1. MATHEMATICAL INDUCTION

1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1 + 2 + 3 +... + k k(k + 1. 2 STEP 3: Prove that (1.1 is true for k + 1, that is We have 1 + 2 + 3 +... + k + (k + 1 ST.2 1 + 2 + 3 +... + k + (k + 1? k(k + 1 2 (k + 1(k + 2. 2 k + (k + 1 (k + 1 2 + 1 (k + 1(k + 2. 2 EXAMPLE 2: Prove that for ay iteger 1. Proof: STEP 1: For 1 (1.2 is true, sice 1 1 2. 1 + 3 + 5 +... + (2 1 2 (1.2 STEP 2: Suppose (1.2 is true for some k 1, that is 1 + 3 + 5 +... + (2k 1 k 2. STEP 3: Prove that (1.2 is true for k + 1, that is 1 + 3 + 5 +... + (2k 1 + (2k + 1? (k + 1 2. We have: 1 + 3 + 5 +... + (2k 1 + (2k + 1 ST.2 k 2 + (2k + 1 (k + 1 2. 1

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 0. 8 3 2 1 (1.4 Proof: STEP 1: For 0 (1.4 is true, sice 8 3 0 1. 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+1 1 3 2k+2 1 3 2k 9 1 3 2k (8 + 1 1 3} 2k {{ 8} + } 3 2k {{ 1 }. div.by 8 St. 2 div.by 8 EXAMPLE 5: Prove that for ay iteger 1. 7 7 (1.5 Proof: STEP 1: For 1 (1.5 is true, sice 7 1 7 1. 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 + 1 7 (k + 1. We have (k + 1 7 (k + 1 k 7 + 7k 6 + 21k 5 + 35k 4 + 35k 3 + 21k 2 + 7k + 1 k 1 k 7 k } {{ } St. 2 div. by 7 + 7k } 6 + 21k 5 + 35k 4 {{ + 35k 3 + 21k 2 + 7k }. div. by 7 2

2. THE BINOMIAL THEOREM DEFINITION: Let ad k be some itegers with 0 k. The! k k!( k! is called a biomial coefficiet. PROPERTIES: 1. 0 1. Proof: We have 0! 0!( 0!! 1! 1,!!(!!! 0!!! 1 1. 2.. 1 1 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. + 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 + 2 + k 1 k k + 1 + 2. k + 1 Proof: By property 4 we have + 2 + k 1 k + k + 1 k 1 + 1 + k + k + 1 k + 1 + k k + 1 + 2. 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 2 +... + a 2 b 2 + ab 1 + b. 1 2 2 1 4

PROBLEM: For all itegers 1 we have + 0 + 1 +... + 2 2. Proof: Puttig a b 1 i the Theorem above, we get (1 + 1 1 + hece 1 1 1 + 1 2 1 + 1 2 1 2 +... + 1 2 1 2 + 1 1 1 + 1, 2 2 1 + 1 therefore by property 1 we get 2 + + 0 1 +... + + + 1, 2 2 1 +... + + + 2 2 1. PROBLEM: For all itegers 1 we have 0 + 1... + ( 1 0. 2 Proof: Puttig a 1 ad b 1 i the Theorem above, we get hece (1 1 1 + 1 1 ( 1 + 1 0 1 + 1 1 2 ( 1 2 +... + 1 ( 1 1 + ( 1, 2 1 2... + ( 1 1 + ( 1, 1 therefore by property 1 we get 0 +... + ( 1 1 + ( 1. 0 1 2 1 5

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

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

PROBLEM 4: Prove that 1 3 2 + 5 is irratioal. Proof: Assume to the cotrary that 1 3 2 + 5 is ratioal, that is where p ad q are itegers ad q 0. The 1 p 2 + 5 3 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

4. DIVISION ALGORITHM PROBLEM: Prove that 3 is irratioal. Proof: Assume to the cotrary that 3 is ratioal, that is 3 p q, where p ad q are itegers ad q 0. Moreover, let p ad q have o commo divisor > 1. The 3 p2 q 2 3q 2 p 2. Sice 3q 2 is divisible by 3, it follows that p 2 is divisible by 3. The p is also divisible by 3 (i fact, if p is ot divisible by 3, the...??? THEOREM (DIVISION ALGORITHM: For ay itegers a ad b with a 0 there exist uique itegers q ad r such that b aq + r, where 0 r < a. The itegers q ad r are called the quotiet ad the remider respectively. EXAMPLE 1: Let b 49 ad a 4, the 49 4 12 + 1, so the quotiet is 12 ad the remider is 1. REMARK: We ca also write 49 as 3 12 + 13, but i this case 13 is ot a remider, sice it is NOT less tha 3. EXAMPLE 2: Let a 2. Sice 0 r < 2, the for ay iteger umber b we have ONLY TWO possibilities: b 2q or b 2q + 1. So, thaks to the Divisio Algorithm we proved that ay iteger umber is either eve or odd. EXAMPLE 3: Let a 3. Sice 0 r < 3, the for ay iteger umber b we have ONLY THREE possibilities: b 3q, b 3q + 1, or b 3q + 2. Proof of the Problem: Assume to the cotrary that 3 is ratioal, that is 3 a b, where a ad b are itegers ad b 0. Moreover, let a ad b have o commo divisor > 1. The 3 a2 b 2 3b 2 a 2. (4.1 9

