Ancient Mathematics

Known empirically in Mesopotamia and China (`gougu` rule), it received a classic deductive area-shearing proof in Book I, Proposition 47 of Euclid's Elements (c. 300 BCE), dropping an altitude from the right angle to split the hypotenuse square into two rectangles equal in area to $a^2$ and $b^2$.

Step 1 (Square dissection / gougu configuration): Arrange four congruent right triangles with legs $a, b$ and hypotenuse $c$ around the perimeter of a large square of side length $a + b$. Because the two acute angles of each right triangle sum to $\alpha + \beta = 90^\circ$, each interior corner formed by adjacent hypotenuses measures $180^\circ - (\alpha + \beta) = 90^\circ$, so the central region is a tilted square of side $c$ and area $c^2$.

Step 2 (Equating total area in two ways): Expanding the area of the outer square algebraically gives $(a + b)^2 = a^2 + 2a b + b^2$. Decomposing the same outer square into the central square plus four right triangles gives $c^2 + 4 \cdot \frac{1}{2}a b = c^2 + 2a b$. Subtracting $2a b$ from both expressions yields $a^2 + b^2 = c^2$.

Step 3 (Euclid's altitude-projection proof, Elements I.47): Drop the altitude from the right angle to the hypotenuse, splitting $c$ into segments $p$ and $q$ with $p + q = c$. By triangle similarity (or Euclid's area-shearing of each leg's square onto the corresponding hypotenuse rectangle), the two squares on the legs have areas $a^2 = c p$ and $b^2 = c q$. Adding the two rectangles gives $a^2 + b^2 = c p + c q = c(p + q) = c^2$.

Euclid considers $N = p_1 p_2 \cdots p_n + 1$. Since $N > 1$, $N$ has a prime factor $q$. No $p_i$ can divide $N$ because dividing $N$ by $p_i$ leaves remainder $1$; hence $q$ is a new prime outside $\{p_1, \dots, p_n\}$.

Step 1 (Euclid's construction, Elements IX.20): Given any finite list of primes $p_1, p_2, \dots, p_n$, define the integer $N = p_1 p_2 \cdots p_n + 1$. Because $p_1 \ge 2$, the product $p_1 p_2 \cdots p_n \ge 2$, so $N \ge 2 + 1 = 3 > 1$.

Step 2 (Existence of a prime divisor, Elements VII.31): Every integer greater than $1$ has a prime divisor $q$ (its smallest divisor greater than $1$ must be prime), so there exists a prime $q$ satisfying $q \mid N$.

Step 3 (Eliminating all primes in the given list): If $q = p_i$ for some $i \in \{1, \dots, n\}$, then $q \mid p_1 p_2 \cdots p_n$. Since we also have $q \mid N$, $q$ must divide the difference $q \mid (N - p_1 p_2 \cdots p_n) = 1$, which contradicts $q \ge 2$. Consequently $q \notin \{p_1, p_2, \dots, p_n\}$, proving that any finite list of primes is incomplete.

In On the Sphere and Cylinder (c. 225 BCE) and The Method of Mechanical Theorems, Archimedes balanced cross-sections of a sphere and a cone against a cylinder on a virtual lever, then supplied a rigorous exhaustion proof. He was so proud of the $\frac{2}{3}$ ratio that he asked for a sphere inscribed in a cylinder to be engraved on his tomb.

Step 1 (Archimedes' slice setup): Place a hemisphere of radius $r$ next to a right circular cylinder of base radius $r$ and height $r$ from which an inverted cone of base radius $r$ and height $r$ (apex at the base center) has been carved out. Cut both solids by a horizontal plane at height $z \in [0, r]$ above the base.

Step 2 (Matching cross-sectional areas): At height $z \in [0, r]$, the hemisphere's cross-section is a circle of radius $\rho = \sqrt{r^2 - z^2}$ by the Pythagorean theorem, so its area is $A_{\text{sphere}}(z) = \pi \rho^2 = \pi(r^2 - z^2)$. At the same height $z \in [0, r]$, the cone's cross-section has radius $z$ by similar triangles, so the punctured cylinder's annular cross-section has area $A_{\text{annulus}}(z) = \pi r^2 - \pi z^2 = \pi(r^2 - z^2)$.

Step 3 (Volume and surface area ratios): Because the cross-sectional areas match at every height $z \in [0, r]$, Cavalieri's principle (and Archimedes' mechanical lever / method of exhaustion) gives the hemisphere volume as cylinder minus cone: $V_{\text{hemi}} = \pi r^3 - \frac{1}{3}\pi r^3 = \frac{2}{3}\pi r^3$. Doubling yields the full sphere volume $V = \frac{4}{3}\pi r^3 = \frac{2}{3}(2\pi r^3)$. Decomposing the sphere into thin cones of height $r$ from the center gives $V = \frac{1}{3} r S$, whence $S = \frac{3V}{r} = 4\pi r^2 = \frac{2}{3}(6\pi r^2)$.