37

It seems evident from infinitely many primitive pythagorean triples $(a,b,c)$ that there are infinitely many rational points $\left(\frac{a}{c}, \frac{b}{c}\right)$ on the unit circle.

But how would one go about, and show that they are dense, in the sense that for two rational points $x$ and $y$ of angles $α$ and $β$ on the unit circle, if $α<β$ there is a third rational point $z$ of angle $γ$ on the unit circle, such that $α<γ$ and $γ<β$.

Is it to expect that this conjecture holds and that it isn't an unsolved number theory problem?

projectilemotion
  • 15,914
  • 14
    For a different argument: $\mathbb{Q}(i) \cap S^1$ is an infinite subgroup of $S^1$, hence $\overline{\mathbb{Q}(i) \cap S^1}$ is an infinite closed subgroup of $S^1$. But the closed subgroups of $S^1$ are 1. $S^1$ itself, and 2. the finite subgroups. Conclusion, $\overline{\mathbb{Q}(i) \cap S^1} = S^1$. – Daniel Fischer Jan 14 '17 at 21:31
  • 1
    @DanielFischer Nice! Although to be fair, your second-to-last sentence does take a bit of proof. – Noah Schweber Jan 14 '17 at 21:43
  • @DanielFischer Nice! Is it easy to show that $\mathbb Q(i) \cap S^1$ is infinite? My first thought was to show that, say, $(3+4i)/5$ has no finite order. I'm not sure how an elementary argument along these lines would go; it's easy with some basic Galois theory. – Dustan Levenstein Jan 14 '17 at 21:44
  • 3
    @DustanLevenstein Find infinitely many Pythagorean triples which correspond to different rational points. – Noah Schweber Jan 14 '17 at 21:45
  • Oh, of course. :) – Dustan Levenstein Jan 14 '17 at 21:45
  • 2
    @DustanLevenstein: Alternately, see here for an elementary proof that $(3+4i)/5$ has no finite order. – Micah Jan 15 '17 at 21:57
  • Technically, you want $\mathbb{Q}(i)^*$. – Steve D Jan 20 '17 at 14:55
  • Maybe you should see 3blue1brown video about pythagorean triples, as it is directly (and surprisingly) related with your question: https://youtu.be/QJYmyhnaaek – Carlos Toscano-Ochoa Sep 17 '18 at 11:35

4 Answers4

46

Consider a nonvertical line through the point $(1, 0)$ of slope $m$. This line meets the circle at exactly one other point, and it's not hard to show that the coordinates of that point are $$P_m=\left({m^2-1\over m^2+1}, {-2m\over m^2+1}\right).$$ As long as $m$ is rational, $P_m$ has rational coordinates; so now think about the lines through $(1, 0)$ of rational slope . . .

ruakh
  • 703
Noah Schweber
  • 260,658
  • Can you show that P_m is on the unit circle? –  Jan 14 '17 at 21:03
  • @j4nbur53 Fixed a (godawful) typo. But yes - $(m^2-1)^2=m^4-2m^2+1$, and $(-2m)^2=4m^2$. So adding the squares of the numerators we get $(m^4+1)^2$. But that's exactly the square of the denominator. – Noah Schweber Jan 14 '17 at 21:05
  • 1
    I think I did it right. The slope is ${y_2-y_1\over x_2-x_1}$. Let $A=y_2-y_1, B=x_2-x_1$. $y_2={-2m\over m^2+1}$ and $y_1=0$, so $A={-2m\over m^2+1}$. $x_2={m^2-1\over m^2+1}$ and $x_1=1$, so $B={m^2-1\over m^2+1}-1={m^2-1\over m^2+1}-{m^2+1\over m^2+1}={-2\over m^2+1}$. Cancelling denominators, we get ${A\over B}={-2m\over -2}=m$. – Noah Schweber Jan 14 '17 at 21:12
  • My bad, was short cutting rational arithmetic. gr8! –  Jan 14 '17 at 21:13
  • 6
    Just for reference this is essentially an inverse stereographic projection. – Derek Elkins left SE Jan 15 '17 at 04:07
  • @DerekElkins And there is a nice illustration in the subsection you link; and we can think of rational points on a straight line, if that is easier to "see" than a bunch of rational slopes. – Jeppe Stig Nielsen Jan 15 '17 at 22:31
22

We are interested in angles $\theta$ such that $(\cos \theta, \sin \theta)$ are both rational. Call such an angle a $\mathbb Q-$ angle.

