# Is the area of a line = 1?

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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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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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 at 14:28