Discrete Mathematics W. Ethan Duckworth Spring 2018, Loyola University Maryland Contents 1 Logic and Proofs3 1.1 Even and Odd Numbers..................................3 1.2 Common Ground: Algebraic Properties.........................3 1.3 Divides...........................................6 1.4 Some Basic Logic......................................7 2 Sets 11 2.1 Introduction to Sets.................................... 11 2.2 Operations......................................... 14 2.3 Algebraic properties of set operations........................... 14 3 Quantifiers and Functions 20 3.1 Quantifiers......................................... 20 2 Chapter 1 Logic and Proofs 1.1 Even and Odd Numbers 1.2 Common Ground: Algebraic Properties 3 CHAPTER 1. LOGIC AND PROOFS 4 Theorem 1.2.1 (Algebraic Properties of R). The following properties hold for any a; b; c 2 R: (a) If a + b = a then b = 0. (b) If a + b = a + c then b = c. (c)0 × a = a × 0 = 0. (d) −(−a) = a. (e) −a = −1 × a. (f) a × (−b) = (−a) × b = −(a × b). (g)( −a) × (−b) = a × b. (h) a × (b − c) = a × b − a × c. (i) If a × b = a × c, and a 6= 0, then b = c. (j) If ab = 0 then a = 0 or b = 0. (k)1 > 0. (l) If a > b > 0 then a2 > b2. (m) If a < b, and c < 0 then a × c > b × c. (n) If a < 0 and b > 0 then a × b < 0. If a < 0 and b < 0 then a × b > 0. (o) If a 2 Z then there is no integer d 2 Z such that a < d < a + 1. (p) There exists some n 2 Z such that n × b > a. Proof. Part (a). Suppose a; b 2 R such that a+b = a. By the additive inverse property −a exists. Add −a to both sides to get −a + (a + b) = −a + a: By the associative property we can rewrite the left hand side to get (−a + a) + b = −a + a: By definition of the additive inverse we have −a + a = 0 and so this becomes 0 + b = 0. (In the future we will combine the last two steps by saying \subtract a from both sides.) By definition of the additive identity this means that b = 0. Part (b). Suppose a; b; c 2 R such that a + b = a + c. By the additive inverse property −a exists. Adding −a to both sides we get −a+(a+b) = −a+(a+c). Using the associative property we have (−a + a) + b = (−a + a) + c. By definition of the additive inverse we have 0 + b = 0 + c. (In the future we will combine the last two steps by saying \subtract a from both sides".) By definition of the additive identity we have b = c. Part (c). Let a 2 R. Note that 0 = 0 + 0 and so 0a = (0 + 0)a. Apply the distributive law to get 0a = 0a + 0a. Subtract 0a from both sides to get 0a − 0a = 0a + 0a − 0a which comes 0 = 0a. Part (d). Let a 2 R. By definition of the additive inverse we have a + (−a) = 0. Subtract (−a) from both sides to get a = −(−a). Part (e). Let a 2 R. Note that 0 = 1 + (−1) and so 0a = (1 − 1)a. Apply the distributive law to get 0a = 1a + (−1)a. Using the previous part, and the defining property of the multiplicative identity, we get 0 = a + (−1)a. Subtracting a from both sides gives −a = (−1)a. Part (f). Let a; b 2 R. By the previous part we have −b = (−1)b and −a = (−1)a and −(ab) = (−1)(ab). Then the following three expressions may be rewritten a × (−b) = a (−1)b ; and (−a) × b = (−1)a b; and − (a × b) = (−1)(ab): All the right hand sides are equal by the associative and commutative properties. Part (g). Let a; b 2 R. Applying part (e) gives (−a) × (−b) = (−1)a × (−1)b Applying the commutative and associative properties gives RHS above = (−1)(−1) ab CHAPTER 1. LOGIC AND PROOFS 5 Applying part (d) gives RHS above = 1ab and applying the defining property of the multiplicative identity gives RHS above = ab: Part (h). Let a; b; c 2 R. Then part (e) gives a(b − c) = a(b + (−1)c) Now we can apply the distributive property to get RHS above = ab + a(−1)c and the associative property and part (f) gives RHS above = ab − ac: Part (m). Let a; b; c 2 R with a < b, and c < 0. Subtract c from