# 1. MATHEMATICAL INDUCTION

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1 1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: ( (1.1 STEP 1: For 1 (1.1 is true, sice 1 1( STEP 2: Suppose (1.1 is true for some k 1, that is k k(k STEP 3: Prove that (1.1 is true for k + 1, that is We have k + (k + 1 ST k + (k + 1? k(k (k + 1(k k + (k + 1 (k (k + 1(k EXAMPLE 2: Prove that for ay iteger 1. Proof: STEP 1: For 1 (1.2 is true, sice (2 1 2 (1.2 STEP 2: Suppose (1.2 is true for some k 1, that is (2k 1 k 2. STEP 3: Prove that (1.2 is true for k + 1, that is (2k 1 + (2k + 1? (k We have: (2k 1 + (2k + 1 ST.2 k 2 + (2k + 1 (k

2 EXAMPLE 3: Prove that for ay iteger 1.! (1.3 Proof: STEP 1: For 1 (1.3 is true, sice 1! 1 1. STEP 2: Suppose (1.3 is true for some k 1, that is k! k k. STEP 3: Prove that (1.3 is true for k + 1, that is (k + 1!? (k + 1 k+1. We have (k + 1! k! (k + 1 ST.2 k k (k + 1 < (k + 1 k (k + 1 (k + 1 k+1. EXAMPLE 4: Prove that for ay iteger (1.4 Proof: STEP 1: For 0 (1.4 is true, sice STEP 2: Suppose (1.4 is true for some k 0, that is 8 3 2k 1. STEP 3: Prove that (1.4 is true for k + 1, that is 8 3 2(k+1 1. We have 3 2(k k k k ( } 2k {{ 8} + } 3 2k {{ 1 }. div.by 8 St. 2 div.by 8 EXAMPLE 5: Prove that for ay iteger (1.5 Proof: STEP 1: For 1 (1.5 is true, sice STEP 2: Suppose (1.5 is true for some k 1, that is 7 k 7 k. STEP 3: Prove that (1.5 is true for k + 1, that is 7 (k (k + 1. We have (k (k + 1 k 7 + 7k k k k k 2 + 7k + 1 k 1 k 7 k } {{ } St. 2 div. by 7 + 7k } k k 4 {{ + 35k k 2 + 7k }. div. by 7 2

3 2. THE BINOMIAL THEOREM DEFINITION: Let ad k be some itegers with 0 k. The! k k!( k! is called a biomial coefficiet. PROPERTIES: Proof: We have 0! 0!( 0!! 1! 1,!!(!!! 0!!! Proof: We have 1 1! 1!( 1! ( 1! 1! ( 1!,! ( 1![ ( 1]!! ( 1! 1! ( 1! ( 1! 1!. 3.. k k Proof: We have k! k!( k!! ( k!k!! ( k![ ( k]!. k 3

4 4. + k k 1 Proof: We have + k k 1 ( + 1 k.! k!( k! +! (k 1!( k + 1!!( k + 1 k!( k!( k + 1 +!k (k 1!k( k + 1!!( k + 1 k!( k + 1! +!k k!( k + 1!!( k + 1 +!k k!( k + 1!!!k +! +!k k!( k + 1!! +! k!( k + 1!!( + 1 k!( k + 1! ( + 1! k!( k + 1! ( + 1! k!( + 1 k! ( + 1 k. PROBLEM: For all itegers ad k with 1 k we have k 1 k k k + 1 Proof: By property 4 we have k 1 k + k + 1 k k + k + 1 k k k k + 1 THEOREM (The Biomial Theorem: Let a ad b be ay real umbers ad let be ay oegative iteger. The (a + b a + a 1 b + a 2 b a 2 b 2 + ab 1 + b

5 PROBLEM: For all itegers 1 we have Proof: Puttig a b 1 i the Theorem above, we get ( hece , therefore by property 1 we get , PROBLEM: For all itegers 1 we have ( Proof: Puttig a 1 ad b 1 i the Theorem above, we get hece ( ( ( ( ( 1, ( ( 1, 1 therefore by property 1 we get ( (

