TITLE: A rope and Pi's irrationality
QUESTION [5 upvotes]: Here is a question which has been puzzling me for some time.
You have a thin rope of an integer length $L$. You can bend it to create a rectangle of perimeter $L$. Fine so far.
Next, through some mechanism, you create a circle with the rope. This circle has a perimeter of $L$. 
But then $L = 2\pi r$.
This suggests that $L$, which is a multiple of $\pi$, must be irrational too.
What am I missing?

REPLY [6 votes]: You’re missing the fact that $r$ can (and in this case must) be irrational. For example, if $L=2$, then $r=\frac1{\pi}$, an irrational number. Rational multiples of $\pi$ are guaranteed to be irrational; irrational multiples are not.