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The Timpson Lyceum · History of Philosophy (2024)

Historical Context and Thales of Miletus – Lecture 2 (History of Philosophy)

Overview

Before meeting the first philosopher, this lecture builds the world he was born into — because philosophy is in large part a response to its environment. It sketches world history up to roughly 500 BC: the two great river civilizations of Egypt (stable, unified, geometry-rich thanks to Nile-flood land surveying) and Mesopotamia (fickle rivers, warring city-states, a base-60 number system and sky-watching that still shapes our clocks and calendars); the Bronze Age collapse (~1100 BC) that erased most of the Mediterranean world but which Egypt and Babylon survived; and the rise of Lydia and then Persia. It then turns to the Greeks: scattered by colonization into an "information highway" of trading city-states, split culturally toward Sparta (a rigid, communal warrior state — the "Godfather" of Greece) and Athens (moving from Draco's harsh code to Solon's foundations of democracy).

Against this backdrop arrives Thales of Miletus (~624–546 BC), traditionally counted the first philosopher and one of the Seven Sages. He is the first of the physikoi — the natural philosophers who tried to explain the cosmos not by myth but by a single underlying principle.

Key Concepts

Worked Example

Thales' theorem — any angle inscribed in a semicircle is a right angle. A clean specimen of the deductive geometry Thales helped inaugurate.

  1. Let ABAB be a diameter of a circle with center OO, and let CC be any other point on the circle. Claim: ∠ACB=90∘\angle ACB = 90^\circ.
  2. Draw radius OCOC. Then OA=OB=OCOA = OB = OC (all radii), so triangles OACOAC and OBCOBC are isosceles.
  3. Isosceles △OAC\triangle OAC gives equal base angles: let ∠OAC=∠OCA=α\angle OAC = \angle OCA = \alpha.
  4. Isosceles △OBC\triangle OBC gives equal base angles: let ∠OBC=∠OCB=β\angle OBC = \angle OCB = \beta.
  5. The angle at CC is ∠ACB=α+β\angle ACB = \alpha + \beta.
  6. The angles of △ABC\triangle ABC sum to 180∘180^\circ:   α+β+(α+β)=180∘\;\alpha + \beta + (\alpha + \beta) = 180^\circ.
  7. Hence 2(α+β)=180∘2(\alpha + \beta) = 180^\circ, so α+β=90∘\alpha + \beta = 90^\circ.
  8. Therefore ∠ACB=90∘\angle ACB = 90^\circ for every choice of CC on the circle. ■\blacksquare

Summary