i.e. could we find a subset $X\subset \mathbb{Q}^2$ such that $\overline{X}=\mathbb{R}^2$ and that for any $x,y\in X$ the distance $|x-y|$ is an irrational number?


I'm considering the following assertion of which I'm not sure :



> 
> Given finite rational points $p\_1,p\_2,\dots,p\_n$ , and an open ball $D$ on the plane, there is a rational point $x\in D$ such that $|x-p\_i|\in \mathbb{R}\backslash \mathbb{Q}$ for $i=1,2,\dots,n$.
> 
> 
> 


But this assertion accounts to be the following seemingly number-theoretic problem:



> 
> Given $n$ pairs $(a\_i,b\_i) (i=1,2,\dots,n)$ of positive integers such that $a\_i^2+b\_i^2$ is not square of any integer. Could we find an integer $N\geq 2$ such that the integral pairs $(Na\_i+1,Nb\_i)$ still satisfy the previous property(i.e. $(Na\_i+1)^2+(Nb\_i)^2$ is not square of any integer).
> 
> 
> 


BTW the distribution of the Pythagorean triples might help.



Is there a dense planar rational point set within which the distance of any two points is an irrational number?
Yes.


Observation: if (x,y) is half-integral, then it is at irrational distance (square root of a number that is 2 mod 4) from all integer points.


Construction: Find a sequence $p\_1,p\_2,\dots$ of the rational points such that each point appears infinitely often, and let $G\_1$ be the integer grid. For each $i$ in sequence, replace $p\_i$ by the nearest half-integral point $q\_i$ of grid $G\_i$, and then subdivide the grid by two to produce a new grid $G\_{i+1}$ in which $q\_i$ is integral. Then the resulting sequence of points $q\_i$ gets arbitrarily close to every rational point, so it is dense, every point is rational (more strongly dyadic rational), and by the observation every point is at irrational distance from all previous points.


