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The Timpson Lyceum · Discrete Math (2025)

Factorial, Sets, and Quantifiers - Lecture 5 (Discrete Math)

Factorial, Sets, and Quantifiers – Lecture 5

Overview

This lecture pulls together the counting tools from the previous week and builds the foundational language of set theory and logic. It introduces factorials and falling factorials as ways to count arrangements, then compact notation for long sums and products (sigma and pi). From there it develops sets — membership, equality, cardinality, subsets, set-builder notation, and the power set — and practices writing rigorous proofs of set equality. It closes with quantifiers (existential and universal), how to read nested quantifier statements, and the mechanical rules for negating them.

Key Concepts

Counting choices (with vs. without repeats)

Factorial

Summation notation (sigma)

Product notation (pi)

Sets

Set equality, cardinality, finiteness

Set-builder notation

Subsets and the power set

Quantifiers

¬ ∀x, P(x) ≡ ∃x, ¬P(x)¬ ∃x, P(x) ≡ ∀x, ¬P(x)\begin{aligned} \neg\,\forall x,\ P(x) \ &\equiv\ \exists x,\ \neg P(x) \\ \neg\,\exists x,\ P(x) \ &\equiv\ \forall x,\ \neg P(x) \end{aligned}

"Not all swans are white" becomes "there exists a swan that is not white."

Worked Example

Claim: The set of even integers E={ x∈Z:2∣x }E = \{\, x \in \mathbb{Z} : 2 \mid x \,\} equals the set of integers expressible as a sum of two odds, F={ z∈Z:z=a+b, a,b odd }F = \{\, z \in \mathbb{Z} : z = a + b,\ a, b \text{ odd} \,\}.

Prove set equality by showing both inclusions: E⊆FE \subseteq F and F⊆EF \subseteq E.

(E⊆FE \subseteq F) Take any n∈En \in E. Being even means n=2mn = 2m for some integer mm. Rewrite it as a sum of two odds:

n=2m=(2m+1)+(−1)=(2m+1)⏟odd+(−1)⏟odd\begin{aligned} n &= 2m \\ &= (2m + 1) + (-1) \\ &= \underbrace{(2m+1)}_{\text{odd}} + \underbrace{(-1)}_{\text{odd}} \end{aligned}

Both summands are odd, so n∈Fn \in F. (Concretely, 8=2⋅4=(2⋅4+1)+(−1)=9+(−1)8 = 2\cdot 4 = (2\cdot 4 + 1) + (-1) = 9 + (-1).)

(F⊆EF \subseteq E) Take any n∈Fn \in F, so n=a+bn = a + b with a,ba, b odd. Then a=2x+1a = 2x + 1 and b=2y+1b = 2y + 1 for integers x,yx, y:

n=a+b=(2x+1)+(2y+1)=2x+2y+2=2(x+y+1)\begin{aligned} n = a + b &= (2x + 1) + (2y + 1) \\ &= 2x + 2y + 2 \\ &= 2(x + y + 1) \end{aligned}

Since x+y+1x + y + 1 is an integer, nn is two times an integer, hence even, so n∈En \in E. (Concretely, 18=15+3=(2⋅7+1)+(2⋅1+1)=2(7+1+1)=2⋅918 = 15 + 3 = (2\cdot 7 + 1) + (2\cdot 1 + 1) = 2(7 + 1 + 1) = 2\cdot 9.)

Both inclusions hold, therefore E=FE = F. ■\blacksquare

Summary