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I tried to teach my son multiplication using a rectangle. (e.g. 3cm * 4cm = 12cm^2). Now I have 12 little squares. But how do I explain where the "little square" came from?

My best guess: If I cut a little square in two, I can get a rectangle 2cm long * 1/2cm deep. Then I cut that rectangle in two, I get 4cm long * 1/4 cm deep. I keep going and say I get a rectangle 100,000cm long by 1/100,000 cm deep.

This is very long and very thin. And I deduce, that the limit would be a line. And so I conclude that the area of a line = 1. (units become irrelevant, infinity takes care of that).

I also like this to explain why the integral of y=0x+1 (for x from 0 to 1) is 1. Note: this should be a 1 unit square, and integrals are an area but we say the answer is 1 not 1 unit^2.

My question: Do you agree? (if not, why not?)

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5 Answers 5

The ‘little squares’ are simply the result of dividing each side into $1$ cm segments and using those to form a grid of $1\times 1$ squares within the $3\times 4$ rectangle. There’s no reason to look any further for an explanation of where they come from. We define the area of a $1\times 1$ square to be $1$ square unit (where unit is whatever linear unit was used to measure the sides of the square), so the obvious way to find the area of some other shape is to divide it up into $1\times 1$ squares $-$ if we can.

Most of the rest is simply wrong, I’m afraid. For starters, a straight line has no area. If you want to explain geometrically/pictorially why a rectangle of dimensions $\frac12\times 2$ has an area of $1$, divide it into $4$ squares of dimensions $\frac12\times\frac12$ and observe that they can be rearranged into a single $1\times 1$ square, which by definition has area $1$ square unit. The same reasoning applies to a rectangle of dimensions $\frac1{100}\times 100$: since $100=10000\cdot\frac1{100}$, it can be divided into $10000$ squares of dimensions $\frac1{100}\times\frac1{100}$, which can then be arranged in $100$ rows of $100$ squares to form a unit square. There is no limiting procedure involved here.

Finally, $\int_0^1 1~dx$ is $1$ square linear unit (e.g., sq. cm.) when it’s interpreted as the area between the $x$-axis and the line $y=1$ over the interval $0\le x\le 1$; it has other units when the integral is given other interpretations.

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I think you are taking my parts and putting them back in a square. By halving the height and doubling the width, would I not eventually end up with a line? length = infinty, height = 1/infinty. –  Thomas Dec 12 '12 at 6:29
@Thomas: (1) Yes, I am: that’s what one needs to do with them in order to explain in simple, intuitive, geometrical terms why the area of those rectangles is $1$. (2) No, you would not; at each stage of the process you would have a rectangle of area $1$. The limiting object is a straight line, but it has no area. –  Brian M. Scott Dec 12 '12 at 6:33
I know you are right. I am aware that a line is defined to have no width. Thanks. Its just that when I think of adding I think of a line, when I think of multiplication I think of an area and I feel there is a sleight of hand going on. I will ponder some more. –  Thomas Dec 12 '12 at 8:53

Sometimes it is possible to formalize an intuition like this and sometimes it isn't. In this case it isn't, because if you were to start with a $2 \times 2$ square instead of a $1 \times 1$ square and then perform the same process, the "limit" would be the exact same line. Therefore your argument, if it were valid, would show that the same line has an area of 1 and also an area of 4.

The problem is that the function assigning area to figures is not continuous in the appropriate sense, so the area of the limit is not the limit of the areas.

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The ambiguity can be cleared by considering the following two points:

1) The idea is to cut and add the square an infinite no. of times so that in the end the length of the rectangle has become x (infinite) and the breadth 1/x (zero). The mistake then, is to multiply the length and breadth i.e. (x)*(1/x) to get 1. Because x has been taken to be infinity we have an ∞/∞ form which is indeterminate.

2) One might argue (against pt.1) that we are doing a limiting procedure where x tends to infinity and is not actually infinite (that's what the definition of limit says) and thus when we take the limit x/x as x tends to infinity we get 1 (by using L'Hospital's rule). Well, the mistake here is that when we are taking the limiting case we are only approaching towards a line but never really get to it (that is what the concept of limit says e.g. when you say that y tends to a that means that y can come closer and closer to a but never actually become a). Actually we are, thus, not calculating the area of a line but of some rectangle with a non-zero width.

The apparent ambiguity is thus resolved no matter which way you want to consider it.

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I disagree with your second point. The concept of limit is not “when you say that y tends to a that means that y can come closer and closer to a but never actually become $a$)“. Both parts (“closer and closer” and „never becoming $a$”) are just wrong, as coming closer and closer would just imply a being $a$ limit point and the limit of the constant function $a$ is $a$, obviously. –  Keba Jan 27 '14 at 14:28

A line cannot have an area, since it is not a shape, but that which makes a shape a shape. Think of it like this, if you draw a line with a highlighter, then you have a small rectangle. Of which you can calculate the area, let us say this breadth is 5mm, and the length is 100mm, the area is simply 500mm^2. Suppose now we reduce the drawing apparatus and you are working with a pen, breadth = 2.5mm, area is now 250mm^2, now with a pencil, breadth reduce to 1mm, so the new area is 100mm^2. All of these can be defined as lines (especially the last 2) so we can draw another line with some other instrument where the breadth is 0.5mm, the area reduces to 50mm^2. This can continue to infinity, and we end up with a breadth = 0.000000000001mm, and you know the result.

The point is all of these lines you draw are lines, which determines a shape. If the area of a line exists then it would alter the area of whatever it is you draw!

For example, if you draw a rectangle with dimensions 4x3, the area is 12. But if the line has an area then it would spill over into the area of the rectangle. Assuming the line is symmetrical (which is a safe assumption in my opinion :D) then 50% of the line would spill into the are of the rectangle, if the line gets more area, then the difference of the areas of the lines would change the area of the drawn shape. This will reduce the area of the initial shape.

Therefore I think lines do not have an area.

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a fractal line may actually have Hausdorff dimension 2 –  mau Mar 26 '14 at 9:11

In my opinion, an area of the point is 1 in unit, an area of the line (irrespective of direction) is its length in unit. The formula to compute an area of a square or a rectangle is height*width, the same formula can be used to compute the area of a point as well as a line. In case of point, height=width=1, so the area is 1. In case of line, map it's length along either x-axis or y-axis, if you map it along x-axis then its y-coordinate (say height) will be 1 and if you map it along y-axis then its x-coordinate (say width) will be 1, in the end, the area of line will be height*1 or 1*width depending upon the mapping along one of the axes, this mapping we would call direction and to the extent, the line goes after mapping, would be called as its length and also its area.

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You should either be more precise with what you're saying (e.g., that points and lines have nonzero area, which is typically understood to mean $2$-dimensional volume) or your opinion is not at all widely held. I think Lebesgue would disagree with your opinion. –  pjs36 Jun 21 at 15:48

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