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 write r =2j for some integer j. Hence r2 =4j2. Substituting in (1.1), we get s2 =2j2. Therefore, s2 is even. We conclude as before that s is even. Thus, bothr 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) andx <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) find the supremum (infimum). Prove your claims. (a) 1+ (−1)n n : n∈N (b) 3n n+4 : n∈N (c) (−1)n + 1 n : n∈N (d) (−1)n − (−1)n n : n∈N 1.6.2 ▶Let r be a rational number such that 0<r <1. Prove that there is n∈Nsuch that 1 n+1 <r ≤ 1 n .
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