Why Gauss wanted a heptadecagon on his tombstone(scientificamerican.com)
scientificamerican.com
Why Gauss wanted a heptadecagon on his tombstone
https://www.scientificamerican.com/article/why-this-great-mathematician-wanted-a-heptadecagon-on-his-tombstone/
66 comments
https://archive.is/a6pQb
A surprising related fact is that 200 years after Gauss and with a vast amount of progress in mathematics, we still don’t know the largest regular polygon with an odd number of sides that can theoretically be constructed in the Euclidean manner. For the curious, this is because the answer reduces to combinations of multiples of Fermat Primes, and nobody knows if there are any Fermat Primes beyond 3, 5, 17, 257, 65537. (See https://en.m.wikipedia.org/wiki/Constructible_polygon)
There's a series of two excellent videos on youtube about this proof:
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
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
I learned about the construction of this shape from a numberphile video a couple of years back. https://www.youtube.com/watch?v=87uo2TPrsl8
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.
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.
>"Can a compass and straightedge construct a line segment of any length? By Gauss’s time, mathematicians knew the surprising answer to this question.
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/CORDIC
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/CORDIC
What privileges the square root over any other fractional power?
Finding intersection points of a circle with a line is equivalent to solving a system of equations, where one equation is that of a circle, (x-a)^2 + (y-b)^2 = r^2, and the second is that of a line, Ax + By = C. To solve it, you’ll be taking square roots, and not other roots. Similarly, to find intersection of two circles, you’ll be taking square roots, and not other roots.
Presumably the hypotenuse of a right angled triangle.
Indeed. Related is the Ammann-Beenker tiling and it's connections to the 'Silver Ratio' of sqrt(2) + 1.
* https://en.wikipedia.org/wiki/Silver_ratio
* https://en.wikipedia.org/wiki/Ammann%E2%80%93Beenker_tiling
* https://en.wikipedia.org/wiki/Silver_ratio
* https://en.wikipedia.org/wiki/Ammann%E2%80%93Beenker_tiling
Hypotenuse of a 1x2 unit right triangle, to be precise. By Pythagoras, the square root of any sum of squares can be drawn trivially with a compass and a straight edge. So 2, 5, 7, 10, 13, 17, etc
I recommend anyone to try to work through a few compass-and-straightedge constructions. It can be really satisfying and meditative.
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.)
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.)
There's also game versions https://www.euclidea.xyz/
Wow thank you for sharing this project. Definitely going to buy a copy.
Does Gauss's headstone actually have a 17-pointed star on the back? I can't find any pictures of this online.
Gauss's tombstone actually has a circle[1]. Gauss wanted the heptadecagon (not stellated, a regular 17-gon) but the mason making the tombstone decided it was similar enough to a circle for it not to matter and doing a 17-gon was too hard so he just did a circle.
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.
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.
The headstone (https://de.wikipedia.org/wiki/Carl_Friedrich_Gau%C3%9F#/medi...) does not (it does feature the star of David, but I couldn't find any notion that Gauß was jewish).
But there's a statue with that star (https://de.wikipedia.org/wiki/Carl_Friedrich_Gau%C3%9F#/medi...)
But there's a statue with that star (https://de.wikipedia.org/wiki/Carl_Friedrich_Gau%C3%9F#/medi...)
Gauß wasn’t jewish and it was a Christian cemetery. The gravestone insinuates the form of a cross and the David star just symbolizes the Old Testament.
FYI Gauss's grave is in Albani cemetery, 37085 Göttingen, Germany. https://en.wikipedia.org/wiki/Albanifriedhof
But per below comments, the star is on the monument in Brunswick/Braunschweig, not on his headstone.
But per below comments, the star is on the monument in Brunswick/Braunschweig, not on his headstone.
The end of the article indicates it does not
Huh? No, the article says the stonemason chose a star because nobody would be able to tell that a 17-gon isn't a circle. I have heard this story repeated before in Gaussian biographies, but I'm surprised I can't find a single picture online that shows that this did or did not happen.
If they made the sides slightly outwardly concave, think an opened bottle cap, it would highlight the vertices and I think that would help people identify it as a polygon rather then a circle --I definitely could be wrong though.
The article says the 17-point star is on a monument to Gauss in his home town of Brunswick Germany, 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.
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.
Image showing the star on the statue https://commons.m.wikimedia.org/wiki/File:Braunschweig_Gauss...
To me, the result is exciting because it shows how algebra, over hundreds of years, came back round to improve Euclidian geometry. Without the background, I wouldn't even know why it was an interesting problem. The motivation is very similar to that of the Langlands program.
