Abstract Algebra. Ch 0: Preliminaries. Properties of Integers

Transcription

1 Abstract Algebra Ch 0: Preliminaries Properties of Integers Note: Solutions to many of the examples are found at the end of this booklet, but try working them out for yourself first. 1. Fundamental Theorem of Arithmetic: (Theorem 0.3:) Every integer n > 1 can be written uniquely as a product of increasing primes raised to positive integer powers: i.e., n = p m pr mr where p 1 < p 2 <... < p r are primes, m 1 1, and this is unique. Some questions to think about: Some huge number you factor and I factor - how do you know that we get the same list of factors. Why do we say the prime factors? 2. Well Ordering Principle: Every nonempty set S of positive integers contains a smallest member. Note: Well Ordering Principle can t be proved from usual properties of arithmetic, so taken as an axiom. You might think the Well Ordering Principle is obvious - there is an element s S, so just take everything less than s that is in S and take the smallest element - there are only finitely many, so why would there be a problem? But what if we took Q + (positive rationals) instead of N (natural numbers)? From the point of view of addition and multiplication Q + and N act the same. But Q + doesn t satisfy the Well Ordering Principle: For example, the set S = {1,1/2,1/3,1/4,...} has no smallest member. So the Well Ordering Principle distinguishes between the positive integers and the positive rationals. Some questions to think about: What does it mean to have a smallest element? Why must S be non-empty? Why must the elements of S be positive? Why must the elements of S be integers? The following version of the Well Ordering Principle is often useful: 3. Lemma: Let S be a non-empty set of non-negative integers. Then S has a smallest element. Proof: If 0 S we are done. If not, then use the Well Ordering Principle. 1

2 4. Definition: Let m,n Z,m 0. Then m divides n if n = qm for some integer q. We write m n. We also say that n is a multiple of m or that m is a divisor of n. We write m n if m does not divide n. 5. Definition: A prime is a positive integer which has exactly two distinct natural number divisors. We usually think of this as: A prime is an integer m > 1 which is divisible only by 1 and m. Question: Why isn t 1 a prime? 6. Theorem 0.1: Division Algorithm: Let a, b Z with b > 0. Then there exist unique integers q and r with the property that a = bq + r, where 0 r < b. This is really something that you have been using since about 3rd grade: here r stands for remainde and q stands for quotient. Note that in the division algorithm, the remainder r is smaller than the divisor b. The proof is hard and typical. Idea: subtract multiples of b until a qb is just < b. Note: need to prove both existence and uniqueness. Proof: Existence of q and r: Let S = {a bk : k Z,a bk 0}. If 0 S, then b a so we get the desired result with q = a/b,r = 0. If 0 / S, we first note that S is non-empty: if a > 0, then a b 0 S if a < 0 then a b (2a) = a(1 2b) S since b > 0. Thus we may apply the Well Ordering Principle (actually the Lemma)to conclude that S has a smallest member, say r = a bq. Then a = bq + r and r 0, so all that remains to be proved is that r < b. If r b, then a b(q+1) = a bq b = r b 0, so that a b(q+1) S. But a b(q +1) < a bq and a bq is the smallest member of S. Contradiction! So r < b. Uniqueness of q and r: Suppose there are integers q,q,r,r such that a = bq + r,0 r < b and a = bq + r,0 r < b. Suppose (wlog) that r r. Then bq + r = bq + r and b(q q ) = r r. So b (r r) and 0 r r r < b. Thus r r = 0 and so r = r and q = q. 7. Example: Use the division algorithm on a = 473, b = 2 to find q, r. 2

