Math 312 Intro to Real Analysis Midterm Exam 1 Solutions Stephen G Simpson Friday February 13 2009 1 True or False 3 points each a Every ordered field has the Archimedean property Answer False b The ordered field axioms imply a b a b for all a b Answer True c If lim an then lim sup an Answer True d For any sequence of real numbers the lim inf and the lim sup always exist and furthermore the lim inf is always the lim sup Answer True e The equation 3x3 2x2 3x 2 0 has a rational solution Answer True A rational solution is x 2 3 f 3 216 is an irrational number Answer False 3 216 6 g The limit of a convergent sequence of negative numbers is negative Answer False For example the sequence 1 n for n 1 2 3 has 0 as its limit h The limit of a convergent sequence of rational numbers is rational Answer False i Every interval contains at least three rational numbers Answer True j Every bounded sequence of real numbers is convergent Answer False k Every convergent sequence of real numbers is bounded Answer True l Every monotone sequence of real numbers is convergent Answer False For example the sequence 1 2 3 is divergent to 1 m If an is a monotone sequence of real numbers then lim an exists and belongs to the interval Answer False Same example as above 2 7 points each a Give an example of a sequence of real numbers such that inf an lim an sup an Answer An example is an and the limit is 0 1 n The inf is 1 the sup is 1 2 n b Give an example of a sequence of real numbers such that lim sup an lim inf an sup an inf an are four distinct real numbers 1 The inf is 2 the sup Answer An example is an 1 n 1 n is 1 1 2 the lim inf and lim sup are 1 and 1 c Give an example of a sequence of real numbers such that lim inf an and lim sup an 2 Answer An example is 1 2 2 2 3 2 4 In other words a2n 2 and a2n 1 n for all n p 3 3 8 points It can be shown that 1 5 is an algebraic number i e it is a solution of some polynomial equation with integer coefficients Find such an equation p 3 1 5 Then 3 1 5 i e 3 1 5 Answer Let hence 6 2 3 1 5 i e 6 2 3 4 0 The desired equation is x6 2x3 4 0 4 8 points Find all candidates for rational solutions of the equation 2x2 ax 5 0 where a is an unspecified integer 5 1 Answer The candidates are 5 1 This is according to the 2 2 Rational Zeros Theorem page 9 in the Ross textbook 5 12 points Use algebra plus limit laws to calculate 2n2 5n lim n 4 2 Answer p 2 5n lim 2n n 4 r 5 2 n lim 4 1 n r 5 lim 2 n 4 lim 1 n 2 0 1 0 5 n 4 lim 1 n r 5 2 lim n 4 1 lim n 2 lim r 2 6 12 points It can be shown that 3 n 5001 1 lim 3 n 1001 Given 0 find an N so large that 3 n 5001 1 3 n 1001 holds for all n N Answer Given 0 we want 3 n 5001 3 n 1001 and by algebra this is equivalent to saying that 4001 3 n 1001 If n 10013 then 3 n 1001 0 and 4001 4001 so the above is equivalent to saying that 4001 3 n 1001 By algebra this is equivalent to 4001 1001 3 n i e n N 4001 1001 3 3 Thus we may take 4001 1001 3