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The Timpson Lyceum · First-Order Predicate Logic (2024)

Axioms of the Real Numbers - Lecture 12 (Logic)

Axioms of the Real Numbers – Lecture 12

Overview

First-order logic gives us the language for building axiomatic systems — the way we encode systematic bodies of knowledge from a small set of starting assumptions. This lecture develops one such system: arithmetic, treating the real numbers R\mathbb{R} as an abstract domain governed by fifteen axioms about two operators (++, ×\times) and one relation (<<). Alongside the axioms, the lecture introduces informal proof — proofs written in ordinary mathematical style rather than as fully formalized derivations. We see how every familiar arithmetic fact (that 1+1≠11+1\neq 1, that 2+2=42+2=4, cancellation laws, x⋅0=0x\cdot 0=0, the sign rules) must be earned as a theorem from the axioms, with nothing about real numbers assumed in advance.

Key Concepts

Components of an axiomatic system.

Operators vs. relations.

The 15 axioms (all variables range over R\mathbb{R}):

Convention on parentheses. Strictly, each operator takes exactly two arguments, so an expression like x⋅y+zx\cdot y+z is ambiguous without parentheses. By the agreed convention "multiplication before addition," x⋅y+zx\cdot y+z means (x⋅y)+z(x\cdot y)+z. This is a notational agreement, not an axiom.

Equality is logical. == is a symbol of logic (first-order logic with identity), so its rules — reflexivity, substitution of equals for equals, and adding/applying the same operation to both sides (a=b⇒a∘c=b∘ca=b \Rightarrow a\circ c=b\circ c for any well-defined deterministic operation ∘\circ) — come for free and need not be re-axiomatized.

Formal vs. informal proof. Formal proofs cite an exact rule at every step and become tedious through repetition. Informal proofs state the key axiom/theorem used at each step in ordinary prose — there are no rigid rules, only style and clarity. Every step must still be justified by an axiom, a prior theorem, or a law of logic.

Building the number system. Starting from only 00 and 11, closure forces new elements into R\mathbb{R}: one proves 1+11+1 differs from both 00 and 11, defines it as 22, then 2+1=:32+1=:3, and so on, giving all the naturals. The additive-inverse definition then yields the integers, and the multiplicative-inverse definition yields the rationals.

Defined symbols.

Representative theorems proved.

What's still missing. These axioms give all of basic algebra but not continuity. The one further axiom needed is completeness — every nonempty set bounded above has a least upper bound — which is what makes 2∈R\sqrt{2}\in\mathbb{R} and underlies all of calculus/real analysis.

Worked Example

Theorem (zero product): for all xx,   x⋅0=0\;x\cdot 0 = 0.

The strategy is to make zero "interact" with multiplication, which only the distributive axiom does. Write 00 as 0+00+0, distribute, then cancel.

x⋅0=x⋅(0+0)(0+0=0, additive identity)=x⋅0+x⋅0(distributivity)x⋅0+0=x⋅0+x⋅0(add identity 0 to the left)0=x⋅0(left cancellation of the x⋅0 term)\begin{aligned} x\cdot 0 &= x\cdot(0+0) && \text{(}0+0=0\text{, additive identity)} \\ &= x\cdot 0 + x\cdot 0 && \text{(distributivity)} \\ x\cdot 0 + 0 &= x\cdot 0 + x\cdot 0 && \text{(add identity }0\text{ to the left)} \\ 0 &= x\cdot 0 && \text{(left cancellation of the }x\cdot 0\text{ term)} \end{aligned}

So x⋅0=0x\cdot 0 = 0 for every xx. ■\blacksquare

Theorem (uniqueness of the additive identity): x+y=x⇒y=0x+y=x \Rightarrow y=0.

Assume x+y=xx+y=x. By the additive-inverse axiom pick ww with x+w=0x+w=0, then add ww to both sides and regroup.

w+(x+y)=w+x(add w to both sides)(w+x)+y=w+x(associativity)(x+w)+y=x+w(commutativity)0+y=0(substitute x+w=0)y=0(additive identity)\begin{aligned} w+(x+y) &= w+x && \text{(add }w\text{ to both sides)} \\ (w+x)+y &= w+x && \text{(associativity)} \\ (x+w)+y &= x+w && \text{(commutativity)} \\ 0+y &= 0 && \text{(substitute }x+w=0\text{)} \\ y &= 0 && \text{(additive identity)} \end{aligned}

So 00 is the only element that behaves like an additive identity. ■\blacksquare

Summary