3 8. Example: Use the division algorithm on a = 591, b = 9 to find q, r. 9. Definition: m is a common divisor of a and b if m a and m b. 10. Example: What are all common divisors of -8 and -12? 11. Definition: The greatest common divisor of two nonzero integers a, b is the largest of all common divisors of a and b. Notation: gcd(a,b). When gcd(a,b) = 1 we say that a and b are relatively prime. (That is, no positive common factor other than 1.) 12. Example: What is gcd( 8, 12)? 13. Example: What is gcd(45, 84)? 14. Example: What is gcd( , )? (OK - so I don t actually expect you to answer this... But this question is to prove to you that the Euclidean algorithm, below, is incredibly useful...) 15. Definition: When a = qb + r where q is the quotient and r is the remainder upon dividing a by b, we write a mod b = r. 16. Lemma: Let a,b N. If a > b then gcd(a,b) = gcd(b,a mod b). Proof: Write a in the form a = qb + r. If c a and c b then c divides b and c divides r = a qb = a mod b. Thus any common divisor of a and b is a common divisor of b and a mod b. Conversely, if c b and c r = a qb = a mod b, then c a = qb + r and c b. So every common divisor of b and r is a common divisor of a and b. In particular, the gcd s have to be the same. (Why? Suppose d = gcd(a,b),d = gcd(b,a mod b). Then d is a common divisor of b,a mod b, so d d. Likewise d d. Thus d = d.) 17. Euclidean algorithm for finding gcd(a,b): Let a,b be positive integers and wlog suppose a > b. Let r 1 = a mod b. If r 1 = 0 then b a so gcd(a,b) = b. If r 1 > 0 then gcd(a,b) = gcd(b,r 1 ) by the lemma. So let r 2 = b mod r 1. If r 2 = 0 then gcd(a,b) = r 1. If r 2 > 0 let r 3 =... The final nonzero remainder r N 1 is the greatest common divisor of a and b. Note: Since the r i s decrease with every step but can never be negative, a remainder r N must eventually equal zero, at which point the algorithm stops. The final nonzero remainder r N 1 is the greatest common divisor of a and b. The number N cannot be 3

4 infinite because there are only a finite number of nonnegative integers between the initial remainder r 1 and zero. 18. Example: Use the Euclidean algorithm to find gcd(22471,3266). 19. Example: Use the Euclidean algorithm to find gcd(243,111). 20. Theorem 0.2: For any nonzero integers a and b, there exist integers s and t such that gcd(a,b) = as + bt. Moreover, gcd(a,b) is the smallest positive integer of the form as + bt. Note: This is an important result! It means that gcd(a,b) is a linear combination of a and b. Proof: Consider the set S = {am + bn : m,n Z,am + bn > 0}. S is nonempty since if some choice of m,n makes am + bn < 0 then replace m,n by m, n. Thus the well ordering principle says that S has a smallest member, say d = as + bt. Claim: d = gcd(a,b). d a: a = dq + r where 0 r < d. We show that r = 0. If r > 0: r = a dq = a (as + bt)q = a asq btq = a(1 sq) + b( tq) S (this is in S because it is of the form as + bt and it is > 0 because it is equal to r > 0.) Thus r S and r < d contradiction. Thus r = 0. d b: similar Thus d is a common divisor of a and b. Now suppose d is another common divisor of a and b and write a = d h,b = d k. Then d = as+bt = (d h)s+(d k)t = d (hs + kt), so d d. Thus among all common divisors of a,b the number d is greatest. 21. Corollary: If a and b are relatively prime, then there exist integers s and t such that as + bt = To find the integers s,t in the previous theorem: Use the Euclidean algorithm (it is convenient to set r 0 = b). Then a = q 1 r 0 + r 1 r 0 = q 2 r 1 + r 2 r n 2 = q n r n 1 + r n r 1 = q 3 r 2 + r 3 r n 1 = q n+1 r n Now start with the next-to-last equation, solve for the remainder in each equation back to the first. Recall that r n is gcd(a,b). 4

