1.4. The Real Numbers 25
Definition 1.4.7. An ordered field is said to have the Archimedean property if,
for every x R, there is a natural number n such that x n. An ordered field
with the Archimedean property is called an Archimedean ordered field.
Theorem 1.4.8. The field of real numbers has the Archimedean property.
Proof. We use the completeness property. Suppose there is an x such that n x
for all n N. Then N is a non-empty subset of R which is bounded above. By the
completeness property, there is a least upper bound b for N. Then b is an upper
bound for N, but b 1 is not. This implies there is an n N such that b 1 n.
Then b n + 1, which contradicts the statement that b is an upper bound for
N. Thus, the assumption that N is bounded above by some x R has led to a
contradiction. We conclude that every x in R is less than some natural number.
This completes the proof.
The Archimedean property can be stated in any one of several equivalent ways.
One of these is: for every real number x 0, there is an n N such that 1/n x
(Example 1.4.9). Another is: given real numbers x and y with x 0, there is an
n N such that nx y (Exercise 1.4.6).
Example 1.4.9. Prove that, in an Archimedean field, for each x 0 there is an
n N such that 1/n x.
Solution: The Archimedean property tells us that there is a natural number
n 1/x. Since n and x are positive, this inequaltiy is preserved when we multiply
it by x and divide it by n. This yields 1/n x, as required.
Another consequence of the Archimedean property is that there is a rational
number between each distinct pair of real numbers (Exercise 1.4.7).
Exercise Set 1.4
1. For each of the following sets, describe the set of all upper bounds for the set:
(a) of odd integers;
(b) {1 1/n : n N};
(c) {r Q : r3 8};
(d) {sin x : x R}.
2. For each of the sets in (a), (b), (c) of the preceding exercise, find the least upper
bound of the set, if it exists.
3. Prove that if a subset A of R is bounded above, then the set of all upper bounds
for A is a set of the form [x, ∞). What is x?
4. Show that the set A = {x : x2 1 x} is bounded above, and then find its
least upper bound.
5. If F is an ordered field, prove that there is a sequence of elements {nk}k∈N, all
different, such that n1 = 1 (the identity element of F) and nk+1 = nk + 1 for
each k N. Argue that the terms of this sequence form a subset of F which
is a copy of the natural numbers, by showing that the correspondence k nk
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