Claim: $\theta$ is a $\mathbb Q-$ angle iff $\tan \frac {\theta}2 \in \mathbb Q$.

Proof:

$\implies$ follows from the half angle formula, $\tan \frac {\theta}2 =\frac {1-\cos \theta}{\sin \theta}$

To see $\impliedby$ it may be easiest to work geometrically. Note that $\frac {\theta}2$ is the angle formed by the line connecting $(-1,0)$ to the point $(\cos \theta, \sin \theta)$. That line has equation $y=h(x+1)$ where $h=\tan \frac {\theta}2$. Solving this and $x^2+y^2=1$ simultaneously we see that we are trying to solve $$x^2+h^2(x^2+2x+1)-1=0\implies x^2+\frac {2h^2}{1+h^2}x+\frac {h^2-1}{1+h^2}=0$$ of course one root is given by $x=-1$ and it follows at once that the other root must also be rational. Thus $\cos \theta \in \mathbb Q$. Now we can use the half-angle formula again to see that this implies that $\sin \theta \in \mathbb Q$.

Note: this is the key point behind the Weierstrass substitution, also known as the $\tan \frac {\theta}2$ substitution.

It is clear that the angles $\theta$ such that $\tan \frac {\theta}2\in \mathbb Q$ are dense so we are done.

