[HN Gopher] Why Gauss wanted a heptadecagon on his tombstone
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Why Gauss wanted a heptadecagon on his tombstone
Author : sohkamyung
Score : 188 points
Date : 2024-09-16 14:14 UTC (1 days ago)
(HTM) web link (www.scientificamerican.com)
(TXT) w3m dump (www.scientificamerican.com)
| jmclnx wrote:
| 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
| ColinWright wrote:
| 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...
| jmclnx wrote:
| Yes, there was a link about it in the article.
| empath75 wrote:
| 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
| extraduder_ire wrote:
| 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.
| sbussard wrote:
| 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
| kennywinker wrote:
| Ok but could we make it a blurry heptadecagon?
| zansara wrote:
| https://archive.is/a6pQb
| calini wrote:
| Are we sure it wasn't just a low poly cirlce? :)
| peter_d_sherman wrote:
| >"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
| lainga wrote:
| What privileges the square root over any other fractional
| power?
| xrisk wrote:
| Presumably the hypotenuse of a right angled triangle.
| gilleain wrote:
| 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
| hinkley wrote:
| 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
| xyzzyz wrote:
| 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.
| jordigh wrote:
| Does Gauss's headstone actually have a 17-pointed star on the
| back? I can't find any pictures of this online.
| sameoldtune wrote:
| The end of the article indicates it does not
| jordigh wrote:
| 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.
| yardshop wrote:
| 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.
| mungoman2 wrote:
| Image showing the star on the statue https://commons.m.wi
| kimedia.org/wiki/File:Braunschweig_Gauss...
| mc32 wrote:
| 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.
| plg94 wrote:
| 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/w
| iki/Carl_Friedrich_Gau%C3%9F#/medi...)
| zengid wrote:
| 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?
| lanternfish wrote:
| It has to do with Fermat primes.
| gus_massa wrote:
| 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
| plg94 wrote:
| For 17, Gauss noticed that cos(360deg/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)
| zengid wrote:
| Very cool, thanks!
| gus_massa wrote:
| 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.)
| paulpauper wrote:
| 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.
| yarg wrote:
| 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.
| dunham wrote:
| 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(p/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.
| yarg wrote:
| 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.
| yarg wrote:
| 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.
| jepler wrote:
| 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
| cooljoseph wrote:
| 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.
| CorrectHorseBat wrote:
| 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.
| wwarner wrote:
| 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.
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