CSE 1400 Applied Discrete Mathematics Number Theory and Proofs

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1 CSE 1400 Applied Discrete Mathematics Number Theory ad Proofs Departmet of Computer Scieces College of Egieerig Florida Tech Sprig 01 Problems for Number Theory Backgroud Number theory is the brach of mathematics that studies properties of the itegers. Prime umbers are a major topic i umber theory. There are may practical applicatios of umber theory, for istace, cryptology, radom umber geeratio, ad check digits i codes. Problems labeled! are importat ad must be mastered. are harder ad cover iterestig advaced material that is ot properly part of this course. are tagetial to the course but cover material that is useful elsewhere. 1.! Aswer the followig True or False. (a) The set of prime umbers is This is True. P {, 3, 5, 7, 11, 13,...} {p N : p has exactly two differet divisors.} (b) The iteger 1 is prime because it is divisible by itself ad 1. This is False. (c) The set of composite umbers is This is True. C {4, 6, 8, 9, 10, 11,...} {c N : c more tha two differet divisors.} (d) Every atural umber ca be writte as the products of prime umbers, that is This is False. Both 0 ad 1 are couterexamples. They are atural umbers that caot be writte as the product of primes. (e) Every atural umber greater tha 1 ca be writte as the products of prime umbers. This is True. It is kow as the Fudametal Theorem of Arithmetic. (f) There are ifiitely may primes, that is, This is True. (g) There is a largest prime umber, that is ( p P)( q P)(q > p) ( p P)( q P)(q p) This is False. The propositio states there is a largest prime umber p, which is ot True.

2 umber theory ad proofs. A repeated product ca be writte usig product otatio (a) Write factorial (!) usig product otatio 1 a k a 0 a 1 a 1! ( 1)( ) 1 (b) Write the biomial coefficiet ( m ) usig product otatio Preted that m m. ( ) m! m!( m)! m 1 m 1 (k + 1) (k + 1) m 1 1 k m+1 1 (k + 1) (k + 1) (k + 1) (k + 1) (c) The atural value to assig to the empty product 1 k is 1. This is True. Problems o Modular Numbers The itegers mod are where each k value represet the set of values 1. Describe the residue (equivalece) classes for (a) Cogruece mod 4. Backgroud Z {0, 1,,..., ( 1)} {a + k : a Z} [0] {4k : k Z} [1] {4k + 1 : k Z} [] {4k + : k Z} [3] {4k + 3 : k Z}

3 umber theory ad proofs 3 (b) Cogruece mod 5. [0] {5k : k Z} [1] {5k + 1 : k Z} [] {5k + : k Z} [3] {5k + 3 : k Z} [4] {5k + 4 : k Z}.! How may residue classes are there mod? There are residue classes, amed [0] through [ 1]. 3.! For the give values of a ad below, compute the quotiet q ad o-egative remaider r that satisfy the quotiet-remaider equatio a q + r, r 0 (a) a 6, 13. (b) a 7, 13. (c) a 7, 13. (d) a 45, Costruct a additio table for the itegers mod Costruct a multiplicatio table for the itegers mod Costruct a multiplicatio table for the itegers mod 7.

4 umber theory ad proofs Solve the followig liear cogruece equatios. (a) 3x 5 (mod 7) x 4 sice (mod 7). (b) 5x 4 (mod 7) x 5 sice (mod 7). (c) x 1 (mod 7) x 4 sice (mod 7). (d) 6x 3 (mod 7) x 4 sice (mod 7). 8. Give the values of p ad q below: What is their greatest commo divisor gcd(p, q)? What is their least commo multiple of lcm(p, q)? Verify the gcd(p, q) lcm(p, q) pq. (a) Let p ad q Use Euclid s algorithm to compute the greatest commo divisors listed below. (a) gcd(19, 8) The greatest commo deomiator is 1. (b) gcd(5, 40)

