# Visually stunning math concepts which are easy to explain

Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.

Do you know of any other concepts like these?

• It looks like mathpop or demand for math entertainment) – rook Apr 2 '14 at 12:59
• @ColeJohnson the 'transcendentality" is the beauty of it! – Guy Apr 4 '14 at 17:07
• There is a considerable overlap with mathoverflow.net/questions/8846/proofs-without-words – Martin Brandenburg Jun 20 '14 at 22:13
• @TheGuywithTheHat That's the reason for the second spike of visits, on August 27. The comment by LTS is from April; back then the traffic was driven by Ycombinator. As a result, this same question made both April 7 and August 27 the two days with most visits to the site. – user147263 Aug 29 '14 at 18:37

One of my favorites - I've seen it somewhere on the web but can't find it again now, so had to reconstruct myself. It is not as pretty but suffices to convey the idea.

It gives good grasp both for $e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$ and for $e^{2k\pi i}=1$

• I like this very much! – Noir Jan 27 '18 at 17:48

Visualisation in ancient times: Sum of squares

Let's go back in time for about 2500 years and let's have a look at visually stunning concepts of Pythagorean arithmetic.

Here's a visual proof of

\begin{align*} \left(1^2+2^2+3^2\dots+n^2\right)=\frac{1}{3}(1+2n)(1+2+3\dots+n) \end{align*}

The Pythagoreans used pebbles arranged in a rectangle and linked them with the help of so-called gnomons (sticks) in a clever way. The big rectangle contains $$(1+2n)(1+2+3\dots+n)$$ pebbles. One third of the pebbles is red, two-thirds are blue. The blue thirds contain squares with

$$1\cdot1, 2\cdot2, \dots,n\cdot n$$

pebbles. Dismantling the blue squares into their gnomons shows that they appear in the red part. According to Oscar Becker: Grundlagen der Mathematik this proof was already known to the Babylonians (but also originated from hellenic times).

This is what happens when you take Pascal's Triangle, and color each entry based on the value modulo 2:

The exact code for this is extremely simple:

def drawModuloPascal(n, p):
for i in range(0, n + 1):
print " " * (n - i) ,
for k in range(0, i + 1):
v = choose(i , k) % p
print '\x1b[%sm ' % (';'.join(['0', '30', str(41 + v)]), ) ,
print "\x1b[0m" # reset the color for the next row


Just provide your own choose(n, r) implementation. The image above is a screenshot of drawModuloPascal(80, 2).

You can also do this modulo other primes, to get even more remarkable patterns, but then it becomes much less "easy to explain."

• I'd also note that it's possible to compute ${i \choose k} \mod p$ without computing $i \choose k$. For large $i$ this would matter. – Michael Lugo Jan 9 '15 at 14:11
• The basic idea is pretty simple: ${i \choose k} = {i \choose k-1} + {i-1 \choose k-1}$, and this recurrence holds $\mod p$ as well. – Michael Lugo Jan 9 '15 at 14:31
• @Coffee_Table: It's literally just the terminal. The code I pasted above write ANSI color codes to the terminal to produce the colored blocks you see above. – Adrian Petrescu Feb 10 '15 at 22:40
• right, thanks. I edited your code to work for Python 3 and I realize now that I made a stupid error when doing so. – Coffee_Table Feb 10 '15 at 23:08
• It is a finite version of the Sierpinski triangle. – Jean Marie Apr 8 '17 at 21:07

Divergent Series can be visual:

from the Wikipedia

showing that $(1-1+1-1+\dots)^2=1-2+3-4+\dots$

This is from betterexplained.com. It's a really cool website with lots of intuitive explanations of maths concepts. This helped me understand Pythagoras' theorem. Actually my go-to website for intuitive explanations of concepts.

These are similar triangles. This diagram also makes something very clear:

Area (Big) = Area (Medium) + Area (Small) Makes sense, right? The smaller triangles were cut from the big one, so the areas must add up. And the kicker: because the triangles are similar, they have the same area equation.