5 r n = r n 2 q n r n 1 r 3 = r 1 q 3 r 2 r n 1 = r n 3 q n 1 r n 2 r 2 = r 0 q 2 r 1 r 1 = a q 1 r 0 Now substitute each succeeding equation into the previous. 23. Example: Write gcd(243,111) as a linear combination of 243 and Example: Write gcd(22471,3266) as a linear combination of and Euclid s Lemma: If p is a prime that divides ab then p a or p b. Proof: Suppose p is a prime that divides ab but does not divide a. (WTS p b). Since p a, there are integers s,t such that 1 = as + pt. (Why? Because p s only divisors are 1 and p so a isn t a divisor of p and p isn t a divisor of a...) Then b = abs + ptb and since p divides the right side of this equation, p also divides b. 26. Definition: The least common multiple of two non-zero integers a and b is the smallest positive integer that is a multiple of both a and b. Notation: lcm(a,b). 27. Example: Find lcm(15, 24). 5

6 Ch 0: Preliminaries Modular Arithmetic 1. Recall that when a = qb + r where q is the quotient and r is the remainder upon dividing a by b, we write a mod b = r. 2. Example: Find 7 mod 3, 31 mod 4,25 mod Note: if a,b Z, n Z,n 2, then a mod n = b mod n iff n (a b). (This is exercise 9 on page 22.) 4. Exercise 11 on page 22: If a, b integers, then for addition and multiplication, we can mod first - i.e. (a + b) mod n = (a mod n + b mod n) mod n and (a b) mod n = (a mod n b mod n) mod n. 5. Example: mod 3 can be figured out by 2 2 mod 3 = 1. 6

7 Ch 0: Preliminaries Mathematical Induction 1. Theorem 0.4: First Principle of Mathematical Induction: Let S be a set of integers containing a. Suppose S has the property that whenever some integer n a belongs to S, then the integer n+1 also belongs to S. Then S contains every integer greater than or equal to a. 2. Example: 2 2n is divisible by 3 for all n 1. Proof: Let S = {n : 2 2n is divisible by 3}. 1 S because = 3. Assume n S, to show that n + 1 S: 2 2(n+1) 1 = 2 2n = n = 2 2 (2 2n 1) Example: If a prime p divides a 1... a n then p a i for some i. Proof: Let S = {n 1 : if prime p divides a 1...a n then p divides some a i.}. That 1 S is obvious. If n S, to show n+1 S : If p (a 1... a n+1 ), then p (a 1... a n ) a n+1 so by Euclid s lemma, p (a 1... a n ) or p a n+1. If p a n+1 then done. If not, p (a 1... a n ). But n S so p a i for some i. Either way, p a i for some i. 4. Example: n = n + 1 for all positive integers Proof : Let S = {n : n = n + 1}. If n S, to show n + 1 S: n = n + 1 implies n + 1 = (n + 1) + 1. What is wrong with this proof? 5. Example: Let P(x) = x(x 1)(x 2) (x ). Then P(n) = 0 for all n N. Proof : Let S{n : P(n) = 0}. Then 1 S,2 S,3 S, etc. So all n in S. What is wrong with this proof? 6. Theorem 0.5: Second Principle of Mathematical Induction: Let S be a set of integers containing a. Suppose S has the property that n belongs to S whenever every integer less than n and greater than or equal to a belongs to S. Then S contains every integer greater than or equal to a. 7. Example: Fundamental Theorem of Arithmetic: Every integer n > 1 can be written uniquely as a product of increasing primes raised to positive integer powers: i.e., n = p m p mr r where p 1 < p 2 <... < p r are primes, m 1 1, and this is unique. Proof of Existence: Let S = {n 2 : n factors into a product of primes}. Note 2 S. Now assume that for some integer n, S contains all integers k with 2 k n. WTS n + 1 S. If n + 1 is prime then n + 1 S by definition. If n + 1 is not prime 7