5 umber theory ad proofs 5 The greatest commo deomiator is 5. (c) gcd(70, 7) gcd(7, 70) 1. (d) gcd(66, 99) gcd(66, 99) 33. (e) gcd(189, 80). (f) gcd(511, 55) gcd(511, 55) Usig your work from questio 9 fid itegers a ad b such that (a) 1 19a + 8b. Fill i the magic table usig the quotiets from the Euclidea algorithm. Start from the iitial two colums ad compute ext colums by the recurrece 1 19 ( 3) where a 3 ad b 7. c qc 1 + c (b) 5 40a + 5b. Fill i the magic table usig the quotiets from the Euclidea. Start from the iitial two colums ad compute ext colums by the recurrece c qc 1 + c

6 umber theory ad proofs ( ) ad, multiplyig this equatio through by 5 gives ( 10) (c) 1 70a + 7b. Fill i the magic table usig the quotiets from the Euclidea Start from the iitial two colums ad compute ext colums by the recurrece c qc 1 + c ( 5) + 7 (13). (d) 33 66a + 99b. Fill i the magic table ( 1) ad, multiplyig this equatio by 33 we get ( 1) Usig your work from questio 10, solve the followig liear cogruece equatios. (a) 8x 5 mod 19. Sice 19 ( 3) , mod 19 it follows that x 7 8x x mod 19. (b) 5x 38 mod 40. (c) 7x 4 mod 70. Sice 70 ( 5) , mod 70 it follows that x 13 7x x mod 70. (d) 66x mod 99. Problems o Proofs Backgroud Statemets have bee proved throughout the course. Now these ideas will be orgaized ito proof methods. Vacuous ad Trivial Proofs Backgroud

7 umber theory ad proofs 7 A statemet of the form is vacuously True. A statemet of the form is trivially True. False p p True 1. Prove that the empty set is a subset of every set. X is a subset of Y if for every x X, x Y. The statemet is vacuously True because x is False.. Prove that less tha is atisymmetric. ( x U)(x xy) There is a vacuous proof: The statemet that less tha is atisymmetric is ( x, y R)((x < y y < x) (x y)), which is True by virtue of its logical form: p q where the premise p (x < y y < x) is always False which makes the coditioal always True. Direct Proofs Backgroud Direct proofs may be the most commo. They rely o the modus poes rule of iferece (p (p q) q That is, if p is True ad p q is True, the q must be True. 1. Let x a mod ad y b mod. Prove that x + y (a + b) mod. If x a mod ad y b mod, the x a c ad y b d for some itegers c ad d. Therefore (x a) + (y b) (x + y) (a + b) c + d (c + d), that is x + y (a + b) mod.. Let x a mod ad y b mod. Prove that xy ab mod. If x a mod ad y b mod, the x a c ad y b d for some itegers c ad d. Therefore (x a)(y b) xy ay bx + ab, that is x + y (a + b) mod. 3. Prove that if the itegers m ad are both odd, the m is odd. If m ad are odd, the m k + 1 ad j + 1 for some itegers k ad j. Therefore m (k + 1)(j + 1) 4kj + k + j + 1 (kj + k + j) + 1 which shows m is odd. 4. Prove that (1 + 5)/ ad (1 5)/ are solutios to the equatio x x + 1. The quadratic formula computes the two solutios to the quadratic equatio x x 1 0 as (1 + 5)/ ad (1 5)/.