Let's call the long side c (5), the middle side b (4), and the small side a (3). Our area equation for these triangles is:

Area = F * hypotenuse^2

where F is some area factor (6/25 or .24 in this case; the exact number doesn't matter). Now let's play with the equation:

Area (Big) = Area (Medium) + Area (Small)

F c^2 = F b^2 + F a^2

Divide by F on both sides and you get:

c^2 = b^2 + a^2

Which is our famous theorem! You knew it was true, but now you know why.

This explains the product rule:

• This depends on the idea that the area of all shapes increases on the order of the square of a scale parameter. It's true but it takes a lot of machinery to prove if you don't already believe it to start with. – enthdegree Aug 11 '16 at 5:04
• @enthdegree That's actually not true. It's true for well behaved 2D shapes (and in fact you can use this to define a type of dimension) but in general it's a factor of $(\Delta x)^d$. I say "well behaved shapes" because when you use this with fractals, you get fractions! – Stella Biderman Apr 6 '17 at 14:06

Ulam Spiral:

Discovered by Stanislaw Ulam, the Ulam Spiral or the Prime Spiral depicts the certain quadratic polynomial's tendency to generate large number of primes.Ulam constructed the spiral by arranging the numbers in a rectangular grid . When he marked the prime numbers along this grid, he observed that the prime numbers thus circled show a tendency to occur along diagonal lines. A 150x150 Ulam Spiral is shown below where the dots represent the occurance of prime numbers. The high density along the diagonal lines can be seen as represented by the darker shade of blue.

This is @Blue's very nice visual proof from trigonography.com that

$$x+\frac{1}{x}\;\geqslant\; 2$$

Two more illustrations from http://www.doubleroot.in

We see $(x+(1/x))^2 \geq 4$:

We know the hypotenuse is always the longest side of a triangle:

• BTW: It's cool (and surprising) that you know me from trigonography.com and not here. Maybe I should update the site more often, now that I know I have an audience. :) – Blue Apr 9 '16 at 1:12
• @Blue Yes, everyone likes a good diagram (a lot) – user311151 Apr 9 '16 at 1:13
• hey, i saw you did some editing to remove my watermark there. but please cite the source atleast? www.doubleroot.in or facebook.com/doubleroot – yomayne Aug 14 '18 at 11:05
• @yomayne the original poster is not active anymore, thus there was no reply initially. I added a link to the site you mentioned; since by now you passed 100 points you can edit Community Wiki posts. Please add more precise reference as you see fit. (It seems the site is under a CC licence, so I suppose you it is fine that it is reproduce in principle; if not we can also remove it entirely.) – quid Aug 27 '18 at 15:08

The Julia set of a complex number $c$ is a fractal (for each $c$ you have one) that has a weird property: they visually look like the Mandelbrot set around that point $c$. This becomes clear in this illustration I made for a school project, which consists of tiny images of Julia sets:

Magically the Mandelbrot set appears...

• Mandelbrot defined his Set as the set of values for which the Julia set is connected; its formula is intimately related to that of a Julia set. – Anton Sherwood Jun 21 '16 at 23:46

The following animation shows how the surface area of a sphere is calculated.

I recently find some stunning visualizations. I preferred to share them all:

$5)$ Mean inequalities [from Proof without words]

$4)$ Streographic projection [by H.Segerman]

$3)$ Farey-Ford Tessellation in non-euclidean geometry [by F.Bonahon]

$2)$ Steiner porism [by Wikipedia]

$1)$ Polynomial roots [by J.Baez]

Aren't them Incredible??

It's not exactly stunning, but it is interesting and visual and simple enough for an elementary school child:

There are only 5 platonic solids.

Numberphile has a great video explaining it: https://www.youtube.com/watch?v=gVzu1_12FUc

In short, the reason is that there are only enough space for 3, 4, or 5 equilateral triangles at a corner; only enough space for 3 squares at a corner; and only enough space for 3 pentagons at a corner; and not even enough space for 3 hexagons at a corner, so there are only 5.

Although I guess it was stunning enough for the ancient Greeks to decide that they were the geometric basis of the five elements of the universe: earth, fire, wind, water, aether.

