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I've a few questions in mind:

  • Why is $\pi$ irrational?
  • If it is, then how can 2 rational quantities (circumference, diameter) can produce irrational number?
  • How are we able to determine digits of $\pi$ accurately?
Zev Chonoles
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mehulmpt
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    The diameter of the unit circle is indeed rational. Now, what makes you think the circumference is also rational? – guest Sep 12 '15 at 19:00
  • Because if you cut the circle (theoretically) and stretch it out, it becomes a line, which is indeed rational. (I think) – mehulmpt Sep 12 '15 at 19:03
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    @MehulMohan: If you agree that there do exist irrational numbers, then obviously you can have lines whose length is an irrational number. – Zev Chonoles Sep 12 '15 at 19:04
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    There you go : https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational – krirkrirk Sep 12 '15 at 19:04
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    See Approximations of $\pi$ – Mauro ALLEGRANZA Sep 12 '15 at 19:05
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    The diagonal of the unit square is already a straight line, no stretching required, and has irrational length by the Pythagorean theorem. – guest Sep 12 '15 at 19:06
  • @ZevChonoles Okay I do agree that irrational lengths exists. But for example, if I take a square with side 4cm and stretch it theoretically to make it a circle, what would happen? – mehulmpt Sep 12 '15 at 19:08
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    If you think the diameter and circumference of a circle are "two rational quantities" then I start to wonder if the meaning of "rational number" is clear in your mind. As to your third question, see this: http://math.stackexchange.com/questions/297/simple-numerical-methods-for-calculating-the-digits-of-pi ${}\qquad{}$ – Michael Hardy Sep 12 '15 at 19:09
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    @MehulMohan : Then you get a circle whose circumference is $16$ centimeters and whose diameter is $16/\pi$ centimeters. The latter is an irrational number of centimeters. ${}\qquad{}$ – Michael Hardy Sep 12 '15 at 19:10
  • @MehulMohan if a circle has circumference $C$ and diameter $D$, then we define $\pi := \frac{C}{D}$. Since it is known that $\pi$ is irrational (see the list of proofs posted above), we conclude that $C$ and $D$ cannot both be rational. – 727 Sep 12 '15 at 19:14
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    So you all are saying that technically, I can never split a circle's circumference into equal parts with each part equal to radius of that circle? Or more precisely if I start with any point on circumference and then jump the distance equal to radius, on circumference, I'd never come back to my original position? Interesting! – mehulmpt Sep 12 '15 at 19:17
  • The problem is not that ... the issue is that, whatever the lenght ot your jump is, provided that is a finite fraction of the radius, you will never succeed. – Mauro ALLEGRANZA Sep 12 '15 at 19:27
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    @MauroALLEGRANZA: the circumference is $\mathbf{2}\pi r$ so you need $7$ jumps to get past the starting point. – Rob Arthan Sep 12 '15 at 19:28
  • 2 rational numbers CANT produce a irrational number. I found a good proof of the irrationality here. http://worldpifederation.org/Fed/irrational.pdf – Ole Petersen Oct 30 '18 at 16:23
  • Part 2 of the question was asked earlier here: Why is π irrational if it is represented as C/D ? – David K Aug 03 '21 at 02:03

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The name "irrational number" has an ancient source...

It was an (implicit) assumption common to all "archaic" Greek mathematics that given two magnitudes, e.g. to segments of lenght $a$ and $b$, it is always possible to find a segment of "unit lenght" $u$ such that it measure both, i.e. such that [using modern algebraic formulae which are totally foreign to Greek math] :

$a = n \times u$ and $b = m \times u$, for $n,m \in \mathbb N$.

From the above instance of the assumption, it follows that :

$a/b = (n \times u) / (m \times u) = n/m$.

The assumption amounts to saying that the ratio between two magnitudes is always a ratio between numbers (i.e. in modern terms : a rational number; but note that for Greek math the only numbers are the natural ones and they are distinguished from magnitudes : a segment, a square, ... which are "measured" by numbers).

The discovery of the existence of irrational magnitudes, through the proof that the case where $a=1$ is the side of the unit square and $b=\sqrt 2$ is the diagonal is not expressible as a ratio between (natural) numbers, leads Greek math to the withdrawal of the "commensurability assumption" and to the axiomatization of geometry.

Those couples of magnitudes was called "incommensurable" (i.e. without common measure).

For the same reason, $\sqrt 2$ is an irrational number, exactly because the ratio "diagonal/side" is not expressible as a ratio between natural numbers.

The irrationality of $\pi$ was proved by Johann Heinrich Lambert in 1761.

Mauro ALLEGRANZA
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