Introduction to Mathematical Analysis I - Second Edition

29 and, hence, 2 s 2 = r 2 . (1.3) It follows that r 2 is an even integer. Therefore, r is an even integer (see Exercise 1.4.1 ) . We can then write r = 2 j for some integer j . Hence r 2 = 4 j 2 . Substituting in ( 1.3 ) , we get s 2 = 2 j 2 . Therefore, s 2 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) find the supremum (infimum). Justify all your conclusions. (a) 3 n n + 4 : n ∈ N (b) ( − 1 ) n + 1 n : n ∈ N (c) ( − 1 ) n − ( − 1 ) n n : n ∈ N 1.6.2 I Let r be a rational number such that 0 < r < 1. Prove that there is n ∈ N such that 1 n + 1 < r ≤ 1 n . 1.6.3 Let x ∈ R . Prove that for every n ∈ N , there is r ∈ Q such that | x − r | < 1 n .