Math 312, Intro. to Real Analysis: Midterm Exam #1 Solutions

1 Math 312, intro . to real Analysis: Midterm Exam #1 SolutionsStephen G. SimpsonFriday, February 13, 20091. True or False (3 points each)(a) Every ordered field has the Archimedean :False.(b) The ordered field axioms imply|a b| |a|+|b|for alla, :True.(c) If liman= then lim supan= .Answer:True.(d) For any sequence of real numbers, the lim inf and the lim sup alwaysexist and furthermore the lim inf is always the lim :True.(e) The equation 3x3+ 2x2+ 3x+ 2 = 0 has a rational :True. A rational solution isx= 2/3.(f)3 216 is an irrational 216 = 6.(g) The limit of a convergent sequence of negative numbers :False. For example, the sequence 1/nforn= 1,2,3, ..has 0 as its limit.(h) The limit of a convergent sequence of rational numbers :False.

2 (i) Every interval contains at least three rational :True.(j) Every bounded sequence of real numbers is :False.(k) Every convergent sequence of real numbers is :True.(l) Every monotone sequence of real numbers is :False. For example, the sequence 1,2,3, ..is divergent to .1(m) If (an) is a monotone sequence of real numbers, then limanexistsand belongs to the interval ( , ).Answer:False. Same example as (7 points each)(a) Give an example of a sequence of real numbers such that <infan<liman<supan< .Answer:An example isan=( 1)nn. The inf is 1, the sup is 1/2,and the limit is 0.(b) Give an example of a sequence of real numbers such thatlim supan,lim infan,supan,infanare four distinct real :An example isan= ( 1)n(1 +1n).

3 The inf is 2, the supis 1 + 1/2, the lim inf and lim sup are 1 and 1.(c) Give an example of a sequence of real numbers such thatlim infan= and lim supan= :An example is 1, 2, 2, 2, 3, 2, 4, .. In otherwords,a2n= 2 anda2n 1= nfor (8 points) It can be shown that3 1 + 5 is analgebraic number, , itis a solution of some polynomial equation with integer coefficients. Findsuch an :Let =3 1 + 5. Then 3= 1 + 5, , 3 1 = 5hence 6 2 3+ 1 = 5, , 6 2 3 4 = 0. The desired equation isx6 2x3+ 4 = (8 points) Find all candidates for rational Solutions of the equation2x2 ax+ 5 = 0whereais an unspecified :The candidates are 5, 52, 1, 12. This is according to theRational Zeros Theorem, page 9 in the Ross (12 points) Use algebra plus limit laws to calculatelim 2n2+ 5nn+ :lim 2n2+ 5nn+ 4= lim 2 +5n1 +4n=lim 2 +5nlim 1 +4n= lim 2 +5nlim 1 +4n= 2 + lim5n1 + lim4n= 2 + 01 + 0= (12 points) It can be shown thatlim3 n 50013 n 1001= >0, find anNso large that 3 n 50013 n 1001 1 < holds for alln > :Given >0, we want 3 n 50013 n 1001 < and by algebra this is equivalent to saying that 40013 n 1001 <.

4 Ifn >10013then3 n 1001>0, and| 4001|= 4001, so the above isequivalent to saying that40013 n 1001< .By algebra, this is equivalent to4001 + 1001<3 ,n >(4001 + 1001) we may takeN=(4001 + 1001)