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 $s^2 &lt; s$, which doesn't make sense to me for the reason above.
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 $s&gt;1$.
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  $a&lt;1$ i.e., $ a &lt;10^0$ units produces an area of magnitude $a^2&lt;a $  and when magnitude of its side is $a&gt;1$ unit, then it produces an area whose magnitude is $a^2&gt;a.$