8 umber theory ad proofs 8 5. Prove that if is eve the is eve ad if is odd the is odd. If k is eve, the 4k is eve. If k + 1 is odd, the 4k + 4k + 1 (k + k) + 1 is odd. 6. Prove that if is a eve iteger, the 4k or 4k + for some iteger k. If is eve, the m for some iteger m. If m is eve, that is, if m k, the m (k) 4k. If m is odd, that is, if m k + 1, the m (k + 1) 4k Prove that ( x R, x 0)( x(x + ) < x + 1). Square the left-had side of the iequality to obtai ( x(x + )) x(x + ) x + x < x + x + 1 (x + 1) which shows the result is True. 8. (Quotiet Remaider Lemma 1) Give itegers a, Z, 0, cosider the set of atural umbers Prove A is ot empty. To see this, let A {a q 0 : q Z} a/ if > 0, that is, a q if a/ q. q a/ if < 0, that is, a q if a/ q. Sice a/ a/ a/b, the first case, whe > 0, q a/ a/ implies a q 0. Ad, i the secod case, whe < 0, a/ a/ q, implies a q 0. I both cases, there is a elemet i A 9. Prove that lg! lg. By the log of a product is the sum of logs rule lg(!) lg ( 1)( ) 1 lg() + lg( 1) + lg( ) + + lg() + lg(1) For each k 1,,...,, lg(k) lg() Therefore lg(!) lg() + lg( 1) + lg( ) + + lg() + lg(1) lg() 10. (Euclid s Lemma) Let p P be a prime umber ad let a, b N be atural umbers. If p divides ab, the p divides a or p divides b. Preted p divides ab ad p does ot divide a. That is, ( c N)(pc ab) (a mod p r 0) By the fudametal theorem of arithmetic, ab ca be factored as a product of primes. Sice p does ot divide a, p is ot oe of the prime factors of a. Therefore, sice the prime factorizatio is uique, p must be a prime factor of b. That is, p divides b.

9 umber theory ad proofs 9 Proof by Cotrapositio Backgroud Proofs by cotrapositio are idirect proofs. To prove p q prove istead. q p 1. Prove that if is eve the is eve. If is odd, the k + 1 ad 4k + 4k + 1 (k + k) + 1 is odd.. Prove that if is odd the is odd. If is eve, the k ad 4k (k ) is eve. Proofs by Cotradictio Backgroud Suppose that p q ad p q are both True. The it must be that p is False, that is, p is True. The cojuctio q q is a cotradictio ad showig is True proves that p is True. p q q 1. If 1 is ot divisible by 8, the is eve. Note that the form of the statemet is which has egatio p q p q p q Assume 1 is ot divisible by 8 ad is odd. The k + 1 for some iteger k ad 1 4k + 4k 4k(k + 1). Sice k ad k + 1 are cosecutive itegers oe of them is eve, so i fact, 4k(k + 1) is divisible by 8.. Prove that 3 is irratioal. By way of cotradictio, preted 3 is ratioal ad let 3 a/b where a ad b are relatively prime itegers with b 0. The 3 a b a3 b 3 b 3 a 3

10 umber theory ad proofs 10 ad so a is a multiple of. Let a k, the b 3 8a 3 b 4a 3 ad so b is a multiple of. This cotradicts that a ad b are relatively prime. 3. Prove that if k < < k + 1 for some iteger k, the is irratioal. Assume is ratioal. a, a, b Z, b 0, gcd(a, b) 1 b b a b a divides a. Let be the prime factorizatio of, ad let a m 1 p e k k j 1 q f k k be the prime factorizatio of a. Sice a each prime factor p e k k of must divide oe of the prime factors of a. Therefore divides a ad a c for some atural umber c. But the b a c ad b c. That is divides both a ad b which cotradicts that a ad b are relatively prime. Proofs by Couterexample A uiversally quatified statemet such as Backgroud ( x U)(p(x)) ca be proved False by givig oe istace (a couterexample) where it is False. 1. (Be able to costruct couterexamples) Give a couterexample to prove that the followig uiversally quatified statemets are false (a) If a eve atural umber the + m is eve for all atural umbers m. Let 0 ad m 1. The is eve ad + m 1 is odd. (b) If atural umber > 0 is a multiple of 3, the 1 is divisible by 4. The statemet is True for 1 ad 3: Both 0 ad 8 are divisible by 4, but for 6, 1 35 is ot divisible by 4. (c) If a ad b are ratioal umbers, the a b is a ratioal umber. Let a ad b 1/. The a b 1/ is a irratioal umber. (d) If a is a multiple of b, the a is a multiple of b. Let a ad b 4. The a 4 is a multiple of b 4, but a is ot a multiple of b 4.

