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Can you produce an example where both the area of a circle and it's radius are integers?

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No, if that were true then $r^2$ would be an integer and then $\pi=\frac{A}{r^2}$ would be rational, which it isn't. – Tom Oldfield May 18 '13 at 18:34
If r = 0 because then A = 0. But that's hardly an example, is it? Otherwise Tom answered it. – imranfat May 18 '13 at 18:36
@Tom: That's an answer, not a comment. – Asaf Karagila May 18 '13 at 18:54
@Asaf I suppose, I wrote it up as one at first but I didn't feel like I had put enough work into it to make it an answer! I was sure that someone else would point out the exact same thing (and I hate it when some questions receive many almost identical answers). If no-one did I would have changed it into an answer at some point for completeness. – Tom Oldfield May 18 '13 at 21:00
there is $1019514486099146/324521540032945$ or $20,000,000/2325^2$ but it isn't pi. – user52413 Aug 23 '13 at 13:07
up vote 3 down vote accepted

$$r\neq 0\;,\;\pi r^2\;,\;r\in\Bbb N\implies \pi\in\Bbb Q\;,\;\text{which is false}$$

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OK, then is it correct to say that at least one among A and r has to be irrational. – kBisla May 18 '13 at 18:54
Yes, it is correct. – DonAntonio May 18 '13 at 18:56

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