Sections 2.4-3.10 (no 3.9)

Definition 3.1.7- Subsets

(A ⊆ B) ⇔ (∀x)(x∈A ￫ x∈B) ⇔ (∀x∈A)(x∈B)

Theorem 3.4.6 (PMI)- Suppose S is a subset of the positive integers with the following properties:

(I1) 1 ∈ S (I2) If n >= 1 and n ∈ S, then n + 1 ∈ S Then S = N (natural numbers)

Theorem 3.5.17 (SPMI)- Suppose S is a subset of the positive integers that has properties:

(K1) 1 ∈ S (K2) If n >= 2 and 1, 2, ..., n - 1 ∈ S, then n ∈ S. Then S = N (natural numbers).

Exercise 3.7.11- Suppose ≡ defines an equivalence relation on a set S. Define F = {[x] : x ∈ S}. That is, F is the set of all equivalence classes generated by letting x take on all values in the set S. Prove that F satisfies properties P2 and P3 from the definition of a partition.

(P1) Pick any [x] ∈ F. Since x ≡ x, it follows that x ∈ [x]. Thus [x] is not empty. (P2) Pick [x], [y] ∈ F and suppose [x]∩[y] is not empty. Then there exists z ∈ [x] ∩[y]. By definition of [x] and [y], we have z ≡ x and z ≡ y. By the symmetric and transitive properties of an equivalence relation, it follows that x ≡ y. Therefore every element equivalent to y is also equivalent to x, and vice versa, which implies that [y]=[x]. (P3) Since every set in F is defined to be a subset of S, we have ⋃x∈S[x] ⊆ S by a previous exercise. So we must show that S ⊆ ⋃x∈S[x]. Let x0 ∈ S. Then x0 ∈ [x0], so x0 ∈ ⋃x∈S[x]. This implies the desired inclusion.

Exercise 3.1.10- Construct a negation of the following statements.

(a) A ⊆ BThe statement A ⊆ B means that for all x, if x ∈ A then x ∈ B. Thus a negation would be: There exists x ∈ A such that x /∈ B. (b) A ⊂ BThe statement A ⊂ B, in our textbook, means that A ⊆ B and B ⊆ A. Thus a negation is: Either A contains an element that is not in B, or every element of B is also in A. (c) A = BA = B means (A ⊆ B)∧(B ⊆ A), so a negation would be: Either there exists x ∈ B −A or there exists x ∈ A − B. Another possibility: A ∆ B = ∅. (d) A ⊆ B ∪ C Negation: there exists x ∈ A such that x /∈ B and x /∈ C. (e) A ∩ B ⊆ C ∪ D Negation: There exists x such that x ∈ A and x ∈ B, but x is not in either C or D. (f) For all sets A, B, and C, if A ∪ C ⊆ B ∪ C, then A ⊆ B. Negation: There exist sets A, B, and C such that A ∪ C ⊆ B ∪ C, but A ⊆ B. (g) For all sets A and B, if A ⊆ B, then A ∩ B = A. Negation: There exist sets A and B such that A ⊆ B but either A∩B ⊆ A or A ⊆ A∩B. Another possibility, because A ∩B is by definition a subset of A: There exist sets A and B such that A ⊆ B but A ⊆ A ∩ B.

Exercise 3.1.13- In this exercise, A and B represent sets, and x is an unspecified element. In each pair of statements below, one is stronger than the other. State which one is stronger.

(a) x ∈ A x ∈ A ∪ B The first statement is stronger: if x ∈ A, then x ∈ A ∪ B. (b) x ∈ A x ∈ A ∩ B The second statement is stronger: if x ∈ A ∩ B, then x ∈ A. (c) x ∈ A x ∈ A − B The second statement is stronger: if x ∈ A − B, then x ∈ A. (d) x ∈ A ∆ B x ∈ A − B The second statement is stronger: if x ∈ A − B, then x ∈ A ∆ B. (e) A ⊆ B A = B The second statement is stronger: if A = B, then A ⊆ B.

Definition 3.1.4- Difference

A - B = A ∩ B^c = {x: x ∈ A ^ x ∉ B}

Definition 3.1.9- If A and B are sets, we say that A=B provided A ⊆ B and B ⊆ A.

A = B ⇔ (A ⊆ B) ^ (B ⊆ A) ⇔ (∀x)(x∈A ￫ x∈B) ^ (∀x)(x∈B ￫ x∈A)

Definition 3.1.5- Symmetric Difference

A Δ B = {x: x ∈ A ∩ B^c v x ∈ A^c ∩ B}

Definition 3.1.3- Intersection of a Set

A ∩ B = {x: x ∈ A v x ∈ B}

Definition 3.1.2- Union of a Set

A ∪ B = {x: x ∈ A v x ∈ B}

Definition 3.1.1- Complement of a Set

A^c = {x : x ∈ U and x ∉ A}

Linear Combination

An expression of the form mb+nc is called a linear combination of b and c.

