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Chapter 1 - 1. Elementary Notions and Notations. Section 1.1 A Proof Primer. A proof is a demonstration that some statement is true. We normally demonstrate proofs by writing English sentences mixed with symbols.

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## Chapter 1 - 1

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**Chapter 1 - 1**Elementary Notions and Notations**Section 1.1 A Proof Primer**• A proof is a demonstration that some statement is true. We normally demonstrate proofs by writing English sentences mixed with symbols. • We’ll consider statements that are either true or false. If A and B be are statements, then • “not A,” “A and B,” and “A or B,” are called negation, conjunction, and disjunction, respectively. “not A” is opposite in truth value from A. “A and B” is true exactly when both A and B are true “A or B” is true except when both A and B are false.**“if A then B” (or “A implies B”) is a conditional**statement with hypothesis A and conclusion B. It’s contrapositive is “if not B then not A” and it’s converse is “if B then A”. Statements with the same truth table are said to be equivalent. The table shows that a conditional and it’s contrapositive are equivalent. Conditionals**Vacuously true and Trivially true**• A conditional is vacuously true if its hypothesis is false. • A conditional is trivially true if its conclusion is true.**Some definitions**• integers: …, -2, -1, 0, 1, 2, … • odd integers: …, -3, -1, 1, 3, … (have the form 2k + 1 for some integer k). • even integers:…, -4, -2, 0, 2, 4, … (have the form 2k for some integer k). • m | n (read m divides n) if m ≠ 0 and n = km for some integer k. • p is prime if p > 1 and its only divisors are 1 and p.**Exhaustive Checking**• Some statements can be proven by exhaustively checking a finite number of cases. • Example 1. There is a prime number between 200 and 220. • Proof: Check exhaustively and find that 211 is prime. QED. • Example 2. Each of the numbers 288, 198, and 387 is divisible by 9. • Proof: Check that 9 divides each of the numbers. QED.**Conditional Proof**• Most statements we prove are conditionals. We start by assuming the hypothesis is true. • Then we try to find a statement that follows from the hypotheis and/or known facts. • We continue in this manner until we reach the conclusion.**Example 3**• If x is odd and y is even then x – y is odd. • Proof: Assume x is odd and y is even. Then x = 2k + 1 and y = 2m for some integers k and m. So we have x – y = 2k + 1 – 2m = 2(k – m) + 1, which is an odd integer since k – m is an integer. QED.**Example 4**• If x is odd then x2 is odd. • Proof: Assume x is odd. Then x = 2k + 1 for some integer k. So we have x2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1, which is an odd integer since 2k2 + 2k is an integer. QED.**Example 5**• If x is even then x2 is even. • Proof: Class do as one minute quiz.**Example 6**• If x2 is odd then x is odd. • Proof: The contrapositive of this statement is “if x is even, then x2 is even,” which is true by Example 5. QED.**Example 7**• If x2 is even then x is even. • Proof: This is the contrapositive of Example 4, which has been shown to be true. QED.**If And Only If (Iff) Proofs**• A statement of the form “A if and only if B” means “A implies B” and “B implies A.” So there are actually two proofs to give. Sometimes the proofs can be written as a single proof of the form “A iff C iff D iff … iff B,” where each iff statement is clear from previous information.**Example 8**• x is even if and only if x2 – 2x + 1 is odd. • Proof: x is even iff x = 2k for some integer k (definition) iff x – 1 = 2k – 1 for some integer k (algebra) iff x – 1 = 2(k – 1) + 1 for some integer k – 1 (algebra) iff x – 1 is odd (definition) iff (x – 1)2 is odd (Examples 4 and 6) iff x2 – 2x + 1 is odd (algebra). QED.**Proof By Contradiction**• A false statement is called a contradiction. For example, “S and not S” is a contradiction for any statement S. • A truth table will show us that “if A then B,” is equivalent to “A and not B implies false.” • So to prove “if A then B,” it suffices to assume A and also to assume not B, and then argue toward a false statement. This technique is called proof by contradiction.**Example 9**• If x2 is odd then x is odd. • Proof: Assume, BWOC, that x2 is odd and x is even. Then x = 2k for some integer k. So we have x2= (2k)2 = 4k2 = 2(2k2), which is even since 2k2 is an integer. So we have x2 is odd and x2 is even, a contradiction. So the statement is true. QED.