Lecture Notes, 1 Dec 2010

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§§8.1; 8.5: Relations

1. Database Theory as the Computational Theory of Relations:
   1. All set operations apply to relations: That's how databases work!

   2. E.g.)
      Let C = {c | c is a course at UB}
      Let S = {s | s is a student at UB}

      Then can define R_i ⊆ S × C such that:

      • R₁ = {(s,c) | student s is taking course c}
      • R₂ = {(s,c) | student s must take course c for the CSE major}
      • Then:
        1. R₁ ∪ R₂ = {(s,c) | (s is taking c) ∨ (s must take c for CSE)}
           • i.e.) returns list of all (s,c) specifying students' actual and required CSE courses.

        2. R₁ ∩ R₂ = {(s,c) | (s is taking c) ∧ (s must take c for CSE)}
           • i.e.) returns list of all (s,c) with current courses that are required

        3. R₁ – R₂ = {(s,c) | (s is taking c) ∧ (s doesn't have to take c for CSE)}
           • i.e.) returns list of all (s,c) with current electives.

   3. E.g.)
      • Let < be defined recursively as we did last time

        Let = ⊆ N×N be defined recursively as follows:

        • Base Case: (0,0) ∈ =
          Recursive Case: (x,y) ∈ =  →  (S(x),S(y)) ∈ =

        Then: ≤  =def  < ∪ =

        (= {(x,y) | (x,y) ∈ <   ∨   (x,y) ∈ =})

   ---

2. Equivalence Relations:
   1. Let ∼ ⊆ A × A.
      Then ∼ is an equivalence relation on A =def
      R is reflexive, symmetric, & transitive.
      • Notation: "a ∼ b" for: (a,b) ∈ ∼
        (read: "a is equivalent to b")

   2. E.g.)
      Let WFPL = {A | A is a well-formed proposition of propositional logic}
      Let ≡ ⊆ WFPL×WFPL s.t. A≡B iff tval(A)=tval(B)
      i.e.) iff (A↔B) is a tautology
      Then ≡ is an equivalence relation on WFPL
      • proof:
        tval(A)=tval(A)
        ∴ A≡A

        tval(A)=tval(B) → tval(B)=tval(A)
        ∴ A≡B → B≡A

        (tval(A)=tval(B) ∧ tval(B)=tval(C)) →
                                                  tval(A)=tval(C)
        ∴ (A≡B ∧ B≡C) → A≡C
        QED

   3. Def:
      Let A₁, …, A_n ⊆ A.
      Then:
      1. The A_i are mutually exclusive =def
         they are "pairwise disjoint"
         • i.e.) (∀A_i, A_j)[A_i ∩ A_j = ∅]

      2. The A_i are jointly exhaustive =def ∪_iA_i = A

      3. Let A₁, …, A_n ⊆ A be mutually exclusive
         & jointly exhaustive.
         Then {A₁, …, A_n} is_def a partition of A.
      • So A looks like this:
        

   4. Def:
      Let ∼ be an equivalence relation on set A.
      Let a ∈ A.

      Then:

      1. the equivalence class of a under ∼
         denoted [a]_∼
         =def

         {a′ ∈ A | a′ ∼ a}

      2. α is a representative of [a]_∼ =def α ∈ [a]_∼.

      ---

   5. Thm:
      Let A be a set.
      Then:
      (∀ equiv relation ∼ ⊆ A×A)(∃ A₁,…,A_n ⊆ A)[{A₁,…,A_n} is a partition of A]
      1. i.e.) ∀ equivalence relation, ∃ partition
         i.e.) every equiv relation "induces" a partition

      2. proof sketch:
         Let ∼ be an equiv relation on A.

         Let P = {[a₁]_∼, …, [a_n]_∼} be the set of all equivalence classes of elements of A under ∼.

         Then ∪_i[a_i]_∼ = A.

         ∴ They are jointly exhaustive.

         And (∀i, j)[[a_i]_∼ ∩ [a_j]_∼ = ∅

         ∴ They are mutually exclusive.

         ∴ P is a partition of A.

         QED.

         

      ---

   6. Thm:

      (∀ partition {A₁,…,A_n} of A)(∃ ∼ that is an equivalence relation on A]

      1. i.e.) ∀ partition, ∃ equivalence relation

      2. proof sketch,
         Let P = {A₁,…,A_n} be a partition of A.
         Let a,b ∈ A.
         Let a ∼ b =def ∃A_i[a,b ∈ A_i].
         Show ∼ reflexive, symmetric, & transitive.
         QED

      ---

   7. Best Example: Q as fractions:

      1. Consider the set of all fractions F = {numerals m/n | m,n ∈ Z}
         1. numerals are symbols that name numbers:
            • "1", "one", "un", "uno", and "I"
              are 5 different numerals
              that all name the same number (namely, S(0)).

         2. "½" ∈ F;
            "2/4" ∈ F;
            and those are two different members of F

         3. As numerals, "½" ≠ "2/4".
         4. As numbers, ½ = 2/4

         5. Better: As numerals, ½ ∼ 2/4,
            where a/b ∼ c/d  =def  ad=bc

      2. The relation between 2 fractions
         when they can be reduced to the same lowest common denominator
         is an equivalence relation that partitions F
         into subsets of equivalent fractions
         
         • Let ∼ be that equivalence relation on F, as above.
           Then [½]_∼ = [2/4]_∼
           & any fraction a/b ∼ ½ can represent any other fraction that = ½.

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