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.

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    $\begingroup$ a square with side $0.5 m$ has area $0.25 m^2$ with 0.25 < 0.5. Where is the problem ? $\endgroup$
    – Jean Marie
    Feb 4, 2018 at 19:44
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    $\begingroup$ @JeanMarie Even worse, the same square with side $50$ cm has area $2500 \gt 50$ ;-) $\endgroup$
    – dxiv
    Feb 4, 2018 at 19:45
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    $\begingroup$ Comparing the area of a square to the length of one of its sides is like comparing liters to seconds. It is not defined whether 3 liters is less than, equal to, or greater than 7 seconds. It's not defined whether the length of a side is greater than the area. $\endgroup$
    – user13618
    Feb 4, 2018 at 22:50
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    $\begingroup$ @DanielC I cannot see your problem. Just because something is clear to you does not make it clear to others. OP has a valid question which also is well asked. People tend to vote for all kind of qualities, not only for how mathematically sophisticated the questions is. It even seems to me that questions like these are more valuable for the site as those which only mathematicians can understand. $\endgroup$
    – M. Winter
    Feb 5, 2018 at 19:13
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    $\begingroup$ @CarlWitthoft That's a pretentious and unconstructive comment. Regarding the purpose of the site, "Mathematics Stack Exchange is for people studying mathematics at any level and professionals in related fields." If you want to stick to research-level mathematics, this is not the right place; see MathOverflow. $\endgroup$
    – Théophile
    Feb 6, 2018 at 2:07

6 Answers 6


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 $s>1$.

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    $\begingroup$ Nailed it with the 2nd point. I realized the units were the sticking point, but didn't think of the classic "multiply by 1 (unit)" to make comparison possible. $\endgroup$
    – pjs36
    Feb 4, 2018 at 19:50
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    $\begingroup$ I'm not sure if the second point makes more sense than OP's question. As in the comments above, it depends on how you define $1$ and if $s$ is $50cm$ or $0.5m$. $\endgroup$ Feb 5, 2018 at 8:11
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    $\begingroup$ @EricDuminil I think most of the people in the comments have completely misunderstood the OP's question, which was why you can't just argue "The sides fit into the square, so the area is larger than the side", and I think you are missing the point of the explanation. If $s$ is $50cm$ and you draw a $50cm \times 1cm$, then the rectangle will fit inside the square, and $s^2$ will be a number larger than $s$; if we switch units so that $s$ is $0.5m$ and draw a $0.5m\times 1m$ rectangle, then the rectangle will not fit inside the square, and $s^2$ will be a number smaller than $s$. $\endgroup$
    – mweiss
    Feb 5, 2018 at 14:10
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    $\begingroup$ Also, you can fit loads of squares of any size into a single segment of any size. $\endgroup$ Feb 5, 2018 at 18:29
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    $\begingroup$ So the square does not fit inside the area, if $s = 1$? $\endgroup$
    – fabian
    Feb 6, 2018 at 10:05

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.

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    $\begingroup$ I knew that you could choose different units of measurements that will give different numbers. What was bothering me was when the edge of the square is less than 1, the length of an edge is a bigger number than the area of the square, despite the fact that the area is clearly a larger region of space than the edge. What I failed to realize, is that area and length are different things and so I cannot compare the length of one thing to the area of another to see which is larger. the AREA of the edge is 0, so indeed it is a smaller region of space. $\endgroup$
    – trynalearn
    Feb 5, 2018 at 2:13
  • $\begingroup$ but there's absolutely [no] need [to give the area in m²] - I think you forgot that word $\endgroup$
    – Rafalon
    Feb 6, 2018 at 10:21
  • $\begingroup$ @Peter, well as it completely changes the meaning of the sentence, I prefer to ask to be sure. Also, I don't know here, but in other stacks, you can't edit questions or answers if you don't change at least a certain number of characters $\endgroup$
    – Rafalon
    Feb 6, 2018 at 15:16

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.

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    $\begingroup$ With feet and inches, it's clearer... :) $\endgroup$
    – Jean Marie
    Feb 4, 2018 at 21:23

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).



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.


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.

enter image description here

$\mathbb{R}^2$, $\mathbb{R}$ 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.

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    $\begingroup$ The observation about space-filling curves does not make sense to me. Yes, if you regard the square as the image of a space-filling curve, you can unfold it to get a linear manifold. But the length of that manifold is not finite -- as your own diagram shows, the total length of the various iterations increases at each stage of the recursive construction. So the square does not fit into the segment. $\endgroup$
    – mweiss
    Feb 6, 2018 at 4:46
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    $\begingroup$ Moreover the comment about cardinality seems to really be missing the point of the question, which was clearly about measure-preserving transformations. Yes, any segment and any square have the same number of points, but that hardly seems relevant when the question was about area and length, not about cardinality. $\endgroup$
    – mweiss
    Feb 6, 2018 at 4:48
  • $\begingroup$ @mweiss: Well, I also think that you missed the point in your answer and there's an opportunity for another approach in another answer. $\mathbb{R}$ and $\mathopen{[} 0, 1 \mathclose{]}$ have the same cardinality, so it's not a problem to distribute every point of your infinite line into a segment. $\endgroup$ Feb 6, 2018 at 11:35

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 $a^2>a.$

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    $\begingroup$ I did not provide the downvote. However, FYI the base 10 has no relevance to this question at all. Numbers exist whether or not they are represented in a base that humans can understand. E.g. if a number and the square of that number are both written down in base 10, and you see that the square is more than the number itself, the square will also be bigger if both numbers are written in some writing only understandable to aliens. Or if they are never written down at all. $\endgroup$ Feb 6, 2018 at 12:45

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