Solutions to Math 347 Practice Problems for the final
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1 Solutios to Math 347 Practice Problems for the fial 1) True or False: a) There exist itegers x,y such that 50x + 76y = 6. True: the gcd of 50 ad 76 is, ad 6 is a multiple of. b) The ifiimum of a set is at most as large as the supremum of the set. True: for all x S, we have if(s) x sup(s). c) The sum of two Cauchy sequeces is a Cauchy sequece. True: we proved this about coverget sequeces, ad a sequece is Cauchy iff it is coverget. d) If oe could fid a algorithm to factor large umbers, oe could crack RSA. True: RSA relies o the fact that someoe ca publicly aouce the product of two large primes without ayoe beig able to figure out the actual primes, so if there were a fast algorithm to factor large umbers, RSA would be cracked. e) If A is a proper subset of B, ad A ad B have the same cardiality, the A ad B must both be ifiite. True: fiite sets have the same cardiality iff they have the same umber of elemets. f) A mootoe subsequece of a bouded sequece coverges. True: every subsequece of a bouded sequece is itself bouded, ad bouded mootoe sequeces coverge. g) The remaider of a 0 divided by b > 0 is the smallest oegative value of a bn where N is a iteger. True: proved i class. h) If s ad t diverge, the s t diverges as well. False: cosider s = t = 1. ) Let b 1,b,...,b be itegers such that b i 0 for some i. We say a is a commo divisor of (b 1,b,...,b ) if a b i for all 1 i. Just as it made sese to defie the gcd of two itegers i class, it makes sese to defie the greatest commo divisor g =gcd(b 1,b,...,b ) as the greatest amog commo divisors of (b 1,b,...,b ). Show that there exist itegers x 1,x,...,x such that g = b 1 x 1 + b x + + b x. We prove this by iductio. The base case is =, which we proved i class. Now, for the iductio hypothesis, suppose that for some we have that for ay -tuple of itegers b 1,...,b (ot all 0) there exist itegers x 1,...,x such that b 1 x b x = gcd(b 1,...,b ). Cosider ow a + 1-tuple of itegers c 1,...,c +1 where, without loss of geerality, c 1 0 (ad hece gcd(c 1,c ) 0). Note that gcd(c 1,...,c +1 ) = gcd(gcd(c 1,c ),c 3,...,c +1 ). By the iductio hypothesis, we have that there exist itegers x 1,...,x such that gcd(c 1,c ) x 1 +c 3 x + +c +1 x = gcd(gcd(c 1,c ),c 3,...,c +1 ). Furthermore, by the base case, we ca write gcd(c 1,c ) = c 1 y 1 + c y for some itegers y 1,y. Combiig these last two equalities, we get that c 1 y 1 x 1 + c y x 1 + c 3 x + + c +1 x = gcd(gcd(c 1,c ),c 3,...,c +1 ) = gcd(c 1,...,c +1 ), where y 1 x 1,y x 1,x,...,x Z. So, by iductio, the statemet is true.
