Homework 1 Solutions. The exercises are from Foundations of Mathematical Analysis by Richard Johnsonbaugh and W.E. Pfaffenberger.

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1 Homewor 1 Solutios Math 171, Sprig 2010 Hery Adams The exercises are from Foudatios of Mathematical Aalysis by Richard Johsobaugh ad W.E. Pfaffeberger Let h : X Y, g : Y Z, ad f : Z W. Prove that (f g h = f (g h. Solutio. Let x X. Note that ((f g h(x = (f g(h(x = f(g(h(x = f((g h(x = (f (g h(x Sice this is true for all x X, we have (f g h = f (g h Let f : X Y ad A X ad B Y. Prove (a f(f 1 (B B ad (b A f 1 (f(a. Solutio. (a By defiitio, f 1 (B = {x X f(x B}. So if y f(f 1 (B, the y = f(x for some x f 1 (B, that is, f(x B. Hece y = f(x B, so f(f 1 (B B. (b If x A, the f(x f(a, ad so x f 1 (f(a. Hece A f 1 (f(a Prove that ( x = x for all x R. Solutio. Note that x + ( x = 0 = ( x + x, by defiitio of ( x. By the uiqueess i Axiom 5, x is the additive iverse of ( x, that is, ( x = x Let x, y R. Prove that xy = 0 if ad oly if x = 0 or y = 0. Solutio. To show, suppose that either x = 0 or y = 0. The, by Theorem 3.4, xy = 0. To show, let xy = 0. Suppose for a cotradictio that x 0 ad y 0. Sice x ad y are ozero, by Axiom 10 their multiplicative iverses x 1 ad y 1 exist. Hece we have 0 = 0(y 1 x 1 = (xy(y 1 x 1 = xx 1 = 1 This cotradicts Axiom 9, which says 0 1. Hece it must be the case that either x = 0 or y = 0. This shows. We have xy = 0 x = 0 or y = Prove that if xy > 0, the either x > 0 ad y > 0 or x < 0 ad y < 0, x, y R. Solutio. Let xy > 0. By Exercise 3.5, x 0 ad y 0. If x > 0 ad y < 0, the Theorem 4.2 says xy < 0y = 0, a cotradictio. If x < 0 ad y > 0, the Theorem 4.2 says 0 = 0y > xy, a cotradictio. Hece it must be the case that x > 0 ad y > 0 or x < 0 ad y < Prove that x 2 + y 2 2xy for all x, y R. Solutio. Note (x y 2 0. Expadig, we get x 2 2xy + y 2 0. Addig 2xy to both sides, we get x 2 + y 2 2xy. 1

2 5.1. Let X be a set of real umbers with least upper boud a. Prove that if ɛ > 0, there exists x X such that a ɛ < x a. Solutio. Suppose for a cotradictio that there is o such x. The a ɛ is a upper boud for X, ad a ɛ < a. This cotradicts the fact that a is the least upper boud for X: see part (ii of Defiitio 5.2. Hece there must exist some x X such that a ɛ < x a Let X ad Y be sets of real umbers with least upper bouds a ad b, respectively. Prove that a + b is the least upper boud of the set X + Y = {x + y x X, y Y }. Solutio. First ote that if z X + Y, the z = x + y with x X ad y Y, so z = x + y a + b. Hece a + b is a upper boud for X + Y. Next, let c be ay upper boud for X + Y. Suppose for a cotradictio that c < a + b. Let ɛ = a + b c > 0. By Exercise 5.1, there exists x X such that a ɛ 2 < x, ad there exists y Y such that b ɛ 2 < y. Hece c = a + b ɛ = (a ɛ 2 + (b ɛ 2 < x + y, cotradictig the fact that c is a upper boud for X + Y. So it must be that a + b c. Thus a + b is the least upper boud of X + Y Prove the biomial theorem: If a ad b are real umbers ad is a positive iteger, the ( ( ( ( (a + b = a a b b = a b 0 =0 where ( =!!(!. Solutio. This is a proof by iductio o. The base case whe = 1 is clear, as ( 1 0 = 1 = ( 1 1. For the iductive step, assume it holds for m. The for = m + 1, (a + b m+1 = (a + b(a + b m = (a + b(a m + b m + =1 ab m + ba m + m =1 m =1 m [( m + =1 m + 1 =1 m+1 ( m + 1 = a m+1 b =0 ( m a m b by the iductive hypothesis =1 a m +1 b + a m +1 b + ( m a m +1 b + =0 m j=1 j 1 =1 ( m a m b +1 a m b +1 by combiig terms a m+1 j b j let j=+1 i secod sum ( ] m a m+1 b by combiig the sums 1 a m+1 b from Pascal s rule as desired. Hece the biomial theorem holds for all positive itegers Prove that if X is a oempty subset of positive itegers which is bouded above, the X cotais a greatest elemet. 2

