*x*and

*y*. The simplest binomial is the sum of those two terms

*x+y*. Powers of this binomial (

*x+y*)

*are ruled by the so-called binomial theorem, which states that the coefficients of the expansion terms*

^{n}*x*are given by the Pascal triangle entries. So we have:

^{p}y^{q}(

*x+y*)

*= 1*

^{0}(

*x+y*)

*= 1.*

^{1}*x*+ 1.

*y*

(

*x+y*)*= 1.*^{2}*x**+ 2.*^{2}*xy*+ 1.*y*^{2}(

*x+y*)*= 1.*^{3}*x**+ 3.*^{3}*x*^{2}*y*+ 3.*x**y**+ 1.*^{2}*y*^{3}*x+y*)

*= 1.*

^{4}*x*

*+ 4.*

^{4}*x*

*+ 6.*

*y*^{3}*x*

^{2}*y*

*+ 4.*

^{2}*x*

*y*

*+ 1.*

^{3}*y*

^{4}But where do these coefficients come from? You can just be satisfied by the result algebraic calculation, or you could look for geometric representations. For instance, a nice geometric explanation is given at the www.mathaware.org site.

Geometric proof of square binomial (original at The Geometry of the Binomial Theorem, Math Awareness Month site) |

Geometric proof of cubic binomial (original at The Geometry of the Binomial Theorem, Math Awareness Month site) |

*x*. For third degree binomials, they emerge as the number of cubes or parallelepipeds with volume given by the expansion terms. And for higher degree binomials, they emerge as the number of hypercubes and hyperparallelepipeds given by the expansion terms, so an Euclidean geometric representation for those higher degree binomials seems impossible.

^{p}y^{q}This suggested impossibility prompted me to have a closer look at geometric representations of binomial expansions. As a matter of fact a number to the fourth power

*x*is just a number, but it is also a square

^{4}*x*= (

^{4}*x*)

^{2}*. And there is no impossibility in representing numbers and squares geometrically, so the suggested impossibility is only an impossibility along the common line of thought, which sees cubic expansions as volumes. Along another line of thought, which seems to have been unnoticed, cubic expansions can be seen as areas, and then geometric representations of higher degree binomials become possible. The following figures illustrate this fact for the special case where*

^{2}*x+y*is normalized to 1 (for arbitrary

*x+y*, one just has to rescale the figure each time, the pattern remains the same).

For the binomial (

*x+y*)

*= 1.*

^{1}*x*+ 1.

*y*, we can divide a unit square into two rectangles, one rectangle of area

*x*and one rectangle of area

*y*, see Figure 1.

For the binomial (

*x+y*)

*= 1.*

^{2}*x*

*+ 2.*

^{2}*xy*+ 1.

*y*

*, we can divide the unit square into two squares, one of area*

^{2}*x*

*and one of area*

^{2}*y*

*, plus two rectangles of area*

^{2}*xy*, see Figure 2.

For the binomial (

*x+y*)

*= 1.*

^{3}*x*

*+ 3.*

^{3}*x*

^{2}*y*+ 3.

*x*

*y*

*+ 1.*

^{2}*y*

*, we can divide each of the previous squares and rectangles into proportions*

^{3}*x*and

*y*, giving eight rectangles, one of area

*x*

*, one of area*

^{3}*y*

*, plus three rectangles of area*

^{3}*x*

^{2}*y*and three rectangles of area

*x*

*y*

*, see Figure 3.*

^{2}For the binomial (

*x+y*)

*= 1.*

^{4}*x*

*+ 4.*

^{4}*x*

*+ 6.*

*y*^{3}*x*

^{2}*y*

*+ 4.*

^{2}*x*

*y*

*+ 1.*

^{3}*y*

*, we can again divide each of the previous squares and rectangles into proportions*

^{4}*x*and

*y*, giving one square of area

*x*

*, one of area*

^{4}*y*

*, four squares and two rectangles (giving six) of area*

^{4}*x*

^{2}*y*

*, four rectangles of area*

^{2}*x*

^{3}*y*and four rectangles of area

*x*

*y*

*, see Figure 4.*

^{3}And we could go further indefinitely, doodling areas of incrementing powers, just in 2D, without any reference to unintuitive hyperspaces.