both sides of c < 0 to get 0 < −c. Multiply both sides of a < b by −c to get a(−c) < b(−c). Use part (f) to rewrite this as −ab < −bc. Add ab to both sides to get 0 < −bc + ab and add bc to both sides to get bc < ab. Part (j). Let a; b 2 R. We will prove \if ab = 0 then a = 0 or b = 0" by contrapositive: We'll assume that a 6= 0 and b 6= 0 and prove that ab 6= 0. There there are four cases: Case 1: a > 0 and b > 0, Case 2: a > 0 and b < 0, Case 3: a < 0 and b > 0, Case 4: a < 0 and b < 0. Of course, cases 2 and 3 are essentially the same, so we will just argue one of them. Case 1. Let a > 0 and b > 0. Apply Real Number Property ?? part ?? and multiply both sides of b > 0 by a to get ab > a0. By Proposition ?? this becomes ab > 0, which shows that ab 6= 0. Case 2. Let a > 0 and b < 0. Using the same results as before, multiply both sides of b < 0 by a to get ab < a0. This becomes ab < 0, which shows that ab 6= 0. Case 3. This is the same proof as case 2, with a and b switched. Case 4. Let a < 0 and b < 0. Apply Theorem 1.2.1 part (m) and multiply both sides of a < 0 by b to get ab < a0. This becomes ab < 0, which means that ab 6= 0. We have shown that there are exactly 4 cases, and in every case we conclude that ab 6= 0. Therefore, we have proven \if a 6= 0 and b 6= 0, then ab 6= 0". By contrapositive, this shows \if ab = 0 then a = 0 or b = 0. Part (i). Let a; b; c 2 R with a × b = a × c, and a 6= 0. Subtract ac from both sides to get ab − ac = 0. Factor (i.e. apply the distributive law) to get a(b − c) = 0. By part (j) we see that either a = 0 or b − c = 0. Since we assumed that a 6= 0 we conclude that b − c = 0. Then add c to both sides to get b = c. Part (k). (Probably we should have assumed this as part of our defining properties of R, but oh well.) Suppose this is not the case. Then by the trichotomy property we have either 1 = 0 or 1 < 0. If 1 = 0 then all natural numbers are 0, which is not the case since we assumed that n + 1 6= n for all natural numbers. If 1 < 0 then apply part (m) and multiply the inequality a < b by 1 to get 1a > 1b i.e. a > b. But this contradicts a < b, and so 1 < 0 is false. Part (l). Let a; b 2 R and suppose that 0 < b < a. Then multiply both sides of b < a by b to get b2 < ab. Multiply both sides of b < a by a to get ab < a2. Combine these inequalities to get b2 < a2. Part (n). Let a; b 2 R. Suppose a < 0 and b > 0. Apply part Definition ?? part ?? and multiply a < 0 by b to get ab < 0. Suppose a < 0 and b < 0. Apply part (m) and multiply a < b by b to get ab > 0. CHAPTER 1. LOGIC AND PROOFS 6 Part (o). Let a 2 Z. Case 1: a > 0. Case 2: a ≤ 0. In case 1, we have that a is a natural number. By definition of the natural numbers, every natural number equals one of the following: 1, 2 = 1 + 1, 3 = 1 + 1 + 1, etc. Furthermore, 1 < 2 < 3 < : : : . As a consequence, 1 is the smallest natural number, i.e. 1 ≤ n for all n 2 Z. Therefore, there is no natural number d such that 1 < d < 2. Because if there were, we would have d = m + 1 for some natural number m, and then m < 1. Similarly, there is no d between 2 and 3, or between 3 and 4, and by induction no d between a and a + 1. Case 2: if a ≤ 0 then apply the result just proven to −a + 1. Part (p). Let a; b 2 R. We first prove this the case where a; b 2 N. Let n = a + 1 and note the following b ≥ 1 ab ≥ a ab + b > a (a + 1)b > a nb > a Now suppose that a; b are positive rational numbers. Then let a = c=d and b = e=f. Then nb > a () n(e=f) > c=d () n(ed) > cf and the list line follows from the case just proven for integers. Now suppose that a; b are positive real numbers. Let a0 and b0 be rational numbers such that b > b0 and a0 > b. Apply the case just proven for rational numbers to get nb0 > a0. Then nb > nb0 > a0 > a: Finally, we leave the case where a or b is negative to the reader.