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?
Best Answer
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 $4$.