This one is only visually stunning in your imagination, but I like it. The derivative of a circle w.r.t. the radius is the circumference. $$\frac{d}{dr}\pi r^2=2\pi r$$ The derivative of a sphere w.r.t. the radius is the area. $$\frac{d}{dr}\frac{4}{3}\pi r^3=4\pi r^2$$ The derivative of a 4-dimensional sphere w.r.t. the radius is the 3-dimensional area. $$\frac{d}{dr}\frac{1}{2}\pi^2 r^4=2\pi^2 r^3$$ This works because the radius is invariant in n-dimensional spheres. Holding a circle, a sphere or a hypersphere requires your hands to be the same distance apart.

• Also, this isn't limited to circles/spheres at all. It holds for squares and cubes and tetrahedra and possibly other shapes, if the formulas are expressed in terms of the radius of the inscribed circle/sphere. – user3932000 Mar 9 '17 at 2:00
• @user3932000 Pls write an answer expanding your comment – dushyanth Nov 19 '17 at 15:25

The sum of the first $n$ squared numbers:

The first 3 triangles are the same, just rotated. Also, notice that

\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}

Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$

Thus,

$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$

Please see here for the original post

and here for a more indepth explanation.

Allow me to join the party guys...

This is another proof of the Pythagorean theorem by The 20th US President James A. Garfield.

A nice explanation about Garfield's proof of the Pythagorean theorem can be found on Khan Academy.

I've built a bunch of interactive explorations over at Khan Academy. A few of my favorites are:

• Derivative intuition. Particularly amazing is seeing how $\frac{d}{dx}e^x=e^x$. (Do a few and it should pop up).

• Exploring mean and median. Light bulbs are twice as likely to burn out before the average lifetime printed on the package. If that statement surprises you, this exploration points out that mean and median aren't the same thing.

• Exploring standard deviation. Standard deviation is a term that gets thrown around a lot. Play around with this to get a more intuitive sense of what it means.

• One step equation intuition. Basic introduction to why you can do the same thing to both sides of an equation to solve it.

• All of these links are broken. Could you provide unbroken links? – Drew Christensen Jan 2 '17 at 17:51

How to convert a function from Cartesian to Polar coordinates:

There's also some really cool art in Polynomiography. Dr. Bahman Kalantari seems to have made really interesting visualizations of polynomials, and considering these functions are everywhere, it might be cool to check them out.

Polynomiography

This one $($via Proof Without Words$)$ is wonderful but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it.

Explanation:
Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$ So \begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}

• I don't get it. I see a circle divided into 14 segments and 3 oddly placed parallelograms. Explain this please. – Nick Feb 22 '15 at 7:17
• @Nick I'll add it in my answer. – Vim Feb 22 '15 at 7:34
• @Nick I've finished. You can take a look at it if you need it. – Vim Feb 22 '15 at 7:45
• @Vim how do we know that, for example, AD and CE are parallel? – Joffan Mar 14 '15 at 14:56
• To my mind, however, this proof makes $$\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}=\frac{1}{2}$$ seem more magical than it really is. One can instead see that the statement is equivalent to $$\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}+\cos\frac{7\pi}{7}+ \cos\frac{9\pi}{7}+\cos\frac{11\pi}{7}+\cos\frac{13\pi}{7}=0,$$ which expresses that the mean horizontal coordinate of seven equally spaced points on the unit circle is $0$, a fact that follows from rotational symmetry. – Will Orrick Dec 6 '15 at 10:01

# Polynomials can describe geometric objects

In high school we learn that some low order polynomials can describe geometric shapes:

Basic shapes we all recognize ( as intro ) $$\begin{array}{llr}y&=kx+m& (\text{ line })\\r^2 &= x^2+y^2 & (\text{ circle })\\y &= x^2+ax+b &( \text{ parabola })\end{array}$$

Cool properties consider the rotation $$\left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{rr}\cos(\phi)&\sin(\phi)\\-\sin(\phi)&\cos(\phi)\end{array}\right]\left[\begin{array}{c}x_{new}\\y_{new}\end{array}\right]$$ and then we substitute each $$x^ay^b$$ and carry out the multiplications and we will still have a polynomial. By similar reasoning we can do scaling and translation and still remain a polynomial. If we rewrite the polynomials to be expressions equal to 0: $$p_a(x,y)= 0, p_b(x,y) = 0$$ then we can multiply them and use the fact that $$b\cdot a = a\cdot b = 0, \forall a \neq 0, \text{iff } b=0$$ This gives us ability to combine shapes into one and the same representation. We can also do something of the opposite: equation systems which can get the intersection. Example is intersection of two lines is an equation system of two lines. The interesection of a sphere and a plane is a point or a circle. This is also where the expression conic section comes from: an intersection between a cone and something!