2 3) a) Show that the umber is irratioal. Suppose Q. The, sice Q is a field, we have that the square of this expressio, Q. Agai, sice Q is a field, we have that subtractig 1 will keep this expressio i Q ad so Q. Squarig ad subtractig 1 agai, we have that 1 + Q, ad squarig ad subtractig 1 agai we have that Q, which we have prove is ot true i class. Hece our origial claim is false ad so is irratioal. b) Show that the umber is ratioal. Cosider the square of this umber. ( ) = ( )(7 4 3) = = 16. There are oly two umbers whose square is 16, ad these two umbers are ±4, which are both ot oly ratioal but itegers. So the umber must be either 4 or 4 ad is thus ratioal. 4) Cosider the set R of real umbers, where multiplicatio ad additio are defied as follows. We defie x + y = x y, where x y is obtaied usig the usual rules of additio ad multiplicatio i the reals; ad we defie x y as it is usually defied i R. Does this make up a field? No, this does ot make up a field. For example, the commutative property of additio fails: for example, 1 + = = = + 1 i this setup. 5) Prove that the sum of the first odd atural umbers is. Fid the sum of the first eve atural umbers ad prove it. We prove the first statemet by iductio. If = 1, the the sum of the first atural umbers is 1 = 1 =. Now suppose that for some 1 we have that =. The the sum of the first + 1 atural umbers is = = ( + 1). Hece, by iductio, the statemet is true. To fid the sum of the first eve atural umbers, ote that the differece betwee the sum of the first eves ad the first odds is ( 1) = times. Hece the differece is, ad by the previous prove claim we have that the sum of the first eve atural umbers is +. 7) Does the sequece a = ( 1) ( 1)( 1)
3 coverge? Note that a = ( 1) ( 1)( 1) = ( 1) ( 1) 1 = ( 1) 1 = 1 if is odd, ad that a = ( 1) ( 1)( 1) = ( 1) ( 1)1 = ( 1) 1 = 1 if is eve. So this is a costat sequece ( 1) which we have prove coverges o a homework assigmet. 8) Suppose give a collectio S ( = 1,, ) of fiite sets of real umbers. Is it ecessarily true that sup{ifs } if{sups } No, this is ot true. Cosider the collectio of sets S = {3,4} for odd, ad S = {1,} for eve. The {ifs } is a set cotaiig 1 s ad 3 s, ad the supremum of such a set is 3. However, {sups } is a set of s ad 4 s, ad so the supremum of this set is 4, ad ) a) Fid the multiplicative iverse of 1 mod 65. I order to do this, we use the Euclidea algorithm to fid X ad Y such that 1X + 65Y = 1, ad X will be the multiplicative iverse we are after. The Euclidea algorithm goes as follows: 65 = = = + 1 = 1 So, we get that gcd(1,65) = 1 ad 1 = 5 = (1 5 ) = ( ) = So the multiplicative iverse of 1 (mod 6)5 is 7, or 38 mod 65. b) Does 6 have a multiplicative iverse mod 83573? Justify your aswer. No: sice the sum of the digits of is 30, which is divisible by 3, we have that 6 ad are ot relatively prime, which is a requiremet for 6 to have a iverse mod c) Fid the greatest commo divisor of 345 ad 134 usig the Euclidea algorithm. The Euclidea algorithm goes as follows: 345 = = 67
4 So the gcd is ) Show that the sets S = { + 1 Z} ad T = {4 + 1 Z} {4 1 Z} are equal. We first show S T. Suppose x S, so x = + 1 for some Z. If is eve, the x = (k) + 1 = 4k + 1 T. If is odd, the x = (k 1) + 1 = 4k 1 T. So S T. Now we show T S. Suppose x T, so x = or x = 4 1 for some Z. I the first case, we have x = () + 1 S, ad i the secod case we have x = ( 1) + 1 S. So T o f S, ad, together with S T from above we have S = T. 11) Prove that the sequece si k coverges for ay k > 0. We use the Squeeze Lemma. Sice 1 si 1, we have that 1 k coverge to 0 as. So, by the Squeeze Lemma, we have that si k 1) a) What is the supremum of the set { si si k 1 k, where both 1 k coverges to 0 as well. N}? Justify (prove) your aswer. The maximum, ad hece supremum of this set is si1. To prove this, we ote that si1 > si π 6 = 1 si for all. This proves that si1 is the maximum of this set. b) What is the ifiimum of the set { +1 N}? Justify (prove) your aswer. ad 1 k We claim that the ifiimum is 0. First of all, 0 is a lower boud, sice each elemet i this set is positive. Next, we must show that 0 is the greatest lower boud. Suppose it is ot, ad L > 0 is a lower boud for the set (hece it is every elemet i the set. However, + 1 = < L for ay > L. Thus there are umerous elemets i our set which are smaller tha L ad we have a cotradictio. Hece 0 is the greatest lower boud as claimed. 13) Show that the sequece defied as s 1 =, ad s +1 = s +1 for 1 coverges. What is its limit? We show that this sequece is decreasig ad bouded below (ad hece coverget). First, we prove by iductio that s > 1 for all. The base case is clear, sice s 1 = > 1. Suppose ow that for some 1 we have s > 1. The s +1 = s +1 > 1+1 = 1 by the iductio hypothesis. So, by iductio, s > 1 for all. To show that (s ) is decreasig, cosider s s +1 = s s +1 = s 1. Sice s > 1 for all, we have that s s +1 = s 1 > 1 1 = 0, ad so s > s +1 for all, ad the sequece is ideed decreasig. Because all bouded mootoe sequeces coverge, we have that s coverges.
