1.3. Integers and Rational Numbers 19

O4. If x ≤ y, then x + z ≤ y + z.

O5. If x ≤ y and 0 ≤ z, then xz ≤ yz.

Remark 1.3.7. Given an order relation “≤”, we don’t distinguish between the

statements “x ≤ y” and “y ≥ x” – they mean the same thing. Also, if x ≤ y and

x = y, then we write x y or, equivalently, y x.

Example 1.3.8. Prove that if F is an ordered field, then

(a) if x, y ∈ F and x ≤ y, then −y ≤ −x;

(b) if x ∈ F, then x2 ≥ 0;

(c) 0 1.

Solution: If x ≤ y, then 0 = x − x ≤ y − x by O4. Using O4 again, along

with A1 through A4, yields −y ≤ (y − x) − y = −x. This completes the proof of

(a).

By O1, if x ∈ F, then 0 ≤ x or x ≤ 0. If 0 ≤ x, then we multiply this

inequality by x and use O4 to conclude that 0 ≤

x2.

On the other hand, suppose

x ≤ 0. Then, by part (a), 0 ≤ −x. As above, we conclude that 0 ≤

(−x)2.

Since

(−x)2

=

x2

(Exercise 1.3.5), the proof of part (b) is complete.

Since

12

= 1, part (b) implies that 0 ≤ 1. By M3, 1 = 0 and so 0 1.

Defects of the Rational Field. The rational number system is very satisfying

in many ways and is highly useful. However, there are real-world mathematic

problems that appear to have real-world numerical solutions, but these solutions

cannot be rational numbers. For example, the Pythagorean Theorem tells us that if

the legs of a right triangle have length a and b, then the length c of the hypotenuse

satisfies the equation

c2

=

a2

+

b2.

However, there are many examples of rational and even integer choices for a and

b such that this equation has no rational solution for c. The simplest example is

a = b = 1. The Pythagorean Theorem says that a right triangle with legs of length

1 has a hypotenuse of length c satisfying

c2

= 2. However, there is no rational

number whose square is 2. We will prove this using the following theorem:

Theorem 1.3.9. If k is an integer and the equation x2 = k has a rational solution,

then that solution is actually an integer.

Proof. Suppose r is a rational number such that

r2

= k. Let r =

n

m

be r expressed

as a fraction in which n and m have no common factors. Then,

n

m

2

= k and so

n2

=

m2k.

This equation implies that m divides n2. However, if m = 1, then m can be

expressed as a product of primes, and each of these primes must also divide n2.

However, if a prime number divides n2, it must also divide n (Exercise 1.3.14).

Thus, each prime factor of m divides n. Since n and m have no common factors,

this is impossible. We conclude that m = 1 and, hence, that r = n is an integer.