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.
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).
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.
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.
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.
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.
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.