3 Solutio. Sice X is bouded above, there exists b P such that x < b for all x X. Hece b x P for all x X. By Theorem 6.10, there exists a least elemet i the oempty set {b x : x X} of positive itegers. That is, there exists some y X such that b y b x for all x X. Subtractig b from both sides ad multiplyig by 1, we get that y x for all x X. Hece X cotais a greatest elemet y Prove the laws of iteger expoets Solutio. (a x m+ = x m x, x R, x 0, m, Z Defiitio 7.3 tells us that x +1 = xx (ad hece x 1 x +1 = x for all P. First we show this is true for all Z. The cases = 0, 1 are clear. Also, if < 1, the we have x +1 = 1 x 1 = 1 x 1 x = 1 1 x 1 = xx x Hece x +1 = xx for all Z. Now, let m Z be arbitrary. We will show that x m+ = x m x by iductio o. The base case = 0 is clear. Suppose the formula is true for. The x m++1 = xx m+ = xx m x = x m x +1 so the formula is true for + 1. This shows the formula is true for all N. Suppose agai the formula is true for. The x m+ 1 = x 1 x m+ = x 1 x m x = x m x 1 so the formula is true for 1. Hece the formula is true for all Z. (b x = 1 x, x R, x 0, Z. If 0, the this is Defiitio 7.3. If > 0, the ote 1 x = 1 1 x = x. (c (xy = x y, x, y R, x 0 y, Z. We prove the formula by iductio o. The base case = 0 is clear. Suppose the formula is true for. The (xy +1 = (xy(xy = xyx y = x +1 y +1 so the formula is true for + 1. This shows the formula is true for all N. Suppose agai the formula is true for. The (xy 1 = (xy 1 (xy = x 1 y 1 x y = x 1 y 1 so the formula is true for 1. Hece the formula is true for all Z. (d (x m = x m, x R, x 0, m, Z Let m Z be arbitrary. We prove the formula by iductio o. The base case = 0 is clear. Suppose the formula is true for. The (x m +1 = (x m (x m = x m x m = x m+m = x m(+1 so the formula is true for + 1. This shows the formula is true for all N. Suppose agai the formula is true for. The (x m 1 = (x m 1 (x m = x m x m = x m+m = x m( 1 so the formula is true for 1. Hece the formula is true for all Z. (e ( x y = x y, x, y R, x 0 y, Z. This follows by replacig y with y 1 i (b. (f If 0 < x < y, the x < y, P 3

4 We prove the formula by iductio o. The base case = 1 is clear. Suppose the formula is true for. The x +1 = xx < xy < yy = y +1 so the formula is true for + 1. This shows the formula is true for all P. (g If < m ad x > 1, the x < x m,, m Z. Let Z be arbitrary. We prove the formula by iductio o m. The base case m = + 1 is clear, as 1 < x implies x < xx = x +1 = x m Suppose the formula is true for m. The x < x m < xx m = x m+1 so the formula is true for m + 1. This shows the formula is true for all m > Prove that o equilateral triagle i the plae ca have all vertices with ratioal coordiates. Solutio. Suppose for a cotradictio that such a triagle exists. Traslate the triagle so that oe vertex is at the origi; ote the ew vertices still have ratioal coordiates. Scale the triagle by the commo deomiator of the ew ratioal coordiates to produce a equilateral triagle with iteger coordiates. Let these coordiates be (0, 0, (a, b, ad (c, d. Let s be the commo legth of each side. We have Expadig the right had side gives so 2(ac + bd = s 2. So Expadig gives s 2 = a 2 + b 2 = c 2 + d 2 = (a c 2 + (b d 2 s 2 = (a 2 + b 2 + (c 2 + d 2 2(ac + bd 4(ac + bd 2 = s 4 = (a 2 + b 2 (c 2 + d 2 4(ac 2 + 8abcd + 4(bd 2 = (ac 2 + (ad 2 + (bc 2 + (bd 2 Cacelig lie terms ad rearragig gives 3(ac + bd 2 = (ad bc 2 The iteger o the left had side has a odd umber of threes i its prime factorizatio, whereas the iteger o the right has a eve umber of threes, which cotradicts uique factorizatio of itegers ito primes. Hece o equilateral triagle i the plae ca have all vertices with ratioal coordiates. (Note: this argumet usig uiqueess of prime factorizatio to show a cotradictio above is essetially the same as the argumet showig 3 is irratioal Prove that every ifiite set has a coutably ifiite subset. Solutio. Let X be a ifiite set. Sice X is oempty, there exists some x 1 X. Note that X \ {x 1 } is oempty, for otherwise X would be a fiite set. Hece there exists some x 2 X \ {x 1 }. For the iductive step, suppose that we have chose {x 1,..., x } X (with the x i disjoit. Note that X \ {x 1,..., x } is oempty, for otherwise X would be a fiite set. Hece there exists some x +1 X \{x 1,..., x }. By iductio, we have costructed a coutably ifiite subset {x 1, x 2, x 3,...} of X Let X be a set. Let S be the set of all fuctios from X ito {0, 1}. Prove that S P (X. How does this exercise coect Example 8.3 ad Corollary 8.5? Solutio. Defie f : S P (X as follows. Give s S, let f(s = {x X : s(x = 1}. Note that f is oe-to-oe, for if f(s = f(s, the {x X : s(x = 1} = {x X : s (x = 1}, ad 4

5 so s = s. Also, f is oto, for if Y P (X, the defie s Y S by s Y (x = 1 if x Y ad s Y (x = 0 otherwise, ad ote that f(s Y = Y. So f is a oe-to-oe fuctio from S oto P (X, givig S P (X. Let X = P, so S is the set of all fuctios from P ito {0,1}, that is, the set of all real sequeces which assume oly the values 0 or 1. By this exercise, S P (P. So S is ucoutable if ad oly if P (P is ucoutable. Hece, Example 8.3 ad this exercise imply Corollary 8.5. Vice-versa, Corollary 8.5 ad this exercise imply Example

2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.

2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F. CHAPTER 2 The Real Numbers 2.. The Algebraic ad Order Properties of R Defiitio. A biary operatio o a set F is a fuctio B : F F! F. For the biary operatios of + ad, we replace B(a, b) by a + b ad a b, respectively.