8 then n = a b where 2 a,b < n + 1 so a S and b S and thus both factor as primes and thus n + 1 S. Proof of Uniqueness: Let Ŝ = {n > 2 : n has more than one factorization into powers of primes.} If Ŝ = we are done. So assume Ŝ and take the smallest integer n Ŝ (can do by well ordering principle). Thus n has two factorizations, but no smaller integer has. Thus n = p 1 m1... p s m s = q 1 n1... q t n t,p 1 < p 2 <... < p s,q 1 <... < q t. Now p 1 n so p 1 divides q 1 n1... q t n t so divides one of the q i by example 3 above. But q i is a prime so p 1 equals one of the q i. But the same argument shows q 1 divides one of the p j. Now p 1 = q i q 1 = p j (note: because q i could be q 1, so p 1 p j, so j = 1. Thus q 1 = p 1. Let n = m p 1 so m = p 1 m p s m s = q 1 n q t n t. But m < n so m / Ŝ so s = t and p i = q i,m i = n i. Contradiction. 8

9 Ch 0: Preliminaries Equivalence Relations 1. Definition: A relation R on a set S is a true statement about ordered pairs of elements of S. (This is an informal definition.) 2. Example: S= set of all people, relation is child of. Then (a,b) R means a is a child of b. 3. Example: S = R, (a,b) R means a b Definition: An equivalence relation is a relation R on a set S such that the following three properties hold: (a) reflexive: (a,a) R for all a S. (b) symmetric: (a,b) R implies (b,a) R (c) transitive: (a,b) R and (b,c) R implies (a,c) R 5. Example: Which of the three properties do the relations of child of and a b 2 defined above have? 6. When R is an equivalence relation on a set S, we usually write arb instead of (a,b) R. Or, more typically, we use a symbol. Then the three properties are a a,a b b a,a b,b c a c. 7. Definition: If is an equivalence relation on a set S and a S then the set [a] = {x S : x a} is called the equivalence class of a. 8. Example: S is all differentiable functions on R, f g if f = g. Find [f]. 9. Example: S = Z, a b iff a = b mod n. Find [a]. 10. Example: S = all people, a b iff a and b live at the same address. Find [a]. 11. Example: Prove that if a b then [a] = [b]. Proof: Suppose a b and let x [a]. Then x a. But a b so by transitive property x b, so x [b]. Now do vice versa. 12. Definition: A partition of a set S is a collection of nonempty pairwise disjoint subsets of S whose union is S. (see picture on page 17.) 13. Example: Partition the integers into two sets. 14. Example: Do the sets {primes}, {composites} partition N? 9

10 15. Theorem 0.6: The equivalence classes of an equivalence relation on a set S constitute a partition of S. Conversely, for any partition P of S, there is an equivalence relation on S whose equivalence classes are the elements of P. Proof: Let be equivalence relation on set S. For any a S, a [a] by reflexivity, so [a] and the union of all equivalence classes is S. Now suppose [a],[b] are distinct equivalence classes but not pairwise disjoint - so let c [a] [b] (to get a contraction). Let x [a]. Then c a,c b,x a. Thus by symmetric property a c and by transitivity x c so by transitivity again, x b. Thus x [a] x [b] so [a] [b]. Similarly, [b] [a] so [a] = [b], so contradiction to distinct equivalence classes. To prove converse: Let P be a collection of nonempty pairwise disjoint subsets of S whose union is S. Define a b if a,b belong to the same subset in the collection. WTS is an equivalence relation. ( This is exercise 55.) 10

11 Ch 0: Preliminaries Functions (Mappings) 1. Definition: A function (or mapping) φ from a set A to a set B is a rule that assigns to each element a A exactly one element b B. The set A is called the domain of φ and the set B is called the range of φ. If φ(a) = b then b is called the image of a under φ. The subset of B comprising all the images of elements of A is called the image of A under φ. The notation is φ : A B. 2. Definition: Let φ : A B,ψ : B C. The composition ψ φ = ψφ is the mapping from A to C defined by ψφ(a) = ψ(φ(a)) for all a A. Note that this book doesn t use for composition (but I sometimes will). 3. Definition: A function φ : A B is called one-to-one if for every a 1,a 2 A, φ(a 1 ) = φ(a 2 ) implies a 1 = a 2. The function is called onto B if for each b B there is an a A such that φ(a) = b. 4. Theorem 0.7: Basic Facts about Functions: Given functions α : A B,β : B C,γ : C D, then (a) γ (β α) = (γ β) α (associativity). (b) If α, β are one-to-one then so is β α. (c) If α,β are onto then so is β α. (d) If α is one-to-one and onto then there is a function α 1 from B onto A such that (α 1 α)(a) = a for all a A and (α α 1 )(b) = b for all b B. (I.e. the function composed with its inverse is the identity and vice versa.) The function α 1 is the inverse function of α. The function α 1 is one-to-one and onto. ( ) Proof of part (a): Let a A. Then γ (β α)(a) = γ β(α)(a) = γ(β(α))(a). On the other hand, (γ β) α(a) = (γ β(α(a)) = γ(β(α(a)). Thus the equality holds. 11

