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$\pi$ is dependent on properties of geometry, assuming that we define it as $C/d$. Could there be a geometry where $\pi$ is a rational number or an integer?

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  • $\begingroup$ You should probably edit into your question that $C,d$ are perimeter and diameter of a circle respectively. $\endgroup$ – Jack's wasted life Feb 8 '16 at 19:38
  • $\begingroup$ You can make $\pi$ equal different values using different metrics. Consider using the $\max$ norm to make $\pi =4$ or even use the taxicab norm. Very relevant: math.stackexchange.com/questions/254620/… $\endgroup$ – Xoque55 Feb 8 '16 at 19:39
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Great question! Let's start by looking at all the data we need to make sense of $\pi$ in the first place.

You may have heard of metric spaces before. These are one of the more basic ways of generalizing geometry: a metric space is just a pair $(X, d)$: a collection $X$ of "points", together with a "distance" function $d$, satisfying some obvious rules. For example, $(\mathbb{R}, \vert x-y\vert)$ is a metric space; indeed, the usual metric space structure on the real numbers.

In a metric space, we can talk reasonably well about circles. The circle of radius $r$ and center $c$ in a metric space $(X, d)$ is the set of points $y$ such that $d(y, c)=r$. However, this isn't enough to define $\pi$, since we need yet one more concept: the length of a circle, or more generally the length of an arbitrary curve.

Calculus (well, limits, to be precise) is our friend here. Suppose $C$ is a circle. Given a finite set of points $A=\{a_1, . . . , a_n\}\subseteq C$, we let the approximation via $A$ be $$\delta_A=d(a_1, a_2)+d(a_2, a_3)+...+d(a_n, a_1),$$ and the mesh of $A$ be $$m_A=\max\{d(a_1, a_2),d(a_2, a_3),...,d(a_n, a_1)\}$$ We now let $circ(C)$ be the limit as $m_A$ goes to zero of of $\delta_A$, if this exists; and the point is, we want to restrict attention to metric spaces where $circ(C)$ always exists.

There are (literally! :P) countless metric spaces for which this limit always exists. However, even in these metric spaces we still can't necessarily make sense of $\pi$ - why should the ration of circumference to diameter always be the same?

So that's the other condition you need to check in a metric space in order to conclude that $\pi$ makes sense there: that the ration of the circumference of a circle to twice the circle's radius does not depend on the circle. That said, there are still countless of these: for instance, consider $\mathbb{R}^2$ with distance function $$d((u,v), (a, b))=\vert u -a\vert+\vert v-b\vert,$$ the so-called "taxicab" metric. A "circle" around a point looks like a square, with sides parallel to the lines $y=x$ and $y=-x$; and the sides are "longer" than they are in the usual Euclidean metric (the line connecting the origin to $(1, 1)$ has length $2$ in the taxicab metric). It's easy to check that here, $\pi$ is indeed $4$.

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  • $\begingroup$ A couple comments about the definition of $circ(C)$: 1. I think the $f$ in the definition of $\delta_A$ and $m_A$ is not needed. 2. We need some condition that in the limit, $A$ fills up $C$. One possibility is to take $m_A=\sup_c \min_i d(c,a_i)$, then define $circ(C)$ to be the liminf of $\delta_A$ as $m_A\to 0$. $\endgroup$ – Julian Rosen Feb 9 '16 at 15:58

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