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 + 1 2 9q 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 + 2 2 9q 2 + 12q + 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

5. GREATEST COMMON DIVISOR AND EUCLID S LEMMA PROBLEM: Prove that 101 is irratioal. Proof: Assume to the cotrary that 101 is ratioal, that is 101 a b, where a ad b are itegers ad b 0. Moreover, let a ad b have o commo divisor > 1. The 101 a2 b 2 101b 2 a 2. Sice 101b 2 is divisible by 101, it follows that a 2 is divisible by 101. The a is also divisible by 101. I fact, if a is ot divisible by 101, the by the Divisio Algorithm there exists q Z such that a 101q + 1 or a 101q + 2 or a 101q + 3 or a 101q + 4...??? DEFINITION: If a ad b are itegers with a 0, we say that a is a divisor of b if there exists a iteger q such that b aq. We also say that a divides b ad we deote this by a b. EXAMPLE: We have: 4 12, sice 12 4 3 4 15, sice 15 4 3.75 DEFINITION: A commo divisor of ozero itegers a ad b is a iteger c such that c a ad c b. The greatest commo divisor (gcd of a ad b, deoted by (a, b, is the largest commo divisor of itegers a ad b. EXAMPLE: The commo divisors of 24 ad 84 are ±1, ±2, ±3, ±4, ±6, ad ±12. Hece, (24, 84 12. Similarly, lookig at sets of commo divisors, we fid that (15, 81 3, (100, 5 5, (17, 25 1, ( 17, 289 17, etc. THEOREM: If a ad b are ozero itegers, the their gcd is a liear combiatio of a ad b, that is there exist iteger umbers s ad t such that sa + tb (a, b. Proof: Let d be the least positive iteger that is a liear combiatio of a ad b. We write d sa + tb, (5.1 11

where s ad t are itegers. We first show that d a. By the Divisio Algorithm we have From this ad (5.1 it follows that a dq + r, where 0 r < d. r a dq a q(sa + tb a qsa qtb (1 qsa + ( qtb. This shows that r is a liear combiatio of a ad b. Sice 0 r < d, ad d is the least positive liear combiatio of a ad b, we coclude that r 0, ad hece d a. I a similar maer, we ca show that d b. We have show that d is a commo divisor of a ad b. We ow show that d is the greatest commo divisor of a ad b. Assume to the cotrary that (a, b d ad d > d. Sice d a, d b, ad d sa + tb, it follows that d d, therefore d d. We obtai a cotradictio. So, d is the greatest commo divisor of a ad b ad this cocludes the proof. DEFINITION: A iteger 2 is called prime if its oly positive divisors are 1 ad. Otherwise, is called composite. EXAMPLE: Numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59... are prime. THEOREM (Euclid s Lemma: If p is a prime ad p ab, the p a or p b. More geerally, if a prime p divides a product a 1 a 2... a, the it must divide at least oe of the factors a i. Proof: Assume that p a. We must show that p b. By the theorem above, there are itegers s ad t with sp + ta (p, a. Sice p is prime ad p a, we have (p, a 1, ad so sp + ta 1. Multiplyig both sides by b, we get Sice p ab ad p spb, it follows that spb + tab b. (5.2 p (spb + tab. This ad (5.2 give p b. This completes the proof of the first part of the theorem. The secod part (geeralizatio easily follows by iductio o 2. COROLLARY: If p is a prime ad p a 2, the p a. Proof: Put a b i Euclid s Lemma. 12

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

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 1 1... p e. Proof: Just collect like terms i a prime factorizatio. EXAMPLE: 120 2 3 3 5. 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

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, 78 2. PROBLEM: Fid (252, 198. 326 78 4 + 14 78 14 5 + 8 14 8 1 + 6 8 6 1 + 2 6 2 3 Solutio: By the Euclidea Algorithm we have therefore (252, 198 18. 252 198 1 + 54 198 54 3 + 36 54 36 1 + 18 36 18 2 15

