32 Chapter 1 • The Real Number SystemTheorem 1.5.3 Every positive element of an Archimedean ordered field can
be located between a unique pair of successive natural numbers. That is,\:/x > 0, :J unique n E N 3 n - 1 :::::: x < n.
*Proof. Part 1 (Existence): Let F be an Archimedean ordered field, and
x > 0. By the Archimedean property, the set S = { n E N : x < n} is nonempty.
Hence, by the well-ordering property (Theorem 1.3.10) the set S has a smallest
element, say no. Then x < no and no -1 ¢ S, since no is the smallest element of
S. Hence, x f:_ no -1. That is, no -1:::::: x. Therefore, we have no -1 :::::: x <no.
Part 2 (Uniqueness): Suppose :J m, n E N 3
n - 1 :::::: x < n and m - 1 :::::: x < m.
That is , x < n :::::: x + 1 and x < m '.s; x + 1. Multiplying through the second
inequality by -1, we havex < n :::::: x + 1 and - x - 1 :::::: -m < - x.
Adding these two inequalities, we have
-1 < n - m < 1.
But there is only one integer between -1 and l; namely, 0. Therefore, n-m = 0.
That is, n = m.
Therefore, there is only one natural number n such that n - 1 :::::: x < n. •Corollary 1.5.4 Every element of an Archimedear: ordered field can be located
between a unique pair of successive integers. That is,\:/x E F, :J unique n E Z 3 n - 1 :::::: x < n.
Proof. Exercise 5. •Corollary 1.5.5 Between any two elements greater than one unit apart in an
Archimedean ordered field, there is an integer. That is, if y - x > 1, then
:J integer n 3 x < n < y.*Proof. Suppose Fis an Archimedean ordered field, x, y E F, and y-x >
- By Corollary 1.5.4, :J integer n 3
n -1:::::: x < n.
Adding 1 to both sides of these inequalities,
n:Sx+l<n+l.
By putting together pieces of these inequalities, we have (justify each st ep)x < n:::::: x + 1 < y.
Thus, :J integer n 3 x < n < y. •