# Visually stunning math concepts which are easy to explain [closed]

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?

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## closed as too broad by Arthur Fischer♦Apr 26 at 14:09

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It looks like mathpop or demand for math entertainment) –  rook Apr 2 '14 at 12:59
@ColeJohnson the 'transcendentality" is the beauty of it! –  Sabyasachi Apr 4 '14 at 17:07
Why has this got 105k views......................? –  LTS Apr 8 '14 at 16:49
@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. –  Mice Elf Aug 29 '14 at 18:37

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.

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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 grey, two-thirds are black. The black thirds contain squares with

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

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

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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 $e^{2k\pi i}=1$

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I made this a while back to explain sine and cosine. http://math.garbl.es/demos/sine-and-cosine

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Great stuff! I'll be using that on my CG classes from now on. –  karlphillip Jan 9 at 11:53

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."

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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 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 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 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 at 23:08

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

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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

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.

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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

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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.

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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.

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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.

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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 $0^0=1, 0^1=0$.

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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

Check out A Mathematical Picture Book at your local library - it has a bit about the Szegő Curve.

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

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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 fulfil 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 fulfil 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.

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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

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}

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Proof that the area of a circle is πr² without words: Proof Without Words: The Circle

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@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

This is how I learned Pythagoras' Theorem:

This diagram makes more sense to me than the other ones posted here.

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This doesn't explain why the theorem is true for any other triangle though, does it? This is basically just "compute $3^2+4^2$ and $5^2$ to see that $3^2+4^2=5^2$" –  Daenerys Naharis Apr 1 '14 at 18:47
The visual is not explaining why (given the three specific squares) the red triangle is a right triangle. –  Johannes Apr 1 '14 at 18:50
I do not thing this answer deserves to be downvoted, given the question wasn't asking for visual proofs. It does demonstrate the truth of Pythagoras' Theorem - albeit for a specific example. Someone who hadn't encountered Pythagoras' theorem before but who understood areas might find it interesting and ask whether it was true for other right-angled triangles. +1 from me. –  Aky Apr 2 '14 at 6:33
Stop downvoting, people. The question asked for a concept. Not a proof. This is perfectly fine. Mnemonics would be fine too. –  smci Apr 3 '14 at 5:51
Clearly this answer requires down-voting. It is a bad answer. The whole raison d'etre of the points are to order the answers by their "goodness" as seen by the community. This is not good. It is not even fair, or mediocre, or ho-hum. It is bad. This is not "easy to explain" because it has a lot of useless information presented (right angle triangle, the a, b, and c, or anything to do with Pythagoras). This doesn't even illustrate that $3^2+4^2=5^2$ except if you count squares, but you can do that with the numbers! And it is certainly not visually stunning... –  ex0du5 Apr 3 '14 at 20:17

## protected by A1DHTHDec 23 '14 at 4:02

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