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Since my understanding of rational number says that it can expressed as fraction of 2 integers.

$Q = \frac{p}{q}$ where, $p,q$ belongs I

What I don't understand is how does PI fail to qualify these conditions?

Since, $\pi$ is an angle and

$\theta = \frac{circumference}{radius}$

So does $\pi$ could be expressed in same terms and we can always make a circle with integral circumference and radius and find the value of $\pi$.

Thus, $\pi$ should be a rational number. What am I missing?

MIB
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    Can you make a circle with both circumference and diameter simultaneously of integer measure?

    That's a rather big assumption on your part.

     – Deepak May 24 '17 at 07:28
  • A circle with integral radius will always have an irrational circumference given by $2\pi r$. – Prasun Biswas May 24 '17 at 07:29
  • There must be a combination which satisfy these conditions. – MIB May 24 '17 at 07:30
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    @MIB Why "must there be"? Otherwise the universe will implode, People will die...or what? – DonAntonio May 24 '17 at 07:31
  • Why can't it be a possibility. Suppose I took a rope of a 100 units to make a circle and then I find its radius. Assume radius is not integral and as a fractional contains n decimal numbers. Then we multiply the size till its decimal numbers become null. – MIB May 24 '17 at 07:38
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    A rather easier example is this: $\sqrt 2$ is equal to $\frac{\text{hypotenuse}}{\text{leg}}$ in a right, isosceles triangle, yet it is well-known to be irrational. That means precicely that there is no right, isosceles triangle with both legs and hypotenuse lengths integers. You can try as much as you want, as large a triangle as you want, and you will never make it work. – Arthur May 24 '17 at 07:38
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    @MIB And what if the number of decimal numbers is infinite? – drhab May 24 '17 at 07:40
  • Thanks @Arthur. And since we can prove sqrt(2) is irrational. Then pi must be too. Thanks everybody else too – MIB May 24 '17 at 07:40
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    @MIB I didn't mean to say that there was any kind of deep connection between $\sqrt2$ and $\pi$... The two examples of circle and right triangle are separate, and both must be proven. But they illustrate the same point, namely that some times, even though you have defined something as a fraction, doesn't mean that it's rational. – Arthur May 24 '17 at 07:41
  • @drhab, it is quite possible but then there will be a circle. Just we might took much time and calculations. Haha. Thanks btw. – MIB May 24 '17 at 07:42
  • Yeah @arthur. Since they relate each other well. So rhere must be a way to prove pi is irrational too. It shouldn't be just hypothetical. – MIB May 24 '17 at 07:43
  • If there is any prove that pi is irrational as with sqrt(2). That will help me. – MIB May 24 '17 at 07:45
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    π was proved irrational by Johann-Heinrich Lambert in 1767. – Bernard May 24 '17 at 07:45
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    In math we will never measure, in physics(real world) pi is rational since you can't measure infinitely precise – kingW3 May 24 '17 at 07:47
  • Thanks @bernard. I will see that proof – MIB May 24 '17 at 07:47
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    Actually, we know: more: a consequence of the Hermite-Lindemann theorem (1882) is that π is trancendent, i.e. it is not a root of any polynomial with integer coefficients (whereas, for instance, $\sqrt 2$ is a root of the polynomial $x^2-2$). – Bernard May 24 '17 at 07:51
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    @kingW3 Pi is not rational in the physical world. Pi is always pi, and it always irrational since it is fixed and immutable. What you probably meant to say is that taking ratios of the circumference to the diameter of circles in the real world will always give you rational numbers because of limited precision in measurement. But these are only estimates of pi and as estimates go, they are naive in method and crude in accuracy. Better estimates (rational approximants of arbitrary precision) are obtainable using deeper mathematics involving infinite series. – Deepak May 24 '17 at 13:45

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