Number z such lint z2 a Modify the argument in Theorem 247

Number z such (lint z^2 = a. Modify the argument in Theorem 2.4.7 to show that there exists a positive real number u such that u^3 = 2.

Solution

Let S = {x: x R , x 0, x³ 2}. We may observe that 1 0 and 13 =1 2. Therefore,1 S. Thus, the set S is non-empty. We claim that 2 is an upper bound of S . If not, then, x S such that x³ > 2. However, this contradicts the definition of S. Therefore, 2 is an upper-bound of S. So, by the completeness axiom (the least-upper-bound property of the real numbers), S has a supremum or least upper bound. Say u = Sup S which must be a positive real number by definition. If u³ = 2, then we have proved that there exists a positive real number u such that u³ = 2. If u³ 2, then either u³ < 2 or, u³ > 2. Let assume that u³ < 2. Then we can find a real number v > u such that v3 < 2 Then u is not the sup S, which is a contradiction. Similarly, there is a contradiction if u³ > 2. Thus, u³ = 2. Hence, there exists a positive real number u such that u³ = 2.