29 ■ Example 1.6.1 Let A={1− 1 n : n∈N}. We claim that supA=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 (ii) we can find n∈Nsuch that 1 n <ε. Then 1−ε <1− 1 n . This proves condition (2’) with a=1−1 n and the result follows. 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 mand a positive integer nsuch 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∈Nsuch that 1<n(y−x). Then ny =nx+n(y−x) >nx+1. By Theorem 1.6.2 (iv), one can choose m∈Zsuch 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 R̸=Q. Proposition 1.6.4 The number √2 is irrational. Proof: Suppose, by way of contradiction, that √2∈Q. Then there are integers r ands with s̸ =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= r2 s2 ,
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