Question

Problem 3: The purpose of this problem is use the Nested Interval Property to show the existence of the square root of a positive number A. Thinking of A as an area, start with a rectangle with sides ao bo such that aoboA. Define b( bo)/2 and aA/b1, so that abi-A. Repeating this process, one creates sequence {an^ni and sbn^n1 defining nested intervals In such that I -VA. The different parts of this problem will guide you through a proof of this statement Part : Show that b>VA. Hint: Takesqares on both sides and recall that A- abo] Part lI: Show that a VA, and that [ai, bi] C lao, bo. Part Il: Show that (bi-a) K (1/2)(bo-a Part IV: Show by induction that an VA bn and that an, b] C[an-1,6-I] Part V: Using Part III and IV, conclude that Eis a trivial interval, consisting on only one point L. Part VI: Show that L = VA.

just part 4,5,6

Answer 1

clearly b_1 > squareroot A, a_1 < squareroot A. Let b_k > squareroot A & a_k < squareroot A b_k + 1 = b_k + a_k/2 & a_k + 1 = A/b_k + 1 b^2_k + 1 = b^2_k + a^2_k/4 + a_k b_k/2 + a_k b_k/2 - a_k b_k/2 > A Doublerightarrow b_k + 1 > squareroot A following similar logics as in II we get a_k + 1 < squareroot A so by induction a_n < squareroot A < b_n It is clear that b_n lessthanorequalto b_n - 1 since b_n = b_n - 1 + a_n - 1/2 lessthanorequalto 2b_n_1 + a_n - 1/2 = b_n & a_n greaterthanorequaltoa_n - 1 so [a_n, b_n] lessthanorequalto [a_n - 1, b_n - 1] From III inductively we have b_k - a_k lessthanorequalto 1/2^k (b_o - a_o). so as k rightarrow infinity b_k - a_k rightarrow 0 so mod I_n rightarrow o as n rightarrow infinity. Thus n^infinity I_n contains just a single point say L since in