Mathistory

Why do we prove things?

Proofs

How “it works” stopped being enough

SIX STOPS

Browse the stops with the arrows or by swiping left and right; the timeline stays at the top.

PROLOGUE · TODAY · A PUZZLE TO START

Is “it works” enough?

I

QuestionHow many examples are enough?

Before we set off into the past, here is a puzzle. Try the expression

n2 + n + 41

for n = 0, 1, 2, 3, … and see what numbers come out.

Do you get a prime for n = 0, 1, 2, 3, 4, 5? How many values of n would you try before believing that the formula always gives primes?

Keep trying

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Reveal the answer – once you have tried it

The formula gives primes for every n = 0, 1, …, 39. But for n = 40 we get 402 + 40 + 41 = 412 = 1681, a composite number. So forty successful checks were not enough.

II

IdeaChecking examples is not a proof.

This is the main idea of the whole journey: even forty successful attempts do not guarantee that the forty-first will turn out the same way.

At the following stops we will see when and why people learned this lesson, and how it gave birth to proof, a way of being certain.

III

FormulaForty successes and one exception.

FORMULA

n2 + n + 41

n = 40:  402 + 40 + 41 = 412

Examples can disprove a claim, but they can never prove it.

STOP I · ROUGHLY 19TH–17TH CENTURY BCE

Mesopotamia: “It works, we can see it”

scribes in Mesopotamia

I

QuestionAre many examples the same as a proof?

Babylonian scribes had procedures that worked, and they checked them on many examples. Is that the same as a proof? What would they need before we could say that they know the formula holds for all p, q?

Something to think about – talk it over in pairs first

They would need a general argument that does not depend on particular numbers. For Pythagorean triples it is enough to multiply out:

(p2 − q2)2 + (2pq)2 = (p2 + q2)2

Try it yourself – it makes a good one-minute exercise.

II

PersonScribes with clay tablets.

About 4000 years ago, in the Old Babylonian period (roughly the 19th–17th century BCE), scribes in Mesopotamia wrote mathematical problems and tables in cuneiform on clay tablets. Thousands of them have survived.

The famous tablet Plimpton 322 (held at Columbia University) contains a table of fifteen rows of numbers. According to today’s interpretation it is a list of Pythagorean triples, that is, triples of natural numbers (a, b, c) for which a2 + b2 = c2. So the Babylonians knew the relation we now associate with Pythagoras more than a thousand years earlier.

Another famous tablet (YBC 7289) gives an approximate value of √2 in base sixty, which corresponds to the number 1.41421296…

What survives on the tablets is mainly procedures and numerical examples, not general justification in today’s sense. The purpose of Plimpton 322 is disputed among historians (a list of triples, a teaching aid, some other explanation). So we do not say that “the Babylonians could not prove things”, only that the surviving texts contain no proof in the later Greek sense. More: Before Pythagoras (ISAW, NYU).

III

IdeaA procedure that works.

TRY IT LIKE A SCRIBE

Check that the triples (3, 4, 5), (5, 12, 13), (8, 15, 17) are Pythagorean. Then try to come up with more. Notice that for any p > q you get the triple

(p2 − q2, 2pq, p2 + q2)

Try it for several values of p, q. Does it always work?

The scribe’s table

choose p and q

Every new example that works makes us more confident. But even a thousand pairs we have tried do not tell us that it works for all of them.

IV

FormulaKnown a thousand years before Pythagoras.

FORMULA

a2 + b2 = c2

(p2 − q2)2 + (2pq)2 = (p2 + q2)2

A procedure that works is not yet a proof.

STOP II · AROUND THE 6TH CENTURY BCE · MILETUS

Thales of Miletus: “Why is it true?”

I

QuestionIsn’t it enough that we can see it?

Draw a circle and its diameter. Is it obvious to you that it divides the disc into two equal halves? And what if someone asked you: Why?

How Thales answered – according to tradition

Proclus reports a tradition that Thales argued like this: fold the disc along the diameter – the two halves must match exactly, because every point of the circle is the same distance from the centre and the diameter passes through the centre.

What matters is not how simple this argument is, but that “we can see it is so” gives way to the question “why must it be so”, and the answer is based on what we already know.

II

PersonOne of the Seven Sages.

