Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When we say the area of a rectangle is the product of the length by the width is it a definition based on geometric intuition or is it a result? I know it is a result that we can find after defining integrals but before that was this a definition?

share|cite|improve this question
Generally, the measure (aka area) of a rectangle $[a,b]\times [c,d]$ is defined to be $|b-a|\cdot |d-c|$. – Alex Becker Mar 30 '12 at 17:59
Yes, they are defined in that way. – Alex Becker Mar 30 '12 at 18:06
If you consider only rectangles with integer-valued dimensions, you can certainly define the area of a rectangle to be the number of unit squares that fit in it, and then prove that this is equal to the product of the length and width. – Rahul Mar 30 '12 at 18:30
Curiously, the usual modern approach defines the area of a rectangle with sides parallel to the axes like this, and derives this property for all other rectangles as a theorem. – Chris Eagle Mar 30 '12 at 18:48
You only need to define the area of the unit square to be 1. – lhf Mar 30 '12 at 20:34

I have no idea how the definition of area originated, but here's my take on it all. Almost none of the story below is true (i.e., if you choose a section of the story at random, it will be wrong with probability $1$).

Very early on, people noticed that it was impossible to fill an area with a line of finite length, so the concept of area was invented. Area had no relation to length, so areas were only defined in terms of other areas. People noticed that you could put two congruent right-angled isosceles triangles together to make a square, so they said that the square had twice the area of each of the triangles. People also noticed the interesting fact that you could take a rectangle with sides of integer length $a$ and $b$ and fill it with $a \times b$ sqares of side length $1$, so they said the rectangle's area was $ab$ times that of the square.

They also noticed that you could split the rectangle into two right-angled triangles, which had the same area, as you could rotate one to put it on top of the other (the notion that area is invariant under rotation matrices and translation maps is implicit here, but nobody noticed) so they said the triangles had areas which were $\frac{ab}{2}$ times that of the square of side length $1$.

They also noticed that any triangle with base $b$ and height $h$ could be split up into two right-angled triangles, one with non-hypotenuse sides of length $c$ and $h$, and one with sides of length $d$ and $h$, such that $c+d=b$. So the total area of the triangle was $\frac{ch}{2}+\frac{dh}{2}=\frac{bh}{2}$ times that of the square of side length $1$.

Spurred on by these discoveries, our fictional early geometers almost without noticing it began to drop the 'times that of the square of side length $1$' and adopt the area of the square of side length $1$ as the basic unit of measurement. So that square had area $1$, a rectangle with sides of length $a$ and $b$ had area $ab$ and a triangle with base $b$ and height $h$ had area $\frac{bh}{2}$.

The geometers quickly realized that they did not have to restrict themselves to polygons with integer side lengths. Most of them were unaware of the existence of irrational numbers, and they realized that they could split their square of side length $1$ into square of side length $\frac{1}{\textrm{the lowest common denominator of all side lengths involved in the question}}$, which they used to show that any rectangle with sides of length $a$ and $b$ had area $ab$, whether or not $a$ and $b$ were integers. Corresponding results for triangles followed easily.

The geometers were by now insatiable - they moved out of the strictly rigorous world of polygons - which can, after all, just be broken up into triangles - and adopted new methods involving infinitesimally small triangles to prove that the area of a circle of radius $r$ was $\pi r^2$. Another important development at around that time was the discovery, by applying the theorem of Pythagoras to at the diagonal of a square, of irrational numbers. Mathematicians agreed that the area of a rectangle with sides of length $a$ and $b$, where $a$ and $b$ were irrational, was still $ab$, and were even able to prove this in special cases. For example, they proved that a $\sqrt{2}$ by $\sqrt{2}$ square had area $2$ by putting four unit squares together to make a large square of side length $2$ and area $4$, and joining the midpoints of the sides of the large square to create a square of side length $\sqrt{2}$ which had exactly half the area - i.e., $2$.

The geometers were unable to solve certain problems, however. One particularly important one was the problem of constructing - using ruler and compasses - a square with area equal to that of a given circle. We now know that this is impossible, but this problem held up the study of area until the end of the golden age of the Greeks and Romans.

No significant advances took place in geometry until Descartes introduced his system of coordinates, and geometry could be stated in terms of algebra. There were a few problems with this - for example, it is easy to give four coordinates that define a unit square, but less easy to give the coordinates of the square when it is rotated through $30^{\circ}$ about the point $(3,-2)$! To get round this problem, concepts such as matrices were introduced, and it was shown that the area of a shape subject to a transformation was equal to its original area multiplied by the determinant of that translation. Those maps which preserved area were those with determinant of $\pm 1$; i.e., the rotation and reflection matrices. The techniques used to do this were not rigorous to the standard of modern algebraic geometry, but they were still correct - it is manifestly obvious that rotating a shape leaves its area constant - the only doubts were in the mathematics used to describe the area of the square.

It was not until the notion of an area integral had been defined - and then redefined to make it rigorous - that a truly formal mathematical definition of area was arrived at. It was quickly decided that the only shape we truly knew the area of was a rectangle - it had area equal to the product of the lengths of its sides - and the definition of the integral and all further development in area theory stemmed from the idea that all shapes could be thought of as being approximately built up from collections of small rectangles - in such a way that we could make the approximations as good as we liked, and the present definition of an area was arrived at. Using the new theory, it became possible to evaluate things like the area of a circle rigorously - not by dividing it into triangles as the ancients had done, but always, always into rectangles.

Why was the rectangle chosen? It was just the shape with the simplest area formula.

Ultimately, we want a definition of area that satisfies the following rules:

  1. Area is always non-negative.

  2. Area is invariant under rotations, reflections and translations.

  3. If a shape is the union of two other shapes, then its area is the sum of the areas of those two shapes.

  4. If a shape $A$ is contained within another shape $B$, then $\textrm{area of } A \le \textrm{area of } B$.

  5. The area of nothing is $0$.

  6. The area of a unit square is $1$.

These rules don't seem too restrictive, but it actually turns out that such a definition of area can never exist, although it is impossible, given a definition of area, to come up with an explicit region which violates any of the rules - only a way to construct such a region using the Axiom of Choice. For all 'nice' (Lebesgue measurable) shapes, though, the standard Lebesgue integral works well for a definition of area.

So... to answer your question, saying that a rectangle with sides of length $a$ and $b$ has area $ab$ is a definition, based on what were once results. All other theorems about area though - including areas of triangles and circles and the invariance of area under rotation - are now classed as results of that definition, whether they seem intuitive or not. It's all part of finding a rigorous foundation for algebraic geometry.

share|cite|improve this answer
The last bit about the non-existence of an ideal area function defined over all sets is very non-intuitive but it does imply that non-measurable sets must exist. – lhf Apr 3 '12 at 1:45

It's a definition based on a geometric intution. The basic intuition behind the area of a rectangle (or hypercube its really the same) is like lhf says the square with length 1, has area one, and when you scale by say, k, one side of a rectangle you scale its area by k. These assumptions alone (along with a notion of orientation which allows scaling by negative numbers) define a unique notion of area. Look up the construction of the determinant for a rigorous proof.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.