Why Gauss wanted a heptadecagon on his tombstone
scientificamerican.com
scientificamerican.com
Detailed description of the problem of constructable regular polygons and a gloss of the proof. https://www.youtube.com/watch?v=EX7U0DGBmbM
A full explanation of the proof: https://www.youtube.com/watch?v=Gdy1u4lsjDw
There's a bit at the end showing the construction used in place of a building number on the front of the Mathematical Sciences Research Institute (MSRI) building at 17 Gauss Way, UC Berkeley.
A length is constructible exactly when it can be expressed with the operations of addition, subtraction, multiplication, division or square roots applied to integers.
[...]
Remarkably, the rudimentary tools that the ancient Greeks used to draw their geometric diagrams perfectly match the natural operations of modern-day algebra: addition (+), subtraction (–), multiplication (x), division (/) and taking square roots (√).
The reason stems from the fact that the
equations for lines and circles only use these five operations
, a perspective that Euclid couldn’t have envisioned in the prealgebra age."
Related:
* https://en.wikipedia.org/wiki/Silver_ratio
* https://en.wikipedia.org/wiki/Ammann%E2%80%93Beenker_tiling
Oliver Byrne made an insanely pretty colourful version of Euclid's Elements, which is available online. Grab a pen, paper, a string to make circles, and the edge of a book to draw straight lines, start with Proposition 1 and go as far as you'd like: https://www.c82.net/euclid/book1/#prop1
(There is also a physical facsimile of Byrne's Elements (ISBN:9783836577380) – it is one of the best additions I've ever made to my library. It is simply gorgeous.)
So arguably the greatest mathematician of all time[2] wanted a particular tribute to something he did while a teenager, felt was one of his greatest achievements (because the problem had been unsolved for over 2000 years) and someone just decided they couldn't be arsed.
The whole thing is described really well here, including showing the full construction https://www.youtube.com/watch?v=EX7U0DGBmbM
[1] Picture here https://www.atlasobscura.com/places/grave-of-carl-friedrich-...
[2] My vote would go for Euler, but a lot of people feel Gauss.
But there's a statue with that star (https://de.wikipedia.org/wiki/Carl_Friedrich_Gau%C3%9F#/medi...)
But per below comments, the star is on the monument in Brunswick/Braunschweig, not on his headstone.
An image search for "gauss monument brunswick germany" on Duck Duck Go includes a picture of the 17-point star at this link:
https://www.braunschweig.de/leben/stadtportraet/stadtteile/n...
I can't go to it to confirm because we are blocked from going to foreign links at my work place. It looks like the star is on the left side of the monument near his right foot.
This can be attributed at least in part to the collapse of the Western Roman Empire and the relative chaos that followed it in Central and Western Europe. Instead, Indian and Middle-Eastern mathematicians took over in the intervening millennium or so. Men like Āryabhaṭa, Brahmagupta, Al-Khwarizmi, et al made significant contributions to modern mathematical understanding.
Oddly, there does seem to be something of a long running fad among popular historians to downplay the achievements of the middle ages. None of that of course is to say that the collapse of the empire in the west wasn’t devastating, especially in the early middle ages.
Later he proved that all n-gons with $n=2^k*p_1…*p_r$ where the p_i are Fermat-primes (2^(2^m)+1 prime, today we only know of 3, 5, 17, 257, 65537) are constructible. The opposite direction, i.e. all other n are not constructible, was only a few years later proved. Look up "Theorem of Gauss-Wantzel". I only skimmed the proof, but it seems to generalize the concept of constructing the cos of the angle with "Galois-Theory".
(edit: or see https://en.wikipedia.org/wiki/Constructible_polygon)
In the complex numbers, the vertices of a pentagon are z^5-1=0. You can factorize it as (z^4+z^3+z^2+z+1)*(z-1)=0. The hard part is solving z^4+z^3+z^2+z+1=0.
Now that equation can't be factorized, and has degree 4. It's important that the solutions have a property that is related to the degree of the equation so they have a property that is 4.
With a compass and a straightedge you can solve only equations of degree 2, that is like taking a square root. If you repeat the process you can solve (some) equations of degree 4. So after a few tricks, you can solve the equation and draw the pentagon.
For 17, the equation is z^16+z^15+...+z+1=0. So the property is 16 and you must use the square root a few times. Each time the solutions double their property, so you get 1 -> 2 -> 4 -> 8 -> 16. Near the bottom of the article is the formula, and it's possible to see a lot of nested and repeated square roots.
