Math 61CM - Solutions to homework 1

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1 Math 61CM - Solutios to homework 1 Cédric De Groote October 1 st, 2018 Problem 1: Use mathematical iductio to check the validity of the formula j 3 = 2 ( + 1) 2 for = 1, 2,.... Note: The priciple of mathematical iductio is oe of the basic properties of the atural umbers N = {1, 2,...}, ad we shall use it frequetly; it says that if a subset P N has the properties (a) 1 P, ad (b) P + 1 P for each = 1, 2,..., the P = N. Slightly more cocretely, that is the same as sayig that if P is a (true or false) propositio for each = 1, 2,... ad if (a) P 1 is true, ad (b) for each we are able to check that P +1 is true if P is true, the P is true for all = 1, 2,.... Let P be the statemet j 3 = 2 ( + 1) 2. I case = 1 both sides of the idetity are equal to 1, so P 1 is true. +1 j 3 = j 3 + ( + 1) 3 ( 2 ( + 1) 2 ) = + ( + 1) 3 (assumig that P is true) = 2 ( + 1) 2 + ( + 1) 3 = ( + 1)2 ( + 2) 2. = ( + 1)2 ( ) Thus, P 1 is true ad P +1 is true if P is true, so P is true for all. Largely copy-pastig from solutios from previous TAs 1

2 Problem 2: Prove that (i) 28 is irratioal (i.e. ot ratioal); (ii) If x, y ratioal, x 0, ad z irratioal, the y + xz, y + x/z are both irratioal. (i) Suppose 28 = p q for some positive itegers p, q with o commo factors. 28 = p q 28q 2 = p 2 p 2 is divisible by 7. Sice 7 is prime, 7 must divide each factor of p 2, so p is divisible by 7, ad we ca write p = 7r for some iteger r. Therefore, 28q 2 = 9r 2 q 2 = 7r 2. This implies that q 2 is divisible by 7: 2 must divide q 2, so 7 divides either or q 2 because 7 is prime, but 7 does ot divide. So as before, q is also divisible by 7. Thus p, q have commo factor 7, a cotradictio to the fact that p ad q were chose to have o commo factors. (ii) Suppose y = k l, x = p q 0 with k, l, p, q itegers, l, p, q 0, ad suppose, cotrary to the claim, that y + xz = r s with r, s itegers, s 0. The z = q p ( r s k l ) = q(lr ks) psl, which is a ratioal, cotradictig the fact that z is irratioal. Similarly if y + x z = r s we get 1 z = q(lr ks) psl which gives i particular that lr ks 0 ad hece z = psl q(lr ks), agai a cotradictio. 2

3 Problem 3: By examiig the proof of the triagle iequality x + y x + y give i lecture (recall that proof bega with the idetity x + y 2 = x 2 + y 2 + 2xcdoty), prove that equality holds i the triagle iequality either at least oe of x, y is 0 or x, y 0 ad y = λx with λ > 0. The proof of the triagle iequality give i lecture/sectio was x + y 2 = x 2 + y 2 + 2x y x 2 + y x y x 2 + y x y = ( x + y ) 2 so equality holds (i.e. x + y = x + y ) if ad oly if equality holds i both the above iequalities. Thus x + y = x + y if ad oly if both x y = x y (i.e. x y 0) ad x y = x y (i.e. equality i the Cauchy-Schwarz iequality), thus equality implies both equality i the Cauchy Schwarz iequality ad also x y 0 so i particular ( ) x y = x y. Recall that i provig the Cauchy Schwarz iequality we established the idetity x 2 y 2 (x y) 2 = y 2 x x y y 2 y for y 0, so, for y 0, equality i the Cauchy Schwarz iequality implies x = y 2 (x y)y = y 1 x y by ( ). Thus we have show i particular that if x, y 0 ad if equality holds i the triagle iequality the x = λy with λ = x y > 0. Thus we have show that that fact that the equality holds i y 2 = x y the triagle iequality implies that either oe of x, y = 0 or else x = λy with λ > 0. O the other had if either oe of x, y = 0 or if x = λy with λ > 0 the equality trivially holds i the triagle iequality, so the required equivalece is proved. 2 Problem : (Aother proof of the Cauchy-Schwarz iequality.) If a = (a 1,..., a ), b = (b 1,..., b ) R, prove the idetity 1 2 i, (a ib j a j b i ) 2 = a 2 b 2 (a b) 2, ad hece prove a b a b. For each i, j = 1,...,, (a i b j a j b i ) 2 = a 2 i b 2 j + a 2 jb 2 i 2a i b i a j b j, hece i=1 (a i b j a j b i ) 2 = i = i ( a 2 i b 2) + i a 2 i j b 2 j + i ( b 2 i a 2) 2 i b 2 i j a 2 j 2 i a i b i j (a i b i (a b)) = 2( a 2 b 2 (a b) 2 ). a j b j Sice (a i b j a j b i ) 2 0 this shows that (a b) 2 ( a b ) 2 ; i.e., a b a b. 3

