I'm looking for some intuition for why does pi equal 3.1416... and not some other number.

I started with the first digit 3. What I've done so far is:

Assume a circle of diameter 1. Assume a square of side 1 touching the circle at four points. The perimeter of the square is 4. The circle lies inside the square and both the circle and the square are convex figures, so the length of the circle must be less than 4. Does this seem right?

Also, the circumference must be greater than 2 because 2 is the shortest journey (forward and return) along the diameter of the circle, so the circle must be longer than that.

Now I can't think of anything to show that the length of the circle must be greater than 3.

Also, it'd help if you could give intuition for the next few digits of pi.

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    $\begingroup$ Try the regular hexagon inside the circle. $\endgroup$ – Michael Aug 4 '18 at 4:08
  • $\begingroup$ @Michael Ok that'll do. I kept thinking of an equilateral triangle but the triangle would come out of the circle lol. Is there some logical method for the next few digits too? $\endgroup$ – Ryder Rude Aug 4 '18 at 4:13
  • $\begingroup$ @RyderRude: Archimedes simply inscribed (and circumscribed) polygons with more and more sides. Do a web search for "archimedes pi" and you'll find lots of references. $\endgroup$ – Blue Aug 4 '18 at 4:22

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There is Archimedes method to find the approximation of $\pi$ by determining the length of the perimeter of a polygon inscribed with in a circle and the perimeter of a polygon circumscribed outside a circle $($which is greater than the circumference$)$. The value of $\pi$ lies between those two length. This might be helpful.


If you know a straight line is the shortest distance between two points this is an excellent way to produce lower bounds for $\pi$. Archimedes used this approach to approximate $\pi$ using polygons with $2^n\cdot 3$ sides. You can compute the side lengths of the polygons using the Pythagorean theorem.

I find the upper bound more problematic. You are claiming that any path between two points of a circle that stays outside the circle is longer than the path following the circle. That seems obvious, but I don't see an easy way to prove it. If you believe this the statement that the circle is shorter than the square is sufficient to show the first digit of $\pi$ is $3$. As Archimedes did, you can get a tighter upper bound by using circumscribing polygons with more sides.

There are many techniques for approximating $\pi$. The Archimedes technique converges rather slowly. There are many more that converge more quickly, with each having a proof that they actually converge to $\pi$. Whether any of them satisfies your definition of intuitive is for you to assess.

  • $\begingroup$ The upper bound arising from circumscribing regular polygons becomes obvious if you consider area rather than arc-length. $\endgroup$ – Lord Shark the Unknown Aug 4 '18 at 5:29

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