11 umber theory ad proofs 11 Mathematical Iductio Backgroud Mathematical iductio is oe of the most importat proof techiques i discrete mathematics. To prove that p() is True for all atural umbers it suffices to show p(0) (p() p( + 1)) is True 1. Prove that the sum of the first atural umbers is ( 1)/. The atural umbers lie i the set The sum of the first atural umbers is N {0, 1,, 3, 4, 5, 6,...} ( 1) Basis for Iductio: For 0 the sum o the left-had side of the equality is empty ad has value zero. Also, the right-had side expressio is 0(0 1)/ is also zero. Iductive Premise: Preted the equality is True for some 0. Iductive Step: If the 1 ( 1) k 1 ( 1) k 1 k k + ( 1) + ( 1) + ( 1) + ( + 1) establishig that if the equality holds for, the it holds for + 1. Coclusio:. Prove the formula for arithmetic sums ( 1) k ( N) ( N)( m, b R)(b + (m + b) + (m + b) + + (m + b) ( + 1) m + b

12 umber theory ad proofs 1 That is, the sum of terms i the arithmetic sequece A b, m + b, b + m,..., b + m is the umber of terms time the average of the first ad last term. For 0, the sum o the left of the equality is equal to b ad the fuctio o the right of the equality is equal to (0 + 1) 0m+b b. Assume that The m + b ( N)( m, b R)(b + (m + b) + (m + b) + + (m + b) ( + 1) m + b b + (m + b) + (m + b) + + (m + b) + (( + 1)m + b) ( + 1) + (( + 1)m + b) ( + 1)m + ( + 1)b ( + 1)m + b) + ( + 1)m + ( + 1)b + ( + 1)m + b ( + )( + 1)m + ( + )b ( + 1)m + b ( + ) 3. Prove that the sum of the first powers of is 1. Basis for Iductio: For 0 the sum o the left-had side of the equality is empty ad has value zero. Also, the right-had side expressio 0 1 is zero also. Iductive Premise: Preted the equality is True for some 0. Iductive Step: If the 1 k 1 1 k 1 k 1 k + ( 1) establishig that if the equality holds for, the it holds for + 1.

13 umber theory ad proofs 13 Coclusio: 4. Prove the formula for geometric sums k 1 ( N) ( N)( r R, r 1)(a + ar + ar + + ar 1 a r 1 r 1 For 0, the sum o the left of the equality is empty ad equal to 0 ad the fuctio o the right of the equality is also equal to 0. Assume that ( N)( r R, r 1)(a + ar + ar + + ar 1 a r 1 r 1 The a + ar + ar + + ar 1 + ar a r 1 r 1 + ar a r 1 r 1 + ar (r 1) r 1 a r 1 + r +1 r r 1 a r+1 1 r 1 5. Prove the fudametal theorem of arithmetic: Every atural umber > 1 ca be writte as the product of primes. (Note that a product may be a sigle factor so that a prime p is itself a product of primes.) Basis: Let. Sice is prime, it is a product of primes. Iductive Hypothesis: Let 3 ad suppose every atural umber from to 1 ca be writte as the product of primes. Iductive Step: Cosider the atural umber. If is prime, the it is the product of primes. If is composite, the a b for some atural umbers a, b <. By the iductive hypothesis, both a ad b ca be writte as the product of primes. Therefore, ca be writte as the product of primes. 6. Prove that ( N, > 0)(3 > ). For 1, 3 > establishig a basis for iductio. Assume that ( N, > 0)(3 > ). The > 3 > Prove that for all atural umbers ad all real umbers a ad b Notice that a b ()(a 1 + a b + a 3 b + ab + b 1 ) a 1 + a b + a 3 b + ab + b 1 1 a 1 k b k

