This question already has an answer here:

Since my understanding of rational number says that it can expressed as fraction of 2 integers.

$Q = \frac{p}{q}$ where, $p,q$ belongs I

What I don't understand is how does PI fail to qualify these conditions?

Since, $\pi$ is an angle and

$\theta = \frac{circumference}{radius}$

So does $\pi$ could be expressed in same terms and we can always make a circle with integral circumference and radius and find the value of $\pi$.

Thus, $\pi$ should be a rational number. What am I missing?


marked as duplicate by J. M. is a poor mathematician, Arthur, drhab, projectilemotion, msm May 24 '17 at 7:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ Can you make a circle with both circumference and diameter simultaneously of integer measure? That's a rather big assumption on your part. $\endgroup$ – Deepak May 24 '17 at 7:28
  • 2
    $\begingroup$ @MIB Why "must there be"? Otherwise the universe will implode, People will die...or what? $\endgroup$ – DonAntonio May 24 '17 at 7:31
  • 1
    $\begingroup$ A rather easier example is this: $\sqrt 2$ is equal to $\frac{\text{hypotenuse}}{\text{leg}}$ in a right, isosceles triangle, yet it is well-known to be irrational. That means precicely that there is no right, isosceles triangle with both legs and hypotenuse lengths integers. You can try as much as you want, as large a triangle as you want, and you will never make it work. $\endgroup$ – Arthur May 24 '17 at 7:38
  • 1
    $\begingroup$ @MIB And what if the number of decimal numbers is infinite? $\endgroup$ – drhab May 24 '17 at 7:40
  • 2
    $\begingroup$ Actually, we know: more: a consequence of the Hermite-Lindemann theorem (1882) is that π is trancendent, i.e. it is not a root of any polynomial with integer coefficients (whereas, for instance, $\sqrt 2$ is a root of the polynomial $x^2-2$). $\endgroup$ – Bernard May 24 '17 at 7:51