30 1.6 Applications of the Completeness Axiom and, hence, 2s2 = r2 . (1.1) It follows that r2 is an even integer. Therefore, r is an even integer (see Exercise 1.4.1). We can then 2 2 write r = 2 j for some integer j. Hence r = 4 j2. Substituting in (1.1), we get s = 2 j2. Therefore, s2 is even. We conclude as before that s is even. Thus, both r and s have a common factor, which is a contradiction. □ The next theorem shows that irrational numbers are as ubiquitous as rational numbers. Theorem 1.6.5 Let x and y be two real numbers such that x < y. Then there exists an irrational number t such that x < t < y. Proof: Since x < y, one has √ √ x − 2 < y − 2 By Theorem 1.6.3, there exists a rational number r such that √ √ x − 2 < r < y− 2 This implies √ x < r + 2 < y. √ Since r is rational, the number t = r + 2 is irrational (see Exercise 1.6.4) and x < t < y. □ Exercises 1.6.1 For each sets below determine if it is bounded above, bounded below, or both. If it is bounded above (below) fnd the supremum (infmum). Prove your claims. (−1)n (a) 1+ : n ∈ N n 3n (b) : n ∈ N n+ 4 1 (−1)n (c) + : n ∈ N n (−1)n (−1)n − (d) : n ∈ N n 1.6.2 ▶ Let r be a rational number such that 0 < r < 1. Prove that there is n ∈ N such that 1 1 < r ≤ . n + 1 n
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