14 umber theory ad proofs 14 I particular, the sum is empty ad equal to 0 whe 0. Therefore, for 0 the left-had side of the equality is a 0 b 0 0 ad the right-had side is () time the empty sum ad equal to 0 also. We ca also assume that a b sice both sides of the equatio are equal to 0 whe a b. Now assume that for some 0. The a b ()(a 1 + a b + a 3 b + ab + b 1 ) a k b k 1 a k b k + b 1 a a 1 k b k + b a a b a a b + b + b a+1 ab + b +1 a+1 b +1 The sum a k b k a 1 + a b + a 3 b + ab + b 1 is called the covolutio of fiite sequeces a 1, a, a 3,, a, 1 ad 1, b, b, b 3,, b 1 8. Let a (1 + 5)/ deote the golde ratio ad let b (1 5)/ be its cojugate. Show that where F is the Fiboacci umber. F a b Sice seedig the Fiboacci recurrece requires values, we eed to establish the basis for both 0 ad 1. First, F 0 0 ad a0 b 0 a b 0 also. Secod, F 1 1 ad a1 b 1 a b 1 also. Now assume that F a b ad F 1 a 1 b 1 for some pair ad 1 where 1. Recall, from problem 4 that (1 + 5)/ ad (1 5)/ are

15 umber theory ad proofs 15 solutios to the equatio x x + 1. The F +1 F + F 1 a b a b + a 1 b 1 + a 1 b 1 a b + a 1 b 1 a 1 (a + 1) b 1 (b + 1) a 1 a b 1 b a+1 b Let > 0 ad prove that the Fiboacci umbers F ad F +1 are relatively prime. 10. Prove that ( N, > 1)( ( 1) 1 ) For, 1/(1 ) ( 1)/ establishig a basis for iductio. Assume that for some. The ( 1) ( 1) + 1 ( + 1) ( 1)( + 1) ( + 1) ( + 1) + 1 ( + 1) + 1 ( + 1) 11. Usig the result i problem 10, give a direct proof that ( N, > 1)( < 1) (Iterestigly, as teds toward ifiity, the sum coverges to π /6.)

16 umber theory ad proofs 16 For ay > 1, > 1 which implies 1/( 1) > 1/. Therefore 1 ( 1) > 1 for all > 1 Sum both sides of this iequality from k to k to obtai < ( 1) 1 < 1 1. Prove that for ay > (3 ) (6 3 1) For 1, 1 1 1(6 3 1)/ establishig a basis for iductio. Assume that for some > 0. The (3 ) (6 3 1) (3 ) + (3 + 1) (6 3 1) + (3 + 1) ( ) ( ) 3( + + 1) ( + 1) 6( + 1) 3 3( + 1) ( + 1) ( + 1)(6( + 1) 3( + 1) 1) 13. Prove that for all < 3 For, < 3 establishig a basis for iductio. Assume < 3 for some. The < 3

17 umber theory ad proofs Prove that for ay N, 3 (mod 3). For 0, 30 (mod 3). Assume 3 (mod 3) for some 0. The ( 3 ) 3 3 (mod 3) 8 (mod 3) (mod 3) 15. Prove that 3 +1 divides for all atural umbers. For 0, which divides establishig a basis for iductio. Assume 3 +1 divides for some 0. We eed to show that 3 + divides By our assumptio 3 +1 divides so to complete the proof we oly eed to show 3 divides ( 3 ) By ( problem 14 3 (mod 3) ad so its square is 3) (mod 3) 4 (mod 3) 1 (mod 3). Therefore ( 3 ) (mod 3) 0 (mod 3). 16. Prove that r 1 divides r 1 for all atural umbers r 1 ad. Let r 1 be fixed but otherwise arbitrary. Usig iductio o : For 0, r 1 divides r If r 1 divides r 1, that is, if (r 1)c r 1 for some atural umber c, the r +1 1 r +1 r + r 1 r (r 1) + (r 1) r (r 1) + (r 1)c (r 1)(r + c) Showig that r 1 divides r Recall the geometric sum formula 1 r k r 1 r 1 for r Prove that For 0 the sum of the left is 0 ( ) 1 + k k ( + k ) 1 k ( ) ad the fuctio o the right is 0 1 establishig a basis for iductio. Assume that ( ) 1 + k k

18 umber theory ad proofs 18 for some 0. Recall Pascal s idetity ( ) ( ) 1 + m m ( ) 1 m 1 The +1 ( ) k k (( k ) ( )) k k