# Can the notion of “squaring” be extended to other shapes?

We all know what squaring is:

$$n^2=n\times n$$

More specifically, I could define it as

$$n^2=\text{ area of a square with side length }n$$

Instead of using normal notation, I wish to say that

$$\operatorname{square}(n)=\text{ area of a square with side length }n$$

Is there any point in having this extend to other shapes?

$$\operatorname{triangle}(n)=\text{ area of a triangle with side length }n$$

$$\operatorname{pentagon}(n)=\text{ area of a pentagon with side length }n$$

etc.

Specifically, is there any good reason for why we would have such things? Secondly, what makes the square so special here that it gets its own operation?

For example, we could've done everything in terms of triangles. Then the area of square would be given as

$$\operatorname{square}(n)=\frac{12}{\sqrt3}\operatorname{triangle}(n)$$

Preferably, I'd like to say $\square(n)$ and $\triangle(n)$, but I can't do $\pentagon(n)$.

# EDIT

My goodness, I completely forgot to include the circle function, the most important of them all! So don't forget to consider that.

A similarly good question is whether or not this has been used before. (I know we use circles/triangles when dealing with polar coordinates)

• The square isn't receiving any special treatment or "operation". mathwords.com/a/area_regular_polygon.htm They all have their own. – Evariste Oct 24 '16 at 22:42
• Well, the square gives the fundamental one. Everybody else is some constant multiple thereof, since area of similar regions scales by the square of the stretching factor. – Ted Shifrin Oct 24 '16 at 22:44
• @TedShifrin Then what makes the square special? We could just as easily make everything in terms of triangles. – Simply Beautiful Art Oct 24 '16 at 22:46
• The geometric square is defined by two orthogonal vectors of the same length. In an equilateral triangle or any other regular polygon the shape is not defined by orthogonal vectors. What make "special" the square is that, as a regular polygon, is defined by it orthogonality. – Masacroso Oct 24 '16 at 22:52
• We do have triangular numbers, pentagonal numbers (see also), and various other figurate numbers. – Rahul Oct 24 '16 at 22:52

Actually, there are so-called polygonal numbers of all sizes. The triangle numbers are 1, 3, 6, 10, 15, ... . They can be arranged in the shape of a (filled in) equilateral triangle. They are formed as

1, 1+2, 1+2+3, ... , so the n'th triangle number is $T_n = n(n+1)/2$.

Similarly, the square numbers can be formed (just using addition) as

1, 1+3, 1+3+5, 1+3+5+7, ..., and the n'th square number is (of course) $S_n = n^2$.

The pentagonal numbers, which form regular pentagons (including the interior points) are 1, 5, 12, 22, ..., which are formed by

1, 1+4, 1+4+7, 1+4+7+10, ... .

And if you study the question of "Which triangle numbers are also square numbers?", you'll be lead to solving the Pell equation $X^2 - 2Y^2 = 1$ and finding infinitely many solutions to $T_n=S_m$, the smallest non-trivial solution being $T_8=S_6=36$. OTOH, I'm not sure if it's known whether there are infinitely many numbers that are simultaneously triangular, square, and pentagonal, or indeed whether there are any such numbers (other than 1).

• Very interesting OTOH. – O. S. Dawg Oct 25 '16 at 1:44
• Wolfram MathWorld says "...there is no other pentagonal square triangular number less than $10^{22166}$" (other than the trivial case) after searching through the first ~10k pentagonal triangular numbers. An (unpublished) analysis using solutions to the simultaneous Pell equations suggests only the trivial solution exists. But very interesting indeed. – Kyle Kanos Oct 25 '16 at 12:18
• How is it not known whether a whole number greater than $1$ is triangular, square and pentagonal? Can't the theory of Pell equations be used to figure it out? – Oscar Lanzi Sep 13 '17 at 10:10
• @OscarLanzi The question is whether there is a number that is simultaneously triangluar, square, and pentagonal. Such a number would lead to a solution of three simultaneous Pell equations. So no, the classical theory of Pell equations doesn't solve the problem, it just tells you about the solutions to each individual Pell equation. – Joe Silverman Sep 13 '17 at 11:16

If you are interested, the area of a regular $n$-polygon with side length $l$ is $$\frac{nl^2}{4}\cot\frac{\pi}{n}$$ Squares/rectangles are fundamental as they are the products of two intervals (set-theoretically): $$[a,b] \times [c,d]$$ It then becomes natural to assign this square/rectangle an area of $(d-c)(b-a)$. Other shapes cannot be expressed in this form.

I think you have the motivation backwards. The function $f(x) = x^2$ is a very useful function in its own right. In fact, it is one of a whole family of functions:

$$1, x, x^2, x^3, x^4, x^5, x^6, \ldots$$

Mathematicians have thought about these functions (called polynomials) for centuries. In fact, the field of classical algebraic geometry is basically all about solving equations involving polynomials.

