PROJECTENGRAM

[REV] CCDSTRU COMPREHENSIVE FINALS REVIEWER

Tags: #◇Computer Science #◇CCDSTRU #◇Discrete Mathematics #◇Reviewer #◇Finals


CCDSTRU Comprehensive Finals Reviewer

◇Overview

This reviewer synthesizes Discrete Structures (CCDSTRU), establishing the rigorous foundations of mathematical logic, set theory, relations and functions, mathematical induction (weak and strong), elementary number theory (Euclidean algorithm, modular arithmetic, Chinese Remainder Theorem), and combinatorics.


◇1. Core Logic & Equivalence Laws

Law NameEquivalence FormNotes
Implication Lawp→q≡¬p∨qp \to q \equiv \neg p \lor qFundamental for contrapositive and deduction proofs
Contrapositivep→q≡¬q→¬pp \to q \equiv \neg q \to \neg pEquivalent truth value; converse q→pq \to p is NOT equivalent
De Morgan's Laws¬(p∧q)≡¬p∨¬q\neg(p \land q) \equiv \neg p \lor \neg q
¬(p∨q)≡¬p∧¬q\neg(p \lor q) \equiv \neg p \land \neg q
Swaps conjunction and disjunction under negation
Biconditional Lawp↔q≡(p→q)∧(q→p)p \leftrightarrow q \equiv (p \to q) \land (q \to p)Proven by showing both directions hold
Distributive Lawsp∧(q∨r)≡(p∧q)∨(p∧r)p \land (q \lor r) \equiv (p \land q) \lor (p \land r)Boolean algebra analog to arithmetic multiplication

◇2. Relations & Properties

Given a relation RR on a set AA:

  • Reflexive: ∀x∈A,(x,x)∈R\forall x \in A, (x, x) \in R.
  • Symmetric: ∀x,y∈A,(x,y)∈R  ⟹  (y,x)∈R\forall x, y \in A, (x, y) \in R \implies (y, x) \in R.
  • Antisymmetric: ∀x,y∈A,((x,y)∈R∧(y,x)∈R)  ⟹  x=y\forall x, y \in A, ((x, y) \in R \land (y, x) \in R) \implies x = y.
  • Transitive: ∀x,y,z∈A,((x,y)∈R∧(y,z)∈R)  ⟹  (x,z)∈R\forall x, y, z \in A, ((x, y) \in R \land (y, z) \in R) \implies (x, z) \in R.

Relational Classifications

  • Equivalence Relation: Reflexive, Symmetric, and Transitive. Partitions AA into disjoint equivalence classes [x][x].
  • Partial Order (Poset): Reflexive, Antisymmetric, and Transitive (e.g., (Z,≤)(\mathbb{Z}, \le), (P(S),⊆)(\mathcal{P}(S), \subseteq)).

◇3. Mathematical Induction Framework

  • Weak Induction: Assume P(k)P(k) holds to prove P(k+1)P(k+1).
  • Strong Induction: Assume P(i)P(i) holds for all n0≤i≤kn_0 \le i \le k to prove P(k+1)P(k+1) (essential for prime factorization, game theory, recurrence).

◇4. Number Theory & Modular Arithmetic Matrix

ConceptDefinition / FormulaAlgorithmic Tool
Divisibilitya∣b  ⟺  ∃k∈Z,b=aka \mid b \iff \exists k \in \mathbb{Z}, b = akDivision Algorithm: a=dq+r,0≤r<da = dq + r, 0 \le r < d
GCD & Bézout's Identitygcd⁡(a,b)=sa+tb\gcd(a, b) = sa + tb for s,t∈Zs, t \in \mathbb{Z}Extended Euclidean Algorithm
Modular Inverseax≡1(modm)  ⟺  gcd⁡(a,m)=1ax \equiv 1 \pmod m \iff \gcd(a, m) = 1Solve as+mt=1  ⟹  a−1≡s(modm)as + mt = 1 \implies a^{-1} \equiv s \pmod m
Linear Congruenceax≡b(modm)ax \equiv b \pmod mHas solutions   ⟺  gcd⁡(a,m)∣b\iff \gcd(a, m) \mid b
Chinese Remainder Theoremx≡ai(modmi)x \equiv a_i \pmod{m_i} where gcd⁡(mi,mj)=1\gcd(m_i, m_j) = 1Unique solution modulo M=∏miM = \prod m_i

◇5. High-Yield Exam Checklist

  • Does converse (q→pq \to p) equal the original conditional (p→qp \to q)? NO.
  • For strong induction, did you verify enough base cases for the recurrence depth?
  • Does a modular inverse exist if gcd⁡(a,m)≠1\gcd(a, m) \ne 1? NO.
  • Did you remember that an equivalence relation requires symmetry, whereas a partial order requires antisymmetry?
  • Is a relation antisymmetric if it has no pairs with x≠yx \ne y? YES (vacuously true).

◇References & Navigation

CONFIRMATION // 理解度の確認

ACTIVE RECALL VERIFICATION

Select an answer to verify conceptual comprehension and reveal the technical rationale.

Q1

What is the negation of the quantified statement ∀x∈R,(x>0  ⟹  ∃y∈R,y2=x)\forall x \in \mathbb{R}, (x > 0 \implies \exists y \in \mathbb{R}, y^2 = x)?

Q2

Let R={(1,1),(2,2),(3,3),(1,2),(2,1)}R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\} on the set A={1,2,3}A = \{1, 2, 3\}. What type of relation is RR?

Q3

Using the Extended Euclidean Algorithm, what is the modular inverse of 7(mod26)7 \pmod{26}?

Q4

In mathematical induction, what is the inductive hypothesis in proving that 1+2+⋯+n=n(n+1)21 + 2 + \dots + n = \frac{n(n+1)}{2}?

Q5

Find the unique solution modulo 15 for the system: x≡2(mod3)x \equiv 2 \pmod 3 and x≡3(mod5)x \equiv 3 \pmod 5.