6 3. RATIONAL AND IRRATIONAL NUMBERS DEFINITION: Ratioal umbers are all umbers of the form p, where p ad q are itegers ad q 0. q EXAMPLE: 1 2, 5 50, 2, 0, 3 10, etc. NOTATIONS: N all atural umbers, that is, 1, 2, 3,... Z all iteger umbers, that is, 0, ±1, ±2, ±3,... Q all ratioal umbers R all real umbers DEFINITION: A umber which is ot ratioal is said to be irratioal. PROBLEM 1: Prove that 2 is irratioal. Proof: Assume to the cotrary that 2 is ratioal, that is 2 p q, where p ad q are itegers ad q 0. Moreover, let p ad q have o commo divisor > 1. The 2 p2 q 2 2q 2 p 2. (3.1 Sice 2q 2 is eve, it follows that p 2 is eve. The p is also eve (i fact, if p is odd, the p 2 is odd. This meas that there exists k Z such that Substitutig (3.2 ito (3.1, we get p 2k. (3.2 2q 2 (2k 2 2q 2 4k 2 q 2 2k 2. Sice 2k 2 is eve, it follows that q 2 is eve. The q is also eve. This is a cotradictio. 6

7 PROBLEM 2: Prove that 3 4 is irratioal. Proof: Assume to the cotrary that 3 4 is ratioal, that is 3 4 p q, where p ad q are itegers ad q 0. Moreover, let p ad q have o commo divisor > 1. The 4 p3 q 3 4q 3 p 3. (3.3 Sice 4q 3 is eve, it follows that p 3 is eve. The p is also eve (i fact, if p is odd, the p 3 is odd. This meas that there exists k Z such that Substitutig (3.4 ito (3.3, we get p 2k. (3.4 4q 3 (2k 3 4q 3 8k 3 q 3 2k 3. Sice 2k 3 is eve, it follows that q 3 is eve. The q is also eve. This is a cotradictio. PROBLEM 3: Prove that 6 is irratioal. Proof: Assume to the cotrary that 6 is ratioal, that is 6 p q, where p ad q are itegers ad q 0. Moreover, let p ad q have o commo divisor > 1. The 6 p2 q 2 6q 2 p 2. (3.5 Sice 6q 2 is eve, it follows that p 2 is eve. The p is also eve (i fact, if p is odd, the p 2 is odd. This meas that there exists k Z such that Substitutig (3.6 ito (3.5, we get p 2k. (3.6 6q 2 (2k 2 6q 2 4k 2 3q 2 2k 2. Sice 2k 2 is eve, it follows that 3q 2 is eve. The q is also eve (i fact, if q is odd, the 3q 2 is odd. This is a cotradictio. 7

8 PROBLEM 4: Prove that is irratioal. Proof: Assume to the cotrary that is ratioal, that is where p ad q are itegers ad q 0. The 1 p q, 3(p 5q 2. q Sice 2 is irratioal ad 3(p 5q q is ratioal, we obtai a cotradictio. PROBLEM 5: Prove that log 5 2 is irratioal. Proof: Assume to the cotrary that log 5 2 is ratioal, that is log 5 2 p q, where p ad q are itegers ad q 0. The 5 p/q 2 5 p 2 q. Sice 5 p is odd ad 2 q is eve, we obtai a cotradictio. 8

10 Sice 3b 2 is divisible by 3, it follows that a 2 is divisible by 3. The a is also divisible by 3. I fact, if a is ot divisible by 3, the by the Divisio Algorithm there exists q Z such that Suppose a 3q + 1, the a 3q + 1 or a 3q + 2. a 2 (3q q 2 + 6q + 1 3(3q 2 + 2q } {{ } q + 1 3q + 1, which is ot divisible by 3. We get a cotradictio. Similarly, if a 3q + 2, the a 2 (3q q q + 4 3(3q 2 + 4q + 1 } {{ } q + 1 3q + 1, which is ot divisible by 3. We get a cotradictio agai. So, we proved that if a 2 is divisible by 3, the a is also divisible by 3. This meas that there exists q Z such that a 3q. (4.2 Substitutig (4.2 ito (4.1, we get 3b 2 (3q 2 3b 2 9q 2 b 2 3q 2. Sice 3q 2 is divisible by 3, it follows that b 2 is divisible by 3. The b is also divisible by 3 by the argumets above. This is a cotradictio. 10

