Why isn't the area of a square always greater than the length of one of its sides? Intuitively, it seems like the area of a square should always be greater than the length of one of its sides because you can "fit" one of its sides in the space of its area, and still have room left over. 
However when the length of a side, $s$, is less than $1$, then the area $s^2 < s$, which doesn't make sense to me for the reason above.
 A: Because you're misunderstanding units. The first assumption you make is that a square with a side of $1$ has an area of $1$ - that assumption is incorrect.
A square with a side of $1000$ $m$ / $1$ $km$ / $0.001$ $Mm$ has an area of $1$ $km^2$, $1000000$ $m^2$, or $0.000001$ $Mm^2$ (square-Mega-meters), depending on how you chose to present it. It's all about presentation, not mathematical properties.
What you need to intuitively understand is that by doubling the length of the side of a square, you get 4 times the area. And by shrinking the side by half you shrink the area to a quarter, regardless of units. 
Once you have that intuitive understanding, it will overrule your current understanding. Knowing that areas shrink "faster" than side lengths, it will be obvious that on a square with a side length of $1$ grok and an area of $1$ grikk, when you reduce the side length the area has to shrink faster than the side length - same is true for a square with a side length of $42$ gruk and an area of $42$ grakk: the area will shrink faster than the side length.
A: Take a square with dimensions $\cfrac 12 ft * \cfrac 12 ft $. The area would be $\cfrac 14$ $ft^2$. This makes perfect sense. I'll show you what I mean.
$\cfrac 12 ft* \cfrac 12 ft$ = $6 in*6 in$
The area is $36$ $in^2$ is equal to $1/4$ $ft^2$.
You can always take a square with the length of sides $x$ being $0<x<1$, but you can convert this $x$ to another unit $>0$. Therefore, the area would now make sense.
A: You're comparing apples with oranges. Lengths and areas are measured in different units. 
The most intuitive way is to consider the units to be part of the measurement. So the length may be three meters (the same as 118.11 inches), but never just three. Area = length squared, which nine square meters (not just nine). Now you can see that 118.11 is much greater than 9, if units are not included as part of the measurement.
To be able to compare them, take the area of both. For example, what is the area of one of the sides of the square? Since the sides are lines, their thickness is zero, and they thus have no area (or zero area).
A: I think your intuition is failing you because you are trying to compare a 1-dimensional object (the length of a side) with a 2-dimensional object (the area of the interior of the square).  You can fit loads of segments into a square of any size -- infinitely many, in fact!  That comparison doesn't really mean anything.
On the other hand, here's a comparison that does make sense:  Set a square of side length $s$ side-by-side with a rectangle whose sides are $s \times 1$.  Now you are comparing area to area.  The rectangle's area will fit inside the square if and only if $s>1$.
A: First physics and then maths. 
You cannot compare an area and length while attempting to make sense. 
If you are talking about areas of different side lengths in the numbering system based on $10$ as the most common scale of notation adopted here then the square whose side has magnitude  $a<1$ i.e., $ a <10^0$ units produces an area of magnitude $a^2<a $  and when magnitude of its side is $a>1$ unit, then it produces an area whose magnitude is $a^2>a.$ 
A: Physics
From a physical point of view, if you consider your length and area as having units (e.g. $\mathrm{m}$ and $\mathrm{m}^2$), your question doesn't make sense : you cannot compare both quantities, just like you cannot compare a length and a time.
Mathematics
When talking about the 2-D plane as $\mathbb{R}^2$, you can compare $x$ and $x^2$. And geometrically speaking:


*

*You can fit any segment into its corresponding square, by "copy-pasting" it along the perpendicular axis.

*You can also fit any segment into another square, even if the diagonal is smaller than the segment. Since the segment is infinitely thin, you can fold it inside the square.

*More surprisingly, you can unfold any square (or cube, or hypercube ...) and it would still fit into any segment. See space-filling curves.



$\mathbb{R}^2$, $\mathbb{R}$ and $\mathopen{[} 0, 1 \mathclose{]}$  have the same cardinality, so you can consider that the square and its side have the same number of points.