(A19) Well Ordering Principle

Any non-empty subset of whole numbers has a smallest element. That is, if A is non-empty, then there is non-empty, then there is some number a e A with the property that a<= x for all x e A.

Collection vs Indexes of Sets

Collection: a "set" of sets Indexes: the subscripts used to address the sets, i.e. for the family {A_1, A-2, A-3, ..., A_10}, the set of indexes would be {1, 2, 3, ... 10}

Theorem 3.5.19 (Fundamental Theorem of Arithmetic)-

Every integer n >= 2 can be written as the product of primes, and this factorization is unique, except perhaps for the order in which the factors are written.

Theorem 3.2.5- For any set A,

For any set A, A ∪ Ø = A

Definition 3.8.2- Rational numbers

For two rational numbers p/q and r/s, we define p/q ≡ r/s provided ps=qr.

Definition 3.8.6- Addition with rational numbers

Given two rational numbers p/q and r/s, we define addition ⊕ in the following way. p/q ⊕ r/s = (ps +qr)/qs

Definition 3.8.9- Multiplication with rational numbers

Given two rational numbers p/q and r/s, we define multiplication Ⓧ in the following way. p/q Ⓧ r/s = pr/qs

Definition 3.1.11 (Disjoint Sets)

Given two sets A and B, we say that they are disjoint provided A ∩ B is empty.

Theorem 3.2.4 (DeMorgan's Law)

If A and B are sets, then (A ∩ B)^c = A^c ∪ B^c

Theorem 3.2.7- If A is any set,

If A is in any set, Ø ⊆ A.

Theorem 3.2.6- If A ⊆ B,

If A ⊆ B, then A - B = Ø

Definition 2.4.1 (Even and Odd Integer)

If n is an integer, we say that n is even provided there exists an integer k such that n=2k. We say that n is odd provided there exists an integer k such that n=2k+1.

Corollary 2.4.5- Ya know

If n is an odd integer, then n^2 is odd.

Corollary 2.5.12

If p | a^2, then p | a.

Definition 3.3.4- Intersection over F

Let F be a family of sets indexed by A. Then the intersection over F is defined by ∩α∈⋏(A_α)={x: ∃α ∈ ⋏)(x ∈ A_α)}

Defintion 3.3.3- Union over F

Let F be a family of sets indexed by A. Then the union over F is defined by ⋃α∈⋏(A_α)={x: ∃α ∈ ⋏)(x ∈ A_α)}

Definition 3.6.1- Equivalence Relations

Let S be a non-empty set, and let "x ≡ y" be a statement for all x, y ∈ S, either x ≡ y or x ≢ y is true. Suppose also that ≡ has the following properties on S. (1) For all x ∈ S, x ≡ x. (2) For all x, y ∈ S, x ≡ y. (3) For all x, y, z ∈ S, if x ≡ y and y ≡ z, then x ≡ z.

Theorem 2.4.7 (Division Algorithm)

Let a and b be integers, where a>0. Then there exist unique integers q and r such that b=aq+r and 0<=r<a.

Theorem 3.5.3- Let a and b be nonzero real numbers and let m and n be nonnegative integers. Then

Let a and b be nonzero real numbers and let m and n be nonnegative integers. Then (a^m)(a^n) = (a^(m+n)) (a^m)^n = (a^(mn)) (ab)^n = (a^n)(b^n)

Exercise 3.2.8- If A ∩ B = A, then A ⊆ B.

Pick x ∈ A, then x ∈ A and x ∈ B since the intersection of A and B is A. Since every element of A must also be an element of B, then A is a subset of B.

Exercise 3.2.9- Prove the following. (b) If A ⊆ B and B ⊆ C, then A ⊆ C.

Pick x ∈ A. Then because A is a subset of B, every element of A is in B. Because B is a subset of C, every element of B is in C. By the transitive property, if every element of A is in B and every element of B is in C, then every element A is in C. Therefore,

Exercise 3.2.9- Prove the following. (a) For all sets A and B, if A ⊆ B, then A ∪ B = B.

Pick x ∈ A. Then, because A is a subset of B, every element of A is in B. Because the union of the sets A and B means an element must be in either set A or B, it follows that the element must always be in set B. It also follows that if an element is in set B, it must be in the union of set A and set B (since every element of A is in B). Therefore,

Exercise 3.2.9- Prove the following. (c) If A ⊆ B, then A^c ⊇ B^c.