PROBLEM: Fid (4361, 42371. Solutio: By the Euclidea Algorithm we have therefore (4361, 42371 7. 42371 9 4361 + 3122 4361 1 3122 + 1239 3122 2 1239 + 644 1239 1 644 + 595 644 1 595 + 49 595 12 49 + 7 49 7 7 + 0, THEOREM: Let a p e 1 1... p e ad b p f 1 1... p f (a, b p mi(e 1,f 1 1... p mi(e,f. be positive itegers. The EXAMPLE: Sice 720 2 4 3 2 5 ad 2100 2 2 3 5 2 7, we have: (720, 2100 2 2 3 5 60. PROBLEM: Let a Z. Prove that (2a + 3, a + 2 1. Proof: By the Lemma above we have (2a + 3, a + 2 (a + 1 + a + 2, a + 2 (a + 1, a + 2 (a + 1, a + 1 + 1 (a + 1, 1 1. PROBLEM: Let a Z. Prove that (7a + 2, 10a + 3 1. Proof: By the Lemma above we have (7k + 2, 10k + 3 (7k + 2, 7k + 2 + 3k + 1 (7k + 2, 3k + 1 (6k + 2 + k, 3k + 1 (k, 3k + 1 (k, 1 1. 16

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 + 1... p l!(p l! (l + 1... p, (p l! therefore p (p l! (l + 1... 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 1 + 1 ( p + p p k p 2 +... + k + 1 k 1 2 p 1 p p k p 1 + k p 2 +... + k. 1 2 p 1 } {{ } div. by p by Lemma 17

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, 25 16 mod 9, 43 27 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 1. 5. 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 +. 6. 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

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. (9.3 2 3... (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

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 2 2 1 mod 3, we get x 1 2 2 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 2 4 1 mod 7, we get x 5 4 6 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 3 3 1 mod 8, we get x 4 3 12 4 mod 8. Problem 4: Fid all solutios of the followig cogruece 2x 5 mod 8. 20

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 2 2 1 mod 3, we get x 1 2 2 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 2 3 1 mod 5, we get x 1 3 3 mod 5. Problem 8: Fid all solutios of the cogruece 9x + 23 28 mod 25. 21

Solutio: We first rewrite this cogruece as 9x 5 mod 25. Note that (9, 25 1. Therefore we ca apply the theorem above. Sice 9 14 1 mod 25, we get x 5 14 70 20 mod 25. Problem 9: What is the last digit of 345271 79399? Solutio: It is obvious that therefore by property 5 we have 345271 1 mod 10, 345271 79399 1 79399 1 mod 10. This meas that the last digit of 345271 79399 is 1. Problem 10: What is the last digit of 4321 4321? Solutio: It is obvious that therefore by property 5 we have 4321 1 mod 10, This meas that the last digit is 1. 4321 4321 1 4321 1 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 2 0 2 0 mod 4. If a 1 mod 4, the a 2 1 2 1 mod 4. If a 2 mod 4, the a 2 2 2 0 mod 4. If a 3 mod 4, the a 2 3 2 1 mod 4. So, a 2 0 or 1 mod 4. Therefore a 2 2 or 3 mod 4. 22

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 4 0 4 0 mod 4. If a 1 mod 4, the a 4 1 4 1 mod 4. If a 2 mod 4, the a 4 2 4 0 mod 4. If a 3 mod 4, the a 4 3 4 1 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 2 0 2 0 mod 10. If a 1 mod 10, the a 2 1 2 1 mod 10. If a 2 mod 10, the a 2 2 2 4 mod 10. If a 3 mod 10, the a 2 3 2 9 mod 10. If a 4 mod 10, the a 2 4 2 6 mod 10. If a 5 mod 10, the a 2 5 2 5 mod 10. If a 6 mod 10, the a 2 6 2 6 mod 10. If a 7 mod 10, the a 2 7 2 9 mod 10. If a 8 mod 10, the a 2 8 2 4 mod 10. If a 9 mod 10, the a 2 9 2 1 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 444444444444444444443 is ot a perfect square. Solutio: The last digit is 3, which is impossible by Problem 13. Problem 15: Prove that 888... 882 is ot a perfect square. Solutio 1: The last digit is 2, which is impossible by Problem 13. Solutio 2: We have 888... 882 4k+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

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 2 5 2 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 2 0 2 0 mod 3. If a 1 mod 3, the a 2 1 2 1 mod 3. If a 2 mod 4, the a 2 2 2 1 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 2 5 1 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 2 2002 24

