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I tried to prove that the area of a rectangle is $ab$ given side lengths $a$ and $b$.

The best I can do is the assume the area of a $1\times1$ square is $1$. Then not the number of $1\times1$ squares that fit in an $a\times b$ rectangle is $ab$. Therefore area is $a\cdot b$. This does not seem rigorous however.

Américo Tavares
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Sidharth Ghoshal
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    Are you assuming that $a$ and $b$ are whole numbers? If so, then I would say that this is the definition of multiplication. The number $a \cdot b$ is total number of boxes when they are arranged in $a$ rows with $b$ boxes in each row. – Sammy Black Apr 03 '13 at 18:58
  • But how can we assume the area of a 1 x 1 square is really 1? It seems obviously true but I can't find a rigorous ( aka axiomatic) justification – Sidharth Ghoshal Apr 03 '13 at 19:01
  • And yes I am using the definition of multiplication in my square counting argument. The fault lies not in the counting but the assumption that 1 x 1 squares have an area of 1. – Sidharth Ghoshal Apr 03 '13 at 19:02
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    I guess it really depends on how rigorous you want to get, what are your current axioms? – Lazar Ljubenović Apr 03 '13 at 19:07
  • Euclid's original set – Sidharth Ghoshal Apr 03 '13 at 19:08
  • It will also depend on what your definition of area is; if $a$ and $b$ are integers then you can do it by counting boxes as @SammyBlack suggests, but you could also use integration for example (although if you're starting from Euclid, this would be a lot of work). – mdp Apr 03 '13 at 19:09
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    @Matt The usual way to go about integration (i.e. Lebesgue measure, ultimately) is to declare $\operatorname{vol}(B) = \prod_i (b_{i,1}-b_{i,0})$ with $B = \prod_i [b_{i,0},b_{i,1})$ an $n$-dimensional half-open rectangle. So that seems like moving the problem. – Lord_Farin Apr 03 '13 at 19:12
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    @Lord_Farin Fair enough, I hadn't thought of that loop. That almost suggests that the sensible thing to do is to define the area of a rectangle to be the product of lengths of sides, but one would hope that this is equivalent to some other natural definitions of area. – mdp Apr 03 '13 at 19:14
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    @frogeyedpeas: Your general question is quite interesting. But as to why the area of a $1\times 1$ square is $1$, the answer is that it is $1$ square unit by the definition of square unit. However, if we measure lengths in yards, and area in acres, as I heard they still do in some parts of the world, then the area of a $1$ yard by $1$ yard square is definitely not $1$ acre. – André Nicolas Apr 03 '13 at 19:14
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    @Sammy It seems that using this as the definition for integer multiplication, we can then proceed to derive the area of squares with side $2^{-n}$ (using that the unit square will be divided into $2^{2n}$ identical such squares (by induction)) and then seemingly all that remains to establish soundness for real $a,b$ is to prove that there is some sound limit concept (which may come down to the elementary theorems in establishing Riemann integration). – Lord_Farin Apr 03 '13 at 19:24

4 Answers4

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This is a good question! On the face of it, it seems that before you can start being rigorous, you have to define precisely what you mean by the area of a rectangle. But this is just what you are trying to prove!

However, there is another approach: you define exactly the properties that you require of a reasonable area metric. Then you adopt these properties as your axioms, and use them to show that an $a \times\ b$ rectangle has area $ab$.

In a Euclidean plane, we would expect the following axioms to hold for a reasonable area metric on $a \times b$ rectangles (with $a, b \ge 0$):

$A1$: The area of a $1 \times 1$ rectangle is $1$.

$A2$: Any two congruent rectangles have the same area.

$A3$: If a rectangle $R$ is the union of disjoint rectangles $S$ and $T$, then the area of $R$ is equal to the sum of the areas of $S$ and $T$.

Given these axioms, I think we can show by induction that for rational numbers $a$ and $b$, the area of an $a \times b$ rectangle is indeed $ab$. But what about rectangles with irrational sides? I think this might require another axiom:

$A4$: If a rectangle $R$ contains a rectangle $S$, then the area of $R$ is not less than the area of $S$.

Otherwise you might be able to construct pathological area metrics using the Axiom of Choice. But I am open to correction on this.