lulu
  • 76,951
  • The intermediate claim is interesting, but the last "it is clear" part might need some explanation. –  Jan 14 '17 at 21:12
  • 2
    @j4nbur53 The map $f:x\mapsto \tan{x\over 2}$ is continuous and surjective onto $\mathbb{R}$, so the $f$-preimage of any dense subset of $\mathbb{R}$ (e.g. $\mathbb{Q}$) is dense in the domain. – Noah Schweber Jan 14 '17 at 21:15
  • 4
    May be worth pointing out that, of course, my construction is exactly the same as that of @NoahSchweber . He projects from $(1,0)$ while I project from $(-1,0)$ but any rational point on the circle would do the job. I just added enough details to show that, in fact, every rational point on the circle arises this way. – lulu Jan 14 '17 at 21:23
  • Indeed (and incidentally I upvoted this answer and your comment). – Noah Schweber Jan 14 '17 at 21:32
  • @NoahSchweber Thanks! Reciprocated. – lulu Jan 14 '17 at 21:36
  • @j4nbur53 Well, you don't need much topology here. And since you want a statement about density you'll need some. All I really need is continuity....nothing too powerful. Alternatively, you can just use the fact that, in my construction, $\tan \frac {\theta}2$ is the $y$-intercept of the line I draw, so all I really need is the density of the rationals on the interval $[0,1]$. – lulu Jan 14 '17 at 21:39
  • Oh, for the most part he did. But don't forget: density wouldn't have occurred to him as a problem....density in what? There was an understanding that things like $\sqrt 2$ weren't rational, though they were constructible, and a suspicion that things like $\pi$ or $\sqrt[3] 2$ might be even worse. But there wasn't a very fleshed out Geometric notion behind such ideas. – lulu Jan 14 '17 at 21:52
  • What are you asking? Euclid certainly used this device, phrased differently, to classify all the Pythagorean triples. – lulu Jan 14 '17 at 22:05
  • I'm coming late to this, and the comments you are responding to are gone, but I find it hard to see this device in Euclid's derivation. It is clearly there in Diophantus's derivation, however. – Will Orrick Jan 15 '17 at 08:37
  • @WillOrrick Oh, you are likely right. My reasoning was based on Euclid's generating mechanism for Pythagorean triples...where did that come from if not from this? But, in truth, I don't know the history. Is there evidence that Euclid also handled rational points on ellipses? – lulu Jan 15 '17 at 12:05
  • The method Euclid gives in Book X of Elements starts with two lines (lengths), $x>y$. Proposition 6 in Book II relates two squares to a rectangle, and in modern, algebraic terms would be $$xy+\left[\frac{1}{2}(x-y)\right]^2=\left[\frac{1}{2}(x+y)\right]^2.$$ If $x$ and $y$ are positive integers of the same parity, then two terms in this expression are perfect squares. Euclid imposes the condition that the remaining term, $xy$, be a perfect square by requiring that $x$ and $y$ be similar plane numbers, that is, areas of similar rectangles with positive integer side lengths. So you... – Will Orrick Jan 16 '17 at 13:16
  • ...might let $x=p(rp)$ and $y=q(rq)$, where $p$, $rp$, $q$, and $rq$ are positive integers. Then you get the triple $$(rpq, \frac{r}{2}(p^2-q^2),\frac{r}{2}(p^2+q^2)).$$ This method is given in a Lemma that is stated in between Propositions 28 and 29 in Book X. Book X contains 115 propositions in total, and its main concern is the classification and relationships of various kinds of irrational numbers. It is believed that much of this material goes back to Theaetetus, who appears in one of Plato's dialogs, but little is known about the origins and motivations for its development. – Will Orrick Jan 16 '17 at 13:40
  • I am not aware of any evidence that Euclid looked at rational points on ellipses, but you certainly shouldn't take that as definitive. There was work, both prior to Euclid, and at the same time or slightly after, by Archimedes, relating to certain Pell equations, which implies knowledge of integer points on hyperbolas, but I have not looked at this, so I can't say what techniques were used or whether the Greeks viewed it in those terms. At any rate, in 300 BCE, the Greeks did not regard what we call rational numbers as numbers, which... – Will Orrick Jan 16 '17 at 13:55
  • ... makes it unlikely that they would have formulated the question as we do today. – Will Orrick Jan 16 '17 at 13:57
  • Diophantus, writing, it is believed, in about 250 CE, was explicitly concerned with rational solutions of equations, and he employed an algebraic notation, apparently of his own invention. The relevant calculation is Problem 8 in Book II of Arithmetica. Diophantus illustrates the method by dividing $4^2$ into two squares. He notes that $16-x^2$ is square and writes it as $(mx-4)^2$. Taking $m=2$ and equating the two gives $x=16/5$ and hence $4^2=(16/5)^2+(12/5)^2$. I am not sure whether Diophantus looked at rational points on ellipses. – Will Orrick Jan 16 '17 at 14:19
  • @WillOrrick Thanks for this. I'm intrigued, and will go read the old texts. Personally, I find it very difficult to orient myself in the classical way of thinking. But of course that's my weakness, not theirs. – lulu Jan 16 '17 at 16:24
2

If the two given points are in different quadrants, then one of the points $(0, \pm1), (\pm1, 0)$ may fairly be said to be between them. So wlog they are both in the first quadrant. Then for any rational point $P$ on the unit circle, there are $t, u\in \mathbb N$ with $t>u$, where $$P=P(t,u)=\left(\dfrac{2tu}{t^2+u^2},\dfrac{t^2-u^2}{t^2+u^2}\right).$$ If $t+u$ is odd and $t$ and $u$ are coprime, that even yields the fractions in their lowest terms.

So given $X=P(t_1,u_1)$ and $Y=P(t_2,u_2)$, $Z=P(t_1+t_2,u_1+u_2)$ is a rational point between $X$ and $Y$ as required.

