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How can $\pi$ be an irrational number if it is a ratio of the circumference over the diameter?



marked as duplicate by Cameron Buie, Andrés E. Caicedo, hardmath, Asaf Karagila, Norbert Nov 5 '13 at 17:26

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    $\begingroup$ Circumference and diameter are not both integers. $\endgroup$ – Daniel Fischer Nov 5 '13 at 17:03
  • $\begingroup$ Please write down the question normally...and why being a ratio would be an obstacle for $\;\pi\;$ to be irrational? $\endgroup$ – DonAntonio Nov 5 '13 at 17:04
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    $\begingroup$ Irrational = Not the ratio of two integers. $\endgroup$ – Asaf Karagila Nov 5 '13 at 17:18

I can write any real number $\alpha$ as a ratio: $\frac\alpha1.$

What makes a number rational is when can be written as a ratio of integers (with the denominator non-zero).

See this comic for all that needs to be said on the subject.

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    $\begingroup$ I don't understand the downvote. This answers the question clearly and concisely. $\endgroup$ – TonyK Nov 5 '13 at 17:27

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