Quiz 2

Advanced Logic: Proof Strategies and Uniqueness

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# Advanced Logic: Proof Strategies and Uniqueness ## 8.1 Proving Universal Statements **Strategy:** "Let $x$ be arbitrary" — choose an arbitrary element and prove the property holds. **Example:** Prove $n^2 \geq n$ for all integers $n \geq 1$.

Advanced Logic: Proof Strategies and Uniqueness

8.1 Proving Universal Statements

Strategy: "Let xx be arbitrary" — choose an arbitrary element and prove the property holds. Example: Prove n2≥nn^2 \geq n for all integers n≥1n \geq 1. Proof: Let nn be an arbitrary integer with n≥1n \geq 1. Then n2−n=n(n−1)≥0n^2 - n = n(n-1) \geq 0 because n≥1n \geq 1 implies n≥0n \geq 0 and n−1≥0n-1 \geq 0. Thus n2≥nn^2 \geq n. □\square

8.2 Proving Existential Statements

Strategy: Find one example (constructive) or show existence without construction (non-constructive). Constructive: "There exists an even prime" — the number 2. Non-constructive: "There exist irrational a,ba,b such that aba^b is rational." Consider 22\sqrt{2}^{\sqrt{2}}. If it's rational, done. If it's irrational, then (22)2=22=2(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^2 = 2 is rational. Either way, such numbers exist.

8.3 Proving Uniqueness

Show existence and uniqueness. Example: Prove there is a unique solution to 3x+5=03x + 5 = 0. Proof:
  • Existence: x=−53x = -\frac{5}{3} satisfies 3(−53)+5=03(-\frac{5}{3}) + 5 = 0.
  • Uniqueness: If 3x+5=03x + 5 = 0 and 3y+5=03y + 5 = 0, then 3x+5=3y+5  ⟹  3x=3y  ⟹  x=y3x + 5 = 3y + 5 \implies 3x = 3y \implies x = y.

8.4 Induction with Inequalities

Example: Prove 3n≥2n+13^n \geq 2^n + 1 for n≥1n \geq 1. Base (n=1n=1): 3≥2+1=33 \geq 2 + 1 = 3. ✓ IH: 3k≥2k+13^k \geq 2^k + 1. Step: 3k+1=3⋅3k≥3(2k+1)=3⋅2k+3=2k+1+2k+3≥2k+1+13^{k+1} = 3 \cdot 3^k \geq 3(2^k + 1) = 3 \cdot 2^k + 3 = 2^{k+1} + 2^k + 3 \geq 2^{k+1} + 1. □\square

8.5 Proof by Minimum Counterexample

Assume the statement is false and let nn be the smallest counterexample. Show this leads to contradiction. Example: Every integer ≥2\geq 2 is a product of primes. Proof: Let nn be the smallest integer ≥2\geq 2 that is not a product of primes. Then nn is not prime (else it would be a product), so n=abn = ab with 2≤a,b<n2 \leq a, b < n. Since a,ba,b are smaller, they are products of primes. Their product gives a prime factorization of nn. Contradiction. □\square Join Discord PreviousModular Arithmetic DeepNextCardinality and Countability
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