# A rope and Pi's irrationality

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?

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## 1 Answer

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.

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So multiplication of 2 irrationals can give an integer? – user13107 Sep 23 '12 at 8:59
Like $\sqrt{2}\sqrt{2}=2$ – Dennis Gulko Sep 23 '12 at 9:00
@user13107: Sure; I just added an example to my answer, and Dennis Gulko just gave another in a comment. – Brian M. Scott Sep 23 '12 at 9:01