13 THEOREM: Let p be a prime. The p is irratioal. Proof: Assume to the cotrary that p is ratioal, that is p a b, where a ad b are itegers ad b 0. Moreover, let a ad b have o commo divisor > 1. The p a2 b 2 pb 2 a 2. (5.3 Sice pb 2 is divisible by p, it follows that a 2 is divisible by p. The a is also divisible by p by the Corollary above. This meas that there exists q Z such that Substitutig (5.4 ito (5.3, we get a pq. (5.4 pb 2 (pq 2 b 2 pq 2. Sice pq 2 is divisible by p, it follows that b 2 is divisible by p. The b is also divisible by p by the Corollary above. This is a cotradictio. PROBLEM: Prove that 101 is irratioal. Proof: Sice 101 is prime, the result immediately follows from the Theorem above. PROBLEM: Prove that if a ad b are positive itegers with (a, b 1, the (a 2, b 2 1 for all Z +. Proof 1: Assume to the cotrary that (a 2, b 2 > 1. The there is a prime p such that p a 2 ad p b 2. From this by Euclid s Lemma it follows that p a ad p b, therefore (a, b p. This is a cotradictio. Proof 2 (Hit: Use the Fudametal Theorem of Arithmetic below. 13

14 6. FUNDAMENTAL THEOREM OF ARITHMETIC THEOREM (Fudametal Theorem of Arithmetic: Assume that a iteger a 2 has factorizatios a p 1... p m ad a q 1... q, where the p s ad q s are primes. The m ad the q s may be reidexed so that q i p i for all i. Proof: We prove by iductio o l, the larger of m ad, i. e. l max(m,. where Step 1. If l 1, the the give equatio i a p 1 q 1, ad the result is obvious. Step 2. Suppose the theorem holds for some l k 1. Step 3. We prove it for l k + 1. Let a p 1... p m q 1... q, (6.1 max(m, k + 1. (6.2 From (6.1 it follows that p m q 1... q, therefore by Euclid s Lemma there is some q i such that p m q i. But q i, beig a prime, has o positive divisors other tha 1, therefore p m q i. Reidexig, we may assume that q p m. Cacelig, we have p 1... p m 1 q 1... q 1. Moreover, max(m 1, 1 k by (6.2. Therefore by step 2 q s may be reidexed so that q i p i for all i; plus, m 1 1, hece m. COROLLARY: If a 2 is a iteger, the there are uique distict primes p i ad uique itegers e i > 0 such that a p e p e. Proof: Just collect like terms i a prime factorizatio. EXAMPLE: PROBLEM: Prove that log 3 5 is irratioal. Proof: Assume to the cotrary that log 3 5 is ratioal, that is log 3 5 p q, where p ad q are itegers ad q 0. The 3 p/q 5 3 p 5 q, which cotradicts the Fudametal Theorem of Arithmetic. 14

15 7. EUCLIDEAN ALGORITHM THEOREM (Euclidea Algorithm: Let a ad b be positive itegers. The there is a algorithm that fids (a, b. LEMMA: If a, b, q, r are itegers ad a bq + r, the (a, b (b, r. Proof: We have (a, b (bq + r, b (b, r. Proof of the Theorem: The idea is to keep repeatig the divisio algorithm. We have: therefore a bq 1 + r 1, (a, b (b, r 1 b r 1 q 2 + r 2, (b, r 1 (r 1, r 2 r 1 r 2 q 3 + r 3, (r 1, r 2 (r 2, r 3 r 2 r 3 q 4 + r 4, (r 2, r 3 (r 3, r 4... r 2 r 1 q + r, (r 2, r 1 (r 1, r r 1 r q +1, (r 1, r r, (a, b (b, r 1 (r 1, r 2 (r 2, r 3 (r 3, r 4... (r 2, r 1 (r 1, r r. PROBLEM: Fid (326, 78. Solutio: By the Euclidea Algorithm we have therefore (326, PROBLEM: Fid (252, Solutio: By the Euclidea Algorithm we have therefore (252,

16 PROBLEM: Fid (4361, Solutio: By the Euclidea Algorithm we have therefore (4361, , THEOREM: Let a p e p e ad b p f p f (a, b p mi(e 1,f p mi(e,f. be positive itegers. The EXAMPLE: Sice ad , we have: (720, PROBLEM: Let a Z. Prove that (2a + 3, a Proof: By the Lemma above we have (2a + 3, a + 2 (a a + 2, a + 2 (a + 1, a + 2 (a + 1, a (a + 1, 1 1. PROBLEM: Let a Z. Prove that (7a + 2, 10a Proof: By the Lemma above we have (7k + 2, 10k + 3 (7k + 2, 7k k + 1 (7k + 2, 3k + 1 (6k k, 3k + 1 (k, 3k + 1 (k,