**Example 10**• If 2 | 5n then n is even. • Proof: Assume, BWOC, that 2 | 5n and n is odd. Since 2 | 5n, we have 5n = 2d for some integer d. Since n is odd, we have n = 2k + 1 for some integer k. Then we have 2d = 5n = 5(2k + 1) = 10k + 5. So 2d = 10k + 5. Solve for 5 to get 5 = 2d – 10k = 2(d – 5k). But this says that 5 is an even number, a contradiction. So the statement is true. QED.**Section 1.2 Sets**• A set is a collection of things. • If S is a set and x is a member or element of S we write x ∊ S. Otherwise we write x ∉ S. • The set with elements x1, …, xn is denoted {x1, …, xn}. • The empty set with no elements is denoted { } or Ø. • A set with one element is called a singleton. e.g., {a} is a singleton. • The set of integers is denoted by Z, the natural numbers {0, 1, …} by N, the rational numbers by Q, and the real numbers by R.**Equal Sets**• Two sets A and B are equal, denoted A = B if they have the same elements. • e.g., {a, b, c} = {c, b, a} (no ordering). • e.g., {a, a, b, c} = {a, b, c} (no repetitions) • Sets can be described by properties that the elements satisfy. If P is a property, then the expression {x | P} denotes the set of all x that satisfy P. • e.g., The set of odd natural numbers can be represented by the following equal sets. {x | x = 2k + 1 for some k ∊ N} = {1, 3, 5, …}.**Subsets**• The set A is a subset of B, denoted A ⊂ B, means every element of A is an element of B. • e.g., N ⊂ Z ⊂ Q ⊂ R. • e.g., S ⊂ S for any set S. • e.g., Ø ⊂ S for any set S.**power set**• The power set of a set S, denoted power(S) is the set of all subsets of S. • e.g., power({a, b}) = {Ø, {a}, {b}, {a, b}}.**Comparing sets**• Example. Let A = {2k + 7 | k ∊ Z} and B = {4k + 3 | k ∊ Z}. • Quiz. Is A ⊂ B? • Answer: No. For example, 9 ∊ A but 9 ∉ B. • Quiz. Is B ⊂ A? • Answer: Yes. Let x ∊ B. Then x = 4k + 3 for some k ∊ Z. But we can write x = 4k + 3 = 4k – 4 + 7 = 2(2k – 2) + 7. Since 2k – 2 ∊ Z, it follows that x ∊ A. Therefore B ⊂ A. QED.**Equality in terms of subsets**• A = B iff A ⊂ B and B ⊂ A. • Example. Let A = {2k + 5 | k ∊ Z} and B = {2k + 3 | k ∊ Z}. Show that A = B. • Proof: First show A ⊂ B. Let x ∊ A. Then x = 2k + 5 for some k ∊ Z. So we have x = 2k + 5 = 2k + 2 + 3 = 2(k + 1) + 3. Since k + 1 ∊ Z, it follows that x ∊ B. Therefore A ⊂ B. Now show the other direction B ⊂ A. …….Class fill in the proof (2 minute quiz). Since A ⊂ B and B ⊂ A it follows that A = B. QED.**Operations on Sets**• Union: A U B = {x | x ∊ A or x ∊ B}. • Intersection: A n B = {x | x ∊ A and x ∊ B}. • Difference: A – B = {x | x ∊ A and x ∉ B}. • Symmetric Difference: A ⊕ B = {x | x ∊ A or x ∊ B but not both} • Note: A ⊕ B = (A – B) U (B – A) = (A U B) – (A n B ) . • Universal Complement: Given a universe U and A ⊂ U, we write A' = U – A.**Example**• For each n ∊ N let Dn = {x ∊ N | x divides n}. So Dn is the set of positive divisors of n. Here are some expressions involving these sets. • D0 = {1, 2, 3, … } = N – {0}, D5 = {1, 5}, D6 = {1, 2, 3, 6}, and D9 = {1, 3, 9}. • D5U D6 = {1, 2, 3, 5, 6} • D5 n D6 = {1} • D9 – D6 = {9} • D5⊕ D6 = {2, 3, 5, 6} • Let N be the universe. Then D0 ' = N – D0 = {0}, and {0}' = D0.**Quiz (2 minutes)**• Draw a Venn diagram for three sets A, B, C with some areas shaded. • Then find an expression to represent the shaded area.**Properties of Set Operations**• Union and intersection are commutative, associative, and distribute over each other. There are many other properties too. For example, • Absorption: A U (A n B) = A and A n (A U B) = A. • De Morgan’s Laws: (A U B)' = A' n B' and (A n B)' = A' U B'.**Counting Sets**• The cardinality of a set S is denoted by |S|. Two useful rules for counting finite sets are: • Inclusion-Exclusion or Union Rule: | A U B | = | A | + | B | – | A n B |. • Difference Rule: | A – B | = | A | – | A n B |.**Quiz (2 minutes)**• Find a rule for the union of three sets: | A U B U C | = ?**Quiz (3 minutes)**• Three programs use a collection of processors in the following way, where A, B, and C represent the sets of processors used by the three programs: • | A | = 20, | B | = 40, | C | = 60, | A n B | = 10, | A n C | = 8, | B n C | = 6. • If there are 100 processors available, what could | A n B n C | be? • Answer: 100 ≥ | A U B U C | = 20 + 40 + 60 – 10 – 8 – 6 + | A n B n C | . So | A n B n C | ≤ 4.**Bags (Multisets)**• Bags (Multisets) are like sets but can contain repeated elements. e.g., [t, o, o, t] = [o, t, t, o]. • Union and intersection can be defined by taking the maximum and minimum occurrences of each element, respectively. • Quiz (2 minutes). Let A = [m, i, s, s, i, s, s, i, p, p, i] and B = [s, i, p, p, i, n, g]. • What are A U B and A n B? • Answer: A U B = [m, i, s, s, i, s, s, i, p, p, i, n, g] and A n B = [s, i, p, p, i].

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