Unit 5: Comprehensive Topics from Exams
I. Set Theory
Core Concepts:
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Set: A well-defined collection of distinct objects.
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Operations:
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Union: $$\displaystyle A \cup B = \{x \mid x \in A \text{ or } x \in B\} $$
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Intersection: $$\displaystyle A \cap B = \{x \mid x \in A \text{ and } x \in B\} $$
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Difference: $$\displaystyle A - B = \{x \mid x \in A \text{ and } x \notin B\} $$
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Complement (w.r.t. universal set $U$): $$\displaystyle A' = U - A $$
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Cartesian Product: $$\displaystyle A \times B = \{(a,b) \mid a \in A, b \in B\} $$
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De Morgan's Laws (Proof via Venn Diagrams):
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$$\displaystyle (A \cup B)' = A' \cap B' $$
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$$\displaystyle (A \cap B)' = A' \cup B' $$
[!TIP] Exam Focus: You may be asked to prove De Morgan's laws using Venn diagrams. Draw two overlapping circles for $A$ and $B$, shade the left side, then show the right side yields the same shaded region.
II. Relations
Properties of a Relation $R$ on set $A$:
| Property | Definition |
|---|---|
| Reflexive | $\forall a \in A, (a,a) \in R$ |
| Symmetric | $\forall a,b \in A, (a,b) \in R \Rightarrow (b,a) \in R$ |
| Transitive | $$\displaystyle \forall a,b,c \in A, (a,b) \in R \land (b,c) \in R \Rightarrow (a,c) \in R $$ |
| Irreflexive | $\forall a \in A, (a,a) \notin R$ |
Equivalence Relation: A relation that is reflexive, symmetric, and transitive. Partitions the set into equivalence classes.
Transitive Closure $$\displaystyle R^+ $$: The smallest transitive relation containing $R$. For common relations on $\mathbb{Z}$:
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$$\displaystyle a+1=b $$ → Transitive closure is $a \le b$ (or $a \le b$ on integers).
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$$\displaystyle a-b=2 $$ → Transitive closure is $a-b$ is even (i.e., $a \equiv b \pmod{2}$).
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$$\displaystyle a^2 - b^2 $$ divisible by 4 → Transitive closure is $a \equiv b \pmod{2}$ (both even or both odd).
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$|a-b| \le 1$ → Transitive closure is $$\displaystyle a=b $$ (since $|a-b|\le1$ and $|b-c|\le1$ does NOT imply $|a-c|\le1$ generally; closure forces equality).
Relation Matrix & Digraph:
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Matrix $$\displaystyle M_R $$: $$\displaystyle [m_{ij}] $$ where $$\displaystyle m_{ij}=1 $$ if $$\displaystyle (a_i, b_j) \in R $$, else 0.
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Digraph: Directed graph with vertices as set elements, edge $a \to b$ iff $(a,b) \in R$.
Domain & Range:
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Domain: $\{a \mid \exists b, (a,b) \in R\}$
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Range: $\{b \mid \exists a, (a,b) \in R\}$
[!TIP] Common Pitfall: For $aRb$ defined by $a$ is multiple of $b$ on $$\displaystyle A=\{1,2,3,4,5,6\} $$, note $a$ is multiple of $b$ means $$\displaystyle a = kb $$. So $(4,2) \in R$ but $(2,4) \notin R$. Domain is all $a$ that are multiples of some $b$—here all of $A$. Range is all $b$ that divide some $a$—here $\{1,2,3,4,5,6\}$ but check: $5$ only divides $5$ in $A$, so yes.
III. Partially Ordered Sets (Posets)
Definition: A set $A$ with relation $\le$ that is reflexive, antisymmetric, and transitive.
Key Terms (for Poset $(A, \le)$):
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Maximal element: $a$ such that no $b \in A$ with $$\displaystyle a < b $$.
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Minimal element: $a$ such that no $b \in A$ with $$\displaystyle b < a $$.
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Greatest element (maximum): $a$ such that $\forall b \in A, b \le a$.
