2
$\begingroup$

Show that there are countably many circles with three rational points.

My interpretation of this question is to prove that there are infinitely countable circles that contain at least three rational points (points x/y where x,y are integers). So far I think one must show that the set of rationals Q has a bijection with the naturals, then a rational point ( QxQ) has a bijection with the naturals, and finally that three rational points (QxQ)^3 has a bijection with the naturals. The bijections will prove that such sets are countable. Note that points cannot be collinear.

TIA

  • 0
    There's one point missing there - what if *more than one* circle passes through the same three rational points? – 2017-02-14
  • 2
    would it not be the same circle then? – 2017-02-14
  • 0
    Yes, it would - but that needs to be stated somewhere in the proof. – 2017-02-15
  • 0
    just as the answer below does i'm asumming? showing C1=C2 if they have the same three rational points – 2017-02-15
  • 0
    what do you mean by "(points x/y where x,y are integers)" ? are you referring to integer points , e.g. (1,2), or rational points, e.g. (1, 2/3) ? – 2017-02-15
  • 0
    rational points (x/y,p/q) where x,y,p,q are integers – 2017-02-15

1 Answers 1

0

For every circle $C$ in the set of circles $S$ such that $C$ contains three rational points, there exists a triplet of rational numbers associated with $C$. That is, there is a map $f:S \to Q^3$.

Now this map must be one-to-one. Suppose that there are two circles $C_1$ and $C_2$ such that $f(C_1)=f(C_2)$. Then they share three points. But three points determine a circle. Hence, $C_1=C_2$.

So the image of $f$, $f(S)$, must be at most countable because it is a subset of a countable set. But it is easy to show there are a countable number of elements in $S$. So $S$ is not finite and exactly countable.