17 8. FERMAT S LITTLE THEOREM Theorem (Fermat s Little Theorem: Let p be a prime. We have p p (8.1 for ay iteger 1. Proof 1: STEP 1: For 1 (8.1 is true, sice p 1 p 1. STEP 2: Suppose (8.1 is true for some k 1, that is p k p k. STEP 3: Prove that (8.1 is true for k + 1, that is p (k + 1 p (k + 1. Lemma: Let p be a prime ad l be a iteger with 1 l p 1. The p p. l Proof: We have p l p! l!(p l! l!(l p l!(p l! (l p, (p l! therefore p (p l! (l p. l Form this it follows that p p (p l!, l p hece by Euclid s Lemma p divides or (p l!. It is easy to see that p (p l!. Therefore l p p. l We have (k + 1 p (k + 1 k p + k p k } {{ } St. 2 div. by p p k p ( p + p p k p k + 1 k 1 2 p 1 p p k p 1 + k p k. 1 2 p 1 } {{ } div. by p by Lemma 17

18 9. CONGRUENCES DEFINITION: Let m be a positive iteger. The itegers a ad b are cogruet modulo m, deoted by a b mod m, if m (a b. EXAMPLE: 3 1 mod 2, 6 4 mod 2, 14 0 mod 7, mod 9, mod 35. PROPERTIES: Let m be a positive iteger ad let a, b, c, d be itegers. The 1. a a mod m 2. If a b mod m, the b a mod m. 3. If a b mod m ad b c mod m, the a c mod m. 4. (a If a qm + r mod m, the a r mod m. (b Every iteger a is cogruet mod m to exactly oe of 0, 1,..., m If a b mod m ad c d mod m, the a ± c b ± d mod m ad ac bd mod m. 5. If a b mod m, the a ± c b ± c mod m ad ac bc mod m. 5. If a b mod m, the a b mod m for ay Z If (c, m 1 ad ac bc mod m, the a b mod m. Proof 2 of Fermat s Little Theorem: We distiguish two cases. Case A: Let p, the, obviously, p p, ad we are doe. Case B: Let p. Sice p is prime, we have (p, 1. (9.1 Cosider the followig umbers:, 2, 3,..., (p 1. 18

19 We have r 1 mod p 2 r 2 mod p 3 r 3 mod p... (p 1 r p 1 mod p, where 0 r i p 1. Moreover, r i 0, sice otherwise p i, ad therefore by Euclid d Lemma p i or p. But this is impossible, sice p > i ad p. So, From (9.2 by property 5 we have (9.2 1 r i p 1. ( (p 1 r 1 r 2... r p 1 mod p (p 1! p 1 r 1 r 2... r p 1 mod p. (9.4 Lemma: We have Proof: We first show that r 1 r 2... r p 1 (p 1!. (9.5 r 1, r 2,..., r p 1 are all distict. (9.6 I fact, assume to the cotrary that there are some r i ad r j with r i r j. The by (9.2 we have i j mod p, therefore by property 6 with (9.1 we get i j mod p, which is impossible. This cotradictio proves (9.6. By the Lemma we have r 1 r 2... r p 1 (p 1!. (9.7 By (9.4 ad (9.7 we obtai (p 1! p 1 (p 1! mod p. Sice (p, (p 1! 1, from this by by property 6 we get hece p 1 1 mod p, p mod p by property 4. This meas that p is divisible by p. COROLLARY: Let p be a prime. The for ay iteger 1 with (, p 1. p 1 1 mod p THEOREM: If (a, m 1, the, for every iteger b, the cogruece ax b mod m (9.8 19

