The number \sqrt{2} is defined as the length whose square is 2. When you first meet this definition, it can feel circular — a length whose square is 2 is a length, and its square is 2, and... that is the definition, yes, but why should such a length exist, and why should its area really come out to exactly 2?
There is a beautiful, hands-on answer that predates algebra by a thousand years. Take a unit square, draw its diagonal, build a new square on that diagonal, and then cut the new square into four right triangles. Rearrange those four triangles, and they tile exactly two unit squares. No measurement. No calculation. The area is 2, and you can see it with your own eyes.
This article is a walkthrough of that rearrangement, with the pieces laid out step by step so you can watch the proof complete itself.
The setup
Start with a unit square — a square of side 1. Its area is 1. Its diagonal has length \sqrt{1^2 + 1^2} = \sqrt{2} by Pythagoras.
Now draw a new, tilted square whose sides are that diagonal. Each side of the tilted square has length \sqrt{2}. The question is: what is the area of this tilted square?
If you trust the formula "area of a square is side squared," you would answer (\sqrt{2})^2 = 2. But that is exactly the claim we want to see, not the one we want to assume. So let us build the tilted square geometrically and then cut it up.
The tile diagram
Counting the areas, one by one
The outer square is 2 \times 2, so its total area is 4 square units.
Inside, the four green triangles are right triangles with legs of length 1 each (half of the outer side). Each has area \tfrac{1}{2} \cdot 1 \cdot 1 = \tfrac{1}{2} square unit. Four of them contribute 4 \cdot \tfrac{1}{2} = 2 square units.
The tilted orange square must therefore have area
Its side length is \sqrt{2} (the diagonal of a unit square), and its area is 2. The claim (\sqrt{2})^2 = 2 has just been verified entirely with scissors-and-rearrange geometry. No algebra was touched.
Why this rearrangement is a proof: the outer 2 \times 2 square is covered exactly once by the four triangles and the tilted square, with no overlaps and no gaps. So the sum of their areas equals the outer area. Triangle areas are known (\tfrac{1}{2} each, so 2 in total); subtracting from 4 pins down the tilted square's area at 2. Every step is arithmetic on integers and halves — no \sqrt{2} needed in the proof itself.
The rearrangement as a puzzle
You can make the proof even more physical. Cut out the four green triangles. Slide them around.
If you take the top-left triangle T_1 and rotate it 180° about its longer edge, it exactly fills the other half of the top-left unit quadrant, making that quadrant a solid unit square. Do the same with T_2 into the top-right, T_3 into the bottom-right, and T_4 into the bottom-left. The four triangles collapse into two unit squares' worth of area — because four half-unit triangles make two whole unit squares.
What is left uncovered inside the 2 \times 2 outer square is the tilted orange square. So the tilted square plus those "two unit squares of triangles" fills the 2 \times 2 grid. The tilted square itself therefore covers 4 - 2 = 2 unit squares. Again: area = 2.
This is what it means for a square to have side \sqrt{2}. Its area is literally made of two unit squares rearranged.
Why this proof mattered historically
The ancient Greeks, especially the Pythagorean school, cared enormously about constructions like this one. Before algebra existed, geometric rearrangement was the only proof technique available. A square's side was a length; a square's area was a region; and if you could show two regions were congruent (rearrangeable from the same pieces), you had proved they had the same area.
The diagonal of a unit square was a famous example of a length that could not be written as a ratio of integers — it is irrational. The same construction that proves (\sqrt{2})^2 = 2 also led the Pythagoreans to discover, through a related proof by contradiction, that \sqrt{2} is irrational. That discovery was so disquieting to the Greek mathematical worldview that legend says the discoverer was thrown overboard to keep the secret.
What the rearrangement above shows is the positive side of the story: \sqrt{2} is a real, constructible length with a visible area of 2. It lives on the number line. It can be drawn with a ruler and a compass. It is not mystical; it is geometric.
Connections to the rest of the curriculum
The same tile-rearrangement style of proof works for many algebraic identities that look like black magic at first.
- (a+b)^2 = a^2 + 2ab + b^2 comes from cutting an (a+b) \times (a+b) square into four pieces: an a \times a square, a b \times b square, and two a \times b rectangles. Rearrangement shows the identity.
- a^2 - b^2 = (a-b)(a+b) comes from cutting an a \times a square with a b \times b corner removed and rearranging into an (a-b) \times (a+b) rectangle.
- (\sqrt{n})^2 = n for any positive integer n generalises the root-2 argument, using an n-by-1 grid rearranged inside a larger square.
Rearrangement proofs are one of the oldest mathematical techniques and remain the cleanest. They show, rather than tell, why an identity must be true.
The takeaway
Square with side \sqrt{2}? Area is exactly 2, and the proof fits on the back of an envelope. Draw a 2 \times 2 grid, draw a tilted square through the midpoints, and note that the four corner triangles occupy two unit squares' worth of area. The tilted square covers the remaining two unit squares. Side \sqrt{2}; area 2.
That is what irrational lengths really are: concrete geometric measurements whose area or volume comes out to a whole number even though the length itself is unwriteable as a fraction. The rearrangement puzzle above is the picture that shows why — no algebra required.
Related: Roots and Radicals · Construction of Root-2 on the Number Line · Spiral of Theodorus: Root-2, Root-3, Root-4, Unfolding Into a Curve · Tile-View Proof of the Three Core Exponent Laws · Number Systems