This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Honors Introduction to Analysis I Homework III Solution February 15, 2009 Problem 1 Let x be a real number. Show that there exists a Cauchy sequence of rationals { x n } representing x such that x n ≤ x n +1 for every n . Solution. We will build two sequences, one increasing, and one decreasing, proving a more general result. If x ∈ Q , consider the constant sequence x n = x , for any n . If x / ∈ Q , choose rational numbers y 1 and z 1 such that y 1 < x < z 1 (we can do this by the completeness of the reals and the density of rationals). Consider m 1 = y 1 + z 1 2 , which is rational. If x > m 1 , define y 2 = m 1 and z 2 = z 1 , and if x < m 1 , define y 2 = y 1 and z 2 = m 1 . Next, we iterate, considering m 2 = y 2 + z 2 2 , and take y 3 = m 2 and z 3 = z 2 if m 2 < x or y 3 = y 2 and z 3 = m 2 if m 2 > x . Similarly we get y 4 ,z 4 from y 3 ,z 3 and so on. We repeat the process of dividing infinitely often, and we will obtain two sequences: { y k } (increasing) and { z k } (decreasing), so that y k ≤ x ≤ z k and z k y k = z 1 y 1 2 k 1 . Both are equivalent Cauchy sequences: given 1 /n , take m large enough so that z 1 y 1 2 m 1 < 1 /n . Then 0 ≤ z m y m < 1 /n and for k ≥ m , [ y k ,z k ] ⊂ [ y m ,z m ], so  y j y k  < 1 /n ,  z j z k  < 1 /n and  y j z k  < 1 /n , for j,k ≥ m . Since the sequences are equivalent Cauchy, they will have a common limit, call it y . But since y k ≤ x ≤ z k , then taking the limit will yeald: y ≤ x ≤ y , so x is their common limit. So we have build an increasing sequence { y n } and a decreasing sequence { z n } , both representing x . Problem 2 Prove that there are an infinite number of rational numbers in between any two distinct real numbers. Solution. Let x < y be the two real numbers. By the Archimedean property, there is an m ∈ N such that y x > 1 /m . By the density of the rationals, there is a rational r 1 such that  r 1 x  < 1 / (3 m ) and r 1 > x (we can make sure r 1 > x by an argument similar to the previous problem) and s 1 such that s 1 < x and  y s 1  < 1 / (3 m ). By the first condition, we know that x < r 1 < s 1 < y . But between any two rationals (in particular for r 1 and s 1 ), there are infinitely many rationals (we can consider a sequence { r k } , defined such that r...
View
Full
Document
This note was uploaded on 09/22/2010 for the course MATH 413 at Cornell.
 '08
 PROTSAK

Click to edit the document details