Several hundred years later, around the 6th century BCE, Thales lived in the Greek city of Miletus. Tradition counts him among the Seven Sages of ancient Greece and credits him with several geometric statements. Little about him is certain – the oldest reports come from later authors, such as Proclus (5th century CE), who quotes an older history of geometry.

According to tradition, he showed among other things that:

  • every diameter divides a disc into two congruent halves,
  • vertical angles are equal,
  • the angle in a semicircle is a right angle (today called Thales’s theorem).

Another tradition says he measured the height of a pyramid from the length of its shadow. Think about why and when that works (hint: similar triangles).

We cannot be sure that Thales proved Thales’s theorem; tradition credits him rather with discovering it. A proof appears in Euclid’s Elements (Book III, Proposition 31). More: MacTutor, University of St Andrews.

III

IdeaDirect proof: step by step from what we know.

THALES’S THEOREM – A FIRST EXAMPLE OF DIRECT PROOF

We are given a circle with centre S, diameter AB and a point C on the circle. Prove that the angle ACB is a right angle.

Hint: the triangles ASC and BSC are isosceles. What do you know about the sum of the angles in triangle ABC?

Solution

The segments SA, SB, SC are radii, so they have the same length. We can therefore write ∠SAC = ∠SCA = α and ∠SBC = ∠SCB = β. The sum of the angles in triangle ABC is

2α + 2β = 180°,  so  α + β = 90° = ∠ACB.

AB CS αβ αβ
The radius SC divides the triangle into two isosceles ones.
For the curious: the converse of Thales’s theorem – the Thales circle

Claim: if a point C (different from A, B) lies so that ∠ACB = 90°, then C lies on the circle with diameter AB. So the set of all points from which AB is seen at a right angle is this circle (without the points A, B).

The proof is indirect, by elimination: the point C (off the line AB) can be inside the circle, outside it or on it. We show that inside gives an obtuse angle and outside an acute one, so a right angle is possible only on the circle.

  • On the circle: ∠ACB = 90° is Thales’s theorem (proof above).
  • Inside: the ray AC meets the circle at a point D beyond C. By Thales’s theorem ∠ADB = 90°. The angle ACB is the exterior angle of triangle CDB at vertex C, so ∠ACB = ∠CDB + ∠CBD = 90° + ∠CBD > 90° (the angle is obtuse).
  • Outside: one of the segments AC, BC (say AC) enters the disc and meets the circle at a point D ≠ A. Then ∠ADB = 90° is the exterior angle of triangle DCB at vertex D, so 90° = ∠DCB + ∠DBC and ∠ACB = ∠DCB = 90° − ∠DBC < 90° (the angle is acute).

So if ∠ACB = 90°, C can be neither inside nor outside, and only the position on the circle remains. The delicate point is the “outside” case: one has to check that at least one of the segments AC, BC really does meet the circle.

A short alternative without cases: if S is the midpoint of AB and r = ½ |AB|, then CA⃗ · CB⃗ = |CS|2 − r2, so the angle is right exactly when |CS| = r, obtuse exactly when |CS| < r and acute exactly when |CS| > r.

IV

FormulaThe angle in a semicircle.

DIRECT PROOF

2α + 2β = 180°  ⇒  ∠ACB = α + β = 90°

From what we know, step by step, to what we want to prove.

STOP III · FROM THE 6TH CENTURY BCE · SOUTHERN ITALY

The Pythagoreans and a discovery that didn’t fit

I

QuestionIs every length a fraction?

The Pythagoreans believed that every length is a ratio of whole numbers. Can √2 be written as a fraction pq? How would you try to find out (by trial? by calculation?), and how could you show that it never works?

Why trying doesn’t help here

There are infinitely many fractions, and no amount of trying can show that none of them works. Here proof becomes a necessity.

II

PersonA school for which all is number.

The next stop is southern Italy, home of the Pythagorean school (from roughly the 6th century BCE). It is linked to the name of Pythagoras of Samos. The Pythagoreans are credited with the claim that all is number, that is, that everything can be expressed using whole numbers and their ratios (fractions).

The theorem a2 + b2 = c2 bears Pythagoras’s name, even though (see Mesopotamia) it was known in practice much earlier. It is not certain whether Pythagoras himself proved it or one of his followers. The difference is that the Greeks began to justify it for all right triangles.

