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The Timpson Lyceum · College Algebra

Lecture 2 (College Algebra)

Overview

This lecture completes propositional logic and builds up to first-order predicate logic, the language in which all of mathematics is written. It reviews the five basic logical operators and the equivalences that let us reword statements, then introduces predicates, quantifiers, and the precise rules for negating quantified statements. It closes with two staple proof strategies, the logical law of identity, and a first taste of axiom-based algebraic proofs. The goal is fluency: being able to read, negate, and prove statements built from propositions, predicates, and quantifiers.

Key Concepts

Proposition (statement). A sentence that is either true or false, but not both. "Proposition" and "statement" mean the same thing. Greek letters such as α,β\alpha, \beta stand in for arbitrary propositions.

The five basic operators. Each is defined by its truth table, which is its exact meaning — the English reading is only an aid to intuition.

Key equivalences.

Predicate. A proposition-valued function of one or more variables: once you supply values for its variables, it becomes a proposition. For example P(x):x>2P(x): x > 2 is not true or false until xx is given — P(5)P(5) is true, P(1)P(1) is false. A predicate of exactly two variables is called a relation (e.g. x<yx < y, often written in infix form rather than lessthan(x,y)\text{lessthan}(x,y)). Every relation is a predicate, but not every predicate is a relation. Predicates can be combined with logical operators to build new predicates.

Quantifiers. Applying a quantifier to a predicate also turns it into a proposition.

Order of quantifiers matters. ∀x ∃y (x+y=0)\forall x\,\exists y\,(x+y=0) is true (pick y=−xy=-x), but ∃x ∀y (x+y=0)\exists x\,\forall y\,(x+y=0) is false (no single xx works for every yy). Swapping the two quantifiers changes the meaning.

Negating quantifiers. Pushing a negation through a quantifier flips it:

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

So the negation of "all men in Georgia are at least six feet tall" is "there exists a man in Georgia who is less than six feet tall."

Two staple proof strategies (read off the implication truth table):

Law of identity. ∀x (x=x)\forall x\,(x = x) — everything is identical with itself. This is a theorem of logic (more basic than any field-specific axiom), used to justify operating on both sides of an equation.

Operating on both sides. If a=ba = b and the operation a∘ca \circ c and b∘cb \circ c are defined, then a∘c=b∘ca \circ c = b \circ c for any operation ∘\circ (add, multiply, etc.). The one forbidden operation is division by zero, because a/0a/0 is undefined.

The two working axioms used in the closing proofs:

Axiom 1:a+b=a+c  ⟹  b=cAxiom 2:(a⋅b=a⋅c and a≠0)  ⟹  b=c\begin{aligned} \textbf{Axiom 1:}\quad & a + b = a + c \implies b = c \\ \textbf{Axiom 2:}\quad & (a \cdot b = a \cdot c \text{ and } a \neq 0) \implies b = c \end{aligned}

Worked Example

Prove: if 3x+4=193x + 4 = 19, then x=5x = 5. Begin by assuming the antecedent, then rewrite each side so the cancellation axioms apply.

3x+4=19(given antecedent)3x+4=15+4(rewrite 19)3x=15(Axiom 1: cancel +4)3x=3⋅5(rewrite 15)x=5(Axiom 2: cancel 3≠0)■\begin{aligned} 3x + 4 &= 19 &&\text{(given antecedent)} \\ 3x + 4 &= 15 + 4 &&\text{(rewrite } 19 \text{)} \\ 3x &= 15 &&\text{(Axiom 1: cancel } +4\text{)} \\ 3x &= 3 \cdot 5 &&\text{(rewrite } 15 \text{)} \\ x &= 5 &&\text{(Axiom 2: cancel } 3 \neq 0\text{)} \quad\blacksquare \end{aligned}

Summary