Math 301 Problem Set 2 Solutions

Math 301 Problem Set 2 Solutions - Math 301/ENAS 513:...

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Math 301/ENAS 513: Homework 2 Solution Triet Le September 21, 2007 1. Show that for any reals x and y with x < y , there are a rational r and an irrational t such that x < r < y and x < t < y . Solution 1. Suppose that x > 0 . Let z = y - x > 0 and let r 0 > 0 be any rational number. By O5 , there exist a positive integer n such that r 0 < nz and r 0 < nx . This implies 0 < r 0 n < z and r 0 n < x . Now let m be the smallest positive integer such that mr 0 n > x . Thus ( m - 1) r 0 n x . We have, x < mr 0 n = ( m - 1) r 0 n + r 0 n x + r 0 n < x + z = y. Therefore, r = mr 0 n is a rational such that x < r < y . Replacing r 0 above by any positive irrational, say 2 , and follow the same construction as above, we would obtain an irrational t such that x < t < y . Now suppose that x 0 , we then replace x and y by x - m and y - m , where m is any integer less than x . Now 0 < x - m < y - m . Follow the same construction as above, we obtain a rational r and an irrational t such that x - m < r < y - m and x - m < t < y - m . Thus r + m and t + m are the rational and irrational that solve the original problem. 2. Show that the Archemedean axiom O5 follows from the Least Upper Bound Property O6 , together with the other axioms for the reals. Solution 2. Let a, b R be both positive. The the set A = { n N : n · a < = b } is either empty or bounded. If A is empty (when a > b ), pick any n N , we have b < n · a . If A is not empty (when a b ), then the least upper bound N = sup { A } exists and in N . Thus ( N + 1) · a > b . 3. True or False: For every ± > 0 there are positive integers m and n such that the inequality | n - m - π | < ± is true. 1
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Solution 3. Suppose n - m π . Let k = n - m . Then k = n - m = ( n - m )( n + m ) π ( n + m π ( π + 2 m ) = π 2 + 2 π m. Let k = [ π 2 + 2 π m ] to be the integer part of π 2 + 2 π m . Then we have | n - m - π | = ± ± ± ± q m + [ π 2 + 2 π m ] - m - π ± ± ± ± ± ± ± ± q ( m + π ) 2 + 1 - ( m + π ) ± ± ± ± = ± ± ± ± ± ² q ( m + π ) 2 + 1 - ( m + π ) ³ p ( m + π ) 2 + 1 + ( m + π ) p ( m + π ) 2 + 1 + ( m + π ) ± ± ± ± ± = 1 p ( m + π ) 2 + 1 + ( m + π ) Now for any ± > 0 , pick m large enough so that 1 ( m + π ) 2 +1+( m + π ) < ± . This implies that | n - m - π | < ±, with n = m + [ π 2 + 2 π m ] .
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Math 301 Problem Set 2 Solutions - Math 301/ENAS 513:...

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