And it is exactly this theorem that leads to a problem: a square with side 1 has a diagonal of length √2.

According to tradition, the discovery that √2 cannot be written as a fraction was made by the Pythagorean Hippasus of Metapontum. Legend says he was punished for it by drowning at sea.

The story of Hippasus’s punishment is most likely a later legend; the old sources do not clearly attribute the discovery of irrationality to Hippasus, and accounts of his fate differ.

III

IdeaProof by contrapositive and proof by contradiction.

The ancient proof is a proof by contradiction: we assume the opposite of what we want to prove and arrive at a contradiction. Aristotle cites this argument about the diagonal of a square as an example of proof by contradiction.

For the proof we will need an auxiliary statement. We prove it in a way other than directly – indirectly, through its contrapositive.

PROOF BY CONTRAPOSITIVE

Prove: if a2 is divisible by three, then a is also divisible by three (a ∈ ℕ).

Hint: doing it directly is awkward. Prove the contrapositive instead: if a is not divisible by three, then neither is a2. What remainders can a leave when divided by three?

Solution

We prove the contrapositive implication (¬B ⇒ ¬A): if a natural number a is not divisible by three, then a2 is not divisible by three either.

A natural number a that is not divisible by three can be written as a = 3k + 1 or a = 3k + 2, where k ∈ ℕ0. We square both expressions:

(3k + 1)2 = 3(3k2 + 2k) + 1
(3k + 2)2 = 3(3k2 + 4k + 1) + 1

In both cases the remainder on division by three is 1, so a2 is not divisible by three.

PROOF BY CONTRADICTION

Using the previous statement, prove by contradiction that √3 is irrational.

Bonus: how would it go for √2? (Use evenness instead of divisibility by three.)

Solution

Suppose that √3 is rational. Then we can write

√3 = pq,   p, q ∈ ℕ,  gcd(p, q) = 1.

Squaring and rearranging gives 3q2 = p2. We see that 3 divides p2, so it also divides p (by the previous exercise). Let p = 3k, where k ∈ ℕ. Then 3q2 = 9k2, so q2 = 3k2. It follows that 3 divides q2, and therefore also q.

This contradicts the assumption that p and q are coprime. So √3 is irrational. The same approach works for √2 (using evenness instead of divisibility by three).

For the curious: the Pythagorean theorem in a picture
c² ab a²b² ab
Both squares have side a + b and contain the same four triangles. What is left over therefore has the same area: c2 = a2 + b2.

Source and more proofs in pictures: Pythagoras.

IV

FormulaA number that is not a fraction.

CONTRAPOSITIVE AND CONTRADICTION

(¬B ⇒ ¬A)  ⇔  (A ⇒ B)

√2, √3 ∉ ℚ

What cannot be tried out must be proved.

STOP IV · AROUND 300 BCE · ALEXANDRIA

Euclid: how to build all of mathematics

Euclid of Alexandria

I

QuestionMust everything be proved?

Why is it important that we accept some statements without proof? What would happen if we wanted to prove every statement from other statements?

Reveal the answer

A proof derives one statement from other statements. If we wanted to prove absolutely everything, we would end up with one of three options:

  1. an infinite chain: every statement would need a proof from another one, that one from yet another, and so on forever, so we could never get started;
  2. a circle: we would prove A using B and B using A, so in fact we would prove nothing;
  3. stopping at a few statements that we accept without proof. These are exactly the axioms.

So the only workable route is the third: agree on a small number of simple basic statements (axioms, postulates) and derive everything else from them by logic. Axioms are chosen to be as simple as possible, not to contradict each other, and to allow a great deal to be derived from them.

II

PersonAuthor of the most influential book in mathematics.

Around 300 BCE, in Alexandria, Euclid wrote the Elements. It is probably the most influential mathematical book in history.

In the Elements we find, for example, the Pythagorean theorem, a proof that there are infinitely many primes, and a proof of irrationality (in a later form). And also the proof of Thales’s theorem from the previous stop (Book III, Proposition 31).

III

IdeaThe axiomatic system.

The new idea was that all of geometry (and a bit more) could be set out like this:

  1. we start from a few definitions and axioms (postulates) that we agree to accept without proof (for example, that exactly one line passes through two points),
  2. we then prove further statements by logical steps from those we already know.

