Is there a rational point in a given open set such that the distance from given rational points to it are all irrational numbers on $\mathbb{R}^2$? On $\mathbb{R}^2$ there is a nonempty open set $A$ and $n$ rational points $a_1,\ldots,a_n$. Is there a rational point $a$ in $A$ such that $\forall i\in\{1,\ldots,n\}, |a−a_i|$ is an irrational number ?
In more detail, rational point $a_1,...,a_n$ and nonempty open set $A$ are given in advance. We don't know exactly where they are. We only know $a_1,...,a_n$ are rational points and $A$ is an nonempty open set. My question is whether there must be a rational point $a$ in $A$ such that all $|a−a_i|$ are irrational numbers.
My idea is to find a measuring method, in which the "length" of all rational points in unit circle is 0 with total-set being Q in unit circle, and then we can solve this question in a way similar to measure theory.
PS: This question is similar to another question I raised, but actually they are different.
 A: The open set contains some (closed) line segment from $(x_0,y)$ to $(x_1,y)$, whose extension hits none of the points $\{a_1,\dots,a_n\}$. Scale and translate so this segment becomes the segment from $(0,0)$ to $(1,0)$.
So, we now have $n$ rational points $(b_1,c_1),\dots,(b_n,c_n)$, and it suffices to find some rational $0\leq x\leq 1$ for which
$$\sqrt{(b_i-x)^2+c_i^2}$$
is irrational for each $1\leq i\leq n$. Select an integer $N\geq n$, and consider $x=k/N$ for $0\leq k\leq n$. If the problem condition doesn't hold, then for each such $k$, there is some $i$ for which $(b_i-k/N)^2+c_i^2$ is the square of a rational number. As there are $n$ possible choices of $i$ and $n+1$ possible choices of $k$, there are some distinct $j<k$ for which
$$(b_i-j/N)^2+c_i^2\text{ and }(b_i-k/N)^2+c_i^2$$
are both squares for the same fixed $i$. Since there are infinitely many possible choices of $N$, there must exist some triple $(j,k,i)$ for which
$$(b_i-j/N)^2+c_i^2\text{ and }(b_i-k/N)^2+c_i^2$$
are both squares for infinitely many positive integers $N>n$. Letting $M$ be the least common denominator of $b_i$ and $c_i$, these squares are each integer multiples of $\frac1{MN}$. In particular,
$$N\left(\sqrt{(b_i-j/N)^2+c_i^2}-\sqrt{(b_i-k/N)^2+c_i^2}\right)$$
is an integer multiple of $1/M$ for infinitely many positive integers $N$, which we'll call $N_1<N_2<\dots$. If these differences corresponding to $N_\ell$ are $x_\ell$, then $(x_1,x_2,\dots)$ is a Cauchy sequence inside $\frac1M\mathbb Z$, and so must be eventually constant. So, there must exist some $\delta$ for which
$$\sqrt{(Nb_i-j)^2+N^2c_i^2}-\sqrt{(Nb_i-k)^2+N^2c_i^2}=\delta$$
for infinitely many positive integers $N$. This implies, using $b=b_i$ and $c=c_i$ for notational simplicity,
\begin{align*}
\sqrt{(Nb-j)^2+N^2c^2}&=\delta+\sqrt{(Nb-k)^2+N^2c^2}\\
(Nb-j)^2+N^2c^2&=\delta^2+2\delta\sqrt{(Nb-k)^2+N^2c^2}+(Nb-k)^2+N^2c^2\\
(k-j)(2Nb-j-k)&=\delta^2+2\delta\sqrt{(Nb-k)^2+N^2c^2}.
\end{align*}
Since $k\neq j$ and $2Nb\neq j+k$ for large $N$, $\delta\neq 0$. So, $\sqrt{(Nb-k)^2+N^2c^2}$ equals $xN-y$ for some fixed $x$ and $y$, and infinitely many $N$. Then
$$N^2(b^2+c^2)-2Nbk+k^2=(Nb-k)^2+N^2c^2=x^2N^2-2xyN+y^2,$$
for infinitely many $N$; this means that the coefficients of the above quadratics in $N$ must be equal. So, $x^2=b^2+c^2$, $y^2=k^2$, and $2xy=2bk$. However, this implies that
$$c^2k^2=x^2k^2-b^2k^2=x^2y^2-x^2y^2=0;$$
so $x^2=b^2+c^2$, $2xy=-2bk$, and $y^2=k^2$. Since $k>j\geq 0$, $k$ is nonzero, which means that $c$ is $0$, and so $(b_i,c_i)$ is on the line $y=0$. However, this contradicts the choosing of our initial segment, since we chose it such that its extension hits none of the $n$ original points. This finishes the proof.
A: Here is another approach:
First, we claim that there is some lattice point $b$ and integer $M$ (relatively prime to each of the denominators of the $a_i$'s) such that $|a-a_i|$ is not a quadratic residue modulo $M$ for each $i$. To see this, just take $p_1,p_2, \ldots, p_n$ to be distinct primes relatively prime to the denominators of the $a_i$'s. Then you can pick points $b_i$ such that $|b_i-a_i|$ are non-quadratic residues (NQRs) modulo $p_i$, and then you can take $M = p_1\cdots p_n$ and $b\equiv b_i\pmod{p_i}$ for each $i$ by the Chinese Remainder Theorem.
For each $i$ $|b-a_i|\equiv |b_i-a_i|$ is an NQR mod $p_i$, so $|b-a_i|$ is an NQR modulo $M$. Then, you can just take $a\in A$ such that $a$ and $b$'s coordinates are the same modulo $M$.
(Note here that for a fraction $c/d$ and modulo $m$ relatively prime to $m$, $c/d\equiv cd^{-1}\pmod m$)