20 has exactly oe solutio where s is such umber that x bs mod m, (9.9 as 1 mod m. (9.10 Proof (Sketch: We show that (9.9 is the solutio of (9.8. I fact, if we multiply (9.9 by a ad (9.10 by b (we ca do that by property 5, we get which imply (9.8 by property 3. ax abs mod m ad bsa b mod m, Problems Problem 1: Fid all solutios of the cogruece 2x 1 mod 3. Solutio: We first ote that (2, 3 1. Therefore we ca apply the theorem above. Sice mod 3, we get x mod 3. Problem 2: Fid all solutios of the followig cogruece 2x 5 mod 7. Solutio: We first ote that (2, 7 1. Therefore we ca apply the theorem above. Sice mod 7, we get x mod 7. Problem 3: Fid all solutios of the cogruece 3x 4 mod 8. Solutio: We first ote that (3, 8 1. Therefore we ca apply the theorem above. Sice mod 8, we get x mod 8. Problem 4: Fid all solutios of the followig cogruece 2x 5 mod 8. 20

21 Solutio: Sice (2, 8 2, we ca t apply the theorem above directly. We ow ote that 2x 5 mod 8 is equivalet to 2x 8y 5, which is impossible, sice the left-had side is divisible by 2, whereas the right-had side is ot. So, this equatio has o solutios. Problem 5: Fid all solutios of the cogruece 8x 7 mod 18. Solutio: Sice (8, 18 2, we ca t apply the theorem above directly. We ow ote that 8x 7 mod 18 is equivalet to 8x 18y 7, which is impossible, sice the left-had side is divisible by 2, whereas the right-had side is ot. So, this equatio has o solutios. Problem 6: Fid all solutios of the followig cogruece 4x 2 mod 6. Solutio: Sice (4, 6 2, we ca t apply the theorem above directly agai. However, cacelig out 2 (thik about that!, we obtai 2x 1 mod 3. Note that (2, 3 1. Therefore we ca apply the theorem above to the ew equatio. Sice mod 3, we get x mod 3. Problem 7: Fid all solutios of the cogruece 6x 3 mod 15. Solutio: Sice (6, 15 3, we ca t apply the theorem above directly agai. cacelig out 3, we obtai 2x 1 mod 5. However, Note that (2, 5 1. Therefore we ca apply the theorem above to the ew equatio. Sice mod 5, we get x mod 5. Problem 8: Fid all solutios of the cogruece 9x mod

22 Solutio: We first rewrite this cogruece as 9x 5 mod 25. Note that (9, Therefore we ca apply the theorem above. Sice mod 25, we get x mod 25. Problem 9: What is the last digit of ? Solutio: It is obvious that therefore by property 5 we have mod 10, mod 10. This meas that the last digit of is 1. Problem 10: What is the last digit of ? Solutio: It is obvious that therefore by property 5 we have mod 10, This meas that the last digit is mod 10. Problem 11: Prove that there is o perfect square a 2 which is cogruet to 2 or 3 mod 4. Solutio 1: By the property 4(b each iteger umber is cogruet to 0 or 1 mod 2. Cosider all these cases ad use property 4(a: If a 0 mod 2, the a 2k, therefore a 2 4k 2, hece a 2 0 mod 4. If a 1 mod 2, the a 2k + 1, therefore a 2 4k 2 + 4k + 1, hece a 2 1 mod 4. So, a 2 0 or 1 mod 4. Therefore a 2 2 or 3 mod 4. Solutio 2: By the property 4(b each iteger umber is cogruet to 0, 1, 2, or 3 mod 4. Cosider all these cases ad use property 5 : If a 0 mod 4, the a mod 4. If a 1 mod 4, the a mod 4. If a 2 mod 4, the a mod 4. If a 3 mod 4, the a mod 4. So, a 2 0 or 1 mod 4. Therefore a 2 2 or 3 mod 4. 22

