The standard proof that \sqrt{2} is irrational is beautiful: assume \sqrt{2} = p/q in lowest terms, square, deduce p and q are both even, contradict "lowest terms." It is short. It works. But it is algebraic — you have to trust the symbol-pushing.

Stanley Tennenbaum's proof does the same thing with shapes. No algebra. Just squares, with integer side lengths, pushed around on a page. This article walks through it.

The algebraic version, in one line

If \sqrt{2} were rational, then p^2 = 2 q^2 would have solutions in positive integers. Equivalently, there would exist positive integers p and q such that

p \times p \;=\; 2 \times (q \times q).

Read that geometrically. It says: you can take a p \times p square and re-cut it into two q \times q squares — same total area. Or equivalently, two q \times q squares can be fit, without overlap or gap, inside a p \times p square of equal area.

Tennenbaum's proof shows that this is geometrically impossible, if p and q are the smallest such positive integers. The picture does all the work.

Setup

Suppose, for contradiction, that there do exist positive integers p and q with

p^2 = 2 q^2.

Choose p and q to be the smallest such pair — the smallest p, and the corresponding q. This is the infinite-descent setup: if we find an even smaller pair (p', q') with (p')^2 = 2(q')^2, we contradict the choice.

Now picture the equation. You have one big square of side p, total area p^2. You have two small squares of side q, total area 2q^2. The equation says these two totals are equal.

A p by p square equal in area to two q by q squares On the left, a large square of side p drawn as a solid outline with label p under it. On the right, two smaller squares side by side of side q each, drawn with matching style and labelled q under each. An equals sign between them, with the word area written underneath in italics. p area p² = q q + q² q²
The hypothesis $p^2 = 2q^2$ in pictures: one $p \times p$ square has the same area as two $q \times q$ squares. The proof now shows that if $p$ and $q$ are the smallest positive integers satisfying this, a smaller pair must exist too — an impossibility.

The packing

Try to fit the two q \times q squares inside the p \times p square. Note first that q < p, because otherwise two squares of side q would have area \ge 2p^2 > p^2 and couldn't fit. Note also that p < 2q, because p = q\sqrt{2} < 2q.

So you can place one of the q \times q squares in the top-left corner of the p \times p square, and the other in the bottom-right corner. Since p < 2q, the two q-squares overlap in the middle — there is a small square where they cover each other, and two small corners near the corners of the big square that are not covered.

Two q by q squares placed in opposite corners of a p by p square, overlapping in the middle A large square of side p drawn with a thin outline. Inside, two smaller squares each of side q are placed — one flush with the top-left corner and the other flush with the bottom-right corner. The two smaller squares overlap in the centre in a square region shaded darker to indicate double coverage. Two small square regions near the top-right and bottom-left corners of the big square remain uncovered by either q-square and are shaded the lightest. Side lengths are labelled: the big square has side p at the bottom; each q-square is labelled q; the overlap region's side is labelled "2q − p"; the uncovered corner region's side is labelled "p − q". p q × q q × q overlap (2q − p)² uncovered (p − q)² uncovered (p − q)²
The two $q$-squares overlap in a central square of side $2q - p$ (purple), while two square regions of side $p - q$ at the opposite corners are not covered by either $q$-square (unshaded). Because the two $q$-squares together have area $2q^2$ equal to the big square's area $p^2$, the overlap and the two uncovered corners must balance exactly.

The balance equation

Total area of the big square: p^2. Total area covered if you just sum the two q-squares' areas: q^2 + q^2 = 2q^2. But we have assumed these are equal! So the overlap counts once too many in the sum, and the uncovered corners are missing from it — the surplus in the overlap must exactly equal the deficit in the uncovered corners:

\underbrace{(2q - p)^2}_{\text{overlap}} = \underbrace{2(p - q)^2}_{\text{two uncovered corners}}.

Check this algebraically as a sanity test:

(2q - p)^2 = 4q^2 - 4pq + p^2, \qquad 2(p - q)^2 = 2p^2 - 4pq + 2q^2.

Subtract:

(2q - p)^2 - 2(p-q)^2 = (4q^2 - 4pq + p^2) - (2p^2 - 4pq + 2q^2) = 2q^2 - p^2.