For 7, the equation is z^6+z^5+...+z+1=0. So the property of the solutions is 6. With the square root you can only double the property, so you get 1 -> 2 -> 4 -> 8 -> 16 -> 32 ... but you can never get a solution which has a property equal to 6.
(There are more technicals details. You can solve some equations of degree 16, for example to draw the 17agon, but you can't solve every one of them.)
For example with 9, you can factorize z^9-1=0 as (z^6+z^3+1)*(z^2+z+1)*(z-1)=0, and now the property to calculate is 6*2 instead of 8, so it's not a power of 2 and the polygon is impossible to construct.
a 17-gon reduces to a 4th-degree polynomial, and a 2nd degree one, which can be solved in radicals, by studying the permutations and multiplicities of the roots of this polynomial, in which the solutions are multiples of n*pi/17.
You can't do it exactly, but you can do it to arbitrary degrees of accuracy; at least about as far as you can go without bumping into the precision limits of a compass and straightedge.
1/7 = 1/8 + 1/64 + 1/512 + 1/4096 + 1/32768... as you can see this will hit the limits of human precision in short order.
In general any fraction 1/(2^n - 1) can be expressed as an infinite sum (or a series that comes infinitesimally close)
1/(2^n - 1) = the sum from x equals one to infinity of 1/(2 ^ (x * n)). And we all know how to section any arch-length into fractions over powers of 2.
So starting with a complete loop, segment take the first piece, then take the second piece, segment and take its first piece... keep on adding all the little pieces together until it's so close enough to 1/7 that you can take a compass measure and use that to resegment the rest of the pie - making sure that you recurse enough that after you've market out 6 additional ones, you get near enough a collision to the first that you're not really worried.
But yeah, I'd be surprised if you could compass and straightedge even to a precision of one part in 4096 - and there's no way in hell that anyone's ever pulling off one part in 32768.
That that the Hilbert Curve covers the totality of the square; but the square contains all bound points of the form [real, real], and you can see from the rational construction of the recursive vertix generator that one of the two values for each co-ordinate pair must necessarily be a rational number (albeit one denominated by an infinite integer exponent of two).
Even if you covered all of [real, rational] + [rational, real] (which you don't), you'd still never reach all of [real, real].
Effectively 100% of the plane is not on the curve and 100% of the plane is within an infinitesimal distance of the curve.
Which I actually think is more interesting than saying that the whole damned thing is in there, which it isn't.
With a Hilbert curve the entire plane becomes a limit.
There are a countably infinite number of rationals between any two rationals, you can even keep splitting up those rational infinitesimal gaps into countably many rationals that are infinitesimal even relative to the earlier infinitesimals.
And you still only end up with a countably infinite set of expressible locations and not the real continuum.
Either x, y, or both are guaranteed to be a number of that form for all values on the curve.
You're looking for a line that is 2*sin(π/7) of the radius. That's 0.86777. The square of that is 0.7530, which is pretty darn close to 0.75 (1 - (1/2) ^ 2).
So make a triangle whose height is half the radius, hypoteneuse is the radius, and the other edge is 0.8660, within 0.001 of the real value and much more accurate than I can possible draw with a straight-edge and compass.
So it quickly turns into a question of perfect tools and other things that don't actually exist.
Somewhat pedantically, if it were archs and lines, I would consider it differently - they are hypothetical constructs and subject to hypothetical boundlessness.
But a straightedge and compass are not imaginary things; they are things of the material world and they are subject to material limitations.
Even one million is nowhere close to infinity, but the sum from x = one to one-million of 1/8^x is so stupidly close to 1/7 that you're most likely getting your toolkit delivered by the Archangel Gabriel himself.
And in less that ten minutes I could write that entire number to file.
Everytime I hear about Yitang Zhang, I cannot help but be amazed by his accomplishment.
https://www.scientificamerican.com/article/prime-number-puzz...
> At just 18 years old Gauss used a heptadecagon to solve a classic problem that had stumped mathematicians for more than 2,000 years.
where the words "stumped mathematicians" are hyperlinked to the article "Prime Number Puzzle Has Stumped Mathematicians for More Than a Century" that you're talking about, even though there's no real connection at all between the problem Gauss solved and the one Yitang Zhang worked on -- they've just linked the two because of the word "stumped" it looks like. (Well prime numbers turn up in both but the problems are not really related beyond that.)