Can I prescribe a V shape to every number on the real number line so that the point of the V is in contact with this number, and no two V shapes intersect? (You may choose any specifically shaped "V" for each number, change the width of the V's, change the length of the legs of the V's, place V's in in the upper half-plane or lower half-plane)

Here's a picture describing the situation

  • $\begingroup$ Is it allowed to change the length of the $V$? $\endgroup$ – Guacho Perez Dec 23 '18 at 22:48
  • $\begingroup$ @GuachoPerez Doesn't matter - it's impossible either way. $\endgroup$ – Patrick Stevens Dec 23 '18 at 22:49
  • $\begingroup$ @GuachoPerez You may change the length of the the V provided that you don't shrink the V to a single point. $\endgroup$ – Drewrl3v Dec 23 '18 at 23:47
  • $\begingroup$ @RobArthan The "V shapes" are pairs of line segments $\endgroup$ – Drewrl3v Dec 24 '18 at 0:33

I'm afraid not. Every V shape (when filled in) contains a rational point, and there are only countably many of those, but there are uncountably many reals, so at least one point is shared between two filled-in V's. Now show that those two V's intersect at an edge.

| cite | improve this answer | |
  • 1
    $\begingroup$ @RobArthan I took it that Patrick Stevens took the problem in $\mathbb{R}^2$, and then noted that $\mathbb{Q}^2$ is dense in $\mathbb{R}^2$. So if the $V$'s were instead solid triangles (this doesn't change the problem) then each triangle will contain a point in $\mathbb{Q}^2$ (a rational point). $\endgroup$ – Drewrl3v Dec 24 '18 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.