What if $\pi$ was an algebraic number? (significance of algebraic numbers)

No such universe is possible, it would be a universe in which $1$ is equal to $2$.

That said, a rational approximation to $\pi$ with error $\lt 10^{-200}$ is undoubtedly good enough for all practical purposes.

Lindemann's proof that $\pi$ is transcendental was a great achievement, but knowing the result has no consequences outside mathematics.

You have to understand that although $\pi$ is a real number, it's not actually a real number. That is, it's in $\mathbb{R}$, but that set does not exist in the physical universe. It's an abstraction, just like the imaginary number $i$ is an abstraction, and one that has found significant use in physics (quantum mechanics and electrical engineering among others). Just like the idea of a number at all is an abstraction: the abstraction of assigning the same description to different quantities that are not directly related.

My point is that the quality of the universe that allows such abstractions to be imagined by intelligent creatures seems not to be separable from the quality that allows intelligent, imaginative creatures to exist at all. It requires only a sufficiently descriptive language, such as the kind considered in mathematical logic, to write down the formal definition of $\pi$ and indeed, of all of our mathematics, which implies all the algebraic and analytic properties of $\pi$ that we have proven because we wrote the proofs in that language!

So no such alternate physical universe can exist. On the other hand, one could imagine basing the definition of $\pi$ on alternate axioms, such as those specifying a particular non-Euclidean geometry, in some of which one does have $\pi = 3$, say. At least for some circles.

Interesting that nobody has mentioned: A practical consequence is that you cannot construct $\pi$ using a compass and a straightedge. This has saved so-so many man-hours; if Lindemann hasn't proved $\pi$ were transcendental we wouldn't have, e.g. caramel macchiato (or more significantly, aircraft).

Just to expand a bit on trb456's answer.

If $\pi$ was algebraic then that could mean there is some reasonably simple polynomial $p(x)$ for which it was a zero. This would be really nice because then all I'd have to do with a mystery number $\xi$ to check if it was $\pi$ is to check if $p(\xi)=0$ and then perhaps do a bit more book-keeping to verify that $\xi$ was really the real $\pi$.

For example, imagine we had some mystery number $\eta$ and we want to check that $\eta = \sqrt{2}$. How to do this? (for the purposes of this hypothetical, suppose calculators are all controlled by evil, self-aware, robot masters, they cannot be trusted, we have to do calculations by pencil and paper to be safe) $p(x)=x^2-2$ has $p(\sqrt{2}) = 0$, but $p(-\sqrt{2}) = 0$ so as a check on the number being $\sqrt{2}$ I'd also need to check on the sign of the number by some method.

So, perhaps you can see the utility of a number being algebraic. There is some finite number of algebraic operators we can perform on a potential candidate to ascertain if it is in fact the algebraic number in question.

In contrast, to show some potential number is $\pi$ we'll have to resort to a deeper mathematics. Some analysis, series, etc... We have convenient notations to hide the sophsitication, but $\sqrt{2}$ is much easier to define than $\pi$.

In any event, it probably should be agreed that there is some sufficiently precise rational number which captures the concept of $\pi$ for the need of any physical problem which involves $\pi$, so the absence of the polynomial check it of little concern. Most of the time $p(x)=x-22/7$ will do just fine.