The question is written like this:
Is it possible to find an infinite set of points in the plane, not all on the same straight line, such that the distance between EVERY pair of points is rational?
This would be so easy if these points could be on the same straight line, but I couldn't get any idea to solve the question above(not all points on the same straight line). I believe there must be a kind of concatenation between the points but I couldn't figure it out.
What I tried is totally mess. I tried to draw some triangles and to connect some points from one triangle to another, but in vain.
Note: I want to see a real example of such an infinite set of points in the plane that can be an answer for the question. A graph for these points would be helpful.
Best Answer
You can even find infinitely many such points on the unit circle: Let $\mathscr S$ be the set of all points on the unit circle such that $\tan \left(\frac {\theta}4\right)\in \mathbb Q$. If $(\cos(\alpha),\sin(\alpha))$ and $(\cos(\beta),\sin(\beta))$ are two points on the circle then a little geometry tells us that the distance between them is (the absolute value of) $$2 \sin \left(\frac {\alpha}2\right)\cos \left(\frac {\beta}2\right)-2 \sin \left(\frac {\beta}2\right)\cos \left(\frac {\alpha}2\right)$$ and, if the points are both in $\mathscr S$ then this is rational.
Details: The distance formula is an immediate consequence of the fact that, if two points on the circle have an angle $\phi$ between them, then the distance between them is (the absolute value of) $2\sin \frac {\phi}2$. For the rationality note that $$z=\tan \frac {\phi}2 \implies \cos \phi= \frac {1-z^2}{1+z^2} \quad \& \quad \sin \phi= \frac {2z}{1+z^2}$$
Note: Of course $\mathscr S$ is dense on the circle. So far as I am aware, it is unknown whether you can find such a set which is dense on the entire plane.