This is how the so-called axiomatic system came about.

An axiom is not “a truth we must never question”, but an agreement on a starting point. Euclid’s fifth postulate (about parallels) later turned out to be independent of the others, and changing it gave rise to other geometries (for example, on a sphere).

DIRECT PROOF

Prove that for all positive real numbers a, b

a + b2 ≥ √ab

Hint: start from the inequality you want to prove and use equivalent transformations to reach a statement that is always true. Then reverse the steps.

Solution

We start from the given statement and try to derive from it, by equivalent transformations, another statement that we know is true. The proof itself would then run in the opposite order.

Since a and b are both positive, squaring the inequality is an equivalent transformation. This gives the inequality (a + b)2 ≥ 4ab. Expanding the left-hand side and rearranging, we arrive at

(a − b)2 ≥ 0,

which certainly holds for all a, b ∈ ℝ+. Here too the proof rests on a general statement, not on trying out pairs of numbers.

IV

FormulaThe arithmetic mean is never less than the geometric mean.

DIRECT PROOF OF AN INEQUALITY

a + b2 ≥ √ab  ⇔  (a − b)2 ≥ 0

If these axioms hold, so does this theorem.

STOP V · 16TH–17TH CENTURY

What next? From the Greeks to induction

Maurolico, Pascal and young Gauss

I

QuestionHow do you prove something for every n?

Work out:

1 + 2 + ⋯ + 100 = ?

You have just computed the sum 1 + 2 + ⋯ + n for n = 100. How would you show that the formula holds for every n? (Trying doesn’t work, as we saw with n2 + n + 41.)

II

PersonMaurolico, Pascal – and young Gauss.

Greek mathematics focused mainly on geometry. To prove properties of all natural numbers we need another tool, so-called mathematical induction. Its beginnings can be found in antiquity, in India and in the Islamic world, but it was explicitly formulated and used only in the 16th–17th century, for example by Francesco Maurolico and Blaise Pascal.

Let us connect this with the story of young Gauss: according to the anecdote, he was told at school to add up the numbers from 1 to 100 and immediately answered 5050.

The Gauss story is an anecdote; it is attested only by tradition and the details vary. More in Dějiny matematiky (in Czech; p. 46, page 28 of the PDF).

III

IdeaA row of falling dominoes.

Imagine a row of dominoes: the first domino falls, and every domino that falls knocks over the next one. Then all of them fall – and we don’t have to look at each one separately. That is exactly how induction works: we prove the first step, and that each step implies the next.

Gauss’s rectangle

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MATHEMATICAL INDUCTION

Prove that for all n ∈ ℕ

1 + 2 + ⋯ + n = n(n + 1)2

Think about what you need to show in the first step and what in the second.

Solution

(a) For n = 1 we get 1 = 1 · 22, which holds.

(b) Suppose that for n we have 1 + 2 + ⋯ + n = n(n + 1)2. We want to prove that it also holds for n + 1:

1 + 2 + ⋯ + n + (n + 1) = (n + 1)(n + 2)2

We substitute the induction hypothesis:

n(n + 1)2 + (n + 1) = n(n + 1) + 2(n + 1)2 = (n + 1)(n + 2)2

The implication P(n) ⇒ P(n + 1) is true, so the original statement holds as well.

IV

FormulaA tool for statements about all natural numbers.

MATHEMATICAL INDUCTION

1 + 2 + ⋯ + n = n(n + 1)2

P(1)  ∧  [P(n) ⇒ P(n + 1)]  ⇒  P(n) for all n ∈ ℕ

The first domino falls – and all the others with it.

What we have seen

Five steps from “it works” to “it must be so”.

  1. IMesopotamia: procedures that work, checked on examples.
  2. IIThales: the question of why it is true; direct proof.
  3. IIIThe Pythagoreans: √2, which we would never have found without proof; proof by contrapositive and proof by contradiction.
  4. IVEuclid: axioms and logical deduction; a direct proof of an inequality.
  5. VInduction: a tool for statements about all natural numbers.

For practice and for the curious

Based on material by M. Vestenická, Proč dokazujeme? Cesta historií důkazů (Why do we prove things? A journey through the history of proof, 2026). Where a historical claim is disputed, the text says explicitly that it is tradition or conjecture.

END OF THE CHAPTER ON PROOFS

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