4 Problem 5: Usig the dot product, prove, for ay vectors x, y R : (a) the parallelogram law: x y 2 + x + y 2 = 2( x 2 + y 2 ); (b) the law of cosies: x y 2 = x 2 + y 2 2 x y cos θ, assumig x, y are o-zero, where θ is the agle betwee x ad y as discussed i lecture; (c) Give a geometric iterpretio of these iequalities (i.e. describe what (a) is sayig about the parallelogram determied by x, y i.e. the parallelogram OACB where OA = x, OB = y ad OC = x+y, ad what (b) is sayig about the triagle determied by x ad y i.e. the triagle OAB, where OA = x, OB = y). (a) (b) by defiitio of the agle θ. x y 2 + x + y 2 = (x y) (x y) + (x + y) (x + y) = x x 2x y + y y + x x + 2x y + y y = 2( x 2 + y 2 ) x y 2 = (x y) (x y) = x x + y y 2x y = x 2 + y 2 2 x y cos θ, (c) The poit (a) is sayig BA 2 + OC 2 = 2 OA 2 +2 OB 2 (= OA 2 + BC 2 + OB 2 + AC 2 ) sice OA = BC ad OB = AC. Thus we have show that the sum of squares of the legths of the diagoals of the parallelogram OACB is equal to the sum of squares of the legths of the sides, which is a classical theorem about parallelograms. The poit (b) is sayig AB 2 = OA 2 + OB 2 2 OA OB cos θ, where θ is the agle <AOB (i.e. the agle betwee the lie vectors OA ad OB). This is the classical Law of Cosies for the triagle OAB, ad shows that the formal defiitio of agle θ give i lecture does ideed coicide with our ituitive geometric uderstadig of agle.

5 Problem 6: Suppose V is a vector space over a field F. (a) Show that if X, Y are subspaces of V the so are X Y = {v : X + Y = {v V : x X, y Y s.t. v = x + y}. v X ad v Y } ad (b) Show that if a ij F, 1 i m, 1 j, the W = {x = (x 1,..., x ) F : i {1,..., m} a ij x j = 0} is a subspace of F. (c) Give a example of a vector space V, ad subspaces X, Y, such that X Y = {v V : v X or v Y } is ot a subspace of V. (a) We oly eed to show that (i) 0 X Y ad 0 X + Y, ad (ii) if v, w X Y ad λ, µ F the λv + µw X Y, while if v, w X + Y ad λ, µ F the λv + µw X + Y. Let s start with X Y : (i) Sice 0 X ad 0 Y as X, Y are subspaces of V, 0 X Y. (ii) Suppose v, w X Y ad λ, µ F. The v, w X ad v, w Y so as X, Y are subspaces of V, λv + µw X ad λv + µw Y, thus λv + µw X Y. This shows that X Y is a subspace of V. Now to X + Y : (i) sice 0 = 0 + 0, ad 0 X, 0 Y by virtue of X, Y beig subspaces of V, 0 ca be writte as the sum of two vectors, oe i X, oe i Y, so 0 X + Y. (ii) Suppose v, w X + Y ad λ, µ F. The there exist x, x X, y, y Y such that v = x + y, w = x + y, so λv + µw = λ(x + y) + µ(x + y ) = (λx + µx ) + (λy + µy ), ad ow o the right had side the first parethetical term is i X as X is a subspace ad the secod is i Y sice Y is a subspace. Thus, λv + µw is writte as the sum of two vectors, oe i X, oe i Y, so it is i X + Y. This shows that X + Y is a subspace of V. (b) It suffices to show that 0 = (0,..., 0) W ad if x, y W, λ, µ F the λx + µy W. But a ij 0 = 0 for all i, showig that 0 W. O the other had, if x, y W the with x = (x 1,..., x ), y = (y 1,..., y ), λx+µy = (λx 1 +µy 1,..., λx +µy ) by the defiitio of vector space operatios i F, so a ij(λx j + µy j ) = λ a ijx j + µ a ijy j = = 0, where the peultimate equality arose from x, y W, shows λx + µy W. This completes the proof that W is a subspace of F. (c) As the vector space, take V = F 2, ad let the two subspaces be X = {(x 1, x 2 ) : x 1 = 0}, Y = {(x 1, x 2 ) : x 2 = 0}. Sice x 1 = 0 is just the statemet 1 x x 2 = 0, by part (b) X is a subspace of F 2, ad similarly Y is a subspace of F 2. But X Y is ot: (1, 0) X X Y, (0, 1) Y X Y, but (1, 0) + (0, 1) = (1, 1) / X Y as either coordiate vaishes sice

