Modify the argument in Theorem 2.4.7 to show that there exists a positive real number u such that u3 = 2. There exists a positive real number x such that x2 = 2. Proof. Let S := s R: 0 s, s2 lt; 2. Since 1 S, the set is not empty. Also, S is bounded above by 2, because if t gt; 2. then t2 gt; 4 so that t S. Therefore the Supremum Property implies that the set S has a supremum in R, and we let x := sup S. Note that x gt; 1. We will prove that x2 =2 by ruling out the other two possibilities: x2 lt;2 and x2 gt; 2. First assume that x2 lt; 2. We will show that this assumption contradicts the fact that x = sup S by finding an n N such that x + 1/n S, thus implying that x is not an upper bound for S. To see how to choose n, note that 1 /n2 1/n so that (x + 1/n)2 = x2 + 2x/n + 1/n2 x2 + 1/n (2x + 1). Hence if we can choose n so that 1/n(2x + 1) lt; 2 - x2, then we get (x + 1/n)2 lt; x2 + (2 - x2) = 2. By assumption we have 2 - x2 gt; 0, so that (2 - x2)/(2x + 1) gt; 0. Hence the Archimedean Properly (Corollary 2.4.5) can be used to obtain n N such that 1/n lt; 2 - x2/2x + 1. These steps can be reversed to show that for this choice of n we have x + 1/n S, which contradicts the fact that .v is an upper bound of S. Therefore we cannot have x2 lt; 2. Now assume that x2 gt; 2. We will show that it is then possible to find m N such that x -1/m is also an upper bound of S, contradicting the fact that x = sup S. To do this, note that (x - 1/m)2 = x2 - 2x/m +1/m² gt; x2 - 2x/m. Hence if we can choose m so that 2x/m lt; x2 - 2, then (x - 1/m)2 gt;x2 - (x2 - 2) = 2. Now by assumption we have x2 - 2 gt; 0. so that (x2 - 2)/2x gt; 0. Hence. by the Archimedean Property, there exists m m N such that 1/m lt; x2 - 2/2x. These steps can be reversed to show that for this choice of m we have (x - 1/m)2 gt; 2. Now if s S, then s2 lt; 2 lt; (x - 1/m)2, whence it follows from 2.1.13(a) that s lt; x -1/m. This implies that x - 1/m is an upper bound for S. which contradicts the fact that x = sup S. Therefore we cannot have x2 gt; 2. Since the possibilities x2 lt; 2 and x2 gt; 2 have been excluded, we must have x2 = 2. Q.E.D.