Pick ∈ C, C being a set not in A or B. If x is not in B, then x is not in A since every element of A is in set B by definition of a subset. Therefore,

Exercise 2.4.8(a)- Let a, b ∈ Z with a > 0. Suppose that b = aq1 + r1 and b = aq2 + r2, where q1,q2,r1,r2 ∈ Z, 0 ≤ r1 < a and 0 ≤ r2 < a. Show the uniqueness of q and r.

Proof: (a) Let q_1 = q_2. Then, b= aq + r_1. aq_1 = b - r_1 q_1 = (b-r_1)/a b = aq_2 + r_2 b = [a(b - r_1)]/a + r_2 b = b- r_1 + r_2 0 = -r_1 + r_2 r_1 = r_2 Let r_1 = r_2. Then, b = aq_1 + r_1 r_1 = b - aq_1 b = aq_2 + r_2 b = aq_2 + (b - aq_1) b = aq_2 + b - aq_1 0 = aq_2 - aq_1 aq_1 = aq_2 q_1 = q_2

Exercise 2.4.6- Prove the following: (a) The sum of two even integers is even. (b) The sum of two odd integers is even. (c) The sum of even and odd integer is odd.

Proof: (a) Suppose m = 2j and n = 2k. Then m + n = 2j + 2k = 2(j + k), which is even. (b) Suppose m = 2j + 1 and n = 2k + 1. Then m + n = (2j + 1) + (2k +1)=2j + 2k +2= 2(j + k + 1), which is even. (c) Suppose m = 2j and n = 2k + 1. Then m + n = 2j + (2k + 1) = 2(j + k) + 1, which is odd.

Exercise 2.4.8(b)- Show that r_1 = r_2 leads to a contradiction

Proof: (b) Assume r_2 > r_1. Then r_2 - r_1 > 0. (b - aq_2) - (b - aq_1) > 0 -aq_2 +aq_1 > 0 a(-q_2 + q_1) > 0 a - r_2 > 0, r_1 >= 0 a - r_2 - r_1 > 0 a > r_2 - r_1 aq_1 + r_1 = aq_2 + r_2 aq_1 + aq_2 = r_2 - r_1 a(q_1 - q_2) = r_2 - r_1 > 0 r_2 - r_1 = a(q_1 - q_2) > a, since q_1 and q_2 must be >= 1 since q_1 and q_2 are integers. So, r_2 - r_1 = a(q_1 - q_2) >= a. We have reached a contradiction.

Exercise 2.5.8(a)- Let a and b be nonzero integers, and let g = m0a + n0b be the smallest element of S = {ma + nb : m, n ∈ Z, ma + nb > 0}. Show g | a

Proof: Assume g does not divide a. Then, a = gq + r. a = q(ma + nb) + r a = qma + qnb + r r = a - qua - qnb r = (1 - qm)a - qnb Let (1-qm) = m_0 and (-qn) = n_0 r = m_0(a) + n_0(b). Therefore, r < g. But we chose g to be the smallest element of S. Therefore we've reached a contradiction, concluding that g is divisible by a.

Exercise 3.4.11- Suppose A is a given set and {Bk}nk=1 is a family of n ≥ 1 sets. Then A ∩ ( ⋃nk=1Bk) = ⋃n(A ∩ Bk). k=1

Proof: First consider the case n = 1: then the left side is A ∩ B1, and so is the right side, so there is nothing to prove. Now suppose that the equality is true for n. By the distributive property of ∩ over ∪, we have A ∩ (n+1⋃k=1Bk) = A ∩ [( ⋃nk=1Bk) ∪ Bn+1] = [A ∩ ( ⋃nk=1Bk)] ∪ (A ∩ Bn+1) = [ ⋃n(A ∩ Bk)k=1] ∪ (A ∩ Bn+1)= n+1⋃k=1(A ∩ Bk). Therefore the equality holds for all n by induction.

Exercise 3.4.9- Prove the following formulas for n ≥ 1. (a) ∑nk=1(−1)kk2 = (−1)n n(n+1)

Proof: First prove for n = 1: Then the left side is (−1)112 = −1, and the right side is (−1)Now 1 assume 1(1+1) 2 = −1. Thus the formula is true for n = 1. the formula is true for n. n+1∑(−1)kk2 = (−1)n+1(n + 1)2 + ∑n(−1)kk2 k=1k=1= (−1)n+1(n + 1)2 + (−1)nn(n 2 + 1) = (−1)n (−(n + 1)2 + n(n 2 + 1)) = (−1)n (n2 + n − 2(n2 − 2n − 1) 2 ) = (−1)n (−n2 − 3n − 2 2 ) = (−1)n+1(n + 1)(n 2 + 2) . Thus the statement is true for n + 1. By induction, the formula is true for all n.