By our assumption p^2 = 2q^2, that right-hand side is zero. So the two sides are equal.

Why the balance equation is the whole trick: it is a new equation of the form "a square equals twice a square," but now with smaller sides. The original equation was p^2 = 2q^2. The new one is (2q - p)^2 = 2(p - q)^2 — the same shape of equation, with p' = 2q - p and q' = p - q.

The smaller pair

Let

p' = 2q - p, \qquad q' = p - q.

Both are positive integers, because q < p < 2q (so p - q > 0 and 2q - p > 0), and both are differences of integers, so they are themselves integers. From the balance equation above,

(p')^2 = 2 (q')^2.

But also, p' = 2q - p < p, because 2q < 2p (since q < p is false — wait, check: q < p implies 2q < 2p, so 2q - p < p). And q' = p - q < q, because p < 2q implies p - q < q.

So (p', q') is another pair of positive integers satisfying the same equation — and p' < p, q' < q.

That contradicts the choice of (p, q) as the smallest such pair. So no such pair can exist. So no rational p/q can satisfy (p/q)^2 = 2. So \sqrt{2} is irrational.

Infinite descent: each contradiction generates a smaller pair of integer squares Three pairs of squares arranged left to right. The leftmost pair is the largest — a p-square paired with two q-squares. An arrow points to a middle pair, labelled p prime and q prime, noticeably smaller. Another arrow points to a further pair, smaller still. The chain of arrows trails off into dots, signifying that if one solution existed, an infinite chain of smaller solutions would exist, which is impossible among positive integers. p q, q → p′ = 2q − p q′ = p − q → p″, q″ … an infinite chain of positive integers, strictly decreasing impossible — there is no infinite descent in ℕ
Each supposed solution $(p, q)$ forces a strictly smaller one $(p', q') = (2q - p, p - q)$. Iterate, and you get an infinite descending chain of positive integers — which cannot exist, because the positive integers have a smallest element. The assumption that $(p, q)$ existed in the first place is wrong.

Why this proof is special

The algebraic proof you have seen is "even-ness percolates." Tennenbaum's proof is geometry. It makes the contradiction visible: you literally see the leftover overlap and the leftover corners, you see that they have to match in area, and you see that this produces a smaller version of the same puzzle.

The technique has a name: infinite descent, first used by Fermat. The structure is always the same. Assume a solution in positive integers exists. Build a smaller positive-integer solution out of it. Since positive integers cannot decrease forever, no solution can exist.

Try it with $p = 7$, $q = 5$ (a near-miss)

7^2 = 49 and 2 \cdot 5^2 = 50. Not equal — 49 \neq 50 — so 7/5 is not \sqrt{2}. But 7/5 = 1.4 is pretty close.

If you did try to pack two 5 \times 5 squares into a 7 \times 7 square, the overlap would be of side 2 \cdot 5 - 7 = 3, so area 9. The two uncovered corners would each be of side 7 - 5 = 2, total area 2 \cdot 4 = 8. Overlap area 9 \neq uncovered area 8. The picture is off by exactly the amount by which 7^2 \neq 2 \cdot 5^2 — namely 1.

If the equation p^2 = 2q^2 were satisfied exactly, the overlap and corners would balance to zero difference, as in the balance equation. The fact that no integer pair actually achieves this balance is the content of \sqrt{2}'s irrationality.

What the picture actually proves

Not just that \sqrt{2} is irrational. The same geometric trick, with slight modifications, proves that \sqrt{n} is irrational for any positive integer n that is not a perfect square. You still build a square, still try to pack smaller squares inside, still get an overlap-vs-corners balance equation, still deduce a smaller pair of integer squares, still infinite-descend.

The version for \sqrt{3}: three q-squares inside a p-square. The version for \sqrt{5}: five. The overlaps get more elaborate but the logic is identical. Tennenbaum's picture generalises to a whole family of irrationality results, all proved not by algebra but by watching shapes misbehave.

This is why the proof is worth memorising. It is short, it is visual, and it is one of the cleanest examples of "the picture is the argument" in all of elementary mathematics.

This satellite sits inside Number Systems.