23 Problem 12: Prove that there is o itegers a such that a 4 is cogruet to 2 or 3 mod 4. Solutio: By the property 4(b each iteger umber is cogruet to 0, 1, 2, or 3 mod 4. Cosider all these cases ad use property 5 : If a 0 mod 4, the a mod 4. If a 1 mod 4, the a mod 4. If a 2 mod 4, the a mod 4. If a 3 mod 4, the a mod 4. So, a 4 0 or 1 mod 4. Therefore a 4 2 or 3 mod 4. Problem 13: Prove that there is o perfect square a 2 whose last digit is 2, 3, 7 or 8. Solutio: By the property 4(b each iteger umber is cogruet to 0, 1, 2,..., 8 or 9 mod 10. Cosider all these cases ad use property 5 : If a 0 mod 10, the a mod 10. If a 1 mod 10, the a mod 10. If a 2 mod 10, the a mod 10. If a 3 mod 10, the a mod 10. If a 4 mod 10, the a mod 10. If a 5 mod 10, the a mod 10. If a 6 mod 10, the a mod 10. If a 7 mod 10, the a mod 10. If a 8 mod 10, the a mod 10. If a 9 mod 10, the a mod 10. So, a 2 0, 1, 4, 5, 6 or 9 mod 10. Therefore a 2 2, 3, 7 or 8 mod 10, ad the result follows. Problem 14: Prove that is ot a perfect square. Solutio: The last digit is 3, which is impossible by Problem 13. Problem 15: Prove that is ot a perfect square. Solutio 1: The last digit is 2, which is impossible by Problem 13. Solutio 2: We have k+2. Therefore it is cogruet to 2 mod 4 by property 4(a, which is impossible by Problem 11. Problem 16: Prove that there is o perfect square a 2 whose last digits are 85. Solutio: It follows from problem 13 that a 2 5 mod 10 oly if a 5 mod 10. Therefore a 2 85 mod 100 oly if a 5, 15, 25,..., 95 mod 100. If we cosider all these cases ad use property 5 is the same maer as i problem 13, we will see that a 2 25 mod 100. Therefore a 2 85 mod 100, ad the result follows. 23

24 Problem 17: Prove that the equatio x 4 4y 3 has o solutios i iteger umbers. Solutio: Rewrite this equatio as which meas that which is impossible by Problem 12. x 4 4y + 3, x 4 3 mod 4, Problem 18: Prove that the equatio x 2 3y 5 has o solutios i iteger umbers. Solutio: Rewrite this equatio as which meas that x 2 3y + 5, x mod 3. By the property 4(a each iteger umber is cogruet to 0, 1, or 2 mod 3. Cosider all these cases ad use property 5 : If a 0 mod 3, the a mod 3. If a 1 mod 3, the a mod 3. If a 2 mod 4, the a mod 3. So, a 2 0 or 1 mod 3. Therefore a 2 2 mod 3. Problem 19: Prove that the equatio has o solutios i iteger umbers. Solutio: Rewrite this equatio as which meas that 3x 2 4y 5 3x 2 4y + 5, 3x mod 4. O the other had, by Problem 11 we have x 2 0 or 1 mod 4, hece 3x 2 0 or 3 mod 4. Therefore x 2 1 mod 4. Problem 20: Prove that the equatio has o solutios i iteger umbers. x 2 y

25 Solutio: By Problem 11 we have x 2 0 or 1 mod 4, hece x 2 y 2 0, 1 or -1 mod 4. O the other had, mod 4. Therefore x 2 y mod 4, Problem 21: Prove that Solutio: We have 11 1 mod 10, therefore by property 5 we get mod 10, which meas that Problem 22: Prove that Solutio: We have therefore by property 5 we get mod 10, which meas that mod 10, Problem 23: Prove that 23 a for ay a Z + with (a, Solutio: By Fermat s Little theorem we have a 22 1 mod 23, therefore by property 5 we get a mod 23, ad the result follows. Problem 24: Prove that 17 a 80 1 for ay a Z + with (a, Solutio: By Fermat s Little theorem we have a 16 1 mod 17, therefore by property 5 we get a mod 17, ad the result follows. Problem 25: What is the remaider after dividig 3 50 by 7? Solutio: By Fermat s Little theorem we have mod 7, therefore by property 5 we get mod 7, therefore mod 7. 25

26 10. PERMUTATIONS DEFINITION: A permutatio of a set X is a rearragemet of its elemets. EXAMPLE: 1. Let X {1, 2}. The there are 2 permutatios: 12, Let X {1, 2, 3}. The there are 6 permutatios: 123, 132, 213, 231, 312, Let X {1, 2, 3, 4}. The there are 24 permutatios: 1234, 1243, 1324, 1342, 1423, , 2143, 2314, 2341, 2413, , 3241, 3124, 3142, 3421, , 4213, 4321, 4312, 4123, 4132 REMARK: Oe ca show that there are exactly! permutatios of the -elemet set X. DEFINITION : A permutatio of a set X is a oe-oe correspodece (a bijectio from X to itself. NOTATION: Let X {1, 2,..., } ad α : X X be a permutatio. It is coveiet to describe this fuctio i the followig way: α. α(1 α(2... α( EXAMPLE: ( ( ( ( ( CONCLUSION: For a permutatio we ca use two differet otatios. For example, are the same permutatios. ( ad 26