And still after all this which is so visually accessible and easy to explain in one sense, still involves challenges in modern math of algebraic geometry has had lots of development even in the last 50 years.

below: $$ax^p + by^p - k^p = 0$$ for $$p=6$$. When $$p$$ grows it will get closer and closer to a rectangle. To the right is the "fifth heart curve" (source: Wolfram Alpha) is an 8th degree polynomial.

A theorem that I find extraordinarily beautiful and intuitive to understand is Gauss' Theroma Egregium, which basically says that the Gaussian curvature of a surface is an intrinsic property of the surface. Implications of this theorem are immediate, starting from the equivalence of developable surfaces and the 2D euclidean plane, to the impossibility of mapping the globe to an atlas. Wikipedia also provides the common pizza eating strategy of gently bending the slice to stiffen it along its length, as a realization

• Sounds interesting but I found that link really impenetrable, as a lay person. – Paul Apr 3 '14 at 13:17
• Yeah, great theorem, bad example ;) – Carsten S Apr 3 '14 at 21:10
• A minor correction: strictly speaking, Gaussian curvature is not a topological invariant. – Michael Apr 3 '14 at 22:48
• @Michael: Yes, you are right. My mistake. I instead had in mind the surface integral of the gaussian curvature over a closed surface, which is a topological invariant (basically the euler characteristic) – surajshankar Apr 5 '14 at 4:59

I would like to add some explorations of the concept asked by the OP of my own:

1. Visualization of the set of real roots of quadratic equations $ax^2+bx+c=0$, for the specific values of the intervals $a \in [-a_i,a_i]$, $b \in [-b_i,b_i]$, $c \in [-c_i,c_i]$, $a,b,c \in \Bbb N$.

By Cartesian coordinates $(x,y)=(x_1,x_2)$. E.g. $a_i,b_i,c_i=75$:

By Polar coordinates $(\theta, r)=(x_1,x_2)$. E.g. $a_i,b_i,c_i=75$:

Due to the symmetries the opposite patterns $(x,y)=(x_2,x_1)$ and $(\theta,r)=(x_2,x_1)$ are similar.

1. The Chaos Game on the metric space $S^{1} \times [0,\infty)$ with the metric $d((\theta_1,x_1),(\theta_2,x_2)) = d_{S^1}(\theta_1,\theta_2) + |x_1-x_2|$. The distance in $S^1$ is given by the smallest angle measure between $\theta_1$ and $\theta_2$ (this is actually a scaled Euclidean metric on the unit circle itself).

In this version, the points are $(\theta, r)$, (the angle in radians and the radius). And the three attractor points are $A=(0,0)$,$B=(\frac{5\pi}{4},1)$ and $C=(\frac{7\pi}{4},10^4)$.

This is another example locating the attractor points in the same axis: $A=(0,0)$,$B=(\pi-\frac{\pi}{8},1)$ and $C=(\frac{7\pi}{4},10^4)$.

1. Contruction step by step of the Voronoi diagram of the points generated by a classic Chaos Game Sierpinski gasket.

1. And my favorite so far, visualization of the $4$-tuples of the extended Euclidean algorithm in a four dimensional tesseract. The projection of the four dimensional points are shown into a $3D$ visualization adding as a reference a tesseract or hypercube:

• @Leila you are welcome, if you want to know more about them, I have linked the original questions on each point. – iadvd Apr 6 '17 at 8:16

Just wanted to point out that The Book of Numbers has a lot of the examples above $($ as well as many others $).$

The beauty of watching graphs being constructed has always mesmerized me; I love how such simple figures can be used to make such complicated pictures. And it's especially satisfying with polar graphs.

Even simpler things like conic sections:

This one might not be as easy to explain, but brachistochrones are wonderful things.