12 Ch 0: Preliminaries Solutions to Examples Properties of Integers Example 7: Use the division algorithm on a = 473,b = 2 to find q,r. Solution: 472= so the quotient q = 236 and the remainder r = 1. Example 8: Use the division algorithm on a = 591,b = 9 to find q,r. Solution: -591=9*-66+3 so q = 66 and r = 3. Remember that r 0. Example 10: What are all common divisors of -8 and -12? Solution: ±1, ±2, ±4 Example 12: What is gcd( 8, 12)? Solution: by example 10 above, it is 4. Example 13: What is gcd(45,84)? Solution: 45 = 3 2 5,84 = so gcd(45,84) = 3. Example 14: What is gcd( , )? Solution: 15 (why? :-)) Example 18: Use the Euclidean algorithm to find gcd(22471,3266). Solution: 22471= = = = = = Thus the gcd is the last non-zero remainder: gcd(22471,3266) =

13 Solutions to Examples Example 19: Use the Euclidean algorithm to find gcd(243,111). Solution: 243= = = =3 2+0 Thus the gcd is 3. Example 23: Write gcd(243,111) as a linear combination of 243 and 111. Solution: We work the solution from example 19 backwards, solving for the gcd of 3 as a linear combination and substituting up, all the while, gathering like terms... 3 = = 21-( ) 3= = ( ) = Thus 3 = Example 24: Write gcd(22471,3266) as a linear combination of and Solution: This does take some practice but isn t hard = = 138-( ) 1= = ( ) 3-391= = ( ) 22= = ( ) = Thus 23 = Example 27: Find lcm(15,24). Solution: 15 = 3 5,24 = 2 3 3, so the lcm is = 120. Modular Arithmetic Example 2: Find 7 mod 3, 31 mod 4, 25 mod 26. Solution: 7 mod 3 = 1, 31 mod 4 = 1,25 mod 26 = 25 13

14 Solutions to Examples Mathematical Induction Example 4: What is wrong with this proof? We didn t check the base case, and when n = 1, = 2. Example 5: Let P(x) = x(x 1)(x 2) (x ). Then P(n) = 0 for all n N. Proof : What is wrong with this proof? We didn t prove that if it is true for GENERAL n then it is true for n + 1, so for example for n > this isn t true. Equivalence Relations Example 5: Which of the three properties do the relations of child of and a b 2 defined above have? Solution: child of is not any of these, a b 2 is reflexive and transitive but not symmetric. (You should make sure to prove this to yourself!) Example 8: S is all differentiable functions on R, f g if f = g. Find [f]. Solution: [f] = {f + C : C is a constant}. Example 9: S = Z, a b iff a = b mod n. Find [a]. Solution: [a] = {a + qn : q Z} Example 10: S = all people, a b iff a and b live at the same address. Find [a]. Solution: [a] = { all people living at the same address as a}. Example 13: Partition the integers into two sets. Solution: Z = {evens} {odds}. Or Z = {n Z : n 5} {n Z : n < 5.23}, etc. Example 14: Do the sets {primes}, {composites} partition N? Solution: No - even though they are disjoint, the number 1 doesn t appear in either set, so Z isn t partitioned. 14