What is the correct definition of Area?

How is the area of a rectangle: length $\times$ breadth?

We know that other areas can be derived from it. Also, the area under curves uses the area of rectangles as a basis.

• I understand why someone downvoted this question, probably due to its lack of context, but I upvoted you back to zero because I remember a time when I really wanted to know the answer to this question but had no idea how to ask it. – David H Nov 11 '14 at 5:40
• Do you mean in a historical or philosophical sense? The ancients knew the area of a rectangle; and in modern math we use rectangles to build up Lebesgue measure. But "correct" is a loaded term, are you asking if there is some other sensible way? Or why the ancients made the choice they did? It's obvious that you can put $lw$ unit squares in a rectangle of length $l$ and width $w$, you can just cut out pieces of paper and convince yourself. This is what the ancients saw. – user4894 Nov 11 '14 at 5:56

Let's imagine that we've never dealt with the concept of area of a shape before and we are the first people to explore this. We might want to quantify how much space certain shapes take up. Also we might want to say that the "area" of a shape is an intrinsic property not depending on how we are looking at it. Finally we might want to define it in a way that would make the most physical sense.

A rectangle has some obvious parameters we can use, namely its lengths - say $a$ and $b$. One could also take a look at the diagonal $d$. An intrinsic property would depend on the lengths $a$ and $b$ and perhaps d but if we know that $d^2 = a^2 + b^2$, then we don't need to consider it. Furthermore, we want an area that will vary in proportion to variations in $a$ and $b$. Let's use a function $A(a,b)$ for this nebulous concept of "area of rectangle".

If we have a rectangle with one of its sides twice the length of one of the sides of the original triangle we would have

$$A(2a,b) = 2A(a,b) \quad \text{ or } \quad A(a,2b) = 2A(a,b)$$

We could also imagine shrinking it

$$A(a/2,b) = A(a,b)/2 \quad \text{ or } \quad A(a,b/2) = A(a,b)/2$$

Why don't we go all the way and imagine shrinking the rectangle's sides by factors of $1/a$ and $1/b$. We have

$$A(a,b) = abA(1,1)$$

whatever this quantity $A(1,1)$ is. We arrive at a formula

$$A(a,b) = kab$$

for some particular constant $k$. If the area of a rectangle is going to depend on its lengths and scale appropriately in some physical sense when we increase/decrease one or both sides' lengths, then it's formula should be as above. It's particularly convenient to choose $k = 1$. This allows us to speak of the relative sizes of different rectangles.

We can imagine taking an arbitrary "nice" shape and subdividing it into rectangles in order to approximately define its area. It then makes sense that stretching the shape from side to side by a factor of $k$ should change its area by a factor of $k$.

This is a sort of "absolute" path where this concept of area is a property of the object itself, invariant under changes in how we observe the shape. It could also be illustrative to imagine going in a "relative" direction where all we are interested in is what the shapes "look like" from our vantage point. That is, an object should look bigger when it's closer or more of it is turned towards us and smaller when further away or seen more edge-on. This leads you towards the idea of angular size and solid angles.

I like this question because it exposes how humans are "dimensionalist", i.e. "bigoted" against certain dimensions (specifically dimensions higher than three). You ask "How is the area of rectangle length×breadth?" The first question I would ask is how do you measure length? I like this example because it deals with two dimensions vs. 1 dimension (where the 1st dimension is somehow "allowed" a priori) and thus we (humans) can reason within it with real world examples. We could even go further to 3D volumes, but I'm happy to stay at 2D here since I can present very nice pictures in 2D--thank the heavens for our 2D screens!

First off, area and multiplication are intertwined. Area is something we must define--it has to be a definition. It's not proper to introduce area without units. Area is very simple: $1 \text{ unit}^2$ equals a square area of $1\text{ unit}$ by $1\text{ unit}$: The area of a region is the number of units of the above that it occupies. So for instance the following occupies $3\times6 = 18$ units squared: So how do we arrive at $18 \text{ units}^2$? We multiply three $1\times1$ squares by six $1\times1$ squares to arrive at eighteen $1\times1$ squares to arrive at $18\text{ units}^2$.

But wait, how did we measure the lengths of the rectangles...what if we had: Is the area of this rectangle $6$ or is it $12$, i.e. $(2\cdot(2\times1))\cdot3 = 12$?

So which of the following is your definition of "length" $1$? I ask because it will affect your definition of area. Furthermore, the definition of area depends on our understanding that both the "x" and the "y" lengths will be measured the same. What if they are not? There is no mathematical basis for saying they will have the same units! And this is often the case such as with $y(t) = y_0 + v_0t + \frac{1}{2}at^2$ where the "x" axis is time and the "y" axis is height--here there is no correlation between the units of the "x" axis (time in this case) and the "y" axis (distance in this case)!

This problem extends further beyond "area" (2D) and "length" (1D) and into "volume" (3D) and 4D volumes, and 5D volumes, and 6D volumes, etc.