > z.

(b) For every x > 0 and for every y ∈ R, there exists an n ∈ N such that nx > y.

(c) For every x > 0, there exists an n ∈ N such that
.

The proof is given in the book. The idea is that (a) is the same as the
Archimedean property

because (a) is essentially the statement that “For every z ∈ R, z is not an
upper bound for

N.” Then, it is fairly easy to see why (b) and (c) follow.

**Theorem (Q is dense in R).** For every x, y ∈ R such
that x < y, there exists a rational

number r such that x < r < y.

Notes: The idea of this proof is to find the numerator and denominator of the
rational

number that will be between a given x and y. To do this, we first find a natural
number

n for which n x and n y will be more than one unit apart (this will require the
Archimedian

property!) Notice that the closer together x and y are, the bigger this n will
need to be!

Picture (assuming x > 0):

Since nx and ny are far enough apart, we expect that there
exists a natural number m in

between nx and ny. Finally, will be the
rational number in between x and y!

**Proof.** Let x, y ∈ R such that x < y be given. We will first prove the
theorem in the

case x > 0. Since y − x > 0, ___∈ R. Then, by the Archimedean property, there

exists an n ∈ N such that n > ___. Therefore, ___< ny. Since we are in

the case x > 0, ___> 0 and there exists m ∈ N such that m − 1 ≤___< m (The

proof that such an m exists uses the well-ordering property of N; see Exercise
12.9.) Then,

ny > ___≥___. Thus nx < m < ny. It then follows that the rational number

satisfies x < r < y.

Now, in the case x ≤0, there exists k ∈ N such that k > |x|. Since k − |x| = k +
x is

positive and k + x < k + y, the above argument proves that there is a rational
number r

such that k + x < r < k + y. Then, letting r' = r − k, r' is a rational number
such that

x < r' < y.

It is also true that for every x, y ∈ R such that x < y, there exists an
irrational number

w such that x < w < y. Combining these facts, it follows that for every x, y ∈R
such

that x < y there are in fact infinitely many rational numbers and infinitely
many irrational

numbers in between x and y!