A visual display that $0^0=1$. The following is a tetration fractal or exponential map with a pseudo-circle shown in orange. The red area is period $1$ and contains $1$. Example is $1^1=1$. The orange pseudo-circle which contains $0$ is period two. Example is $0^0=1, 0^1=0$.

• 0^0 = 1 is not even proven, it was just defined that way to make things easier, right? – Michael Apr 4 '14 at 18:59
• It depends on the context, but in the grand scheme of mathematics, it is considered undefined. This is because there are completely logical steps that point to it being zero, and equally logical ones that point to it being one. We can't call it both zero and one, so we call it undefined. – recursive recursion Apr 4 '14 at 21:41
• There is one problem with this. I have no idea how you make that image mean what you say it means... Visually stunning? Yes. Easy to explain? Maybe. Explained? No. – daviewales Apr 5 '14 at 12:31
• So, er, how does this show that $0^0=1$? – David Richerby Apr 5 '14 at 17:55
• $x^0 = 1$ because the multiplicative identity is one, $0^x = 0$ because zero is the multiplicative fixed point. That is, if and only if we never, ever, ever tussle with fractional powers; then and only then exponentiation is shorthand for repeated multiplication and therefore $0^0 = 1$. If we at any point use fractional powers, then exponentiation is shorthand for it's natural definition: $a^b = e^{b \ln a}$ and we all know that $\ln 0$ is undefined. There is no dispute, once we make the assumptions explicit the problem goes away. – Karl Damgaard Asmussen Apr 13 '14 at 9:16

Proof that the area of a circle is $\pi r^2$ without words: Proof Without Words: The Circle

• but he has to prove $C=2\pi r$ first – zinking Apr 8 '14 at 6:23
• @zinking No, $\pi$ is defined to be the constant that goes in that place in that equation to make it hold. However, I was rather dissatisfied with this "proof". There's lots of distortion involved with deforming the area of the circle into the area of the triangle, that one would have to know calculus to understand why the distortion doesn't matter (at which point, why not just calculate the integral). This is more of a memorization technique to remember the formula. – Travis Bemrose Apr 13 '14 at 20:02

A nationwide math contest in Germany recently came up with a task that I found beautiful to explain, because of two points.

1. You can get an idea, what the proof is, without applying mathematically accurate theory and this intuitive proof is most likely the right way.

2. At any given point of this intuitive proof, you can chime in and ask yourself: But how would I say this in mathematical terms? When you find these terms, eventually you get the proof you were looking for.

So here you go: Lea gets the task to write down 2014 numbers. These numbers have to fulfill a specification. For every set of three numbers from that whole set, the arithmetic average of these three must also be within the whole set of $2014$ numbers.

Your task is to proof, that Lea has to write down the same number $2014$ times. Every set of 2014 numbers with any variation in it would not fulfill the specifications.

So since we are talking about layman maths here, I'll go with the intuitive way. We have to find a reason, why choosing a set with different numbers would violate the specifications and we have to proof that always taking the same number would not violate them. The later one is rather easy. Take any arbitrary number three times. The arithmetic average will be the same number, which is in your set already. That wasn't too bad, right?

But what about sets with not all the same numbers? We are mathematicians, so we'll just do what we always do: Chop the problem into pieces we can solve. The first piece is where we have two equal numbers and one other number in our set. Let's assume, the single number is bigger than the two equal numbers. What will that do to our arithmetic average? Right, it will be below the middle between the bigger and the smaller number. We can write that arithmetic average down and specifications are ok. But now we have created another set of three numbers. The single, big number (I'll call it a), one of the two equal numbers (that would be b) and the arithmetic average of a, and b (I'll call that one c). So now we would have to also add the arithmetic average of a, b and c. A quick sketch will show you, that this new number is also slightly below the middle between a and b.

And like that we will always have to add a new number. The arithmetic average of a, b and the new number will never reach the middle. Something, that you can also verify with a few sketches. So we would have to add infinitely more numbers, but we wanted only 2014. Apparently, no two numbers can be equal.

So what if all numbers are different? There is one special case. Let's call our numbers a, b and c again. If b is equally far away from a and c (so b could be 3, a could be 1, then c would be 5). In that case, b is the arithmetic average. But we have to have 2014 different numbers. As soon as we add a fourth number d, it's spoiled. d could be 7, to be still in a distance of 2 to c, but then the set a, b and d would not contain its own arithmetic average. So we know, that within a set of 2014 numbers, we would have sets of three, where these three numbers don't include their own arithmetic average, no matter what we do.