Now, it just so happens that $x^2$ has the special property that it is equal to the area of the square with side length $x$. Mathematicians thought this was a pretty nice property, so they decided to name this function the "square" function. Likewise, $x^3$ is the "cube" function, and if we lived in higher dimensional space, we would likely have a geometric name for the function $x^4$ as well.

Summary: The function $x^2$ came first, and the name "squaring function" came second.

• Huh, really? I would've thought areas of squares came first, followed by the $x^k$ stuff. – Simply Beautiful Art Oct 24 '16 at 22:48
• Yes, mathematicians thought about squares (the geometric objects) long before anyone wrote down the function $x^2$. What I'm saying is that mathematicians were thinking about the function $x^2$ before anyone called it the "squaring function." The motivation for the function $x^2$ doesn't come from geometric squares. It comes from algebra. – Alex G. Oct 24 '16 at 22:52
• What? When are you trying to say that the function $x^2$ was created/discovered? Euclid states the Pythagorean theorem in terms of areas, and the Greeks pretty famously got stymied by their lack of conceptualizing math as distinct from measurement. By the time Chinese were thinking about the hockey stick identity, they must surely have had some way of understanding multiplication as a unitless operation, but even they preferred to state things in "quasi-practical" terms, suggesting their original motivation, at least, came from the geometry. – Eric Stucky Oct 24 '16 at 23:01
• @EricStucky OP is specifically asking about the name "squaring function" for the function $x^2$. This name did not exist until someone wrote down the function $x^2$, which was long after the ancients as you point out. All I am saying is that the function $x^2$ was studied before the name "squaring function" was given to it. – Alex G. Oct 24 '16 at 23:06
• @AlexG. You really need a citation for this. Geometers have spoken of the area of a square being a function of a side length for a very long time, probably at least back to Hippias. The old style of "function" is for some aspect of a diagram being a function of another. "Completing the square" precedes $x^2$ notation for quadratics, and descriptions of a quadratic formula involve "squaring" and "square root" functions. Polynomials surely precede higher powers of $x$ as functions: a polynomial is what you get when you add and multiply unknown/indeterminant quantities and coefficients. – Kyle Miller Oct 25 '16 at 2:56

There is also some geometrical justification coming from the physical world/space we live in. Squares have all their angles $90^\circ$. Compared to other $2\pi/n$ the angle $2\pi/4$ occurs naturally. Corners of walls of buildings (unless you work in US defence establishment whose name starts with P). have this angle. Road junctions angle of turns are preferred this way.

Among all shapes circle is the most significant. Square has a special relationship with the circle in the following way: The ratio of areas of a circle of radius $r$ with a square of side length $r$ seems to come up all over calculus. (as opposed to trianglular area's ratio)

• I wouldn't consider buildings and road junctions "naturally occurring" phenomena... – Rahul Oct 24 '16 at 22:59
• Oh, right! So instead of all these shapes, I should define everything in terms of circles... – Simply Beautiful Art Oct 24 '16 at 23:00
• I said "preferred this way" (by human beings, not by nature) for road intersections. Possibly steering to turn by $2\pi/4$ radians is natural in mechanical engineering. – P Vanchinathan Oct 24 '16 at 23:02
• @Rahul For a more "natural" example, trees usually grow orthogonally to the ground. – Federico Poloni Oct 25 '16 at 21:23
• Trees are real good examples. – P Vanchinathan Oct 25 '16 at 23:24

Regarding using triangles to model 2nd powering, or any m x n, check here: https://youtu.be/2B1XXV2Eoh8

The idea extends to a tetrahedron as a model of 3rd powering. http://www.rwgrayprojects.com/synergetics/s09/figs/f9001.html

Note that numeric results don't change, only the shapes used to represent the results. To the best of my knowledge, the author making the most use of this alternative model was R. Buckminster Fuller.

Another reason the square is special is because of its direct application in the Pythagorean Theorem, which in modern mathematics leads to the Euclidean metric.

Technically, the area-sum relationship of the Pythagorean Theorem also works for triangles, pentagons, half-circles, and other similar figures, and there have been theorems proven for them also. But the square version appeared first by a wide margin.

• Doesn't your second paragraph contradict the first? – Federico Poloni Oct 25 '16 at 21:24

Another guess is that area is homogeneous of degree 2 (that is for any $x, \lambda \in \Bbb R$, $nagon(\lambda x) = \lambda^2 nagon(x) = square(\lambda)nagon(x)$). Thus $square()$ would be more suited to deal with geometric problem concerning other shapes than other polygonomial function ?

The square is special because its side span a 90° angle. Which make them orthogonal, which is required for area calculation. (A triangles area is also calculated by 2 orthogonal components of it: base*height/2

So area calculation (in euclidean geometry) is basically always some kind of a*b and units are normed to 1 (the same length) so it becomes a*a for the unit itself, the same formula as the area of the square. They share these features and I guess that's why they became interchangeable.