5 To fid the limit, ote that we kow that lims = a for some real umber a by the above, ad lims = lims +1 = lim s +1. Usig limit theorems ad deotig lims = a, we have a = a+1 ad so a = a + 1 ad a = 1. Hece lims = 1. 14) Show that the set of eve itegers ad the set of itegers divisible by 3 have the same cardiality. Deote the set of eve itegers by Z, ad the set of itegers divisible by 3 by 3Z. Let f : Z 3Z be defied by f (x) = 3x/. We show that this map is ijective ad surjective, ad hece the two sets have the same cardiality. To show ijectivity, ote that if f (x) = f (y) the 3x/ = 3y/ ad so x = y. To show surjectivity, let y 3Z. Hece y = 3 for some iteger. Let x =. The f (x) = 3 / = 3 = y. So every y 3Z has a preimage uder f i Z, ad thus f is surjective. 15) Compute (mod 7). Compute (mod 11). By Fermat s little theorem, sice 7 is prime ad 4 is ot a multiple of 7, we have that = 4 4 (4 6 ) = 56 4 (mod 7). Sice 11 is prime ad 3 is ot a multiple of 11, by Fermat s little theorem agai, we have that = 3 9 (3 10 ) = (3 3 ) = 15 4 (mod 11). 16) Describe why the RSA cryptosystem is a secure way of commuicatig, ad explai how to sed a message (say, a umber less tha 55) give the key (55,3) (ad what the ecrypted message would be). How would you decode such a message? Check that the decoded message is correct. The RSA works as follows: Bob fids two large primes p,q ad takes their product, = pq. He the chooses some umber e which is relatively prime to φ() = (p 1)(q 1), ad uses the Euclidea algorithm to fid e s multiplicative iverse, r, mod φ(). He publishes ad r for the world to see, but o oe but he ca figure out p,q,r give the factorig algorithms we have to day (i less tha a billio years, that is). Now, we demostrate how RSA works i the give example, where = 55 ad e = 3. Alice wats to sed the message 9. She computes 9 3 (mod 55) ad gets (mod 55). She seds the ecrypted message 14. The Bob computes 14 r = 14 7 = (14 3 ) 9 ( 6) (mod 55) 17) Prove that if p abc the p divides oe of a, b, or c. First of all, this eed ot be true if p is ot prime (ad the problem should specify that p is prime). I class, we proved that if p AB the p A or p B. Now, lettig A = ab ad B = c, we have that p abc meas p AB ad so p A or p B. If p B the p c sice B = c. If p A the we have p ab, ad by the theorem from class metioed above we have p a or p b, as desired.
6 18) a) Compute φ(6). Sice 6 is a product of two distict primes, we have that φ(6) = (13 1)( 1) = 1. b) Fid a 4 such that φ() = φ( + 1). Let = 15. The φ() = 4 = 8, ad φ( + 1) = φ(16) = 8 as well, sice the atural umbers less tha 16 that are relatively prime to 16 are 1,3,5,7,9,11,13,15, of which there are 8. 19) a) Determie, with proof, whether the followig series is coverget or diverget: =0 ( 1) + 1 We use the compariso test. ( 1) + 1 = < 1 for all. Sice 1/ coverges, so does the series above by the compariso test. b) Determie, with proof, whether the followig series is coverget or diverget: 6 =1 1/ We use the root test. We have lim 6 = lim 6 = 0 < 1 ad so the series coverges by the root test. c) Determie, with proof, whether the followig series is coverget or diverget: = / We use the compariso test. We have 1 + 1/ > 1 for all, ad, sice 1 = 1 1 diverges, we have that our series above diverges as well.
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