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If I think about squaring as working out the area of an actual square then the following statement seems highly intuitive to me

The area of the square will always be bigger than the length of any individual side.

1) $2*2 = 4$ follows this rule

2) $\frac{1}{2} * \frac{1}{2} = \frac{1}{4}$ doesn't

When you make each side of the second square 3 times longer, the statement holds again. So there is a specific subset of small squares for which my statement doesn't work.

I'm hoping this is logical to someone, who can push me into the intuitive insight behind this. I understand it in pure numbers. But is my intuition so absurd given a drawing of a square?

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Your intuition is telling you that $x^2 \gt x$ (it is not when $0\le x \le 1$), or that the parabola $y=x^2$ is above the diagonal line $y=x$ (they intersect twice). In this case your intuition is wrong. –  Henry Nov 20 '11 at 11:48

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The correct insight here is that area and length, or more generally any two different types of quantities, cannot be compared. How much "bigger" is 10 meters than 1 square meter? Such questions don't make sense. You have been leaving off the units in your computations, which almost always leads one astray. The correct statements would be $$2\text{ units }\times 2\text{ units } = 4\text{ square units}$$ $$\frac{1}{2}\text{ units }\times \frac{1}{2}\text{ units } = \frac{1}{4}\text{ square units}$$ where the units could be whatever you want (feet, meters, inches, etc.) It certainly makes sense to say that $$\frac{1}{2}\text{unit}<2\text{ units}$$ and that $$\frac{1}{4}\text{square units}<4\text{ square units},$$ but it doesn't make sense to talk about whether $2\text{ units}$ is bigger or smaller than $4\text{ square units}$. They're apples and oranges.

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You are, of course, correct that disparate units cannot be compared. However, it's not at all unreasonable to ask how "the number of these units" compares to "the number of those units". Isn't that, in fact, what area/volume/etc formulas are all about? Calculating the number of some units (say, cubic inches) from the number of other units (say, linear inches)? –  Blue Nov 20 '11 at 9:23
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Not really, no. For instance, volume of a cube is equal to the side cubed. So if we have a 2m cube, it has a volume of (2m)^3 = 8m^3. A cubic meter is a meter cubed; you don't need to strip off the units, plug in the numbers, and then put the units back on. If the formula is dimensionally correct, the dimensions will work out. Every time you strip off units and tack them back on afterward, you're depriving yourself of a valuable check (unless you know it's already correct). –  Harry Altman Nov 20 '11 at 9:27
    
Essentially by definition, if one wants to specify a number of cubic inches, one must somewhere have taken three linear inch quantities and multiplied them. The numbers, of course, are incomparable; so I wouldn't view a volume formula as "calculating" a cubic inch quantity from three linear inch quantities, but rather we simply need three linear inch quantities in order to give a cubic inch quantity. –  Zev Chonoles Nov 20 '11 at 9:39
    
@Harry: I'm not saying anyone needs to strip-off or tack-on anything in doing a computation. I really just wanted to make the point that the OP's question makes sense as a comparison of numbers. As for what "is a meter cubed" ... I think of it geometrically: the unit such that a 1m-by-1m-by-1m cube has volume "1-of-that-unit"; I bear the burden of showing that this geometric definition plays well with unit arithmetic. You define it as the formal third-power of "meter"; you must show that this plays well with the geometry. Either/or. Neither helps much in non-Euclidean space: no cubes! ;) –  Blue Nov 20 '11 at 10:02
    
As you say, it does of course make sense to compare the numbers (though of course only if a unit length, unit time, etc. have been fixed; otherwise while you can do so it's meaningless). I was just pointing out that we should think of such formulae as having dimensioned quantities, not necessarily numbers, as their input and output. –  Harry Altman Nov 20 '11 at 21:11

It doesn't really make sense to compare an area to a length. We measure both with the same kind of numbers, but that is mostly just because it's the numbers we have. So your statement is not quite right to begin with.

What we mean by the length of a line is the number of times a certain, chosen unit length goes into it. And what we mean by the area of a shape is the number of times the unit area goes into it. For convenience (because it makes computations relatively easy and is not obviously less practical than any other choice) there's a convention that the unit area is that of a square whose side is a unit length.

But that doesn't really mean that the unit square is "the same as" the unit length, even though we represent both by the number $1$.

If we have a line with the length $1/2$, it is shorter than the unit length. The canonical shape with the area $1/2$ is a rectangle with side lengths $1/2$ by $1$. If we compare that to a square of length $1/2$, we need to make one of the sides of the rectangle shorter while the other stays the same. Therefore the number for the area of the square must be less than the number for the length of its side.

On the other hand, if the length if the side is $2$, then the canonical shape with area $2$ is a rectangle with sides $1$ by $2$. Here we have to make one of the sides longer in order to get a square of side length $2$, so the number for the sqare's area is larger than the number for its side's length.

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