Prove that $Q$ is rational if and only if $t$ is rational

Question

In the accompanying diagram from "The Language of Mathematics: Making the Invisible Visible" by Keith Devlin (Figure 6.21), he states that "It is then an easy exercise in algebra and geometry to verify that the point $Q$ will have rational coordinates if, and only if, the number $t$ is rational." Note that the circle is the unit circle, and so the coordinates of $P$ are $(-1,0)$.

He gives certain examples, such as if $t = 1/2$, "a little computation shows that the point $Q$ has coordinates $(3/5, 4/5)$. Similarly, $t = 2/3$, leads to the point $(5/13, 12/13)$."

I understand these coordinates yield Pythagorean Triplets, but how are they obtained algebraically (and/or geometrically)? and how is the general case he gives above—regarding $Q$ being rational if and only if $t$ is rational—proven?

Answer 1

The equation of the line is $$y=t(x+1)$$ and the circle's, $$x^2+y^2=1$$ The first equation implies that $x,y\in\Bbb Q\implies t\in\Bbb Q$.

Merging both of them we obtain $$x^2+t^2(x+1)^2=1$$ But if $t$ is rational, this is a quadratic equation on $x$ with rational coefficients and one rational root (namely, $-1$). So the other solution is rational, too. This implies that $t\in\Bbb Q\implies x,y\in\Bbb Q$

Answer 2

The line $PQ$ is given by $$ y=tx+t $$ Inserting this into the circle equation $$ x^2+y^2=1 $$ we get $$ x^2+(tx+t)^2=1\\ (t^2+1)x^2+2t2x+t^2-1=0\\ x^2+\frac{2t^2}{t^2+1}x+\frac{t^2-1}{t^2+1}=0 $$ This is the equation that gives the $x$-coordinates of $P$ and $Q$. However, we already know the $x$-coordinate of $P$. So Vieta's formulas (or polynomial division, or actually solving the quadratic) tells us that the $x$-coordinate of $Q$ is $\frac{1-t^2}{1+t^2}$. Using the equation for the line, that gives a $y$ coordinate of $$ t\cdot\frac{1-t^2}{1+t^2}+t=\frac{t-t^3+t+t^3}{1+t^2}\\ =\frac{2t}{1+t^2} $$ So, clearly, if $t$ is rational, the $x$ and $y$ coordinates of $Q$ are both rational.

On the other hand, if the $x$ and $y$ coordinates of $Q$ are both rational, then the slope of the line (using the coordinates of $P$ and $Q$ to calculate it) is rational. And this slope is $t$.

Answer 3

It follows from your post, but you didn't define it, that that is the unit canonical circle $\;x^2+y^2=1\;$ and thus $\;P(-1,0)\;$. Call $\;(0,t)\;$ the point where that cord intersects the $\;y\,-$ axis, that so by basic trigonometric we have that the slope of $\;PQ\;$ is just $\;t\;$ , and if we call $\;Q(a,b)\;$ the coordinates of $\;Q\;$ , then

$$t=m_{QP}=\frac b{a+1}\implies b=t(a+1)$$

and since $\;Q\;$ on the unit circle, we get:

$$1=a^2+b^2=a^2+a^2t^2+2at^2+t^2\implies(1+t^2)a^2+2t^2a+t^2-1=0$$

The discriminant of the above quadratic in $\;a\;$ is

$$\Delta=4t^4-4(t^2+1)(t^2-1)=4\implies a_{1,2}=\frac{-2t^2\pm2}{2(t^2+1)}=\begin{cases}-1\\{}\\-\cfrac{t^2-1}{t^2+1}\end{cases}$$

Now just prove that the above means $\;a\;$ (and thus also $\;b\;$) are rational iff $\;t\;$ is...

Answer 4

A high-level handwavy explanation of why this must be so:

The $x$-coordinates of the two intersection points between the line and the circle are solutions to a quadratic equation whose coefficients can be computed from the known points $(-1,0)$ and $(0,t)$ by purely rational operations.

Thus, the quadratic equation has rational coefficients. Such a quadratic equation has either both roots rational or both roots irrational. (This is clear from the quadratic formula).

Since $P$ has a rational $x$-coordinate, $Q$ therefore also has.

Similarly for the $y$ coordinates.

Answer 5

The function $f(x)=\dfrac{1-x}{1+x},\; x\neq -1$ is its own inverse.

The slope of the line $PQ$ is $t$ and also $\dfrac y{1+x}=\sqrt{\dfrac{1-x}{1+x}}$.

Then, it is clear that if $x,y$ are rational, then $t$ either is.

Also, $t^2 =f(x)$, and hence $f(t^2)=x$. Then, if $t$ is rational, $x$ and $y$ either are.