Proving the Existence of Rational Points on a Circle

[jing]

Techno - I agree with Matt Grime Gonzo never changed his mind about what he wanted proof of. However there was a lack of clarity (for me) about the statement.

For any positive integer k, you can find k points on a circle such that each point is a rational distance from every other point.

I took this to mean

For any positive integer k, you can find k points on any circle such that each point is a rational distance from every other point.

There followed a discussion with Gonzo about unit circles and such that led to an agreed restatement

So is the question now

'given k, a positive integer, is it possible to find a circle with k points on the circumference of the circle such that the distance between any two of these points is rational?' ?

which meant that I should have read the OP as

For any positive integer k, you can find k points on some circle such that each point is a rational distance from every other point.


Reading posts in general on the site has led me to conclude that:

Communicating in posts can be difficult as we know what we mean and so expect others to know exactly what we mean when we write the posts but often do not write with sufficient clarity to put over our exact meaning.

So extra care must be taken to read all posts carefully, to ask for clarification when needed and to give clarification when required.

Gonzo - I am sure everyone's posts have been helpful in clarifying your thoughts and in finding the proof you required, which I am pleased you have done.

 

[gonzo]

Thanks jing. But like I said, I can't take credit for it. Wish I could since it was really obvious and simple when I saw it. But I can't.

 

[tehno]

The proof I have for the unit circle (and it's not my proof, I can't take credit for it) doesn't directly extend to circles with non-perfect square radii. However, there is another proof I've seen that might.

Why do you ask?

Probably becouse I came up with solution for all circles.
OTOH,I'm not satisfied with it.
I'm almost certain there must be more elegant ,shorter,and
elementar proof of it.
It would be intereting to see other proof for c=1 case you saw.
Can you post it?Maybe it can be extended to the all cases.
Here I'll present an "ugly" proof.
Firstly ,about terminology and some definitions I will use.

For collection of points $C \subset \mathbb{R}^2$ I say to be at r-distance if the distance between every pair of points is rational.Let's call such collection of points
a rational distance set.Note that rationals form a rational
distance subset of the reals.Particularly,if $C$ is a line
in $\mathbb{R}^2$,we say $C$ is a dense
set of points at r-distance.
Prior to the proof two known lemmas I will reffer to (without proof).

Lm1.

A triple (x,y,z) of naturals is a primitive pythagorean triple
if $x^2+y^2=z^2$ holds and if x,y,z are coprime.
All primitive pythagorean triplets where y is even are given by:
$x=m^2-n^2,y=2mn,z=m^2+n^2;m>n$
m,n are relatively prime and one of them is even.

Lm2.

Affine-rational solutions to $\alpha^2 =\beta^2 + 1$

are parameterized by:

$$\alpha=\frac{m^2+1}{2m},\beta=\frac{m^2-1}{2m};m\in\mathbb{Q}$$
.

_____________________________________________________________
With terminology given above your proposition can be formulated now in this way:

Any circle contains a dense set of points at r-distance.
_____________________________________________________________
Proof:

Let's identify the complex plane,as is often done,with $\mathbb{R}^2$.
Two points in the complex plane $z_{1},z_{2}$ we can always choose that
they are at r-distance and have rational lenght.Then,becouse of the identity:
$$\frac{||z_{1}-z_{2}||}{||z_{1}||\cdot ||z_{2}||}=||\frac{1}{z_{1}}-\frac{1}{z_{2}}||$$

Points [itex]1/z_{1}[/tex] and $1/z_{2}$ are at r-distance too.

Consider a vertical line I in the complex plane.
By Lm1 & Lm2 ascertain that we can always parameterize all points on I whose
lenghts and imaginary parts are rational.This set of points is dense in I.
Furthery ,this set points is mapped, as long as I isn't imaginary axis,
by the complex map$f(z)=z^{-1}$ to a dense ,rational distance subset of a
circle.Since I can be translated at will through all vertical lines,we can obtain
circless of all radii.
QED

That's a cumbersome proof of something geometrically obvious (to me).
Ugly and quite probably overkill to the problem.
But it's a proof.
[EDIT: That I can be parameterized in described manner make sure like this
Vertical line is x=a.We seek points (a,y) on it such that y is rational and also
(a^2+y^2)^(1/2}=z is rational.We get pythagorean equation a^2+y^2=z^2 or 1+(y/a)^2=(z/a)^2.By Lm2 we have y= a[(m^2-1)/(2m)],for $m\in\mathbb{Q}$ .The parametrization occurs ]
If need be I'll post the proof for lemma 2.It's not long.
But that's is only thing what I'm willing to do more ( lack of time and nerves for a chit-chat on this subject).

but that is the only question i asked, and one you utterly failed to answer until now.

Ok.How about posting your solution instead?

Communicating in posts can be difficult as we know what we mean and so expect others to know exactly what we mean when we write the posts but often do not write with sufficient clarity to put over our exact meaning.

So extra care must be taken to read all posts carefully, to ask for clarification when needed and to give clarification when required.

Well said.
Quite sometimes,I read to quickly.Somewhere around half of this thread I saw the title
"Rational points on the circle" .Didn't care before that about rational points,and I don't know why I thought he was interested in both distances and points being rational.
he wanted both.

 

[gonzo]

I already posted the basis of my proof, the rest is trivial.

Pick two rational points (points with rational coordinates) on the unit circle:

$$e^{i \theta_1}$$ and $$e^{i \theta_2}$$

Then it is really easy to show that the distance between the points with double the angles $$e^{2i \theta_1}$$ and $$e^{2i \theta_2}$$ is rational.

Since there are infinite rational points, you can always find what you need.

 

[matt grime]

l
Ok.How about posting your solution instead?

did I say I had one? I merely have very strong objections to your misuse of mathematics (M-W theorems, and invoking elliptic curves: x^2+y^2=-1 is a perfectly good conic (over C - birational, isomorphic as a curve over C, in fact, to x^2+y^2=1) and has precisely one Q-rational point contradicting your previous assertions, for example).

 

[tehno]

I pleadge gulity as charged for invoking elliptic curves (and for misreading problem conditions as the thread developed as I already said).
I reffered to elliptics only in comparative purposes to conics,but that was completely unnecessary complication to the problem I agree. I accept your criticism as concerns that matter.
I never said x²+y²=-1 has infinitely many rational points. x²+y²=1 in Cartesian plane has that property.

 

[matt grime]

I never said x²+y²=-1 has infinitely many rational points. x²+y²=1 in Cartesian plane has that property.

I meant x^2+y^2=0, not -1, a degnerate case - but what differentiates it from x^2-y^2=0? Or for that matter, what about x^4=3y^4? That has 1 Q-rational point, and is a quadratic, so what's special about conics? Or x^2-1=sqrt(3)y^2? Another curve with only one Q-rational point, although it isn't defined over Q.