**Problem Statement:**Find the least upper bound and the greatest lower bound of \[ S = \left\{ \frac{2pq+1}{pq}, p \in \mathbb{N}, q \in \mathbb{N} \right\} \]and prove that they are the least upper bound and the greatest lower bound of \( S \).

The given set is S = {2pq+1pq, p∈ℕ, q∈ℕ}. Now, we know that for any p∈ℕ, q∈ℕ. pq < 2pq⇒pq < 2pq + 1⇒1 < 2pq+1pq So, 1 is a lower bound of the set S. Now, we have to show that 1 is the greatest lower bound if the set S.Suppose z < 1 is the greatest lower bound of S. Then, by definition, there must exist an element a∈S, such that z<a<1.Since a∈S, a must be of the form 2p1q1+1p1q1, p1,q1∈ℕ. From a<1, we get 2p1q1+1p1q1<1⇒2p1q1+1<p1q1⇒p1q1<-1which is a contradiction as both p1,q1 are positive integers, and hence their product can not be less than -1.Therefore, 1 is the greatest lower bound of S. To find the upper bound of S, consider the following.2pq+1pq=2 +1pqNow, 1pq attains its maximum value for pq = 1. Because if pq > 1 ⇒1 >1pq. So, 2pq+1pq is maximum when pq = 1. 2pq+1pq<2 + 1pq<2+1 = 3 Thus, 3 is an upper bound for S.Now, we need to prove that 3 is a least upper bound for S. Suppose, on contrary, that 3 is not an upper bound for S. Let u > 3 be an upper bound for S.Then, by definition, there exists, b∈S, such that 3<b<u. Since b∈S, b must be of the form 2p2q2+1p2q2, p2,q2 ∈ℕ.From 3 < b, we get, 3<2p2q2+1p2q2 ⇒3<2+1p2q2⇒1<1p2q2⇒p2q2 < 1 This is a contradiction as p2,q2 being positive integers, their product cannot be less than 1. Hence, 3 is the least upper bound of S.

Answered: Find the least upper bound and the…

Math

Advanced Math

Find the least upper bound and the greatest lower bound of S = { 2pg+1 PEN, q E N pq and prove that they are the least upper bound and the greatest lower bound of S.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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