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, 2002 2 mod 4. Therefore x 2 y 2 2002 mod 4, Problem 21: Prove that 10 11 10 1. Solutio: We have 11 1 mod 10, therefore by property 5 we get 11 10 1 10 1 mod 10, which meas that 10 11 10 1. Problem 22: Prove that 10 101 2003 1. Solutio: We have therefore by property 5 we get 101 1 mod 10, which meas that 10 101 2003 1. 101 2003 1 2003 1 mod 10, Problem 23: Prove that 23 a 154 1 for ay a Z + with (a, 23 1. Solutio: By Fermat s Little theorem we have a 22 1 mod 23, therefore by property 5 we get a 22 7 1 7 1 mod 23, ad the result follows. Problem 24: Prove that 17 a 80 1 for ay a Z + with (a, 17 1. Solutio: By Fermat s Little theorem we have a 16 1 mod 17, therefore by property 5 we get a 16 5 1 5 1 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 3 6 1 mod 7, therefore by property 5 we get 3 6 8 1 48 1 mod 7, therefore 3 50 9 2 mod 7. 25

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, 21. 2. Let X {1, 2, 3}. The there are 6 permutatios: 123, 132, 213, 231, 312, 321. 3. Let X {1, 2, 3, 4}. The there are 24 permutatios: 1234, 1243, 1324, 1342, 1423, 1432 2134, 2143, 2314, 2341, 2413, 2431 3214, 3241, 3124, 3142, 3421, 3412 4231, 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: 1 2... α. α(1 α(2... α( EXAMPLE: ( 1 2 2 1 ( 1 2 3 1 2 3 ( 1 2 3 2 3 1 ( 1 2 3 4 1 4 3 2 ( 1 2 3 4 5 3 5 4 1 2 CONCLUSION: For a permutatio we ca use two differet otatios. For example, 24513 are the same permutatios. ( 1 2 3 4 5 2 4 5 1 3 ad 26

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 2 (1 1 cycle 1 2 1 2 (12 2 cycle 2 1 1 2 3 (13 2 cycle 3 2 1 1 2 3 (123 3 cycle 2 3 1 1 2 3 4 (1423 4 cycle 4 3 1 2 1 2 3 4 5 (13425 5 cycle 3 5 4 2 1 1 2 3 4 5 (125 3 cycle 2 5 3 4 1 1 2 3 4 5 is ot a cycle 2 5 4 3 1 REMARK: We ca use differet otatios for the same cycles. For example, 1 2 3 1 2 3 (1 (2 (3, (123 (231 (312. 1 2 3 2 3 1 WARNING: Do ot cofuse otatios of a permutatio ad a cycle. For example, (123 123. Istead, (123 231 ad 123 (1. 27

Compositio (Product Of Permutatios Let The α ( 1 2... α(1 α(2... α( ( α β ( β α ad β 1 2... α(β(1 α(β(2... α(β( 1 2... β(α(1 β(α(2... β(α( ( 1 2... β(1 β(2... β(,.. WARNING: I geeral, α β β α. EXAMPLE: ( 1 2 3 4 5 Let α 5 1 2 4 3 α β β α ( 1 2 3 4 5, β 4 2 5 1 3 ( 1 2 3 4 5 5 1 2 4 3 ( 1 2 3 4 5 4 2 5 1 3 ( 1 2 3 4 5 4 2 5 1 3 ( 1 2 3 4 5 5 1 2 4 3. We have: ( 1 2 3 4 5 4 1 3 5 2 ( 1 2 3 4 5 3 4 2 1 5,. REMARK: It is coveiet to represet a permutatio as the product of circles. EXAMPLE: ( 1 2 3 4 5 6 7 8 9 3 2 6 9 5 7 1 8 4 (1367(49(2(5(8 (1367(49 REMARK: Oe ca fid a compositio of permutatios usig circles. EXAMPLE: 1 2 3 1. Let α (123, β 2 3 1 ( 1 2 3 2 1 3 α β (123(12 (13(2 (13 β α (12(123 (1(23 (23 28 (12(3 (12. We have: ( 1 2 3 3 2 1 ( 1 2 3 1 3 2,.

2. Let We have: 1 2 3 4 5 α (1532(4 (1532, 5 1 2 4 3 1 2 3 4 5 β (14(2(35 (14(35. 4 2 5 1 3 α β (1532(14(35 (1452(3 (1452 β α (14(35(1532 (1324(5 (1324 ( 1 2 3 4 5 4 1 3 5 2 ( 1 2 3 4 5 3 4 2 1 5,. 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: 1 2 3 4 5 6 7 Let α (15724(36. Fid α 5 4 6 1 7 3 2 1. 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 (12345 12345 (e (12(34 (123(234 (k (12345 23451 (f (12(34 (123(234(341 (l (23451 23451 29

2. Factor the followig permutatios ito the product of cycles: 1 2 3 4 5 6 7 8 (4 5 1 2 3 5 4 6 7 8 1 2 3 4 5 6 7 8 9 10 11 12 (1 5 11 8 9(2 3 10(6 12 7 5 3 10 4 11 12 6 9 1 2 8 7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (3 12 10(4 7 8(5 9(6 14 13 11 1 2 12 7 9 14 8 4 5 3 6 10 11 13 15 3. 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

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

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