Math 14A Elemetary Aalysis Homewor Questios 1 1 Itroductio 1.1 The Set N of Natural Numbers 1 Prove that 1 2 2 2 2 1 ( 1(2 1 for all atural umbers. 2 Prove that 3 11 (8 5 4 2 for all N. 4 (a Guess a formula for 1 3 (2 1 by evaluatig the sum for 1, 2, 3, ad 4. [For 1 the sum is simply 1] (b Prove your formula usig mathematical iductio. Prove that 11 4 is divisible by 7 for all N. 8 The priciple of mathematical iductio ca be exteded as follows. A list P m, P m1,... of propositios is true provided (i P m is true, (ii P 1 is true wheever P is true ad m. (a Prove that 2 > 1 for all itegers 2. (b Prove that! > 2 for all itegers 4. [Recall that! ( 1 2 1.] 1 Prove (2 1 (2 3 (2 5 (4 1 3 2 for all N. 11 For each N, let P deote the assertio 2 5 1 is a eve iteger. (a Prove that P 1 is true wheever P is true. (b For which is P actually true? What is the moral of this exercise? 12 For N, let! deote the factorial fuctio. Also let! 1 ad defie the biomial coefficiet!!(! for, 1,..., The biomial theorem asserts that, for all N, (a b a a 1 b 1 a 2 b 2 ab 1 2 1 (a Verify the biomial theorem for 1, 2, ad 3. (b Show that ( ( 1 (1 for 1, 2,...,. (c Prove the biomial theorem by mathematical iductio. b a b 1
1.2 The Set Q of Ratioal Numbers 1 Show that 3, 5, 7, 24, ad 31 are ot ratioal umbers. 2 Shwo that 2 1/3, 5 1/7, ad 13 1/4 are ot ratioal umbers. 3 Show that (2 2 1/2 is ot ratioal. 4 Show that (5 3 1/3 is ot ratioal. 5 Show that (3 2 2/3 is ot ratioal. Discuss why 4 7b 2 must be ratioal if b is ratioal. 1.3 The Set R of Real Numbers 1 (a Which of the Properties A1 A4, M1 M4, D, O1 O5 fail for N? (b Which of these properties fail for Z? 4 Prove ( ad (m of Theorem 1.8 from class ((v ad (vii of Theorem 3.2 i boo. (a Prove that a b c < a b c for all a, b, c R. Hit: Apply the triagle iequality twice. Do ot cosider eight cases. (b Use iductio to prove a 1 a 2 a a 1 a 2 a for ay a 1,..., a R. 7 (a Show that b < a a < b < a. (b Show that a b < c b c < a < b c (c Show that a b c b c a b c 8 Let a, b R. Show that if a b 1 for every b 1 > b, the a b. 2
1 Itroductio Math 14A Elemetary Aalysis Homewor Aswers 1 1.1 The Set N of Natural Numbers 1 Let A N be the set of such that 1 2 2 2 2 1 ( 1(2 1 The 1 A, sice 1 1 1 (1 1 (2 1 1. Now suppose A for some. The 1 2 2 2 2 ( 1 2 1 1 ( 1(2 1 ( 12 ( 1 (2 1 ( 1 whece 1 A. Therefore A N ad we have the result. 2 Let P be the statemet: 3 11 (8 5 4 2 P 1 is simply 3 4 1 2 1, which is clearly true. Now suppose P is true for some. The 1 ( 1 ((2 1 ( 1 1 ( 1(22 7 1 ( 1( 2(2 3 1 ( (( ( 1 1 1 2( 1 1, 3 11 (8 5 (8( 1 5 4 2 8 3 4 2 7 3 ( ( 1(4 3 ( 1 4( 1 1 4( 1 2 ( 1, whece P 1 is true. Therefore P is true for all by iductio. 4 (a 1 gives 1 2 gives 4 3 gives 9 4 gives 1 It therefore appears that 1 3 (2 1 2. (b Let P be 1 3 (2 1 2. P 1 is true, as checed i part (a. Suppose P is true for some. The 1 3 (2 1 (2( 1 1 2 2 1 ( 1 2, whece P 1 is true. Therefore P is true for all by iductio. 3
Let A { N : 11 4 is divisible by 7}. Clearly 11 1 4 1 7, so 1 A. Now suppose A for some. The 11 7 4λ for some iteger λ. Thus 11 1 7 1 11 11 7 1 11(4 7 7 1 11 4 7 (11 7 4(11 7, which is clearly divisible by 4. Hece 1 A. Thus A N ad the claim is true by iductio. 8 (a P 2 is true, sice 2 2 > 2 1. Assume P is true for some 2. The ( 1 2 ( 1 2 2 1 1 2 > 1 >, whece ( 1 2 > 1 ad so P 1 is true. Therefore 2 > for all 2 by iductio. (b P 4 is true, sice 4! 24 > 1 4 2. Assume P is true for some 4. The ( ( 1! ( 1 2 ( 1! ( 1 > ( 1( 2 1. Usig the quadratic formula we see that the RHS is positive for all > 1 5 2. I particular, for 4. Thus ( 1! > ( 1 2 ad so P 1 is true. Therefore! > 2 for all 4 by iductio. 