Exercise 3.4.9- Prove the following formulas for n ≥ 1. (b) ∑nk=1(2k − 1) = n2

Proof: First prove for n = 1: Then the left side is 2(1) − 1 = 1, and the right side is 12 = 1. Thus the formula is true for n = 1. Now assume the formula is true for n. n+1∑k=1(2k − 1) = ∑n(2k − 1) + (2(n + 1) − 1) k=1= n2 + 2n + 2 − 1 = n2 + 2n + 1 = (n + 1)2 Thus the statement is true for n + 1. By induction, the formula is true for all n.

Exercise 3.4.9- Prove the following formulas for n ≥ 1. (c) ∑n1 k=1 k(k+1) = n+1 n

Proof: First prove for n = 1: Then the left side is Thus the formula is true for n = 1. 1 1(1+1) = 12, and the right side is 11+1 = 12. 22 Now assume the formula is true for n. n+1∑k=1 nk(k 1 + 1) = ∑k=1 k(k 1 + 1) + (n + 1)(n 1 + 2) = n + n 1 + (n + 1)(n 1 + 2) = (n n(n + +2)+1 1)(n + 2) = (n n2 + + 1)(n 2n + + 1 2) = (n + (n 1)(n + 1)+ 2 2) = n n + + 1 2. Thus the formula is true for n + 1. By induction, the formula is true for all n.

Exercise 3.4.15- Suppose n is a positive integer and a, b1,b2,...,bn are real numbers. Then a( ∑nk=1bk) = ∑n(abk). k=1

Proof: For n = 1, the equality becomes ab1 = ab1, and there is nothing to prove. Suppose that the equality is true for n real numbers bk. Then given real numbers a, b1,...,bn+1, the distributive property of multiplication over addition implies a(n+1∑k=1 bk) = a( ∑nk=1bk + bn+1) = ∑n(abk) + abn+1 = n+1∑k=1k=1(abk). Thus the equality is true for all n by induction

Exercise 3.4.12- Suppose A1,A2,A3,... are sets with the property that An ⊆ An+1 for every n ≥ 1. Use induction to show that A1 ⊆ An for all n.

Proof: For n = 1, we have A1 ⊆ A1, which is true. Now suppose that A1 ⊆ An. We already know that An ⊆ An+1. By Exercise 3.2.9(b), we conclude A1 ⊆ An+1. Therefore A1 ⊆ An for all n by induction.

Exercise 2.2.7(f)- If a and b are integers such that ab = 1, then a = b = ±1. (This last chain of equalities should be interpreted as saying that either a and b are both 1, or a and b are both −1.) [Hint. There are five cases: a = −1, a = 0, a = 1, a > 1, and a < −1. Three are impossible. Remember that 1 is the smallest positive integer.]

Proof: If a, b ∈ Z and let c ∈ Z and c > 1. (1) Let a = -1. Then (-1)(b) = 1, and (-1)(-1)(b) = (-1)(1) b=-1. (2) Let a <-1. Then (-a)(b) = -1, and ab = 1 b = a^-1. (3) Let a = 0. Then (0)(b) = 1. 0 =/= 1. (4) Let a = 1. Then (1)(b) = 1. b = 1. (5) Let a > 1. Then (a)(b) = 1. b = a^-1. Since case 3 is impossible and case 2 and 5 present non-integers, only case 1 and 4 hold true.

Exercise 2.5.5- If a and b are nonzero integers such that a | b and b | a, then a = ±b. [Hint. Use Exercise 2.2.7(f).]

Proof: If a, b, k_1, k_2 ∈ Z, then a | b is b=ak_1 and b | a is a=bk_2. b = ak_1 b = (bk_2) k_1 b = b(k_1k_2) b = b(1), so k_1k_2 = 1. Since we proved if ab=1, then k_1 = k_2 = ± 1. Therefore, a = ±1.

Exercise 3.6.19- Construct the set consisting of all integers that are equivalent (modulo 6) to 14. [That is, devise a way to explicitly produce all such integers.]

Proof: If x ≡ 14 (mod 6), then x − 14 = 6k for some integer k. Rewriting this equation gives us x =14+6k. Thus the set of all integers that are equivalent to 14 modulo 6 can be produced by adding multiples of 6 to 14.

