Introduction to Mathematical Analysis I - Second Edition

28 1.6 APPLICATIONS OF THE COMPLETENESS AXIOM Example 1.6.1 Let A = sup { 1 − 1 n : n ∈ N } . We claim that sup A = 1. We use Proposition 1.5.1 . Since 1 − 1 / n < 1 for all n ∈ N , we obtain condition (1’). Next, let ε > 0. From Theorem 1.6.2 (b) we can find n ∈ N such that 1 n < ε . Then 1 − ε < 1 − 1 n . This proves condition (2’) with a = 1 − 1 n and completes the proof. Theorem 1.6.3 — The Density Property of Q . If x and y are two real numbers such that x < y , then there exists a rational number r such that x < r < y . Proof: We are going to prove that there exist an integer m and a positive integer n such that x < m / n < y , or, equivalently, nx < m < ny = nx + n ( y − x ) . Since y − x > 0, by Theorem 1.6.2 (3), there exists n ∈ N such that 1 < n ( y − x ) . Then ny = nx + n ( y − x ) > nx + 1 . By Theorem 1.6.2 (4), one can choose m ∈ Z such that m − 1 ≤ nx < m . Then nx < m ≤ nx + 1 < ny . Therefore, x < m / n < y . The proof is now complete. We will prove in a later section (see Examples 3.4.2 and 4.3.1 ) that there exists a (unique) positive real number x such that x 2 = 2. We denote that number by √ 2. The following result shows, in particular, that R 6 = Q . Proposition 1.6.4 The number √ 2 is irrational. Proof: Suppose, by way of contradiction, that √ 2 ∈ Q . Then there are integers r and s with s 6 = 0, such that √ 2 = r s . By canceling out the common factors of r and s , we may assume that r and s have no common factors. Now, by squaring both sides of the equation above, we get 2 = r 2 s 2 ,