Show that a quarter-circle always contains a point $(x,y)$ with rational $x$ and irrational $y$

Question

Let $r\in\mathbb{R}^{+}.$ Let $\mathbb{D}_r=(0,r)\cap\mathbb{Q}.$ That is, $\mathbb{D}_r$ is the set of all rationals strictly greater than $0,$ and strictly less than $r.$ Let $f_r:\mathbb{D}_r\to\mathbb{R}$ be a function given by $f_r(x)=\sqrt{r^2-x^2}.$ Show that the range of $f$ contains atleast one irrational number.

Case $1:$

Let $r\in\mathbb{Q}.$ Then, $\frac{r}{2}\in\mathbb{Q},$ and since $0<\frac{r}{2}<r,$ we have that $\frac{r}{2}\in\mathbb{D}_r.$ Moreover, $f_r\left(\frac{r}{2}\right)=\frac{r\sqrt{3}}{2}.$ This must be irrational as $\sqrt{3}$ is irrational.

Case $2:$

Let $r\in\mathbb{R}\setminus\mathbb{Q},$ and let $r^2\in\mathbb{R}\setminus\mathbb{Q}$ as well. Let $a\in\mathbb{D}_r.$ Then, $f_r(a)^2+a^2=r^2.$ Let $f_r(a)$ be rational. Then, $r^2,$ as a sum of two rational is also rational. This contradicts the irrationality of $r^2.$ Hence, $f_r(a)$ must be irrational.

Case $3:$

This is the case where $r$ is irrational but $r^2$ is rational. How to deal with this case?

Answer 1

Suppose for a contradiction that all numbers in the range of $f$ are rational. Pick any $x\in\mathbb D_r$. Then $r^2-x^2$ is rational, so $r^2$ is rational too. Then there exist natural $p$ and $q$ such that $r^2=\tfrac pq$. Then $\frac 1{2q}\in\mathbb D_r$. Thus $r^2-\left(\frac 1{2q}\right)^2$ is a square of a rational number. Then $4pq-1$ is a square of a natural number, that is impossible because each square of a natural number is $0$ or $1$ modulo $4$.

Answer 2

This answer deals with case 3 where $r$ is irrational but $r^2$ is rational $\frac qp$ where $p,q$ are positive integers satisfying $\gcd(p,q)=1$.

• If $q$ is odd, then taking an even integer $a$ larger than $r$, we get $$f_r\bigg(\frac{r^2}{a}\bigg)=\sqrt{\frac qp--\frac{q^2}{a^2p^2}}=\frac{1}{ap}\sqrt{a^2pq-q^2}$$ which is irrational since $a^2pq-q^2\equiv 3\pmod 4$.

• If $q$ is even of the form $2^st$ where $s,t$ are positive integers with $t\equiv 1\pmod 2$, then taking $a$ of the form $2^{s+1}u$ larger than $r$ where $u\equiv 1\pmod 2$, we get $$\begin{align}f_r\bigg(\frac{r^2}{a}\bigg)&=\sqrt{\frac{a^2pq-q^2}{a^2p^2}} \\\\&=\sqrt{\frac{2^{2s+2}u^2p\cdot 2^st-2^{2s}t^2}{2^{2s+2}u^2p^2}} \\\\&=\sqrt{\frac{2^{s+2}u^2pt-t^2}{4u^2p^2}}\\\\&=\frac{1}{2up}\sqrt{2^{s+2}u^2pt-t^2}\end{align}$$ which is irrational since $2^{s+2}u^2pt-t^2\equiv 3\pmod 4$.