29 ■ Example 1.6.1 Let A = {1− 1 : n ∈ N}. We claim that supA = 1. n 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 (ii) we can fnd n ∈ N such that 1 < ε. Then n 1 1− ε < 1 − . n This proves condition (2’) with a = 1 − 1 and the result follows. n 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 (iii), there exists n ∈ N such that 1 < n(y− x). Then ny = nx + n(y− x) > nx + 1. By Theorem 1.6.2 (iv), 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 x2 = 2. We denote that number by √ 2. The following result shows, in particular, that R ̸ = 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 ≠ 0, such that √ r 2 = . 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 2r 2 = s2 ,
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