Exercise 2.5.15- If n is an integer such that n ≥ 2 and 3 | (n − 1), then 3 | (n2 − 1).

Proof: In order for 3 | (n-1), then (n-1) = 3k, for some k ∈ Z. So n = 3k + 1. Let n = 2, so (n^2 - 1) = 3j, for some j ∈ Z. Then (2^2 - 1) = 3j. (4 - 1) = 3j 3 = 3j j = 1, therefore 3 is a divisor for n^2 - 1 when n = 2. Let n > 2, so [(3k + 1)^2 - 1] = 3j 9k^2 + 24k + 16 - 1 = 3j 9k^2 + 24k + 15= 3j 3k^2 + 8k + 5= j, therefore 3 is a divisor for n^2 - 1 when n > 2.

Theorem 2.5.7- Suppose a and b are nonzero integers. Then there exists a unique positive integer g with the following properties: (D1) g | a and g | b (D2) If h is any positive integer such that h | a and h | b, then h | g also.

Proof: Let a and b be nonzero integers, and define S= {ma + nb : m, n ∈ Z, ma + nb > 0} That is, S is the set of all positive integer linear combinations of a and b. First, S is not empty, for depending on the signs of a and b, we may let m, n =±1 to produce some positive value of ma+nb. By the WOP, S contains a smallest element g, which may be written in the form g= m_0a +n_0b for some integers m_0 and n_0. If g_1 and g_2 are both positive integers that have properties D1 and D2, then g_1=g_2.

Theorem 2.5.10- Suppose a is a nonzero integer and p is prime. Then either a and p are relatively prime or p | a.

Proof: Let a be a nonzero integer, and let p be prime. Since the only positive integer divisors of p are 1 and p, then either gcd(a, p) = 1 or gcd(a, p) = p. In the former case, a and p are relatively prime. In the latter case, p satisfies property D1, so that p | a.

Exercise 3.3.14- Suppose F = {A} is a family of sets, and suppose D is a set for which D ⊆ A for every A ∈ F. Show that D ⊆ ⋂F A.

Proof: Let x ∈ D. Then by assumption x ∈ A for all A ∈ F. This means that x ∈ ⋂F A, which is what we needed to show.

Exercise 3.3.13- Suppose F = {A} is a family of sets, and suppose C is a set for which A ⊆ C for every A ∈ F. Show that ⋃F A ⊆ C.

Proof: Let x ∈ ⋃F A. Then there exists some particular A ∈ F such that x ∈ A. By assumption, this implies x ∈ C. We have shown that every element of ⋃F A is also an element of C, and so the assertion is proved.

Exercise 3.3.12- Suppose F1 = {Aα}α∈A and F2 = {Bα}α∈A are two families of sets with the property that Bα ⊆ Aα for every α ∈ A. Then ⋃α∈A Bα ⊆ ⋃α∈A Aα

Proof: Let x ∈ ⋃α∈A Bα. Then there exists α ∈ A such that x ∈ Bα. By assumption, then, x ∈ Aα, and so x ∈ ⋃α∈A Aα. This proves the assertion. (b) ⋂α∈A Bα ⊆ ⋂α∈A Aα Let x ∈ ⋂α∈A Bα. Then x ∈ Bα for all α ∈ A. This implies x ∈ Aα for all α ∈ A, and so x ∈ ⋂α∈A Aα.

Exercise 3.6.7- Let S be the xy-plane; that is, S = {(x, y) : x, y ∈ R}. For two points (x1,y1) and (x2,y2), define the two points to be equivalent, written (x1,y1) ≡ (x2,y2), provided y2−y1 = 3(x2−x1). Show that ≡ defines an equivalence relation on S.

Proof: Name three elements of S that are equivalent to (2,1).We must show that the three properties of an equivalence relation are satisfied: • ≡ is reflexive: y − y = 3(x − x), because both sides are equal to 0, so (x, y) ≡ (x, y). • ≡ is symmetric: suppose (x1,y1) ≡ (x2,y2), so that y2 − y1 = 3(x2 − x1). Then, multiplying both sides by −1 produces the equation y1 − y2 = 3(x1 − x2), which means (x2,y2) ≡ (x1,y1). • ≡ is transitive: suppose (x1,y1) ≡ (x2,y2) and (x2,y2) ≡ (x3,y3) so that y2 − y1 = 3(x2 − x1) and y3 − y2 = 3(x3 − x2). Then adding the respective sides of these equations produces the new equation (y2 − y1)+(y3 − y2) = 3(x2 − x1) + 3(x3 − x2) (y3 − y1)+(y2 − y2) = 3(x3 − x1) + 3(x2 − x2) y3 − y1 = 3(x3 − x1), so that (x1,y1) ≡ (x3,y3). For an element (x, y) ∈ S to be equivalent to (2,1), we must have y − 1 = 3(x − 2), or y = 3x − 5. For example, (0,−5) and (3,4) are equivalent to (2,1).