6 Problem 7(a): (Simo s book, Exercise 1.1 page 89) Usig oly properties F1-F6, prove (i) a 0 = 0 a R; (ii) ab = 0 either a = 0 or b = 0; (iii) a b + c d = ad+bc bd a, b, c, d R with b 0, d 0. (i) Note that a 0 = a (0 + 0) by F, which is equal to a 0 + a 0 by F3. Addig by F5 the additive iverse of a 0, we get a 0 = 0. (ii) Assume that ab = 0, ad that both a ad b are o zero. The by F6 they have iverses a 1 ad b 1, both of which ca ot be zero because otherwise 1 = 0 by poit (i). Multiplyig the equality ab = 0 o both sides by a 1 (this uses F2 ad F6), we get b = 0 (agai, usig poit (i)). Multiplyig both sides by b 1 (agai, usig F2 ad F6), we get 1 = 0, which is ot possible. (iii) We have to prove that (ad + bc)(bd) 1 = ab 1 + cd 1. Notice that both (bd) 1 ad d 1 b 1 are iverses of bd, hece by the uicity of iverses (see i class) we have (bd) 1 = d 1 b 1. So (ad + bc)(bd) 1 = (ad + bc)(d 1 b 1 ), which by F2, F3 ad F6 simplifies to ab 1 + cd 1. Problem 7(b): (Simo s book, Exercise 1.2 page 90) Usig oly properties F1-F6 ad O1-O2 (ad the agreed termiology), prove the followig: (i) a > 0 0 > a, i.e. a < 0; (ii) a > 0 1 a > 0; (iii) a > b > 0 1 a < 1 b ; (iv) a > b ad c > 0 ac > bc. (i) If a > 0, the by O2 we would have 0 = a + ( a) > 0, a cotradictio. If a = 0, the a = 0, also a cotradictio. Therefore, by O1, we must have 0 > a. (ii) If 1 1 a < 0, the by O2 we would have 0 = a 0 > 0 a = 0, ot possible. If 1 a = 0, the 1 = a 1 a = 0 by problem 7(a)(i), which is ot possible. Therefore, by O1, we must have 1 a > 0. (iii) If a > b > 0, the by the previous questio we kow that 1 a ad 1 b are both > 0. Multiplyig a b > 0 by these two umbers ad usig O2, we get (a b) 1 1 a b > 0. Distributig (usig F1, F2 ad F6), this yields 1 b 1 a > 0, which is 1 a < 1 b. (iv) Similarly to the previous questio: a b > 0, hece (a b)c > 0 by O2, so ac bc > 0 usig F1, F2 ad F6. Therefore, ac > bc. 6

7 Problem 8: (Simo s book, Exercise 1.7 page 90) Prove that every positive real umber has a square root. (That is, for ay a > 0, prove there is a real umber α > 0 such that α 2 = a.) Let S = {x R : x > 0 ad x 2 < a}. To show that S is o-empty, simply take N such that > 1/a; the 2 > 1/a, hece a > 1/ 2 ad so 1/ S. To show that S is bouded above, we ote that a > 0 a < (a + 1) 2 ad x S = x 2 < a = x 2 < (a + 1) 2 = x < a + 1. Therefore, by the supremum axiom (axiom C), there exists a least upper boud α for S. If α 2 a, either α 2 > a or α 2 < a. If α 2 > a, the for N, we have (α 1 )2 = α α α 2 2α = a + (α 2 a) 2α. By takig large eough so that 2α < α2 a (i.e. > 2α α 2 a ), we have (α 1 )2 > a. By takig larger if ecessary, we ca guaratee 1 < α (i fact > max{ 2α α 2 a, α} 1 esures both iequalities are satisfied). Therefore, for all x S, x 2 < a < (α 1 )2 so, i fact, for all x S, α 1 > x. Therefore, α 1 is also a upper boud for S cotradictig the fact that α is the least upper boud. Similarly, if α 2 < a, the for N, we have (α + 1 )2 = α α α 2 + 2α + 1 = a + 2α + 1 (a α 2 ). By selectig such that 2α+1 < a α 2 (i.e. > 2α+1 ), we get (α + 1 a α 2 )2 < a. This implies α + 1 S, cotradictig that α is a upper boud for S. 7

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