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
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
On the other hand, suppose
x 0. Then, by part (a), 0 −x. As above, we conclude that 0
(Exercise 1.3.5), the proof of part (b) is complete.
= 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
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
= 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
= k. Let r =
be r expressed
as a fraction in which n and m have no common factors. Then,
= k and so
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.
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