Theorem 2.4.2- Let m and n be integers, at last one of which is even . Then mn is even.

Proof: Pick integers m and n, suppose (without loss of generality) that n is even. Then there exists integer k_1 such that n=2k_1. Thus we have that mn=m(2k_1)=2(mk_1), so that we may let k_2=mk_1 (which is an integer) to have mn=2k_2. Thus mn is even.

Corollary 2.4.3- If n is an integer, then n^2 is even.

Proof: Pick integers m and n, suppose that m=n and that n is even. Then there exists an integer k_1 such that n=2k_1. Thus we have that nn=(2k_1)(2k_1)=4(k_1)^2=2(2k_1)^2, so that we may let k_2=(2k_1)^2 (which is an integer) to have n^2= 2k_2. Thus n^2 is even.

Exercise 3.2.9- Prove the following. (f) If A and B are disjoint, then A ∆ B = A ∪ B.

Proof: Pick x ∈ A. Because A and B are disjoint, x ∉ B if x ∈ A. Accordingly, if x ∈ B, then ∉ A. the symmetrical difference of two sets is subtracting any elements that are in both sets A and B. Since A ∩ B = Ø, the symmetric difference would just be the union of A and B. Therefore,

Exercise 2.5.11- Suppose a and b are nonzero integers, p is prime, and p | ab. Then either p | a or p | b.

Proof: Since p is prime, that means p and a are relatively prime, so pm + an = 1, which is equivalent to bpm + ban = b. Let j ∈ Z and ab = jp. It follows that bpm + jpn= b. So (bm +jn)p = b. Let (bm + jn)= k and k ∈ Z. Then kp = b. Therefore p | b.

Theorem 3.2.1- If A ⊆ B, then A ∩ B = A.

Proof: Suppose A ⊆ B. To show that A ∩ B= A, show that A ∩ B ⊆ A and A ⊆ A ∩ B. (A ∩ B ⊆ A): Pick x ∈ A ∩ B. Then x ∈ A, we have that A ∩ B ⊆ A. (A ⊆ A ∩ B): Pick x ∈ A. Then, since A ⊆ B, we have that x ∈ B also. Therefore, since x ∈ A and x ∈ B, it follows that x ∈ A ∩ B. Thus A ⊆ A ∩ B. Since we have shown that A ∩ B ⊆ A and A ⊆ A ∩ B, we have demonstrated that A ∩ B = A.

Exercise 3.4.19- Suppose a1,a2,...,an are nonzero integers and p is a prime number. Suppose also that p | a1a2 ···an. Then there exists some k (1 ≤ k ≤ n) such that p | ak.

Proof: Suppose a1,a2,...,an are nonzero integers and p is a prime number. Suppose also that p | a1a2 ···an. Then there exists some k (1 ≤ k ≤ n) such that p | ak. For n = 1: if p | a1, then tautologically p | a1. Thus the statement is true for n = 1. Now suppose we are given n + 1 integers a1,a2,...,an,an+1 such that p | a1a2 ···anan+1. We can rewrite the divisibility condition as p | (a1a2 ···an)an+1. By Exercise 2.5.11, either p | a1a2 ···an or p | an+1. • If the latter is true, then we are done; we can choose k = n + 1. • If the former is true, then by the induction assumption, there exists 1 ≤ k ≤ n such that p | ak, and we are done. In either case, there exists 1 ≤ k ≤ n + 1 such that p | ak. By induction, the statement is true for all n.

Exercise 2.5.4- If a, b, and c are nonzero integers such that a | b and a | c, then a | (mb + nc) for all integers m and n.

Proof: Suppose b = aj and c = ak. Then mb + nc = m(aj) + n(ak) = a(mj + nk), so a | mb + nc.

Exercise 2.5.8(c)- Let a and b be nonzero integers, and let g = m0a + n0b be the smallest element of S = {ma + nb : m, n ∈ Z, ma + nb > 0}. Show that if g_1 and g_2 are positive integers with properties D1-D2, then g_1 = g_2.

Proof: Suppose both g1 and g2 satisfy D1. Putting g1 in the role of h and g2 in the role of g in D2, we get g1 | g2. Similarly, g2 | g1. By Exercise 2.5.5, we conclude g1 = ±g2. Because g1 and g2 are both positive, the negative sign is not possible. Therefore g1 = g2.