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Least element (minimum): $a$ such that $\forall b \in A, a \le b$.
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Lattice: A poset where every pair of elements has a least upper bound (join) and greatest lower bound (meet).
Hasse Diagram:
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Draw vertices as elements.
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Omit reflexive loops and transitive edges.
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Place higher elements above lower ones.
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Connect with edges only for covering relation ($a$ covers $b$ if $$\displaystyle a > b $$ and no $c$ with $$\displaystyle a > c > b $$).
Common Divisor Sets:
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$$\displaystyle D_{15} = \{1,3,5,15\} $$: 1 at bottom, 15 at top, 3 and 5 in middle, edges: $1\to3$, $1\to5$, $3\to15$, $5\to15$.
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$$\displaystyle D_{24} = \{1,2,3,4,6,8,12,24\} $$: Build stepwise: 1 covers 2,3; 2 covers 4,6; 3 covers 6; 4 covers 8,12; 6 covers 12; 8 covers 24; 12 covers 24.
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$$\displaystyle A=\{1,2,3,4,6,8,9,12,18,24\} $$: Similar to $$\displaystyle D_{24} $$ but missing 16? Actually includes 9,18. Check divisibility: 1 covers 2,3; 2 covers 4,6; 3 covers 6,9; 4 covers 8,12; 6 covers 12,18; 8 covers 24; 9 covers 18; 12 covers 24; 18 covers 24.
[!TIP] Exam Tip: For Hasse diagram, always list elements in order of divisibility. Draw from bottom (1) to top (largest). Ensure no crossing edges if possible.
IV. Propositional Logic
Basic Connectives & Truth Tables:
| Connective | Symbol | Example |
|---|---|---|
| NOT | $\neg p$ | $$\displaystyle \neg T = F $$ |
| AND | $p \land q$ | $$\displaystyle T \land F = F $$ |
| OR | $p \lor q$ | $$\displaystyle T \lor F = T $$ |
| Implies | $p \to q$ | $$\displaystyle T \to F = F $$ |
| IFF | $$\displaystyle p \leftrightarrow q $$ | $$\displaystyle T \leftrightarrow F = F $$ |
Tautology: Always true (e.g., $p \lor \neg p$).
Contradiction: Always false (e.g., $p \land \neg p$).
Logical Equivalence: $p \equiv q$ iff $$\displaystyle p \leftrightarrow q $$ is a tautology. Prove via truth tables or laws.
Normal Forms:
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DNF (Disjunctive Normal Form): OR of ANDs (minterms). Example: $(p \land \neg q) \lor (\neg p \land q)$.
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CNF (Conjunctive Normal Form): AND of ORs (maxterms). Example: $(p \lor q) \land (\neg p \lor q)$.
Conversion to CNF:
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Eliminate $$\displaystyle \leftrightarrow $$, $\to$.
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Move $\neg$ inward (De Morgan, double negation).
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Distribute $\lor$ over $\land$.
[!EXAMPLE]
$p \land (p \to q)$
Step 1: $p \land (\neg p \lor q)$
Step 2: $(p \land \neg p) \lor (p \land q)$
Step 3: $F \lor (p \land q) \equiv p \land q$ (already CNF? Actually $p \land q$ is a conjunction of literals, so it's CNF).
V. Graph Theory
Basic Definitions:
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Graph $$\displaystyle G = (V,E) $$: $V$ vertices, $E$ edges.
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Simple graph: No loops, no multiple edges.
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Weighted graph: Edges have weights/costs.
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Directed graph (digraph): Edges are ordered pairs.
Representations:
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Adjacency Matrix $A$: $$\displaystyle a_{ij} = \text{number of edges from } v_i \text{ to } v_j $$.
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Adjacency List: For each vertex, list adjacent vertices.
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Incidence Matrix: Rows = vertices, columns = edges; entry = 1 if vertex incident to edge.
Eulerian Graph:
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Euler's Theorem: A connected graph has an Eulerian circuit iff every vertex has even degree.
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**Eulerian path