27 DEFINITION: Let X {1, 2,..., } ad α : X X be a permutatio. Let i 1, i 2,..., i r be distict umbers from {1, 2,..., }. If α(i 1 i 2, α(i 2 i 3,..., α(i r 1 i r, α(i r i 1, ad α(i ν i ν for other umbers from {1, 2,..., }, the α is called a r-cycle. NOTATION: A r-cycle is deoted by (i 1 i 2... i r. EXAMPLE: 1 (1 1 cycle (1 1 cycle (12 2 cycle (13 2 cycle (123 3 cycle ( cycle ( cycle (125 3 cycle is ot a cycle REMARK: We ca use differet otatios for the same cycles. For example, (1 (2 (3, (123 (231 ( WARNING: Do ot cofuse otatios of a permutatio ad a cycle. For example, ( Istead, ( ad 123 (1. 27

28 Compositio (Product Of Permutatios Let The α ( α(1 α(2... α( ( α β ( β α ad β α(β(1 α(β(2... α(β( β(α(1 β(α(2... β(α( ( β(1 β(2... β(,.. WARNING: I geeral, α β β α. EXAMPLE: ( Let α α β β α ( , β ( ( ( ( We have: ( ( ,. REMARK: It is coveiet to represet a permutatio as the product of circles. EXAMPLE: ( (1367(49(2(5(8 (1367(49 REMARK: Oe ca fid a compositio of permutatios usig circles. EXAMPLE: Let α (123, β ( α β (123(12 (13(2 (13 β α (12(123 (1(23 (23 28 (12(3 (12. We have: ( ( ,.

29 2. Let We have: α (1532(4 (1532, β (14(2(35 (14( α β (1532(14(35 (1452(3 (1452 β α (14(35(1532 (1324(5 (1324 ( ( ,. THEOREM: The iverse of the cycle α (i 1 i 2... i r is the cycle α 1 (i r i r 1... i 1. EXAMPLE: Let α (15724(36. Fid α We have: I fact, ad THEOREM: α 1 (42751(63 α α 1 (15724(36(42751(63 (1 α 1 α (42751(63(15724(36 (1. Every permutatio α is either a cycle or a product of disjoit (with o commo elemets cycles. Examples 1. Determie which permutatios are equal: (a (12 12 (g (124(53 (53(124 (b (1 12 (h (124(53 (124(35 (c (1(2 (1 (i (124(53 (142(53 (d (12(34 (1234 (j ( (e (12(34 (123(234 (k ( (f (12(34 (123(234(341 (l (

30 2. Factor the followig permutatios ito the product of cycles: ( ( (2 3 10( ( (4 7 8(5 9( Fid the followig products: 4. Let α (135(24, β (124(35. We have: (a αβ (143 (b βα (152 (c β 1 (421(53 (d α 2004 (1 (12(34(56(1234 (24(56 (12(23(34(45 (12345 (12(34(56 (12(34(56 (123(234(345 (12(45 30

31 11. GROUPS DEFINITION: A operatio o a set G is a fuctio : G G G. DEFINITION: A group is a set G which is equipped with a operatio ad a special elemet e G, called the idetity, such that (i the associative law holds: for every x, y, z G, x (y z (x y z; (ii e x x x e for all x G; (iii for every x G, there is x G with x x e x x. EXAMPLE: Set Operatio + Operatio Additioal Coditio N o o Z yes o Q yes o for Q \ {0} R yes o for R \ {0} R \ Q o o EXAMPLE: Set Operatio + Operatio Z >0 o o Z 0 o o Q >0 o yes Q 0 o o R >0 o yes R 0 o o 31

32 EXAMPLE: Set Operatio + Operatio {2 : Z} yes o {2 + 1 : Z} o o {3 : Z} yes o {k : Z}, where k N is some fixed umber yes o {a : Z}, where a R, a 0, ±1, is some fixed umber o yes { p 2 : p Z, Z 0} yes o EXAMPLE: Set Operatio R >0 a b a 2 b 2 o R >0 a b a b o 32

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