Exercise 3.2.9- Prove the following. (e) For all sets A and B, (A ∪ B)^c = A^c ∩ B^c.

Proof: We must prove both inclusions: (A ∪ B)^c ⊆ A^c ∩ B^c and A ∩ BC ⊆ (A ∩ B)C. Let x ∈ (A∪B)C. Then x /∈ A∪B, which means x /∈ A and x /∈ B (because the negation of (x ∈ A) ∨ (x ∈ B) is (x /∈ A) ∧ (x /∈ B)). Therefore x ∈ AC ∩ BC. Now let x ∈ AC∩BC. Then, reversing the above reasoning, we have x ∈ AC and x ∈ BC, which means x /∈ A and x /∈ B. Thus x /∈ A ∪ B, so x ∈ (A ∪ B)C.

Exercise 3.6.23- Suppose x and y are integers and n is a positive integer. Then x ≡n y if and only if x and y have the same remainder when divided by n according to the Division Algorithm.

Proof: We need to prove both directions of this "if and only if" statement. First, suppose x and y have the same remainder when divided by n. This means that x = pn + r and y = qn + r for unique integers p,q,r with 0 ≤ r<n. By substitution: x − y = (pn + r) − (qn + r)=(p − q)n, and so n | (x − y). Therefore x ≡n y. For the other direction, suppose that x ≡n y, which means that x− y = kn for some integer k. By the Division Algorithm, there exist unique integers p, q, r, s with 0 ≤ r < n and 0 ≤ s<n such that x = pn + r and y = qn + s. Without loss of generality, assume r ≥ s (otherwise, switch the roles of x and y). From previous equations, we know that kn = x − y = (pn + r) − (qn + s)=(p − q)n + (r − s), and so r − s = (k − p + q)n, which means n | (r − s). But because 0 ≤ s ≤ r<n, we must have 0 ≤ r − s<n. From this string of inequalities and the fact that n divides r − s, we conclude r − s = 0, or r = s. Therefore x and y have the same remainder when divided by n.

Exercise 3.4.9- Prove the following formulas for n ≥ 1. (d) ∑nk=1 k3 = (∑nk=1 k)2

Proof: When n = 1, both sides equal 1. So assume the formula is true for n. For n + 1, on the right we get (n+1∑k=1 k)2 = ((n + 1) + ∑nk=1k)2 = (n + 1)2 + 2(n + 1) ( ∑nk=1k)2 . By assumption the last term equals first two terms equals (n + 1)3. ∑nk=1 k3. We know that ∑So nk=1 we k need = to n(n+1) 2 show that the sum of the , so (n + 1)2 + 2(n + 1) ∑nk=1k + ∑nk=1k = (n + 1)2 + 2(n + 1)n(n 2 + 1) = (n2 + 2n +1)+(n3 + 2n2 + n) = n3 + 3n2 + 3n +1=(n + 1)3. This shows that the formula is true for n + 1. By induction, it is true for all n. Another approach to the inductive step: adding (n + 1)3 on the left produces n+1∑k=1 k3 = ∑nk3 + (n + 1)3 k=1= (n(n + 1) 2 )2 + (n + 1)3 = n4 + 2n4 3 + n2 + n3 + 3n2 + 3n + 1 = n4 + 6n3 + 10n4 2 + 12n + 4 . Meanwhile, the right side becomes (n+1∑k=1 )2 k= ((n + 1)(n + 2) 2 )2 = (n2 + 3n 4 + 2)2 = n4 + 6n3 + 10n4 2 + 12n + 4 . The two sides are equal, so the formula holds for n + 1.

Exercise 3.4.18- For all n ≥ 1, we have 3 | (4n − 1).

Proof: When n = 1, the statement becomes 3 | (41 − 1), or 3 | 3, which is true. Now assume that 3 | (4n−1), which means that there exists some k ∈ Z such that 4n−1=3k. We want to show that 3 | (4n+1 − 1). We have: 4n+1 − 1=4 · 4n − 1 = 4(4n − 1 + 1) − 1 = 4(3k + 1) − 1 = 3(4k + 1) and so 3 | (4n − 1), as desired. By induction, 3 | (4n − 1) for all n ≥ 1.

Exercise 2.5.8(b)- Let a and b be nonzero integers, and let g = m0a + n0b be the smallest element of S = {ma + nb : m, n ∈ Z, ma + nb > 0}. Show that if h is any positive integer with the properties that h | a and h | b, then it must be that h | g.

Proof: Let k_1, k_2 ∈ Z, and a = hk_1 and b = hk_2. Then g = m_0(hk_1) + n_0(hk_2) = h(m_0k_1 + n_0k_2). Let (m_0k_1 + n_0k_2) = k_3, so g = hk_3. Therefore, h | g.