For example,

$X=P(2,1)=(\frac45,\frac35)=(.8,.6), Y=P(4,3)=(\frac{24}{25},\frac7{25})=(.96,.28), Z=P(6,4)=(\frac{48}{52},\frac{20}{52})=(\frac{12}{13},\frac5{13})\approx(.92,.38).$

Rosie F
  • 3,231
  • It looks like a geometric interpretation of $P(t,u)$ might help clarify the claim that $P(t_1+t_2,u_1+u_2)$ "lies between" $P(t_1,u_1)$ and $P(t_2,u_2)$, given both the latter are in the first quadrant. – hardmath Jan 15 '17 at 13:38
  • From t1u2 < t2u1 it follows (t1+t2)u2 < t2(u1+u2) and t1(u1+u2) < (t1+t2)u1. Hence for m1=t1/t2, m2=t2/u2, m=(t1+t2)/(u1+u2) and m1<m2 it follows m1<m & m<m2. This means the same geometry as in the answer by Noah Schweber or lulu can be used. –  Jan 15 '17 at 16:39
1

It's super easy to prove. For any complex number on the unit circle, there exists a sequence of complex numbers with integer parts such that if you replace each term with itself divided by its magnitude, the new sequence is a sequence of points on the unit circle that approaches the square root of the chosen complex number. If you replace each term of the original sequence with its square, then every term in the sequence will have an integral magnitude. If after that, you replace each term of the new sequence with itself divided by its magnitude, the new sequence is a sequence of complex numbers on the unit circle with rational parts that approaches the chosen complex number.

Timothy
  • 860
  • I described a second operation to the sequence that replaces each number with that number divided by its magnitude creating a new sequence which is a sequence of numbers on the unit circle with rational parts. – Timothy Jan 15 '17 at 23:40
  • 1
    I fixed my answer to give a more direct proof that doesn't have any missing parts that are as hard to figure out what are. – Timothy Jan 16 '17 at 05:38
  • 1
    I believe I have given a legitimate proof and you just can't understand it. I know it's not a formal proof that completely proves it. – Timothy Jan 16 '17 at 23:44
  • @j4nbur53 This argument does indeed work, although it takes a couple steps to make clear. We want to show that, given $z$ on the unit circle there are rational points on the circle arbitrarily close to $z$. It would be enough, then, to show that there's a sequence of rational points on the circle that converge to $z$. Well, our first step is to pick a complex $w$ such that $w^2=z$ (think of $w$ as "halfway around" the circle to $z$). Now, we can find (exercise) a sequence $p_i$ of points in the plane with integer coordinates, such that ${p_i\over\vert p_i\vert}\rightarrow w$. (Cont'd) – Noah Schweber Jan 17 '17 at 01:29
  • Let $q_i={p_i\over\vert p_i\vert}$. Good news: $q_i$ is on the unit circle! Bad news: $q_i$ (probably) doesn't have rational coordinates! Moreover, the sequence $q_i$ converges to $w$, not to our desired $z$. But we can fix these two problems at once: let $v_i={(x_i^2, y_i^2)\over\vert p_i\vert ^2}$, where $p_i=(x_i, y_i)$. Since $x_i, y_i$ are integers, $\vert p_i\vert ^2$ is rational (indeed an integer), so $v_i$ has rational coordinates. Moreover, it's not hard to see that the $v_i$s are on the unit circle and approach $z$. So we're done. – Noah Schweber Jan 17 '17 at 01:32
  • The key here is the phrase "square root of" in the post; easy to miss (I did on my first reading), but that reveals the point: we're building a sequence of "coming-from-integer-points" points that converge to the square root of what we want; the squares of these points then converge to what we want, and moreover are rational. – Noah Schweber Jan 17 '17 at 01:33
  • There's no problem that my answer is the way it is because this question has another answer by Noah Schweber at the top that's really good. – Timothy Jan 17 '17 at 03:57