If you read most of the mathematics articles you will be forgiven to think that there is no contributions made by the Middle Ages mathematicians ever. For some unknown reasons the writers will always mention and not missing Greek mathematicians contributions in this case Euclid, and then conveniently and ignorantly skipped about a thousand years by going straight to the Renaissance mathematicians in this particular case Gauss, who's the main character of the article.
> you will be forgiven to think that there is no contributions made by the Middle Ages mathematicians ever
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.
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.
The thing is that’s not really true. Fibonacci is one you’ve heard of, but there are plenty of other great medieval European mathematicians, such as Oresme, de Nemore, Wallingford, and Bradwardine.
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.
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.
This is really interesting.. does anyone who knows more about Gauss's proof know why you can construct a 5 sided polygon with ruler and compass, but not a 7 or 11 sided polygon? Why do some primes work and others not?
If you are interested and have the time you can watch the 2 videos from the YouTube channel Another Roof[1] on that subject. He spends some time on easy stuff too to allow the general public to relatively understand the bases, so don't be surprised if the videos are quite long.
[1]: https://youtube.com/@anotherroof
[1]: https://youtube.com/@anotherroof
This is described in quite an approachable way in the video I posted in another comment. https://www.youtube.com/watch?v=EX7U0DGBmbM
It has to do with Fermat primes.
This was downvoted. I guess someone though it was a joke. But it's part of the correct answer. More details in https://en.wikipedia.org/wiki/Fermat_number
For 17, Gauss noticed that cos(360°/17) can be written only with elementary operations, see https://www.heise.de/imgs/18/2/1/2/3/3/6/4/siebzehneck-b95b5...
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)
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)
Very cool, thanks!
I can give a very fast and incomplete explanation, but you must trust me.
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.)
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.)
Interesting, so in general you could construct polygons with 2^n+1 sides?
Yes, but only if that number is a prime.
I agree. More details:
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.
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.
cyclotomic polynomials and Galois theory
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.
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.
Seven sided never seemed that problematic to me?
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.
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.
This actually reminds me of another claim that I think is wrong for the opposite reason;
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.
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.
You're right that the hilbert curve only visits certain points in the unit square, and never a (non-rational,non-rational) point. While the Wikipedia article doesn't seem to mention it, other sources like [1] mention that the definition of a space-filling curve is one that comes arbitrarily close to any point within its space. I think you would be able to see that the iteration of the hilbert curve does get arbitrarily close to (say) the point (sqrt(2)/2, sqrt(2)/2).
[1]: https://people.csail.mit.edu/jaffer/Geometry/PSFC
[1]: https://people.csail.mit.edu/jaffer/Geometry/PSFC
The Hilbert curve does contain every point in the unit square. It is a limit of curves, and so can contain points even not in the intermediate constructions. This is similar to how the limit of 1/x as x -> infinity can be 0, even though 1/x never equals 0.
Also, a curve which gets arbitrarily close to every point in the unit square actually touches every point in the unit square. This is because (by definition) a curve is a continuous map from a compact space (the unit interval) to a Hausdorff space (R^2), and so its image is compact, and thus closed. A closed set contains every point that it is arbitrarily close to.
If I travel one half of the distance from where I am to the finishing line an infinite number of times, I reach the finishing line but still never finish the race.
With a Hilbert curve the entire plane becomes a limit.
With a Hilbert curve the entire plane becomes a limit.
This doesn't seem to fly with the inductive fact that 1/2 of a power of two is always one over a power of two no matter how many times you perform the iteration.
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.
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.
Sure you can do that, but exact is exactly what they're after. If you allow infinite series then you can approximate anything using Taylor series.
heptagon is not "constructable", but it's easy to draw. I played around with this back in college.
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.
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.
As I said, it's approximable to arbitrary degrees of accuracy.
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.
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.
[deleted]
Ok hear me out - headstones are installed after someone dies. It’s still after his death, so the problem can still be corrected for future generations
Ok but could we make it a blurry heptadecagon?
[deleted]
Nice read and explains what Yitang Zhang did in a manner I can almost understand!
Everytime I hear about Yitang Zhang, I cannot help but be amazed by his accomplishment.
https://en.wikipedia.org/wiki/Yitang_Zhang
Everytime I hear about Yitang Zhang, I cannot help but be amazed by his accomplishment.
https://en.wikipedia.org/wiki/Yitang_Zhang
Zhang's work is not mentioned in this article, so I'm assuming you clicked through to this one:
https://www.scientificamerican.com/article/prime-number-puzz...
https://www.scientificamerican.com/article/prime-number-puzz...
Yes, there was a link about it in the article.
It's really weird. This article contains the sentence
> 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.)
> 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.)
djoldman(3)