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Problem 10.1.2. Let (xn), (yn) be sequences as in the NIP. Show that for all п, т € N, zn < Ут-

Problem 10.1.2. Let (xn), (yn) be sequences as in the NIP. Show that for all п, т € N, zn < Ут-

Advanced Engineering Mathematics
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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math analysis, please help me solve problem 10.1.2

Axiom 10.1.1. Nested Interval Property of the Real Number System (NIP). Suppose we have two sequences of real numbers (xn)and (yn) satisfying the following conditions: 1. x1 < x2 < x3 <... (this says that the sequence, (xn), is non-decreasing) 2. y1 2 Y2 > Y3 >... (this says that the sequence, (Yn), is non-increasing) 3. Vn, xn < Yn 4. lim (yn –- xn) = 0 n00 Then there exists a unique number c such that xn < c Yn for all n. Geometrically, we have the following situation. X1 X2 X3 Уз y2 y1 ... ... Notice that we have two sequences (x,) and (yn), one increasing (really non- decreasing) and one decreasing (non-increasing). These sequences do not pass each other. In fact, the following is true: Problem 10.1.2. Let (xn), (yn) be sequences as in the NIP. Show that for all п, т € N, хn < Ут- They are also coming together in the sense that limn+ (Yn – xn) = 0. The NIP says that in this case there is a unique real number c in the middle of all of this In <c< yn for all n e N. X1 X2 X3 Уз y2 y1 ... ... If there was no such c then there would be a hole where these two sequences come together. The NIP guarantees that there is no such hole. We do not need to prove this since an axiom is, by definition, a self evident truth. We are taking it on faith that the real number system obeys this law. The next problem shows that the completeness property distinguishes the real number system from the rational number system.
Transcribed Image Text:Axiom 10.1.1. Nested Interval Property of the Real Number System (NIP). Suppose we have two sequences of real numbers (xn)and (yn) satisfying the following conditions: 1. x1 < x2 < x3 <... (this says that the sequence, (xn), is non-decreasing) 2. y1 2 Y2 > Y3 >... (this says that the sequence, (Yn), is non-increasing) 3. Vn, xn < Yn 4. lim (yn –- xn) = 0 n00 Then there exists a unique number c such that xn < c Yn for all n. Geometrically, we have the following situation. X1 X2 X3 Уз y2 y1 ... ... Notice that we have two sequences (x,) and (yn), one increasing (really non- decreasing) and one decreasing (non-increasing). These sequences do not pass each other. In fact, the following is true: Problem 10.1.2. Let (xn), (yn) be sequences as in the NIP. Show that for all п, т € N, хn < Ут- They are also coming together in the sense that limn+ (Yn – xn) = 0. The NIP says that in this case there is a unique real number c in the middle of all of this In <c< yn for all n e N. X1 X2 X3 Уз y2 y1 ... ... If there was no such c then there would be a hole where these two sequences come together. The NIP guarantees that there is no such hole. We do not need to prove this since an axiom is, by definition, a self evident truth. We are taking it on faith that the real number system obeys this law. The next problem shows that the completeness property distinguishes the real number system from the rational number system.
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