Why PI is an irrational number? [duplicate]

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, msmMay 24 '17 at 7:50

• 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. – Arthur May 24 '17 at 7:38
• 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 $x^2-2$). – Bernard May 24 '17 at 7:51