Exercise 2.5.14 If n ≥ 3 is an odd integer, then 8 | (n2 − 1). [Hint. If k is an integer, then either k or k + 1 is even.]

Proof: Let n = 2k + 1, where k ∈ Z. Let c ∈ Z, then: If n = 3, then k =1. So (3^2 - 1) = 8c (9-1) = 8c 8 = 8c c = 1, therefore 8 is a divisor for n^2 - 1 if n =3. If n > 3, then k > 1. So (2k + 1)^2 - 1 = 8c 4k^2 + 4k + 1 -1 = 8c 4k^2 + 4k = 8c k^2 + k = 2c k(k + 1)= 2c. Either k or k + 1 must be even, therefore 8 is a divisor for n^2 - 1 if n > 3.

Exercise 3.2.13- Show that union and intersection do not allow for cancellation by providing coun- terexamples to the following statements. (b) If A ∩ C = B ∩ C, then A = B.

Proof: truth table and say its a contradiction ugh

Exercise 3.2.13- Show that union and intersection do not allow for cancellation by providing counterexamples to the following statements. (a) If A ∪ C = B ∪ C, then A = B

Proof: truth table, idgaf just say it's not a tautology

Theorem 3.4.10- Suppose A is a given set and {B_k} is a family of n >= 1 sets, where {k : 1< k< n}. Then

Suppose A is a given set and {B_k} is a family of n >= 1 sets, where {k : 1< k< n}. Then A ∪ (B_1 ∩ B_2 ∩ ... ∩ B_n) = (A ∪ B_1) ∩ (A ∪ B_2) ∩ ... ∩ (A ∪ B_n).

Theorem 3.3.7 (DeMorgan's Law)- Suppose F is a family of sets. Then

Suppose F is a family of sets. Then [∪_F(A)]^c = ∩_F(A)^c.

Theorem 3.3.11- Suppose F_1 is a family of sets, and F_2 is a subfamily of F_1.

Suppose F_1 is a family of sets , and F_2 is a subfamily of F_1. Then: (1) ∪_(F_2)A ⊆ ∪_(F_1)A (2) ∩_(F_1)A ⊆ ∩_(F_2)A

Definition 3.7.1- Partition of a Set

Suppose S is a set and F = {A} is a family of subsets of S. Then F is said to be a partition of S if (1) Every A in F is non-empty; (2) If A, B ∈ F and A ∩ B is non-empty, then A = B; (3) ∪_(⋏∈F) A = S.

Theorem 3.5.1 (PMI)- Suppose S is a subset of integers that has properties J1- J2.

Suppose S is a subset of integers that has properties J1- J2. Then S = {j, j+1, j+2, ...} = {n ∈ Z : n >= j}.

Definition 2.5.6 (Greatest Common Divisor)

Suppose a and b are nonzero integers, and suppose g is a positive integer with the following properties: (D1) g | a and g | b (D2) If h is any integer such that h | a and h | b, then h | g. Then g is called the greatest common divisor of a and b, and is denoted gcd(a, b).

Definition 2.5.1 (Divisibility)

Suppose a and b are nonzero integers. We say that a divides b, written a | b, provided there exists an integer k such that b=ak. We also say that a is a divisor of b. If a | b where a ∉ {±1, ±b}, we call that a proper divisor of b. If a does not divide b, we write a ∤ b.

Theorem 3.4.16- Suppose m and n are positive integers and let a, b_1, b_2, ..., b_n are real numbers.

Suppose m and n are positive integers and let a, b_1, b_2, ..., b_n are real numbers. Then [Σ(j=1, m) a_j][Σ(k=1, n) b_k] = Σ(j=1, m) [Σ(k=1, n) a_jb_k].

Definition 3.7.5 (Equivalence Class)

Suppose ≡ defines an equivalence relation on a non-empty set S. For an element x ∈ S, define [x] = {y ∈ S : y ≡ x}. That is, [x] contains all elements of S that are equivalent to x. This subset of S is called the equivalence class of x.

Theorem 3.7.10- Suppose ≡ defines an equivalence relation on a set S. Define F = {[x] : x ∈ S}. That is, F is the collection of all equivalence classes generated by letting x take on all values in S.

Suppose ≡ defines an equivalence relation on a set S. Define F = {[x] : x ∈ S}. That is, F is the collection of all equivalence classes generated by letting x take on all values in S. Then F is a partition of S.

Relatively Prime

Two numbers are relatively prime if their greatest common factor is 1