Irrationality of $\pi$ and circumference to diameter ratio. [duplicate]

This question already has an answer here:

How is $\pi$ actually defined? If it is defined as the ratio of the circumference of a circle to its diameter then from this definition itself either of the circumference and diameter has to be irrational. The circumference and diameter both are finite but $\pi$ is irrational. An irrational number is a number with never ending numbers after decimal, till now no recurrence has been discovered.

Yet there are methods to represent them in a number line which means they have finite length. So, if we take a finite length and make a circle around it as diameter, does it mean circumference is irrational but finite? In our everyday life, do we always take rounded off values of circumference or diameter?

marked as duplicate by J. M. is a poor mathematician, Misha Lavrov, tilper, Lord Shark the Unknown, JMPMay 25 '17 at 0:09

• Yes, if you take a diameter which is an integer, say, the circumference must be irrational, and vice versa. You can also notice this with triangles: if you take a right angle triangle with the two orthogonal sides having length $1$, the length of the hypotenuse will be $\sqrt{2}$, which is also irrational. – user139388 Apr 17 '14 at 6:32
• Note that $\frac13$ also has "never ending numbers after decimal". This doesn't contradict the fact that $\frac13$ is finite. – Hagen von Eitzen Apr 17 '14 at 6:41
• Watch this. – Shaun Apr 17 '14 at 11:23
• "Finite" and "irrational" are not mutually exclusive. Having an infinite decimal expansion and being infinite are not the same thing. $\pi$ has infinitely many digits after the decimal point, but $\pi$ is still a finite number, just like the circumference and the diameter are finite numbers. What do you mean by "recurrence" when you say "no recurrence has been discovered"? – tilper May 24 '17 at 16:15
• Ah, didn't realize how insanely old this question is... – tilper May 24 '17 at 16:16

• A better way to write that sum would be $\sum_1^{\infty} \frac {4(-1)^r }{2r-1}$ – imulsion Apr 22 '15 at 20:29