1 Could do this by iductio, or alteratively use the aswer to questio 4: [ ] (2 1 (2 3 (4 1 1 3 (2(2 1 1 3 (2 1 as required. 11 (a If 2 5 1 is eve, the (2 2 2 3 2 ( 1 2 5( 1 1 2 2 1 5 5 1 ( 2 5 1 2, which is clearly eve. Thus P P 1. (b P is ever true: 2 5 1 ( 5 1 which is always odd! Sice P 1 is false, we caot use iductio to claim that P 2 is true, etc. Iductio argumets always require the first step to be explicitly checed! 12 (a 1: (a b 1 a b ( 1 a (1 1 b. 2: (a b 2 a 2 2ab b 2 ( 2 a2 ( 2 1 ab (2 2 b2. 3: (a b 3 a 3 3a 2 b 3ab 2 b 3 ( 3 a3 ( 3 1 a2 b ( 3 2 ab2 ( 3 3 b3. (b ( as required. ( 1!!(!! ( 1!( 1!! ( 1!(!! ( 1!(! 1 ( 1! ( 1!(! ( 1! 1!( 1!, 4 [ 1 1 1 ( 1 1 ]
(c Let A N be the set of for which the biomial theorem holds. Clearly 1 A, by part (a. Assume A for some. The, (a b 1 (a b(a b (a b a b a 1 b a b 1 a 1 b 1 a 1 j b j (Let j 1 j 1 j1 a 1 a 1 b a 1 b 1 1 1 a 1 [ ] 1 a 1 b b 1 1 1 1 a 1 1 1 a 1 b b 1 1 1 1 a 1 b. 1 Thus 1 A ad so A N, whece the biimial theorem holds by iductio. 1.2 The Set Q of Ratioal Numbers 1 b 1 (j 1 1 The relevat polyomials are x 2 3, x 2 5, x 2 7, x 2 24, ad x 2 31. Appealig to the Ratioal Roots Theorem, if x p q is a solutio, the, i all cases q 1, while ad p divides the costat coefficiet. It is easy to see that o such p s exist. 2 2 1/3 satisfies x 3 2 whece, if x p q is a ratioal solutio, the q 1 ad p ±1, ±2. Noe of the four combiatios satisfy the polyomial, hece there are o ratioal solutios. 5 1/7 ad 13 1/4 are similar with x 7 5 ad x 4 13 respectively. 3 (2 2 1/2 satisfies x 2 2 2, or x 4 4x 2 2. If x p q is a ratioal solutio, the q 1 ad p ±1, ±2, oe of which satisfy the polyomial. Alteratively, ote that if x was ratioal, the x 2 2 would be, ad so 2 would have to be ratioal. A cotradictio. 4 (5 3 1/3 satisfies x 3 5 3. If x was ratioal, the x 3 5, ad thus 3, would be ratioal. A cotradictio. 5 (3 2 2/3 satisfies x 3 (3 2 2 11 2, or equivaletly x3 11 2. If x was ratioal, the x3 11, ad thus 2, would be ratioal. A cotradictio. Suppose that b p q is ratioal. The 4 7b 2 4 7p2 q 2 4q2 7p 2 q 2 5
is ratioal. Alteratively, the set Q is closed uder additio ad multiplicatio (ay sums ad multiples of ratioal umbers are ratioal, hece 4 7b 2 must be. 1.3 The Set R of Real Numbers 1 (a A4 fails: e.g. 3 / N. M4 also fails: e.g. 3 1 1 3 / N. Strictly A3 also fails, sice / N, this axiom is meaigless (this is very picy!. (b Oly M4 fails for Z. 4 ( By (j, < 1 2 1 1 1. (m By (l, the iequalities ivolvig zero are clear. Now suppose that < a < b but that a 1 b 1. The, multiplyig though by ab > we see that a 1 ab b 1 ab b a, a cotradictio. (a a b c (a b c a b c a b c (b Let P be: a 1 a a 1 a for all choices of real umbers a 1,..., a. The P 1 reads a 1 a 1, which is trivially true. Suppose P is true for some. The a 1 a 1 (a 1 a a 1 a 1 a a 1 a 1 a a 1 (Usual -iequality (Sice P true Hece P 1 is true, ad so P is true for all by iductio. 7 (a b < a a >. Now cosider two cases. If b, the b < a b < a. Sice a >, clearly also have a < b. If b, the b < a b < a a < b. Sice a >, clearly also have b < a. Puttig both halves together gives the result. (b By part (a, a b < c c < a b < c b c < a < b c. (c The iequality part is covered i (b. For equality, a b c c ±(a b, i.e. c a b, or c b a. Equivaletly, a b c or a b c, as required. 8 Suppose, for cotradictio, that a b 1 for all b 1 > b, ad that a > b. But the cosider b 1 ab 2. Beig half way betwee a ad b, we have b < b 1 < a, which cotradicts a b 1. Thus a b as required.