Math 312 Intro to Real Analysis Homework 1 Solutions Stephen G Simpson Wednesday January 21 2009 The assignment consists of Exercises 1 4 1 7 1 11 2 4 3 4 3 7 in the Ross textbook Each problem counts 10 points 1 4 a 1 1 1 3 4 1 3 5 9 1 3 5 7 16 This pattern suggests that the sum of the first n odd numbers is always n2 In other words 1 3 5 2n 1 n2 for all n and this is our guess b Let Pn be the proposition 1 3 5 2n 1 n2 We shall prove that Pn holds for all n For n 1 we have P1 because 1 1 If we assume that Pn holds for a particular n then for n 1 we have 1 3 5 2 n 1 1 1 3 5 2n 1 2n 1 n2 2n 1 in view of Pn n 1 2 so this gives us Pn 1 By the Induction Principle it follows that Pn holds for all n 1 7 Let Pn be the proposition that 7n 6n 1 is divisible by 36 The base of the induction is P1 which says that 7 6 1 0 is divisible by 36 which is obvious Next assuming Pn for a particular n we need to prove Pn 1 i e we need to prove that 7n 1 6 n 1 1 is divisible by 36 An algebraic manipulation shows that the expression in question is equal to 36n 7 7n 6n 1 Obviously the first term is divisible by 36 Moreover Pn implies that the second term is divisible by 36 Thus the whole expression is divisible by 36 We have now shown that Pn implies Pn 1 Therefore by the Induction Principle Pn holds for all n 1 11 a Assuming that n2 5n 1 is even we prove that n 1 2 5 n 1 1 is even An algebraic manipulation shows that the latter expression is equal to n2 5n 1 2 n 3 Since n2 5n 1 and 2 n 3 are even it follows that the whole expression is even Q E D b In actuality n2 5n 1 is never even 1 To see this consider two cases If n is even n2 and 5n are even hence n2 5n 1 is odd If n is odd n2 and 5n are odd hence n2 5n is even hence n2 5n 1 is odd In both cases n2 5n 1 is odd The moral is that in inductive proofs we cannot omit the base step If we omit the base step we may obtain incorrect conclusions Part a may be viewed as an inductive proof that n2 5n 1 is even for all n This conclusion is incorrect The proof was faulty because the base step n 1 was omitted 2 4 Let 5 3 1 3 Then 3 5 3 i e 3 5 3 hence 3 5 2 3 i e 6 10 3 22 0 By the Rational Zeros Theorem 2 2 page 9 the only candidates for a rational solution of the equation x6 10x3 22 0 are x 1 2 11 22 It is easy to check that none of these candidates is actually a solution of the equation For example with x 2 we have x6 10x3 22 64 10 8 22 6 6 0 Since is a solution but is not among the candidates for a rational solution it follows that is not rational 3 4 Part v By part iv we have 0 a2 for all a In particular with a 1 we have 0 12 By M3 12 1 1 1 Thus 0 1 One of our axioms stated in class was that 0 6 1 Hence 0 1 Q E D Part vii Assume 0 a b By O3 it follows that 0 b also Applying part vi to a and b it follows that 0 a 1 and 0 b 1 Multiplying the inequality a b by a 1 and using O5 we get 1 a 1 b Multiplying by b 1 and using O5 again we get b 1 a 1 This completes the proof Note Some of our axioms and parts of theorems were stated with but we have applied them with This is justified because in each case it is easy to prove that the version with implies the version with 3 7 a First assume b a By 3 2 i a b By 3 5 i b 0 hence b 0 b by 3 2 i Thus we have a b b a Since b is equal to either b or b it follows that a b a Now conversely assume a b a By 3 2 i we have a b a also Since b is equal to either b or b it follows that a b a This completes the proof b By part a we have a b c if and only if c a b c By O4 this is equivalent to b c a b c c We can redo parts a and b with replacing throughout Then the redone version of b says that a b c if and only if b c a b c 2