And now we look back at our idea about the set with two equal numbers. We see: As soon as we have a bigger and a smaller number and the number in between those is not exactly in the middle, we can once again start with our endless construction of arithmetic averages. We always replace the number between the bigger and the smaller one by the arithmetic average of the three and we can never reach the middle, but it will always get closer to the middle (thus be another number).

And as I said, making this proof mathematical will not alter it. It will be all the same, but with more equations and sequences. Since we excluded the option of making anything infinite, it is correct as it stands here. This one made me realize: Proofs are not the miracles or the magic they seemed to be for me during high school. Of course, there are hard proofs (and things you can't proof, there is a proof for that), but often you only have to think clearly and to chop the problem into the right pieces.

• Hmm, I don't think "induction by contradiction" ("contradiction by induction"?) is valid even though our intuition would like it. In other words, you've shown that you can't inductively build a set with the desired property, having smaller sets along the way that also have the property. But this doesn't exclude the possibility that the property might not hold for all smaller sets, yet 2014 is the first time you have a set large enough for the property to hold. – Travis Bemrose Apr 13 '14 at 20:36
• Sort the list in non-decreasing order as $a_1, a_2, \ldots, a_{2014}$. Since the average of $a_1$, $a_2$, and $a_3$ is among these, it must equal $a_2$, so $a_2 = a_1 + d$ and $a_3 = a_1 + 2d$ for some $d$. Since the average of $a_2$, $a_3$, and $a_4$ is among the list, is must equal $a_3$. Proceeding in this manner we see that $a_k = a_1 + (k-1)d$ for some $d$. Now note that the average of $a_1$, $a_2$, and $a_4$, which is $a_1 + \frac{4}{3} d$, is also in the list. It follows that $d = 0$. – Daniel McLaury Aug 28 '14 at 1:36

Check out the "Proofs Without Words" gallery (animated) here:

http://usamts.org/Gallery/G_Gallery.php

And the related proofs here:

http://www.artofproblemsolving.com/Wiki/index.php/Proofs_without_words

Many of these are similar to the ones already listed here, but you get a bunch in one place.

A connection between Mathematics and Love: the story goes that a very shy mathematician had fallen in love with a girl but did not dare to tell her. In stead he wrote her a letter with only the following formula: $$y=\pm \sqrt{25-x^2} -\frac{3}{|x|+1}$$ If she was really interested he counted on her drawing the graph of the formula. How the story ended no one knows ...

So how does one construct such a formula? The first part of the formula (the square root without the fraction) is an ellipse: $(3x)^2+(5y)^2=15^2$. Now to get a heart shape the top and bottom of this ellipse must be somewhat lowered and this is accomplished by adding the fraction which is really an adjusted orthogonal hyperbola, $y=\frac{1}{x}$, in such a way that it connects to the ellipse. Desmos is a fantastic tool to illustrate this and hence explaining graphs, functions and and a bit of analytical geometry.

The (otherwise also easy to prove) fact that $\sum_\limits{k=1}^n k=\frac{n(n+1)}{2}$ in one picture:

Source of the picture

I made this earlier this year in Blender after having spent a few days trying to think of a visual proof of $$a^3-b^3 = (a-b)(a^2+ab+b^2)$$ so that I could make myself a nice paperweight. I think it's quite clear but I'll explain it anyways,

When it's put together you see the cube $$a^3$$ with the piece $$b^3$$ cut out of it. This lets you recognize that each block has a dimension of $$(a-b)$$ somewhere, so then I pull the pieces apart and lay them next to each other.

When it's laid down you can see how all of them have the same height $$(a-b)$$ with the red block having a base area of $$a^2$$, the blue block having a base area of $$b^2$$ and the green block having a base area of $$ab$$.

So that shows how $$a^3-b^3 = (a-b)a^2 + (a-b)b^2 + (a-b)ab$$, which is nicer factored as,

$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$

## protected by DeepSeaDec 23 '14 at 4:02

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