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Given,

$$b\in \Bbb Q\implies \exists \;p,q\in \Bbb Z:\:b=\frac pq \wedge q \ne 0 $$

Any line passing through $(0,b)$ and $(1,0)$ such that $\forall \;b \in \Bbb Q$, intersect the unit circle at a rational point $(x,y)$ such that the slope of the line through $(0,b)$ and $(1,0)$ is $m=-\frac pq$. Since we want points on the unit circle we know $x^2+y^2=1$. So we want the rational points $(x,y)$ where boths graphs intersect, hence I tried to set both equations equal to each other $$y-mx-b=x^2+y^2-1$$ $$y+\frac pq \left(x-1\right)=x^2+y^2-1$$ I always run into counter example of irrational $x$ and $y$ having rational squares. Therefore, I wonder if my intuition is wrong.

So instead of trying to find the intersecting rational points on the unit circle and the line when the $y$-intercept is $b \in \Bbb Q$, I tried another approach. Basically, what I'm seeing is that since $x^2+y^2=1$ the hypotenuse $h$ of any triangle of sides $x$ and $y$ is equal to 1. The angle $\phi$ between the $x$-axis and the hypotenuse is $\tan^{-1}\left(\frac b1\right)=\tan^{-1}\left( \frac yx\right)$ from similar triangles. Does it follow that $$\tan\phi = b = \frac yx \in \Bbb Q \implies \exists\;r,s,t,u\in\Bbb Z : x=\frac rs \wedge \frac tu \wedge s,u \ne 0$$ is true? This would mean that $\cos \phi = x \in \Bbb Q:x = \frac rs$ and $\sin \phi = y \in \Bbb Q : y = \frac tu$ and I would claim that $$ (i) \quad \forall \; b \in\Bbb Q \implies \exists \; p,q \in \Bbb Z : q \ne 0 \wedge b = \frac pq = \tan \phi = \left( \frac yx \right)$$ $$(ii) \qquad \exists \; r,s,t,u, \in \Bbb Z : x = \frac rs \wedge y = \frac tu \wedge s,u \ne 0$$ $$\therefore \quad (i) \implies (ii)$$

Is this correct? Does this show that for the infinitely many points on the unit circle, those points $(x, y)$ intersected by a line passing through $(0, b\in\Bbb Q)$ and $(1, 0)$ are rational such that $\forall \;b \in \Bbb Q \implies \left(x = \frac rs, y = \frac tu \right)$? If not, am I on the wrong track?

2 Answers 2

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The easiest proof I can think of follows your intuition, via trigonometry: Let the line make an angle of $\theta$ with the $x$-axis. Then it intersects the circle in two points: $(1, 0)$ itself, and $(-\cos 2\theta, \sin 2\theta)$. We will show that both coordinates are rational. (Don't read any further if you don't want more than that hint—I couldn't tell if you don't want to be spoiled.)

enter image description here

First, we see that $\tan \theta = b$. Then

$$ \cos \theta = \frac{1}{\sqrt{1+b^2}} $$

and

$$ \sin \theta = \frac{b}{\sqrt{1+b^2}} $$

Since $\cos 2\theta = 2\cos^2 \theta-1$ and $\sin 2\theta = 2 \sin \theta \cos \theta$, we have

$$ x = -\cos 2\theta = -\frac{1-b^2}{1+b^2} $$

and

$$ y = \sin 2\theta = \frac{2b}{1+b^2} $$

The above image depicts $0 < b < 1$, though that needn't be the case.

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    For the second and third quadrant I understand how $x$, and $y$ are rational and dependent on $b$ since I make use of double angles formulas which involves squares, yet in the first and fourth quadrant I use $y= \sin \theta$ and $x= 1- \cos \theta$, how can I get rid of the radical $\sqrt {1+b^2}$ ? – 2017-01-31
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    How did you get $x = 1-\cos\theta$? – 2017-01-31
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    I used the same angle $\theta$ between the $x$-axis and the line passing through $(1, 0)$ – 2017-01-31
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    You should still obtain $x = \cos\theta$; it's just on the other side of the $y$-axis. – 2017-01-31
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    I thought that $x= \cos \theta$ would be the case of the angle between the $x$-axis and a line passing through the origin $(0, 0)$, since $x$ is the distance from the origin and I take the angle between the $x$-axis and the point $(1, 0)$ I would have subtracted the $\cos \theta$ from $1$ to get $x$ – 2017-01-31
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    How would $x =\cos \theta =\frac 1{\sqrt {1+b^2}}$ and $y = \sin \theta = \frac b{\sqrt {1+b^2}}$ imply that $x$ and $y$ are rational from knowing that $b \in \Bbb Q$? – 2017-01-31
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    It wouldn't, but I'm having trouble visualizing what it is you're describing. Let me read it more carefully when I have more time and see if I can see what you're seeing. – 2017-01-31
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    I got it! It was a silly mistake, I stopped using the angle $2\theta$ when $b \gt 1$ because from there $2\theta \gt \frac \pi2$, so indeed $x= \pm\cos 2\theta $ is correct. I think that $x=\pm\cos2\theta = 1-cos \theta$ is also correct its just that the double angle formula gets rid of the radical. – 2017-02-01
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For $b\ne 0:$ A point $(x,y)$ on the line satisfies $x+y/b=1,$ that is $$x=1-y/b.$$ A point $(x,y)$ on the circle satisfies $$|x|=\sqrt {1-y^2}.$$ So if both of these hold then $$(1-y/b)^2=x^2=1-y^2$$ which simplifies to $$-2y/b+y^2/b^2=-y^2.$$ Now this holds when $y=0$ iff $x=1-y/b=1-0/b=1$. For the other point, if $y\ne 0$ and $-2y/b+y^2/b^2=-y^2$ then $$-2/b+y/b^2=-y,$$ which is equivalent to $$y=2b/(b^2+1),$$ implying $$x=1-y/b=(b^2-1)/(b^2+1).$$ If $x,y$ are both rational it is necessary that $b$ is rational because $b\ne 0\implies y=2b/(b^2+1)\ne 0$ which implies (from $x+y/b=1$) that $1/b=(1-x)/y.$

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    If $A\ne 0\ne B$ the equation of the line thru $(A,0$) and $(0,B)$ is $x/A+y/B=1.$ – 2017-01-31