Proof — Three Visual Arguments
Theorem I — Euclid, ~300 BCE
There are infinitely many prime numbers.
Proof by contradiction. Assume the opposite and derive an impossibility.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 … N the integers primes in amber p₁ , p₂ , p₃ , … , pₙ assume this is ALL the primes — a complete finite list { } Q = p₁ × p₂ × p₃ × … × pₙ the product of all primes in our list multiply together N = Q + 1 one more than the product N ÷ pᵢ always leaves remainder 1 Contradiction! N is divisible by no prime in our list. So the list was wrong.
Step 1 of 5
The natural numbers stretch infinitely. Among them, the primes — divisible only by 1 and themselves — appear in amber: 2, 3, 5, 7, 11, 13…
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The proof does not exhibit a new prime. It shows that any complete finite list is self-refuting — the integer Q+1 must have a prime factor absent from the list.
Source: Euclid, Elements, Book IX, Proposition 20. This is one of the oldest proofs still taught unchanged.
Theorem II — Pythagorean School, ~500 BCE
√2 is irrational.
Proof by infinite descent. If √2 = p/q in lowest terms, we derive a smaller equal fraction — an impossibility.
1 1 1 √2 A unit square. Its diagonal has length √2. By Pythagoras: 1² + 1² = d² ⇒ d = √2 √2 = p / q Assume p, q are integers, with no common factors (lowest terms). gcd(p, q) = 1 2 = p² / q² ⇒ p² = 2q² So p² is even ⇒ p is even ⇒ p = 2k square (2k)² = 2q² ⇒ q² = 2k² So q² is even ⇒ q is even. sub p=2k Contradiction! Both p and q are even. So gcd(p,q) ≥ 2. This contradicts our assumption that p/q was in lowest terms. ∴ √2 cannot be rational.
Step 1 of 5
A unit square has sides of length 1. By the Pythagorean theorem, its diagonal has length √2. The question: is this length a ratio of two whole numbers?
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Legend holds that Hippasus proved this on a ship and was thrown overboard by fellow Pythagoreans. The integers could not contain the diagonal of their most perfect shape.
The proof generalizes: √n is irrational for any non-perfect-square integer n.
Theorem III — Visual / Classical
1 + 3 + 5 + … + (2n−1) = n²
A proof without words. The sum of the first n odd numbers is always a perfect square.
1 1 = 1² n = 1: sum = 1 = 1² One square cell. Side = 1. +3 1+3 = 4 = 2² n = 2: sum = 1+3 = 4 = 2² Add an L-shaped border of 3 cells. The shape remains a perfect square. +5 1+3+5 = 9 = 3² n = 3: sum = 1+3+5 = 9 = 3² Each odd number adds one more L-shaped gnomon. Squares grow from squares. n² n n ∑ₖ₌₁ⁿ (2k−1) = n² Each successive odd number (2k-1) adds exactly one new L-shaped layer (gnomon) of k cells on the top and k-1 on the side. The geometry is the proof.
Step 1 of 4
The first odd number is 1. It forms a 1×1 square. Conjecture: adding successive odd numbers always produces a perfect square.
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This is a proof without words — the diagram itself constitutes the full argument. No algebra is needed; the visual structure is the demonstration.
Attributed to Nicomachus of Gerasa (~100 CE). The L-shaped pieces are called gnomons after the shadow-casting part of a sundial.