If anybody can suggest corrections or improvements to this axiom set, feel free to post them as a separate answer, so they don't get lost in the comments.

TonyK
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  • You only define area for rectangles. While this appears to be enough for the question at hand, a proper set of axioms should probably rephrase your A4 for arbitrary shapes. That way you could bound a non-rectangular shape by a sequence of boxes from within and without, both converging to the same area, which then my definition is the area of the shape itself. – MvG Apr 03 '13 at 20:45
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    @MvG By $A1$-$A3$ we can calculate for rational sides as mentioned. This will allow us to establish that $A1$-$A4$ define a unique so-called pre-measure on the set $R_{\Bbb Q}$ of rectangles with rational sides (it even suffices to have this only for rectangles aligned with the axes). Carathéodory's Theorem ensures the existence of a unique extension of this pre-measure on $R_{\Bbb Q}$ to the generated $\sigma$-algebra $\sigma(R_{\Bbb Q})$ which can be shown to equal the (usual) Borel $\sigma$-algebra on $\Bbb R^2$; this extension is called Lebesgue measure. So expanding $A4$ is unnecessary. – Lord_Farin Apr 03 '13 at 21:21
  • @frogeyedpeas My above comment has as a further consequence (by a huge mountain of work taking dozens of pages) that the Lebesgue measure, so defined, agrees with Riemann integration on $\Bbb R^2$. This serves as a further confirmation that we have a proper notion of volume and I hope that the above $A1$-$A4$ and my comments have shed sufficient light on the matter for you. (+1 for the answer) – Lord_Farin Apr 03 '13 at 21:24
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    What I like about the question and the answer is that it shows that not everything in maths is proving: a good part of any real maths involves finding the right axioms that are useful to the problem at hand (+1 to both) – carlosayam Aug 12 '14 at 07:12
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    @Lord_Farin I realize this is 3 years old at this point, but do you happen to have a source for that "huge mountain of work taking dozens of pages"? –  May 18 '16 at 21:18
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    @Bye_World It takes a good 25-30 pages in the book I learned measure theory from, Schilling's Measures, Integrals and Martingales. – Lord_Farin May 18 '16 at 21:32
  • @Lord_Farin I'll take a look. Thanks. –  May 18 '16 at 21:37
  • Ok, without axioms, demonstrations don't exist. Mathematics is relative to axioms the same way the area of a square is relatif to the area of a unit square. – Quade Aug 29 '20 at 23:30
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One can also start from the following axioms:

$A1$: The area of a $a \times a$ square is $a^2$.

$A2$: Any two congruent shapes have the same area.

$A3$: If a shape $R$ is disjoint union of shapes $S$ and $T$, then the area of $R$ is equal to the sum of the areas of $S$ and $T$.

With these axioms at hand one readily finds the area of a rectangle with sides $a$ and $b$: enter image description here

On the sides of the rectangle $ABCD$ build the squares $ABC_1D_1$ and $A_1BCD_2$. Let $D_3= (A_1D_2)\cap (C_1D_1)$.

Obviously the rectangles $ABCD$ and $A_1BC_1D_3$ are congruent. Therefore from our axioms one obtains: $$\begin{align} 2[ABCD]&=[DD_1D_3D_2]-[ABC_1D_1]-[A_1BCD_2]\\ &=(a+b)^2-a^2-b^2=2ab. \end{align} $$

user
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In order to interpret lengths and areas as numbers, you have to fix a unit length and a unit area. The former is usually done by defining a coordinate system, whereas the latter usually defines the square of the unit length as the unit of area. But that square is a product, so the definition already assumes that the area of the $1\times 1$ square is $1$, which seems to be something you want to prove.

If you accept this kind of setup, then you can consider transformations of your rectangle which turn it into another rectangle of the same area but with the unit length as the length of one edge. In that case you can prove that the other edge will be the product of the original edge lengths. You can use a von-Staudt construction to express this product geometrically, based on the unit length.

MvG
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If the length and breadth are any real numbers, then divide the whole rectangle into small squares of sides having length $ =(1/lb)$ ,so there will be $lb^2$ squares along the breadth and $bl^2$ squares along the length, so total area $= (1/lb)~(1/lb)~(lb^2)~(bl^2)= lb $ square units

Larry
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Avinashreddy Pakala
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